Niche Modeling

232 responses so far ↓

• 1 Niche Modeling » Greenhouse Effect Physics // May 24, 2008 at 11:18 pm

  [...] Figure: The major relationships between fluxes in the description of Miskolczi’s atmospheric theory. [...]

• 2 Nick Stokes // May 25, 2008 at 12:42 am

  David,
  A quick review of how energy balance equations are derived. You nominate a region of space, and equate
  Flux out - Flux in = Nett Internal source.
  M does that for the Earth and the atmosphere, to get (M1) and (M2).

  You said on an earlier thread (GHE) that (M7) is for Earth and atmosphere combined. But you can’t get an independent equation that way. It’s just the sum of the parts. And M7 isn’t that. So what region is it?

  M’s explanation doesn’t help: “the second term (Ed-Eu) maintains the surface energy balance.” What surface? Ed is from the bottom of the atmosphere, Eu from the top.

• 3 admin // May 25, 2008 at 1:04 am

  Nick, The surface is the earths surface. The energy balance is

  energy in = sum of energies = energy out

  isn’t it? BTW I don’t think M1 and 2 are independent. If you combine the simplified versions:

  F=Eu
  F+Ed=Su

  There is no constraint on the values of Ed and Su. Another equation is needed and M7 (or Ed=F(1+τ)) supplies that, doesn’t it?

• 4 jae // May 25, 2008 at 2:55 am

  Snip as expedient, BUT: This old chemist thinks that all these radiative “games” and “cartoons” are completely ignoring the other 98% of the atmosphere (O2, N2). The discussion seems to presume that the Earth is composed of only GHGs and all that matters is radiative balance? What about the other 98% if the atmosphere? This component of the atmosphere has a big role in energy balance (and temperature) and it seems to be totally ignored in all these (simplistic?) analyses. Just how do all these other IR-inactive gases get warmed up each day and cooled off each night? Where does the energy come from?

• 5 admin // May 25, 2008 at 3:20 am

  Sure they absorb IR http://www.iop.org/EJ/abstract/0022-3700/10/3/018.

• 6 Nick Stokes // May 25, 2008 at 4:10 am

  David, energy in = sum of energies = energy out
  No, I don’t really know what that means, but I can’t recognize it as a usable energy balance equation. You and M both mix up energy and flux. All the things M deals with in this model are actually fluxes. To get an energy balance equation, you need to argue that energy is conserved; in a steady state (as here) it can’t be allowed to pile up anywhere, so for every volume in space, if there is a difference between flux in and flux out, it must be balanced by an internal source or sink.

  To keep things clear I prefer to always balance over volumes. You add up fluxes over all the surfaces. Here one volume, Earth, has only one surface and no internal source (if P=0), so it becomes possible to regard it as a surface balance of fluxes. It’s still only one equation, though.

  On your last observation, yes, the equations are independent; they just don’t match the number of unknowns. Unfortunately, in mathematics there is no law that says the supply of equations will meet the demand. So M7 might meet the need, but you have to establish that it is true.

• 7 Nick Stokes // May 25, 2008 at 4:14 am

  ps I forgot to query the identification of the surface balanced by Ed=Eu as that of the earth. How does this work when Eu is a flux at the top of the atmosphere?

• 8 admin // May 26, 2008 at 2:18 am

  #6 you need to argue that energy is conserved;
  This is what is being assumed. Are you saying it is not?
  It is obviously usable, as the post has shown, and it is also useful,
  particularly for looking at alternative solutions to the system of linear equations.
  Still it seems your objection is more editorial that substantive.

  #7 Ed=Eu in the case of the ’steel shell’ idealized model only,
  where the fluxes propogate up and down from opposite sides of the shell
  and there is a thermal discontinuity (e.g. vacuum) between earth and shell.

• 9 Nick Stokes // May 26, 2008 at 5:53 am

  Ed=Eu - my typo, I meant Ed-Eu, as in M’s statement “the second term (Ed-Eu) maintains the surface energy balance.”. The question remains; if “surface” means the Earth’s surface, how can it be balanced by a flux (Eu) at the top of the atmosphere.

  My other objection is far from editorial. We have an equation for conservation of energy where it is still not satisfactorily defined over which region the conservation is tested. If we could figure that out, then it would be necessary to explain how in F= Su-F + Ed-Eu (M7) Su and Ed are being added, when they are fluxes in the opposite direction. This is a vital equation in his logic, and we still don’t have a shred of argument to say that it is true. Conservation equations aren’t hard - it must be possible to step through and identify which fluxes are supposed to balance where and why.

• 10 Niche Modeling » Models of Greenhouse Effect // May 26, 2008 at 7:04 am

  [...] Nick Stokes @ Radiative Equilibrium (Miskolczi Part 4) [...]

• 11 Jan Pompe // May 26, 2008 at 2:25 pm

  Hi NIck,

  “The question remains; if “surface” means the Earth’s surface, how can it be balanced by a flux (Eu) at the top of the atmosphere.”

  The top of the atmosphere is the only place that the radiation leaving that atmosphere leaves to space the difference between that and E_D, which is the same at equilibrium with the surface as that entering the atmosphere by absorption (A_A), is what remains in the atmosphere to maintain the balance and provides the energy that gets converted to potential energy via convection. To maintain the energy balance at the surface there is also convection and the energy it takes to maintain hydrostatic equilibrium there is some kinetic energy left over and that goes to space as E_U.

• 12 Pat Cassen // May 26, 2008 at 10:25 pm

  Greetings David and Nick –

  I see, David, that you are still striving mightily (and creatively!) to find meaning in Dr. Miskloczi’s paper, where Nick and I see none. I thought that you might grapple with Appendix B and Figure 3, the subjects of my correspondence with Dr. Miskloczi, but since it appears that you are not taking these issues up, I might put my two cents in again.

  So for now, let’s disregard the facts that we disagree on the validity of Miskloczi’s eqn 7, his version of Kirkhoff’s law, his application of the virial theorem, and their collective consequences. In Appendix B he derives what he claims is a valid solution of the radiative transfer equations which has no discontinuity between air and surface temperatures.

  I will refer to your Georgia Tech notes (by Irina Sokolik) for convenience. As you say, she provides the standard, semi-infinite solution, with the Eddington closure assumption. Note that her equation
  B(tau) = Ione*tau + B(0)
  is the same as Miskloczi’s
  B(tau) = (3/4*pi)H*tau + Bo,
  with the obvious correspondence of constants. Since this equation is derived from the primitive radiative transfer equations under the Eddington assumption, Miskloczi has at this point also adopted the Eddington assumption. Now Sokolik goes on to derive in a very straightforward and unambiguous manner the values of Ione and B(0) from the conditions that (a) there is no downward flux at tau = 0 (no LW radiation from space) and (b) the net outward LW radiation must equal the total absorbed solar radiation. It is easy to see that these conditions demand, in Miskloczi’s terminology,
  Bo = (3/2)*Bo
  Fo = OLR,
  the standard result.

  But, by seeking a maximum in Bo(tau), Miskloczi claims to derive a somewhat complicated expression (his eqn 20) for Bo, which, he argues, does away with the discontinuity at the ground and which differs from the standard result. [He would derive the same result had he simply set Bg = B(tauA) in his general expression B(tau) (eqn 21)].

  Miskloczi then goes on to derive the consequent expression for Bg(tauA) [eqn B-11]. But when this relation (demanded by Misklozci’s solution) is used in eqn 20, one obtains the very simple
  Bo = (OLR/2*pi)*[1 + exp(-tauA)]
  Note:
  1.This expression is different from the standard result derived in Sokolik’s notes, so the general B(tau) cannot possibly match the boundary conditions at tau = 0, which must apply to Miskloczi’s model as well.

  2. Bo is a monotonic function of tauA! It has no local maximum. The only way Miskloczi can find a zero derivative for Bo is by incorrectly holding Bg constant, despite his own eqn B-11 for Bg(tauA). Apparently he has also held Bg constant to derive Figure 3, and in applying the upper boundary conditions.

  In our correspondence, Miskloczi “explained” things thusly: “…Bg and tauA are a boundary condition parameters…You may not interpret Eq. B11 as a Bg(tauA) function…..The important relationship here is between the Bg and olr….?”

  How am I to take B-11, if not as Bg(tauA)? What else is it?? How can I specify Bg as a boundary condition? — it’s what I’m trying to calculate! His explanation makes absolutely no sense to me.

  In short, aside from all the other stuff, I don’t believe Miskloczi’s “solution” in Appendix B is valid in any sense whatsoever. Wrong math, wrong physics.

  (I was going to go on about the discontinuity at the surface that bothers Miskloczi so much, but this is getting too long So suffice it to say that the math is telling us something, and nature deals with it just fine.)

  Thanks for reading this.

• 13 Pat Cassen // May 26, 2008 at 11:08 pm

  I should have mentioned in the above post that Miskloczi has apparently incorporated his (incorrect, IMO) eqn 8 to fix both Bg and tauA. In other words, in Appendix B, he has assumed, a priori, that the temperature at the ground is independent of tauA. Aside from rendering the structure of the atmosphere moot, it seems to require a feedback in which, for the Earth, water vapor must decrease as temperature increases (p. 23): “…in case the increased CO2 is compensated by reduced H2O…”, a very strange conclusion. But that gets into the whole question of how this alleged balance is supposed to be maintained.

  Finally, note that his solution is claimed to be general. That is, it applies to any planetary atmosphere (nothing unique to the Earth in the derivation). The implication would be that all planetary surface temperatures depend only on insolation, with the optical depth adjusting, no matter what the nature of the atmosphere…

• 14 admin // May 27, 2008 at 12:01 am

  Hi Pat, Thanks for your detailed reply. I would like to know if the paper is right, so I appreciate attempts to examine it in this way. I have been working through it bit by bit, and it takes time as there are a lot of interesting side-tracks. Though I am not the best person for the job, I’ll keep at it piece by piece and get around to the important issue you raise eventually. I was very glad to find the lecture notes by Irina Sokolik to get the received view.

  I think some of the concerns you and Nick have might have been addressed in http://hps.elte.hu/zagoni/idoj.....08_No4.pdf, a more empirical study of these issues involving the exact distribution and contribution of far-IR etc to greenhouse effects. Certainly I think there is a more extensive justification of Kirchhoff relationship there, as is the CO2/H20 compensation relationship. It really is a Part I to the Part II under discussion here.

  On your final point concerning F as the only variable, the balance of Mars examined in detail in the paper where it is claimed to show that hydrodynamic equibrium and the conditions for the virial theorem do not hold, which leads to a alternative system solution. So he has identified two different possible atmospheres, and Venus may yet a third solution. So the situation is different for Mars.

• 15 admin // May 27, 2008 at 2:59 am

  The question remains; if “surface” means the Earth’s surface, how can it be balanced by a flux (Eu) at the top of the atmosphere.

  Are you saying top of atmosphere flux cannot contribute to a balance a fluxes at the surface? Surely the radiative and thermal mechanisms can.

• 16 Nick Stokes // May 27, 2008 at 4:02 am

  No, I’m not saying that. Again, you don’t seem to have any appreciation of how an energy flux balance equation is created. You add fluxes crossing a surface until you have accounted for the energy budget. You can’t just throw in remote fluxes on the basis that some of the heat may have contributed by some pathway. You’re supposed to be resolving the pathway.

  But anyway, this may be all moot. I see Zagoni has put up what looks like M’s slides from the March 2008 conference here. The eq (M7) is now not included among the energy balance equations. Instead, it is now claimed to be a momentum balance equation (!). This is seriously weird.

• 17 admin // May 27, 2008 at 4:56 am

  Nick, As I understand your position was that the equation was wrong because it was not a flux equation. My position is that I want to know if it is correct at all, not whether it is a flux equation or not. Now that we have a possible derivation I am glad to stop guessing and try to determine whether it is correct or not.

  Perhaps Ms references to pressure of radiation in the paper will make more sense now. On first glace it would make sense, as most solutions of physical system involve both a conservation of energy and a conservation of momentum equation.

• 18 Nick Stokes // May 27, 2008 at 5:32 am

  No, the momentum version makes no sense at all. Radiation pressures are tiny - the Sun’s radiation exerts on Earth a pressure of 4.6 micropascals. If this comes down to a force balance then you have to include all the forces and accelerations. That would include the Sun’s gravitational attraction and the earth’s acceleration in orbit. Any radiation imbalance can be compensated by the earth shifting its orbit by a few microns.

  Secondly, the forces are vectors, and don’t all have the same directions. Most are radial from the Earth’s centre, but solar radiation is not, and you can’t just add them without allowing for direction.

  Thirdly, M can’t just keep changing his story like this if any credibility is to remain. In the paper he clearly says: “The principle of conservation of energy dictates that … Eq M7″. Oops, try momentum? How long can this go on?

• 19 Jan Pompe // May 27, 2008 at 12:31 pm

  “No, the momentum version makes no sense at all.”

  I disagree if we are talking about radiation pressure, due to random motion of particles, then it makes more sense talking about conservation of momentum than conservation of energy. Miklos Zagoni has just taken a somewhat different approach than via Kirchoff’s law as Ferenc Miskolczi did and arrives at the same conclusion.

• 20 admin // May 27, 2008 at 4:20 pm

  you can’t just add them without allowing for direction.

  From wikipedia

  It may be shown by electromagnetic theory, by quantum theory, or by thermodynamics, making no assumptions as to the nature of the radiation, that the pressure against a surface exposed in a space traversed by radiation uniformly in all directions is equal to one third of the total radiant energy per unit volume within that space.

  Greenhouse effect = G = Su/3 = OLR/2

  M can’t just keep changing his story like this if any credibility is to remain.

  From page 9.

  Formally, in the presence of a solid or liquid surface, the radiation pressure of the thermalized photons is the real cause of the greenhouse effect, and its origin is related to the principle of the conservation of the momentum of the radiation field.

• 21 Pat Cassen // May 27, 2008 at 5:05 pm

  typo in my 12, above.
  Bo=(3/2)*Bo should be:
  Bo = (3/2)*(H/3*pi) = H/(2*pi)

  sorry.

  Incidentally, one can prove that Miskolczi’s “solution” is not what he claims it to be, by direct substitution. He claims that Appendix B presents a solution that satisfies both the upper boundary conditions (no downward LW radiation and total LW radiation = total absorbed solar radiation) AND B(tauA) = Bg. Substitution of his solution into these conditions yields a transcendental equation for tauA that is satisfied only in the limit tauA goes to infinity, NOT the finite value Miskolczi claims. In other words, his solution does NOT satisfy all the boundary conditions his model requires.

• 22 Pat Cassen // May 27, 2008 at 5:14 pm

  David -

  As Nick says above, the radiation pressure in the atmosphere is irrelevant; it is totally negligible compared to all other forces. The conservation of momentum equation is satisfied to a high degree of accuracy by the hydrostatic condition. Even in convective zones. Read a textbook. Do the numbers.

• 23 Nick Stokes // May 27, 2008 at 6:03 pm

  Well, David and Jan, I stick to all my objections above to switching to momentum conservation as the justifying principle. But let’s just deal with this one first. If M7 is derived from cons mom, it is then a force balance equation. As such it must include all forces. These include, for example, atmospheric pressure (in kilopascals).

• 24 Jan Pompe // May 28, 2008 at 6:35 am

  “As such it must include all forces. These include, for example, atmospheric pressure (in kilopascals).”

  Perhaps Nick you can quantify it for us and tell us the impact of atmospheric molecular momentum on the momentum of the photons radiated upward from the surface. Does it actually cause the photon momentum not to be conserved?

• 25 Nick Stokes // May 28, 2008 at 7:16 am

  OK, Jan. Conservation of momentum obviously does not require the momentum of any one entity to be invariant (eg billiards). The flux of upward photons that are absorbed will provide an upward pressure on the atmosphere of one or two micropascals. That may reduce the downward pressure of the atmosphere from about 100 kilopascals to …

  The escaping photons provide an effective force on the earth. These might budge the orbit by a nanometre or so, except that the force is outward in all directions, so, no nett force.

  Because these huge forces can so easily absorb radiant imbalances, there is no constraint that says variation of one radiant flux has to be balanced by variation of another.

• 26 Jan Pompe // May 28, 2008 at 8:24 am

  Nick please try to answer what I actually asked. For instance: “The escaping photons provide an effective force on the earth.” While interesting and true as well as insignificant and a red herring it’s hardly relevant to my question and conservation of the momentum of those photons, so I’ll be a bit more specific. What has happened to the momentum of those escaping photons due to interaction with the atmosphere say for instance the ones than comprise S_T? It’s as good a place to start.

• 27 Nick Stokes // May 28, 2008 at 9:00 am

  Jan, the answer was here: “The flux of upward photons that are absorbed will provide an upward pressure on the atmosphere of one or two micropascals.” As I said, momentum is a vector, and because it is symmetric about the Earth’s axis, for S_T it sums to zero anyway. But it does provide a tiny pressure.

  The photons that escape the Earth do have momentum and provide a tiny force. The photons that hit the Earth from the sun pack a total force of about 500 million newtons. About the weight of a medium size ocean liner. This could be balanced by a sub-micron orbit variation. The outgoing IR is again, as a vector sum, a tiny fraction of this, because of symmetry.

• 28 Jan Pompe // May 28, 2008 at 10:27 am

  Nick you are still obfuscating, but thanks for the response anyway.

  “As I said, momentum is a vector, and because it is symmetric about the Earth’s axis, for S_T it sums to zero anyway.”

  Ok so how do they do this?

  “The escaping photons provide an effective force on the earth. These might budge the orbit by a nanometre or so”

  You are making it up as you go by the look of it. Now consider a column through which the photons of S_T are passing through the atmosphere they have a momentum what happens to that momentum due to interaction with the particles in the amosphere on the way out? Try to stay focused on the question.

• 29 Nick Stokes // May 28, 2008 at 11:44 am

  Jan, I think I’ve said, and I don’t think it’s getting to the goal, which is, how can M7 be a force balance at the surface (or anywhere) when radiation forces are tiny and so many big forces are missing?

• 30 Jan Pompe // May 28, 2008 at 12:14 pm

  Nick “how can M7 be a force balance at the surface”

  It’s a pressure balance there is a subtle difference. The incoming is absorbed (theoretically 100% for the putative black earth) so the pressure due to it is zero because all the energy is converted to heat the collision with the surface is perfectly inelastic and all the momentum is absorbed. Now the net pressure of a small parcel of atmospheric is the same in all directions but the net momentum of particles in that same parcel is zero, so even a small net momentum due to the upward radiation, which obviously isn’t isotropic, is quite significant compared to that.

  Now how about that question?

• 31 admin // May 28, 2008 at 9:36 pm

  Hi Guys,

  I consulted a couple of text book as Pat suggested. Thermodynamics of the Atmosphere: A Course in Theoretical Meteorology By Wilford Zdunkowski, Andreas Bott and one by Collins. Both talk about radiation pressures, and use momentum of the photons in equations of solution of atmospheres. I also read up a bit on topics of thermalized photons, photon gases from wikipedia and other sources. This discussion really needs a good background in that material to serve the purpose of reviewing these concepts in Miskolczi view of greenhouse.

  What I think zagoni is saying is that the photons are thermalized in the
  blackbody flux of the earth/atmosphere forming a photon gas that is
  bound by the momentum constraints etc to have, in conjunction
  with the other gases a certain distribution. Departures from that distribution
  are departures from thermal equilibrium. It doesn’t work so well on Mars
  because the atmosphere does not reflect back to the surface enough,
  to give the 1/3 ‘back-pressure’ to set up the ‘container’ for the
  thermalized photon conditions. This is why the 1/3 overpressure is
  attributed to the greenhouse effect. This should be equivalent to Kirchhoff’s law,
  which is really about thermalization of photons in cavities.

  Also I note that M does not necessarily subscribe to this approach by necessity and finds that conservation of energy and radiative equilibrium relations can reach the same result without invoking Kirchhoff’s law.

  Now this is just hand-waving, and I admit no expertise in this area, the
  main goal is to understand what they are saying. Nowhere do they say the radiation pressure has a important effect on the atmosphere directly. But it would seem the text books provide support for the use momentum to derive relationships between fluxes, and M provides empirical support for this.

  Answering directly your question:

  how can M7 be a force balance at the surface (or anywhere) when radiation forces are tiny and so many big forces are missing?

  Its not how big it is but what you do with it. The books talk about the radiation field being decoupled from the hydrostatics. Because of this they can be used to describe relationships that hold irrespective of the external forces.

• 32 Eli Rabett // Jun 28, 2008 at 2:33 pm

  WRT 4 and 5. Yes, there is a very small amount of collision induced absorption by N2 (and O2) in the far IR. OTOH, the cross-section is tiny (see the paper) and the pressures needed to observe this are high (7000 atm). In short, N2 and O2 contribute just about as much to radiation trapping in the atmosphere, as the radiation does to the observed pressure.

  Since there is a lot of atmosphere, N2 can be observed to absorb IR light, via quadrupole absorption. See Demoulin, Farmer, Risland and Zander JGR 96, 13003 (1991). The optical depth from the top of the atmosphere to the top of the Alps is about 10% absorption. O2 has both quadrupole and magnetic dipole lines which are about as strong (Balasubramanian, D’Cunha, Rao, JMS 144 374 (1990). Also Rinsland, et al., JQRST 48 693 (1992). This means that the optical depth of the atmosphere for those lines is longer than the atmosphere is high, e.g. that out of 10 photons emitted only 1 is absorbed in the atmosphere. By comparison the optical depth at CO2 wavelengths is about 10 m.

• 33 admin // Jun 28, 2008 at 7:00 pm

  Eli, thanks for the clarification. I think I deleted the other of your posts in moderation. Please repost if you wish.

• 34 Doug Moseley // Jun 28, 2008 at 8:58 pm

  In your calculation of the steel greenhouse effect on the earth,s temperature you do not take into account the fact there is a IR window. This mans that not all the IR emitted by the surface is intercepted by the greenhouse, and thus the temperature should be lower than what you calculated.

  The earth’s atmosphere cannot be replicated by a single layer greenhouse because the energy that is captured by a greenhouse gas molecule and subsequently released in an upward direction is not guaranteed to escape to space. In the earth’s atmosphere, unless the photon that is emitted upwards has a wavelength that allows it to escape through a transparent window it is almost certain to be intercepted by another GHG molecule unless it is emitted high in the atmosphere. The absorbed energy is subsequently re-emitted and thus has to be represented by a second shell.

  The equation Su=2F does not apply to the earth’s surface, as energy is removed from the surface by conduction and subsequent convection, and also as latent heat due to the evaporation of water. Su is thus considerably less than 2F for a single shell greenhouse. For multiple shell greenhouse effects Su may be more or less than 2F depending on the exact properties of the shells. For a greenhouse of n shells and no windows Su=(n+1)F. No real atmosphere will obey this equation, because there will always be transfer of energy by conduction. Conduction, convection, and the transfer of latent heat create windows in the shells.

  You cannot assume radiative equilibrium at the earth’s surface because the earth’s surface loses energy by means other than radiation. Since the surface is losing energy by other means, radiative balance would result in decreasing temperatures.

• 35 admin // Jun 29, 2008 at 3:27 am

  Doug, These equations do not represent real earth atmosphere, just different conclusions under different assumptions — idealized. One is a fully active greenhouse effect with no IR window.

• 36 David L. Hagen // Jun 29, 2008 at 5:25 am

  Nick et al.
  On “Radiation pressures” etc., I think this is an English as a second language problem of translation. I don’t think he means literal photon radiation momentum, but rather energy fluxes from radiation, conduction, etc. See assumption (a) “The available SW flux is totaly absorbed in the system. In the process of thermalization Fo is instantly converted to isotropic upward and downward LW radiation.”

  M applies the Virial theorem to equate half the gravitational potential energy with internal thermal energy. etc. Thus the fluid use of terms combined with translation to English. (If only my Hungarian, Russian or Chinese were anywhere as close!)

  I find foundational his starting from the detailed line by line modeling of atmospheric radiation Ed and absorption Aa by HARTCODE, from which he observes the their equality and thus identifies an average Kirchhoffs law for the atmosphere. Then applying equilibrium and energy conservation. etc.
  ——————-
  David/Admin
  Could you please clarify
  “Convection is included in the Ed=Su constraint.”

  I see “Aa = Ed” equation (4).
  For Su (=Sg) to equal Ed ie to equal Aa, that would mean that
  St = 0. Did he make that assumption? if so where?

  As you commented, M’s semi-transparent model provides remarkable improvement in fit to the measured source functions shown in Fig. 5 compared to the Semi-infinite model and USST-76, while being very parsimonious.

• 37 Nick Stokes // Jun 29, 2008 at 9:47 pm

  David H,
  Yes, there may be language issues, but somehow someone has to be able to figure out the underlying meaning. What do you make of this sentence from the paper?
  “Formally, in the presence of a solid or liquid surface, the radiation pressure of the thermalized photons is the real cause of the greenhouse effect, and its origin is related to the principle of the conservation of the momentum of the radiation field.”
  The English seems OK.
  And yes, there may be a language problem with describing the basis for the virial theorem. But it’s turned into mathematics and used. Can you describe what the justification is?

• 38 Neal J. King // Jul 4, 2008 at 8:06 am

  I believe M’s application of the Virial Theorem is incorrect. It’s a long argument but it boils down to two points:
  - He uses it in the form 2*avg(KE) = avg(PE), where PE = mgz. But the VT doesn’t work that way: either it should be applied in the form 2*avg(KE) = - avg(PE), where PE = - GMm/r (realistic 3-dimensional case); or it has to be applied taking into explicit account the containing boundaries (the ground at z = 0). When you take the second approach, you only find out that the gas obeys the perfect-gas law, and also the usual equation of hydrostatic equilibrium relating pressure gradient to mass density. You do NOT get a relationship between avg(KE) and avg(mgz).

  - In particular, if the VT were applicable to the case PE = mgz, it would apply to the model case of the adiabatic atmosphere, which can be solved completely. But when you do the calculation explicitly, you find that 2*avg(KE) = 3*avg(PE), which doesn’t fit either the n=1 (mgz) or the n=-1 (-GMm/r) power-law representation of gravity.

  These points are different from, and logically prior to, the previously raised issues of how to take the “results” of the VT and apply it to radiation and other energy fluxes.

• 39 Jan Pompe // Jul 4, 2008 at 12:53 pm

  “He uses it in the form 2*avg(KE) = avg(PE)”

  Don’t think so average KE of a gas relates to it’s temperature virial theorem is not dependant on it’s temperature but the total average kinetic energy. More importantly orbital factors play an important part remember a particle ‘at rest’ on the equator has a tangential velocity of 1000 miles per hour so you need to add that to the total. Now take two masses one heavy (mass M) and one light one (mass m) gravitational potential energy is given by U = - GmM/r where G is gravitation constant and r the distance between the centres and the gravitational force F_g = -GmM/r^2 for the smaller particle to maintain it’s distance it needs a centrifugal force F_c = mv^2/r i.e.

  F_g = F_c = GmM/r^2 = mv^2/r

  Dividing both sides by 2r gives us the energy relation
  GmM/2r = 0.5mv^2

  GmM/2r = U/2 = .5 mv^2 = KE

  U/2 = KE.

  This is where it comes from.

  Now what we usually refer to as the average kinetic energy, that according to kinetic theory is the temperature, is measured in the same moving frame of reference so only serves to add to the total KE the major portion of which is actually determined by the orbital speed.

  You can check out wikipedia for the generalisation to collections of gravitationally bound collection of particles.

• 40 Neal J. King // Jul 5, 2008 at 6:31 pm

  Jan Pompe,
  a) Average KE vs. Total KE: In this application, M is using the VT to relate kinetic energy (KE) to potential energy (PE). You’re talking about the relationship between the total average kinetic energy and the virial, and I’m talking about the relationship between the spatial-average kinetic energy and the spatial-average virial. The difference between the respective quantities is an overall constant factor (the spatial volume), which does not change the relationship. So the distinction you draw makes no difference.
  b) Orbital motion saves application of Virial Theorem: What you are saying would save the application of the VT if the speed due to rotating with the Earth were comparable to the speed an object would have orbiting the Earth. But it’s not:
  – orbital speed = sqrt(GM/R_e) = 7910 (m/s)
  – rotational speed = 2.pi*R_e/day = 463 (m/s)
  – ratio of rotational speed to orbital speed = 463/7910 = 0.0586

  That means that the actual motion of the gas molecules does not begin to approach the speeds necessary to meet the Virial Theorem, if the form you used were to apply:
  2*KE = - U
  (And you forgot the minus sign! This is very important, in distinguishing whether you are talking about a 3-dimensional Earth (U = -GMm/r) or (as M seems to be doing) a flat Earth (U = mgz).)
  Does that mean that the VT is wrong? No, but it means you have to use it more carefully. The VT says that:
  2 * time-averaged(total KE) = - time-averaged(Sum(r_i . F_i))
  If you apply this to a atmosphere shaped like a spherical shell, the LHS gives the total KE as usual; but the RHS, when converted into an integral, has two terms: the bulk volume term (which gives something proportional to the potential energy) and the surface term, which comes from the normal force acting at ground level. So if you separate the molecular motion into random thermal motion and rotational motion, and plug it into the above, you find (suppressing the time-average):
  KE_rotational + KE_thermal = GM_e*M_a/R_e – (4.pi.(R_e)^3)(ground-level pressure)

  As calculated above, the KE_rotational is negligible compared to the gravitational PE term, and the KE_thermal is also negligible (it corresponds to having a speed of about the speed of sound, which is a bit less than the rotational speed). So what the VT says in this case is that the ground-level pressure is what is supporting the atmosphere from collapsing towards the center of the Earth.

  If we go to your case of the orbiting molecule, the KE_rotational would equal the gravitational PE (with a – sign), and the KE_thermal and pressure terms would be zero. But that’s not the case that fits with M’s paper.

• 41 Jan Pompe // Jul 5, 2008 at 11:28 pm

  “orbital speed = sqrt(GM/R_e)’

  this does not make sense you leave out of that the mass of the orbiting body which should be in the numerator.

  So orbital speed is sqrt(m^2G/(m+M)r)

  Your error is a far more serious one than my leaving out the sign given that we are mainly interested in the magnitudes of the energies that balance each other.

  “the KE_rotational would equal the gravitational PE (with a – sign)”

  NO *Total* average KE = - total average gravitational PE.

  “KE_rotational + KE_thermal = GM_e*M_a/R_e – (4.pi.(R_e)^3)(ground-level pressure)”

  4*pi*R_e^3 is neither an energy nor a pressure term so you can’t subtract it from the orbital term you are subtracting what exactly from potential energy?

  The reason we talk about a sum of potential energies because we are talking about an ensemble of particles that have no constraints on the the distance between them or from the the centre of gravity apart from their kinetic energies so it does not make sense to convert the sum to an integral.

• 42 Neal J. King // Jul 6, 2008 at 1:50 am

  Jan Pompe,

  Orbital speed:
  Your own equation was:
  GmM/2r = U/2 = .5 mv^2 = KE
  so v^2 = GM/r , implying:
  v = sqrt(GM/r)
  and this is exactly my equation (except I set r = R_e).
  If you want to worry about the motion of of the Earth in orbit with the molecule, it makes an infinitesimal difference because the ratio of masses is so small.

  But this equation is certainly wrong:
  sqrt(m^2G/(m+M)r)
  As you take the limit as m => 0, this expression vanishes, but you should remember that the path of a particle through a gravitational field becomes independent of the mass; so that should be a hint that this equation is incorrect.

  The correct equation is: v = sqrt(G*M^2/(r*(M + m)), which correctly => sqrt(G*M/r) as m => 0.

  So I am afraid you are in error.

  Virial Theorem: I am basing my analysis on the derivation you can find at, for example, http://en.wikipedia.org/wiki/Virial_theorem, in the section “Time averaging and the virial theorem”: the point is that you have to do an Sum of r_i.f_i. When the mass of particles is held together by mutual gravitation, you can turn this into a volume integral; but when one “particle” is the Earth, which has appreciable size from which it is excluding the other particles, it makes more sense to treat it as an external potential. Then the Sum turns into a volume integral over the available space (outside the Earth) and a surface integral over the Earth’s surface. The 4.pi.R_e^3 term is the surface term part of Sum(r_i.f_i).

  I don’t understand your reluctance to turn a Sum into an integral: It helps when your dealing with a macroscopic number of molecules.

  And there are certainly constraints on the distances: The molecules are excluded from the Earth.

• 43 Neal J. King // Jul 6, 2008 at 1:58 am

  Correction of error above:
  “The 4.pi.R_e^3 term is the surface term part of Sum(r_i.f_i).”
  should be:
  “The 4.pi.R_e^3*(ground-level pressure) term is the surface term part of Sum(r_i.f_i).”

• 44 Neal J. King // Jul 6, 2008 at 2:32 am

  Maybe something that was unclear in 40:

  “KE_rotational + KE_thermal = GM_e*M_a/R_e – (4.pi.(R_e)^3)(ground-level pressure)”

  This is the equation for the total amounts for the gas:
  - KE_rotational: the “moving along” KE of the atmosphere for keeping up with the Earth’s surface
  - KE_thermal: the thermal KE for the atmosphere
  - M_a: the mass of the atmosphere

• 45 Jan Pompe // Jul 6, 2008 at 4:04 pm

  “But this equation is certainly wrong:
  sqrt(m^2G/(m+M)r)”

  I agree it was a slip of the shift key I hadn’t noticed i had done it i apologise for the time you had to waste on it.

  “The correct equation is: v = sqrt(G*M^2/(r*(M + m)), which correctly => sqrt(G*M/r) as m => 0.”

  Fair enough however you take this relation but bear in mind it is an approximation for orbiting bodies of similar mass.

  orbital speed = sqrt(GM/r)
  and rotational speed = 2.pir/day

  a geostationary satellite must be placed at a distance where the rotational speed = orbital speed that is:
  2*pi*r/day = sqrt(GM/r)

  Solving for r
  r= (86400^2*G*M/4*pi^2)^(1/3) plugging in G=6.67E-11 m^3 kg^-1 s^-2 and M = 5.9736E24kg we get an orbital radius of 9,099 which is ridiculous geostationary satellites can have only one orbit and that is at an altitude of 35,768 km.

  I don’t think I’ve done something wrong you might like to check it.

  “The 4.pi.R_e^3*(ground-level pressure) term is the surface term part of Sum(r_i.f_i).”

  Using dimensional analysis
  “4.pi.R_e^3″ has dimension L^3 or volume pressure has dimension M L^-1 T^-2 so it does not make sense.

• 46 Neal J. King // Jul 6, 2008 at 4:35 pm

  Jan Pompe:

  1)v = sqrt(G*M^2/(r*(M + m))
  No, this formula applies with two arbitrary masses M and m. It’s derived from the formulae:
  m*r_S = M*r_L
  r = r_S + r_L
  m*v_S^2/r_S = GMm/r^2 = M*v_L^2/r_L

  In the limit m => 0, v_S = sqrt(GM/r)
  In the limit m = M, v_S = sqrt(GM/(2r))

  2) Orbital vs. rotational speed
  My point is that you were originally saying that the atmosphere would satisfy the VT because the speed that the molecules would have from just “keeping up” with the ground would be enough to do that; and then you calculated the orbital speed (which indeed IS fast enough to satisfy the VT). What I pointed out was the disconnect: the “keeping up” speed is far below the orbital speed, at ground level and for some considerable distance above it. wrt the geostationary orbit: This is exactly why the VT does not apply in the simplistic manner to the atmosphere. The atmosphere is NOT orbiting the Earth.

  My solution is that the VT applies when you understand it correctly: which means, take into account the fact that the actual atmosphere is supported by the ground. The ground is why the atmosphere does not collapse to a much smaller (and much hotter) ball near the geometrical center of the Earth, as it would if the Earth were actually a point mass of the same size,
  6 *10^24 kg.

  But when you take the ground into account, you have to recognize that the r.F terms at the ground give rise to a surface integral, additional to the volume integral. Since the surface is defined as r = R_e, and the pressure is the force/unit area, the integrated r.F is
  R_e*(area)*(ground-level pressure) = R_e * (4.pi.R_e^2)*(glp)
  = 4.pi.R_e^3 * (ground-level pressure)

  3) “4.pi.R_e^3″ has dimension L^3
  Yes, that is why that part is multiplied by the ground-level pressure, which has dimensions (ML/T^2)/L^2 = M/(LT^2). So the entire term has dimensions ML^2/T^2, energy.

• 47 Neal J. King // Jul 6, 2008 at 5:25 pm

  In #46:
  “if the Earth were actually a point mass of the same size,
  6 *10^24 kg.”
  SHOULD BE
  “if the Earth were actually a point mass of the same mass,
  6 *10^24 kg.”

• 48 Jan Pompe // Jul 6, 2008 at 11:19 pm

  “My point is that you were originally saying that the atmosphere would satisfy the VT because the speed that the molecules would have from just “keeping up” with the ground would be enough to do that;”

  No that isn’t quite what I was saying What I actually said is that it needs to be taken into account as well as thermal energy which was the point of the reminder that your “2*avg(KE) = avg(PE)” was not quite enough the word ‘total’ is a very necessary one. Perhaps I wasn’t being clear enough.

  My apology for comment on this again it was late I didn’t notice the ‘*’ included or I would have known that you were multiplying it with (pressure at ground level) and not simply describing it. However writing 4.pi.R_e^3 * p_g (p_g =ground level pressure) would have been a lot clearer.

  “If we go to your case of the orbiting molecule, the KE_rotational would equal the gravitational PE (with a – sign), and the KE_thermal and pressure terms would be zero.”

  I’m not going to accuse you of mounting a strawman because possibly I was clear enough but total KE = rotational energy + thermal kinetic energy + pressure terms and any other KE as yet not thought of. Because that really is a mess is I suspect the motivation behind Clausius’ development of the theory.

  Now he question I have for you is are you trying to make sense of it or are you just trying to tear it down?

• 49 Neal J. King // Jul 7, 2008 at 12:22 am

  Jan Pompe,

  My point of view is expressed in #38:
  - M’s attempt to apply the VT is not correct. When applied correctly, along with the perfect gas law, to the current situation, the VT only implies the equation of hydrostatic equilibrium. It does not relate the total KE to the total PE in any obvious way; what it will mostly tell you is that the pressure on the ground is what is supporting the atmosphere from falling through the ground.
  - The approach would work if the Earth could be treated as a point mass; but then the atmosphere would be a much smaller, hotter cloud of gas, not a spherical shell of gas.
  - A separate point is that the way he relates energy fluxes to the potential and kinetic energy totals is very unclear.

  So, in trying to make sense of it, I think I’ve come across some conceptual problems in M’s paper. I may just write it up more fully (this interface doesn’t allow me to write equations as clearly as I would like; and anyway it takes a lot longer to explain the steps than to state the results) and send it to him. He has engaged people at other blogspots (e.g., RealClimate), so he may be amenable to dealing with email.

• 50 Jan Pompe // Jul 7, 2008 at 4:01 am

  ” I may just write it up more fully (this interface doesn’t allow me to write equations as clearly”

  Then do it I agree it’s a most difficult medium.

  ” what it will mostly tell you is that the pressure on the ground is what is supporting the atmosphere from falling through the ground.”

  This is where I think you are going wrong. All orbital and PE calculations are done as if the larger solid or the aggregate of particles combined or not exist as a point mass at the barycentre (that rock resting on the ground has a net PE to begin with). I’m pretty sure it makes no difference to a parcel of gas if the the balancing upward force is provided by as solid surface or a denser layer of gas beneath it.

• 51 Neal J. King // Jul 7, 2008 at 11:15 am

  Jan Pompe,

  I’ve tried to post substantive responses now twice, but they haven’t shown up. I’m not sure what’s going on, but I’m not going to write anything substantive again until it starts working.

• 52 admin // Jul 7, 2008 at 11:21 am

  Hi Jan, Neal, Sorry, I know one thing that can happen is that more than 2 links gets deleted. I can raise that limit. Otherwise, there are plenty of long posts that get through, so I don’t know what could stop them showing up.

• 53 Neal J. King // Jul 7, 2008 at 12:44 pm

  OK, I’ll try again:

  There is a very important difference between supporting the atmosphere by gas or by ground:
  - When you support it by ground (including the ground of a spherical planet), the virial = Sum(r_i.F_i) only includes the molecules in a spherical shell, consisting of molecules between R_e and R_e + atmospheric_thickness.

  - When you support it by gas (meaning that the Earth is turned into a point-particle of equal mass), all the gas below the previously designated “atmosphere” is really of a piece with the “atmosphere”. It is not separable, because they mix together over time, and you have to do a time-average for the VT to apply anyway. Because they mix together continually, the virial sum has to include ALL the gas molecules in the system, not just the outer ones.

  - So if you want the same amount as before of atmosphere between R_e and R_e + atmospheric_thickness, you have to add a whole lot of gas beneath it. That changes the virial by adding many times more molecules than you had before to the count. You are then talking about a dwarf-star, of which the outer shell (R_e, R_e + a_t) is just an arbitrarily designated subset. The virial is going to be very different.

  - Or, if you want to keep the total number of gas molecules the same, they will not remain at that large distance R_e: they will rush down towards the point-particle and orbit tightly around it.

  - Either of these approaches will satisfy the VT,

  with total-KE = - total-PE

  But neither of them apply to the situation in M’s paper.

  - If you want to understand this point any better, I can’t do much better than to say: Calculate the virial Sum(r_i.F_i) directly, assuming any reasonable distribution of molecules n(r)
  (say, n = 1 for r in (R_e, R_e + a_t),

  with individual mass_i = m,

  and F_i = -(GMm/r_i^3)r_i for gravity; but there is also the normal force pushing upwards at the ground level,
  N/area = Pressure r/r
  so don’t forget that part of the calculation.

• 54 Jan Pompe // Jul 7, 2008 at 2:49 pm

  “- When you support it by gas (meaning that the Earth is turned into a point-particle of equal mass), all the gas below the previously designated “atmosphere” is really of a piece with the “atmosphere”.”

  Actually that mass of gas below the any layer of interest forms as well as the any layer above it will make up the aggregate mass and if the total mass is the same as the total of earth + sea + atmosphere I don’t see that it will make any difference at all. We will somewhere in the gaseous sphere find a level where the pressure and density is that same as that at the surface and from there the pressure profile will be similar to what we have now.

  Rather that pure supposition I prefer too look at a congruous situation to see if it holds water.

  Look at Venus which has far more gas in the atmosphere and at an altitude of 50 km it’s pressure is ~10^3 mbar at 70 km its ~120 mbar earth’s atmospheric pressure is 125mbar at 18KM. This is a fairly similar change in pressure over a similar distance and certainly does not suggest that the layer at 50 km and above is in any way affected by the fact that it’s held up by 50 km of super critical fluid and gas. (BTW the temperature at that level fairly close too and has a lapse rate of -5.5K/km not outrageously different either)

• 55 Neal J. King // Jul 7, 2008 at 4:10 pm

  - Once you have a big ball of gas, you cannot apply the VT to just a small portion of it: “total” means ALL of the gas, not just an arbitrary outer layer. You will be applying the VT to a gas planet or brown dwarf: either held together by mutual gravitation of the gas molecules or by gravitational attraction to a central core.

  - If you replace the hard Earth by gas of equal mass, the density of the gas will be much less, so it cannot be confined to the same volume. Your original shell of gas will be blown out to a much farther distance (and will still be mixed with all the other gas anyway). So your original shell of gas will not be at the same radius, will not have the same pressure, etc. Even if you were to try to apply the VT to this artificial subset, the conditions it would have would be different.

  - Frankly, I am working on putting my calculation in a readable form to send to M for comment. It is based on doing the actual virial calculation, which you still seem not to have tried. If you want to try that, I’ll be interested to see if there are conceptual problems in doing it. But the topics we seem to be discussing right now are kind of off the point, from my perspective; so I will not give a lot of priority to that until I’m through with the note.

• 56 Jan Pompe // Jul 8, 2008 at 1:26 am

  ” So your original shell of gas will not be at the same radius, will not have the same pressure, etc.”

  Radius will be different but the pressure will be the same and quite frankly I can’t see any way of telling the atmosphere that the upward pressure is due to solid ground rather than a denser layer of gas. IOW there is no discontinuity in in pressure at the surface, though the gradient will change, and I see no valid reason to make such an assumption.

• 57 Neal J. King // Jul 8, 2008 at 6:55 am

  When you calculate the virial, if the gas is sitting on gas, the integral/sum has to include ALL of the gas, from r = 0 to the top of the atmosphere, because the “atmosphere” will be mixing it up with the other gas. In other words, it’s all just gas.

  If the gas is sitting on the Earth, the virial integral/sum only covers the atmosphere itself, because that’s all the gas there is; and there is a surface term not present in the first case, because there is no surface in the first case.

• 58 Neal J. King // Jul 8, 2008 at 7:31 am

  As a “practical” application of using this approach to the Virial Theorem, I can calculate that the totals of KE and PE are generally related as:
  [KE] = (3/2)[PE]

  in the “flat-Earth” case that U(z) = mgz
  (m = mass of a molecule, g = local acceleration of gravity)

  This can also be explicitly calculated for the constant-temperature model of the atmosphere, and for the adiabatic atmosphere (see http://farside.ph.utexas.edu/t.....ode56.html), by straight-forward integration.

• 59 Neal J. King // Jul 8, 2008 at 10:08 am

  I don’t know whether that calculation disagrees with M’s result or not, because I’m not absolutely sure whether he’s modeling the atmosphere as a spherical shell or as a vertically layered system.

  His claim that
  [KE] = [PE] , with no minus-sign, suggests he’s using the flat-Earth approach; in which case he’s wrong, because of the actual factor of (3/2)

  On the other hand, he might have meant:
  [KE] = -[PE] , with the minus-sign, implying the spherical shell approach. In that case, he’s wrong for a different reason, having to do with the ground-support term. But I’m not quite ready to finalize my analysis on that case, because I don’t quite yet see how the first case comes out (as it should) as a limiting case when the Earth radius R_e goes to infinity, while g = GM/R_e^2 remains fixed. It’s not yet coming out properly.

• 60 Nick Stokes // Jul 8, 2008 at 12:04 pm

  Neal,
  Before following this too far, I think you should figure out how any statement about these energies can be related to IR fluxes. I don’t believe they can.

• 61 Neal J. King // Jul 8, 2008 at 2:24 pm

  Nick,

  I entirely agree with you. In any event, he doesn’t make much effort to be explicit about it.

  But I’m having some fun educating myself about the Virial Theorem. As I mentioned, I have the flat-Earth case pretty well figured out, but I’m having trouble figuring out how to take the limit of the spherical-Earth case to the flat-Earth case to get an equivalent answer. Once I’ve done that, I’ll be convinced my spherical-Earth solution is also valid.

  Based on that, I will have isolated some points that I believe to be provably in error. My other questions are in the line of, “I don’t follow how you got this from that.”

  We get our fun where we can find it…

• 62 Jan Pompe // Jul 8, 2008 at 3:10 pm

  “as a limiting case when the Earth radius R_e goes to infinity, while g = GM/R_e^2 remains fixed. It’s not yet coming out properly.”

  What remains fixed? if G & M are constants the limiting value as R_e goes to infinity is zero. BTW the Virial theorem is = 1/2 even if he left out the sign (which I don’t think is such a big deal if all that we want to compare is the magnitude)

  “I think you should figure out how any statement about these energies can be related to IR fluxes.”

  Temperature is proportional to the average kinetic energy of a gas if there are radiatively active species in the gas they will radiate with a flux density proportional the 4th power of the temperature and the emissivity (or if you prefer the B21 coefficient for the species) it’s obvious they can be related.

• 63 Neal J. King // Jul 8, 2008 at 3:53 pm

  Jan Pompe,

  a) The limit has to be taken in the case that g = GM/R^2 and the thickness of the atmosphere, h, are both fixed. I would then expect most results to convert into the flat-Earth result.

  b) “The VT gives 1/2″: Correct, but the actual calculable answer, in the flat-Earth case, for total(KE)/total(PE) is (3/2), as can be found out from simple integration. Try it: take fixed T, and n(z) = n(0)*exp(-mgz/(kT)). Calculate total(PE) = integral(n(z)mgz) to infinity, and total(KE) = integral((3/2)*n(z)kT) to infinity. You get the same ratio for the adiabatic atmosphere, but the integration is harder.

  c) Flux relationship: Feel free to spell out in detail how to get E-sub-U = S-sub-U/2. Don’t just cite the text: Explain the reasoning. It doesn’t seem to be there. Plus, you talk about “average KE”, but M talks about the “total KE and PE”.

• 64 Neal J. King // Jul 8, 2008 at 5:39 pm

  Jan Pompe,

  re: #62, #63

  Going over a print-out, I see that in #59, I did incorrectly say that M said [KE] = (+ or -)[PE], whereas it should say
  [KE] = (+ or -)(1/2)[PE]

  So you were correct on that. Basically, a typo.

  Because, as stated in #58 and #63, the flat-Earth case gives:
  [KE] = (3/2)[PE] - which is not what one expects of the VT.

• 65 Jan Pompe // Jul 8, 2008 at 10:32 pm

  “Because, as stated in #58 and #63, the flat-Earth case gives: [KE] = (3/2)[PE] - which is not what one expects of the VT.”

  Flat earth would have us sailing or flying over the edge too I really could not be bothered with it.

  “c) Flux relationship: Feel free to spell out in detail how to get E-sub-U = S-sub-U/2. Don’t just cite the text: Explain the reasoning.”

  It’s observed.

• 66 Neal J. King // Jul 8, 2008 at 10:54 pm

  Jan Pompe,

  The reason I bother with flat-Earth (aside from being easier to clarify than spherical) is that I can’t tell whether M is using a flat-Earth or spherical-Earth model. That’s the reason I’m bothered about the minus-sign in the [KE]-[PE] equation: If I knew which he really meant, I could tell which model he was using.

  The point of M’s paper is to present a theoretical basis for an alternative understanding of the enhanced greenhouse effect. It makes sense to compare the results with observations after having conducted a valid chain of reasoning, leading from the assumptions to the results. If the results match the observations, that confirms the assumptions made in the beginning.

  But if there is no clear chain of reasoning, there’s no connection, no confirmation, and no point to the exercise.

  Example: “My theory is that there are exactly 13 solutions to the diophantine equation x^n + y^n = z^n , and that 13 is prime. Ah, but we’ve “observed” that 13 is prime! That confirms my theory!”

  Not so much.

• 67 Jan Pompe // Jul 9, 2008 at 2:26 am

  “That’s the reason I’m bothered about the minus-sign in the [KE]-[PE] equation: If I knew which he really meant, I could tell which model he was using.”

  I think you are looking for crocodiles under the bed and hoping to find them. Your flat earth model, which is something I don’t think will enter into a physicists head to use for the VT, has put you out by a factor of 3 so I’m not even interested in how you went about it. VT can only apply to spherical systems the centre of gravity does not change for spherical, planar or linear objects if you move a particle along an arc distance r from the CoG the potential energy does not change if you move a particle along a line parallel to a plane the distance from CoG and hence the PE will change.

  In his paper he does NOT actually mention the KE = -PE/2 equation directly only the S_u and E_u fluxes. Both upward and outward so the sign if you can think of it as having a sign so the only thing we are interested in is the magnitude of the relationship. It’s not an energy but a flux balance equation in which the sign is irrelevant. Perhaps you understand it better if written like E_u is proportional to || and S_u is portional to n or E_u/E_ = /n = 1/2. where n = |- sum_i(F_i.r_i)| = absolute value of the total potential energy.

  Now it seems to me you are imputing an error to Miskolczi he did not make. Now we come to a rare point of agreement with Nick:

  ” I think you should figure out how any statement about these energies can be related to IR fluxes.”

  That is the interesting part. E_u proportional to is not a problem, S_u is proportional n might be though the empirical data certainly supports it this is the interesting one to work on.

  You need to understand also that the relationship between a gas’s internal energy, kinetic energy and the rate it radiates at is well established by observation and measurement I don’t think that we are going to come up with a way to determine a priori the Einstein coefficients for each material in nature without observation and measurement. I invite you to try Nick might like to have a go too if you succeed and we can do away such libraries of data like the HITRAN, you won’t be well loved for putting all the maintainers of those databases out of work but you will be quite famous. Your use of that diophantine equation in this context is a nonsensical red herring.

• 68 Nick Stokes // Jul 9, 2008 at 2:33 am

  Neal,
  I think M is talking about a general planet. So if it wasn’t flat in the model, he’d have to specify the radius. I think you can assume it is flat.

  I’ve contended elsewhere that the virial theorem here just relates potential P-V (elastic) energy to the kinetic energy of molecules - T. Presto - PV=nRT. Along with mgh. That’s all there is.

  There’s no point in worrying about energy stores unless they actually change as a result of the processes you are considering (radiative transfer). That’s why there’s no point in including planet rotation KE (or KE of orbiting Sun, or whatever) in virial considerations.

  And indeed, not so much.

• 69 Neal J. King // Jul 9, 2008 at 10:19 am

  #67, Jan Pompe; #68, Nick Stokes:

  - The VT certainly applies for a flat-Earth model, but not in the form that is usually quoted: 2*[KE] = n*[PE], where PE is proportional to r^n. The VT applies as long as the system is bounded in space and momentum, it need not be spherical; but the results differ from the simplest cases, when there are imposed boundaries, as in these cases.

  - As Nick points out, it is far from clear whether M is talking about a flat Earth or a spherical. Further, the flat-Earth model is easy to understand and calculate; and the spherical-Earth model must reduce to the same result in the appropriate limits. If it does not, something is wrong. I am working on that now, and actually I am making progress: It’s getting clearer as to how a
  “-[PE]_se” term is turning into a “+[PE]_fe” term, with the change of zero-point of V. I am still working on some of the other terms, but I think they will work out. Dollars gets you donuts: It will not result in 2*[KE] = -[PE].

  - The point about moving a particle relates to another useful fact: The value of the virial must be invariant under a change of origin, because a change of origin doesn’t affect the [KE]. This turns out to imply that the total force acting on the system must vanish: [F_i] = 0. This gives an additional equation in both cases, although in the flat-Earth case it’s more straight-forward: It leads to the conclusion that
  P(z=0) = g*M_atmosphere/area
  In the spherical case, I am still sorting it out, because it entails integrals over the boundaries. Of course, if you integrate over the entire sphere, everything vanishes by symmetry. But if you integrate over half the sphere (for simplicity), it still must be true that the total force vanishes, but now you get boundary terms that are constrained. (In the flat-Earth case, the boundary terms were proportional to [KE], so that’s why you get the factor of (3/2). In the spherical case, it doesn’t look so neat - yet.)
  - Nick, application of the VT is not quite as unrewarding as you indicate. It’s true that you can use it to get PV = nKT; but you can also use the perfect-gas law in the context of VT to get the result I mentioned before: [KE]/[PE] = (3/2) in the flat-Earth case. I first found this out the hard way, by integrating the adiabatic atmosphere; and after I stumbled across this in the VT calculation, I verified it with the easier exponential atmosphere. But it’s not a self-evident result, in my opinion.

• 70 Neal J. King // Jul 12, 2008 at 10:41 am

  OK, I’ve finally finished the calculations. It took some time, because I had a persistent “-” sign to clear up. A good thing, too, because it was hiding a more subtle error! Anyway, here are the results:

  For the atmosphere as a spherical shell, the application of the general Virial Theorem,

  2[KE] = [-r.F] ,

  yields:
  [KE] = -(1/2)[U_s] - (2.pi.R^3).Pressure(r=R)

  This is clearly different from what one would expect using the VT naively:
  “[KE] = -(1/2)[U_s]” (where U_s = the gravitational (1/r) potential)

  Taking the limit that the height of the atmosphere, h, is small relative to Earth’s radius, R, my result leads to:
  [KE] = (3/2)[-U_s] - (6GMmR)*Integral(number_density(r)dr

  Noting that g = GM/R^2 is the local gravitational acceleration, this result can also be re-expressed in terms of the molecule’s
  height = r-R as:
  [KE] = (3/2)[mgz] = (3/2)[U_f] ,

  which is exactly what I got in the flat-Earth case, and verified by integrating the constant-temperature and the adiabatic atmospheres. Hurrah!

  (The tricky part of the calculation was showing that
  (3/2)[-U_s] - (6GMmR)*Integral(number_density(r)dr => (3/2)[U_f]
  when h/r => 0 while GM/R^2 = g.)

• 71 sadunkal // Jul 17, 2008 at 12:48 am

  So what is it then?
  Is IPCC more correct than Miskolczi is? Were the models used right all along?
  Don’t make me study climatology and mathematics, just tell it in normal words please…

• 72 Neal J. King // Jul 17, 2008 at 10:40 am

  sadunkai,

  The case for the IPCC and for Miskolczi’s paper are not comparable:
  - The IPCC have reviewed and organized a tremendous amount of work by thousands of scientists to create an understanding based on many parallel lines of evidence and reasoning, and constrained by all aspects of atmospheric physics.

  - Miskolczi has created one line of reasoning, that has to be understood on its own. His colleague Zagoni has provided commentary that purports to overview this paper, but I haven’t found it useful in understanding the detailed reasoning and mathematics in the actual paper.

  - My current conclusion is that there assumptions made in the paper which are not correct; and there are logical/mathematical steps that, at very least, need much more explicit explanation (and could be simply wrong). These are in the early part of the paper, so I haven’t felt it useful to study carefully the reasoning on the later part of the paper: Why be concerned with a fortress built on sand?

  - I have sent my detailed discussion to Dr. Miskolczi, who says he may be able to get to it in August (he’s on travel at present). If anyone else wants to look at it, I’m willing to provide it; but it’s reasonably mathematical.

  My bet: I think the paper is incorrect. But I’m willing to be convinced otherwise, if someone can “fill in the gaps”.

• 73 sadunkal // Jul 26, 2008 at 4:53 pm

  I disagree with your first point. As many advantages as it may have to be a big organization, it also brings great difficulties to reach a scientifically correct conclusion. Check Richard Feynman’s Challenger Investigation for example. And that was years ago, today science is even more far away from being pure. The political and financial influences are bigger than ever.

  If Miskolczi is right, then he is right. It doesn’t matter if he reached his conclusions without tremendous help from others. And if you think he’s wrong then it’s good that you really contact him to discuss about it instead of being confident in your own conclusions and telling everybody that he is wrong.

  Anyway I’m excited for his response to you then… It wouldn’t be of any use to me to read your arguments, but I hope that the entire discussion will be made easily available for others to examine. This is an important matter after all…

  And a little suggestion: Maybe studying the rest of the paper carefully might give you a new perspective on the early part. I wouldn’t count on it but it’s not impossible I suppose…

• 74 Neal J. King // Jul 26, 2008 at 8:48 pm

  sandunkal,

  - My point about the IPCC is not that it’s got a lot of people contributing to it, but that the whole field of climate science, and all the conclusions and methods applied to it, are taken into account. On that basis, they find many ways of estimating temperature trends, many ways of trying to evaluate the impact of molecular species in the atmosphere, etc. The greenhouse effect and the general view of the role of CO2 fits into all this pretty well.

  What Feynman’s story on the Challenger shows is that when an engineering organization places meeting its goals above evaluations that point to problems, you can expect the chickens to come home to roost eventually. But the IPCC are not trying to meet goals, they are trying to evaluate what is going on. Feynman is no longer around to ask about climate science. However, as someone who had personal intellectual contact with him for an academic year, I think he would evaluate their effort as sincere and focused upon the evidence, not upon the consequences.

  As mentioned before, I did send my detailed comments to Miskolczi, and he did say he might get back to me in August. I also sent a copy to David, our host, but I don’t know what he did with it.

  And I have certainly looked at the remainder of the paper, but I can’t make much of it.

• 75 Raven // Jul 26, 2008 at 9:02 pm

  Neal J. King says:
  “But the IPCC are not trying to meet goals, they are trying to evaluate what is going on. ”

  The IPCC is a political organization that exists to demonstrate that AGW is a problem which requires government intervention. It is a mistake to assume that the IPCC scientific conclusions are not biased by its desire to justify its existance.

• 76 sadunkal // Jul 27, 2008 at 2:11 am

  I tend to agree with Raven and I suspect that Feynman would also do so. He always disliked politics anyway…

• 77 Neal J. King // Jul 27, 2008 at 9:41 am

  Raven,

  Please feel free to point out the errors in the IPCC report: http://ipcc-wg1.ucar.edu/wg1/wg1-report.html

  sadunkal,

  Feynman’s attitude was, “Do the science, and the hell with who it upsets.” He would definitely have gone up in flames at the Bush administration’s “editing” of the NASA report on the implications of global warming: He talked about that sort of thing as being a very strong reason he would never work for a governmental body. (This was prior to the Challenger issue, which was a very special case.)

  I had many interesting discussions with Feynman over a period of a year. We never discussed global warming: that issue had not emerged as a science-related controversy of general interest.

• 78 sadunkal // Jul 27, 2008 at 12:23 pm

  About IPCC:
  http://climatesci.org/2007/06/20/documentation-of-ipcc-wg1-bias-by-roger-a-pielke-sr-and-dallas-staley-part-i/
  http://climatesci.org/2007/06/.....re-trends/

  I agree with what you said about Feynman. But opposing big oil Bush doesn’t mean supporting IPCC.

• 79 Neal J. King // Jul 27, 2008 at 1:57 pm

  sadunkal,

  As far as I know, no one asked Feynman what he thought about the global-warming question. And no one is in a position to state what his would be.

  But I have spent a certain amount of time reading textbooks on atmospheric physics, and their discussions of radiative transfer, and so on. There are lots of complexities in the studies, and the modeling is very complicated. But the basic principles that they are working from are not that complicated or obscure. My definite impression is of people who are honestly trying to grapple with complexity, not of people who are trying to “avoid experiment” (another one of Feynman’s dislikes).

  And the issue I have with the Bush administration on this point is not whether it is pro or con “big oil”: The issue is that a political appointee who has not even graduated from college should not be editing NASA reports to downplay the consequences of global warming; and even to downplay the implications of recent findings on Big-Bang cosmology (”because it could offend the religious sensitivities of some groups”).

  On this last point, I am really sorry that he was not around for this outrage: I am sure he would have gone through the roof on that, and his opinion would have been heard. As it was, it was a one-day or two-day story: So far from being “plausible” that no one remembers it.

• 80 sadunkal // Jul 27, 2008 at 4:26 pm

  I’d argue that the basic principles used for modeling should be a lot more complicated if the models are supposed to make really correct global predictions. Some reasons can be found here:
  http://climatesci.org/2007/05/23/a-short-summary-of-why-skillful-climate-prediction-is-much-more-difficult-than-skillful-weather-prediction/

  Also you might be interested to know what Freeeman Dyson, an old physicist friend of Feynman, has to say about the current climate models:
  http://www.youtube.com/watch?v=JTSxubKfTBU

• 81 Raven // Jul 27, 2008 at 5:00 pm

  Neal,

  One can be biased without being technically wrong. For example, I could truthfully say that dihydrogen oxide is an output of almost all industrial processes and that it can be lethal to humans. Such a statement would likely get most people worried about this ‘dihydrogen oxide’ stuff - until they figure out I am talking about water.

  By playing down or ignoring non-GHGs influences on climate the IPCC has committed numerous errors of omission.

• 82 Neal J. King // Jul 27, 2008 at 5:22 pm

  Raven,

  As I suggested before, you are welcome to point out the IPCC’s errors of commission or omission.

• 83 Neal J. King // Jul 27, 2008 at 5:33 pm

  sandunkal:

  1) Which is easier to calculate, the average takings of gamblers in a casino, or the takings on 31 July 2008?

  2) Dyson doesn’t challenge the basic physics behind the GW issue (enhanced greenhouse effect due to increased CO2), he argues that:
  a) Stratospheric cooling, which results from it, will take effect sooner and be more serious, as it depletes the ozone layer above the Arctic.
  b) Plants can be “designed” to suck CO2 out of the atmosphere faster.

  For b): No one’s proposed any particular method yet. In the meantime, I would promote the protection and encouragement of rain forests.

  I think Dyson might be surprised to find that his words are interpreted as a reason to forget about the CO2 issue: People like to hear his “GCMs are not to be trusted” message, but somehow don’t hear his “stratospheric cooling is happening sooner, more measurably, and is more serious” message. Frankly, I think he’s a bit naive: He wonders why the public gets excited about “global warming” but not about “stratospheric cooling”. Maybe this is only something that a non-”rocket scientist”, who doesn’t live at the Institute for Advanced Study, can understand.

• 84 Raven // Jul 27, 2008 at 5:53 pm

  Neal,

  The IPCC has largely ignored the effect of natural variability. For example, the attribution studies suggest that the planet would have cooled over the last 30 years without CO2 induced warming despite the fact that the PDO was in its warm phase.

  Further evidence for this blindness can be found by comparing the model distributions used for the attribution studies to the model distributions used to explain the 10 years of non-warming. In the former case the model distributions are very narrow and suggest that random variations could not explain the warming from 1980-2000. In the latter case the distributions are wide enough to allow the modellers to claim that cooling is consistent with the model predictions of 2 degC/century warming.

• 85 Sadun Kal // Jul 27, 2008 at 6:32 pm

  @83 Neal,

  1)I suggest that you read the comments and discuss that over there. I think our solar system is more complicated than a casino and I think that plays an important role.

  2) I also think that he’s a bit naive but he raises some important issues about the models and I see him as a more credible person than Al Gore is. The stratospheric cooling stuff shouldn’t be an excuse to keep on following the IPCC as it is now. If the greenhouse-science could lose some attention, then the energy flowing into it can be shifted to understanding the stratospheric cooling better.

  By the way, he signs petitions/letters against the GW alarmism, so I don’t think he would be that surprised.

• 86 Neal J. King // Jul 27, 2008 at 10:39 pm

  Raven,

  There are a lot of dynamics that complicate the exact behavior. But the basic argument for AGW is pretty simple:
  1) Increased CO2 gives rise to an enhanced greenhouse effect
  2) For each doubling of CO2 in the atmosphere, there is an addition 3.7 Watts/m^2 of radiative forcing: imbalance between incoming solar radiation absorbed and outgoing infrared radiation that escapes to space.
  3) With that imbalance, SOMETHING has got to happen. Roughly, one would expect warming overall. But it doesn’t have to be (and won’t be) monotonic or uniform over the Earth at one time. So the differential effects will create weather events, both hot and cold, which are due to moving air and water masses.

  As I see it, that’s the basic story. I haven’t seen anything that poses a serious challenge to that picture. It is complicated by the feedback loops: oceanic heating, increased water vapor, cloud formation, etc.

  With regards to the detailed attribution studies: You can only compare the different studies for the different time periods directly when the parameters used are the same. However, if there is a serious difference between the basic story for the two studies, when the parameters are changed, then I’m sure some graduate student or postdoc will be all over it, trying to get his shot at fame & glory.

  Otherwise, if you want me to try to respond in detail, it would be appropriate to provide a link to a specific study.

• 87 Neal J. King // Jul 27, 2008 at 11:01 pm

  Sadun Kal,

  1) I’m not hunting for new battlefields: There are endless forums on the internet.

  2) It doesn’t make sense to focus on stratospheric cooling while downplaying the greenhouse effect: This is like emphasizing the “tail” side of the coin in favor of the “head” side. They are part of each other: The reason the stratosphere cools is because the infrared escaping from the troposphere is reduced, and this is exactly the greenhouse effect!

  For the very same reason, it is hopelessly naive of Dyson to decry “GW alarmism” while worrying about the danger of stratospheric cooling: This is like someone who attacks a forest-fire prevention campaign focused on protecting trees, because you think it should be motivated by protecting animals - forgetting that the forest fires will burn both! I suspect one of two things:
  a) The signature on the anti-alarmism list precedes his present concern for stratospheric cooling; or
  b) He thinks that the ultimate extent of the global-warming effect will not be as great as some people think. This is possible; although all that he can say, as a non-expert in the field, is that the uncertainties are perhaps larger than people generally realize. But in that case, the effects may equally well be larger than expected, as happened with respect to the melting of Arctic ice as recently as the year or so.

  There is also a less charitable interpretation: Murray Gell-Mann (a truly great elementary-particle physicist, both a collaborator and a worthy competitor of Feynman’s) made a snide remark about a certain type of British scientist: That they would rather be clever than right. Dyson has certainly been clever all his life, but a number of his ideas have struck others as being far from right: Project Orion (space travel propelled by nuclear bombs), possible origin of life in molecules of clay, etc. He’s certainly gotten a lot of mileage out of being an iconoclast. Why should he switch to “mainstream” mode now? But it doesn’t mean that he’s right.

• 88 Raven // Jul 28, 2008 at 1:01 am

  Neal says:
  “With that imbalance, SOMETHING has got to happen. Roughly, one would expect warming overall. But it doesn’t have to be (and won’t be) monotonic or uniform over the Earth at one time.”

  The basic physics of GHGs are not in dispute and even skeptics agree that adding CO2 to the atmosphere will result in some warming. However, the problem with the IPCC projections is they go beyond the basic radiative physics and assume that water will amplify the CO2 induced warming by a factor of 3 or more. IOW, you can’t defend the IPCC claims by reducing them down to the basic physics - the problem is whether the climate models accurately represent the real atmosphere and I have seed no convincing evidence that they do.

  The IPCC used a subset of the models for the attribution studies which resulted in a much narrowed range. We don’t know whether the subset choose was accidental or deliberate. But we do know that the climate modellers change their estimates of weather noise depending on what they want to show. If they want to show that the CO2 effect is larger than the weather noise then they use a small number of models that produce little noise. If they need to explain away a decade long cooling trend they use many models that produce a lot of noise.

  Roy Spencer’s latest paper tackles the assumptions built into the models and demonstrates that random internal variations can produce decadal trends even if there is no external forcing. His research suggusts that the true sensitivity of CO2 is much closer to what would be predicted with the basic physics.

• 89 Pat Cassen // Jul 28, 2008 at 2:21 am

  Raven –”Roy Spencer’s latest paper…”

  Do you have a link? I don’t seem to get a paper on a google search.

• 90 admin // Jul 28, 2008 at 3:48 am

  Pat, You can get a preprint from here:

  http://landshape.org/stats/pot.....iagnosis1/

• 91 Pat Cassen // Jul 28, 2008 at 5:05 am

  Thanks, David. Got it.

• 92 Jan Pompe // Jul 28, 2008 at 9:45 am

  “With that imbalance, SOMETHING has got to happen. Roughly, one would expect warming overall”

  Indeed and that will lead to increased clouds due to extra water vapour thus increasing albedo and reducing incoming thereby restoring the balance.

  Isn’t wonderful how negative feedback works. Nevertheless an increased concentration of absorbers is also an increase in concentration of emitters so it’s difficult to see that such an increase will cause an imbalance in the first place.

• 93 Neal J. King // Jul 28, 2008 at 10:08 am

  Jan Pompe,

  - What you promote is indeed Lindzen’s hopeful “iris effect”, which Spencer also promotes. But the theoretical justification for it from Lindzen seems to be “It is possible that…”; from Spencer it is “I believe that…”; and from Spencer’s claimed observational support is “not incompatible with…” In other words, it’s not really more than a much-hoped-for alternative scenario. No one has even the calculations to show that it is LIKELY to turn out that way.

  - Your suggestion that “an increased concentration of absorbers is also an increase in concentration of emitters” is based on an incorrect understanding of the greenhouse effect, which is not based on “absorbing” IR photons, but on radiative transfer. If you want, I can try to give a pocket summary, but I point out that even skeptics like Lindzen and Spencer, who are still professional climatologists, don’t question this basic mechanism. I believe that is why your comrade-in-arms, Raven, admitted as much in #88: “The basic physics of GHGs are not in dispute and even skeptics agree that adding CO2 to the atmosphere will result in some warming.”

• 94 Neal J. King // Jul 28, 2008 at 10:46 am

  Raven,

  As far as I can tell, the preprint of Spencer & Braswell indicates the possibility of bias, but don’t provide a quantitative correction factor or decrement. In other words, it seems to amount to: “Due to certain issues not previously taken into explicit account, the numbers could be somewhat larger than they should be.” But it doesn’t seem to give an estimate of what S & B would expect to be an accurate result.

  In any event, comprehension of the entire discussion depends on having some background understanding of the literature which they review: It’s not a straightforward calculation (as Miskolczi’s article partially is), so to fully appreciate the worthiness of their arguments and comparisons would, I believe, require good familiarity with the methods and results they cite: Forster & Gregory; Monterey & Levitus; Hasselman; Stephens; and likely others well-known to the intended audience.

  As this is a preprint, I assume the paper is currently undergoing the peer-review process. Until it has successfully survived that process, it seems to me to be premature for amateurs, like myself, to try to take its considerations into account.

• 95 sadunkal // Jul 28, 2008 at 12:13 pm

  I tried send this yesterday but I had a technical problem, it still seems relevant so here you go:
  http://www.nasa.gov/centers/goddard/news/topstory/2003/0530earthgreen.html

  Things like this are the reason why I think the models should be much more complicated than they currently are.

  Neal,

  I don’t want to keep talking about Dyson, you might be right about his “style”, but I just want to point out something:
  #87,
  1)You shouldn’t see it as a battlefield: People are discussing things to get closer to the truth. You can learn from others and test your own ideas against them. And that particular website is one of the more decent ones.
  2)You’re right. I should have said model-based warming-prediction-”science” instead of greenhouse.
  a)Possible.
  b) I think the probability isn’t equal for things to proceed normally or get dramatically worse. The second scenario can be considered an anomally and is more unlikely. And you might want to check this out about the arctic:
  http://strata-sphere.com/blog/index.php/archives/5589
  http://dotearth.blogs.nytimes......c-sea-ice/

  I also think that Dyson’s thoughts about the models are probably more right than the stratospheric cooling. Simply because one is based on his past experience and the other is not.

• 96 sadunkal // Jul 28, 2008 at 12:15 pm

  I tried send this yesterday but I had a technical problem, it still seems relevant so here you go:
  http://www.nasa.gov/centers/goddard/news/topstory/2003/0530earthgreen.html

  Things like this are the reason why I think the models should be much more complicated than they currently are.

  Neal,

  I don’t want to keep talking about Dyson, you might be right about his “style”, but I just want to point out something:
  #87,
  1)You shouldn’t see it as a battlefield: People are discussing things to get closer to the truth. You can learn from others and test your own ideas against them. And that particular website is one of the more decent ones.
  2)You’re right. I should have said model-based warming-prediction-”science” instead of greenhouse.
  a)Possible.
  b) I think the probability isn’t equal for things to proceed normally or get dramatically worse. The second scenario can be considered an anomally and is more unlikely.

  (And you might want to check this out about the arctic:
  http://strata-sphere.com/blog/.....hives/5589)

  I also think that Dyson’s thoughts about the models are probably more right than the stratospheric cooling. Simply because one is based on his past experience and the other is not.

• 97 sadunkal // Jul 28, 2008 at 12:27 pm

  Here you can read a lot of opinions about the volcanoes under arctic:
  http://dotearth.blogs.nytimes.com/2008/07/01/whats-up-with-volcanoes-under-arctic-sea-ice/

  And here’s another version of the story:
  http://www.canada.com/topics/n.....8c0dc90304

  You can read more about that in the journal Nature apparently. Couldn’t include these in the previous comment.

• 98 Jan Pompe // Jul 28, 2008 at 12:33 pm

  “What you promote is indeed Lindzen’s hopeful “iris effect”,”

  On the contrary the positive feedback effect is the hopeful one that has been put forward to explain away the long CO2 lag to temperature change in both directions. Fact of the matter the evidence in the paleoclimate record is that global temperatures track insolation variation due to orbital factors very well and have done so for 1/2 billion years at least. This is a very unlikely scenario where there is net positive feedback.

  “Your suggestion that “an increased concentration of absorbers is also an increase in concentration of emitters” is based on an incorrect understanding of the greenhouse effect, which is not based on “absorbing” IR photons, but on radiative transfer. ”

  Yes half the radiative transfer the absorption half. In Taking Greenhouse Warming Seriously he comments that the emission bit is conveniently ignored in some models. I think he’s being polite I haven’t seen it effectively taken into account in any except the one we are discussing here.

• 99 Neal J. King // Jul 28, 2008 at 7:27 pm

  #96, sadunkal:

  1st link: I don’t get the connection between enhanced plant growth and complexity of GCMs.

  2nd link: In order for the hot water to melt the ice, it would have to get to it, or warm up all the water in between. I recall seeing someone at RealClimate examine some temperature profiles, and concluding that this wasn’t happening.

  1) Yes, Pielke’s site is better than most, although there is a funny juxtaposition of high-level expertise with what I would consider borderline-unprofessional snarling by some of the guest webloggers. I guess that happens to some extent at RealClimate, as well, although the actions they lambaste tend to be more focused on information-manipulation than on what I would consider differences in technical point of view.
  2) Stratospheric cooling: It’s not clear to me what else needs to be studied about stratospheric cooling. If you believe Dyson, it’s a clear implication of excess CO2, so the effort should be moved directly to CO2 reduction. I can second that! But you will see that that is entirely not a climate-science problem: It’s involved energy production, or, if Dyson’s goal works out, high-intensity horticultural engineering. It’s not the same people or the same pot of money. (By the way, this is as good a place as any to point out that, if the climate scientists were really trying to live “high on the hog” with all this funding, they wouldn’t be saying, as most of them are, that the jury is back and the verdict is in: They would be saying, “We need to study this more, we don’t know if it’s a problem or not.” But instead, the mainstream of climate scientists is saying, “The problem is real, we have to go fix it.” If that advice is taken to heart, that moves money into the solutions area - which is not climate science. So it is actually not in a (selfish) climate scientist’s interest to say that the problem has been identified as CO2. So, do they not get that? Or could it be that they actually believe the results of their work, and they’re not just going along to get another year of funding?

  wrt Dyson’s two attitudes about modeling vs. stratospheric cooling: If you don’t trust Dyson’s scientific judgment, why bother citing him at all? He’s not a real expert in either area: He’s a dilettante - a really really good dilettante, but still a dilettante. If you want to pick just the opinion of his that you like, then who is making the judgment: Him or you? And if it’s you, why mention Dyson?

• 100 Neal J. King // Jul 28, 2008 at 7:34 pm

  #96, sadunkal:

  Interestingly, both the articles you cite concerning the under-sea volcanoes dismiss the idea that they could be responsible for melting icebergs, the idea that your previous citation was promoting:

  - from http://www.canada.com/topics/news/story.html?id=81bb2fd3-63f1-476f-b0be-f48c0dc90304 :
  “The scientists say the heat released by the explosions is not contributing to the melting of the Arctic ice, but Sohn says the huge volumes of CO2 gas that belched out of the undersea volcanoes likely contributed to rising concentrations of greenhouse gases in the atmosphere. How much, he couldn’t say.”

  - http://dotearth.blogs.nytimes......c-sea-ice/ :
  “There was an eruption of assertions in recent days that the increasing summer retreats and thinning of Arctic Ocean sea ice might be a result not of atmospheric warming but instead all the heat from the recent discovered volcanoes peppering the Gakkel Ridge, one of the seams in the deep seabed at the top of the world.

  Several experts said it was not plausible from the get-go, but for the sake of due diligence, I queried a heap of the Arctic oceanographers and climate and ice experts I’ve gotten to know since my North Pole journey in 2003. They uniformly reject the idea that heat from the bottom — either from the general geothermal activity beneath the seabed or the occasional outbursts of lava or vents — could have a significant impact on the veneer of floating, drifting ice on the surface.

  The deep saltier, warmer water is largely isolated from the cold, fresher waters near the surface, they say.”

• 101 Neal J. King // Jul 28, 2008 at 8:31 pm

  #97, Jan Pompe:

  - Net positive feedback does not mean that the thermometer will forever be pegged at max. Over periods of about 1000 years, CO2 is removed from the atmosphere by biological and, to a lesser extent, chemical processes.

  - The feedbacks identified by the studies cited by the IPCC are specific mechanisms, the magnitudes of which can be estimated from the identification and analysis of the mechanism. It would indeed be remarkable if all of these were to somehow cancel out.

  - Whereas Lindzen’s mechanism, last I checked, has not been estimated: It is essentially postulated to have the magnitude necessary to tamp down GW. That’s why I describe it as “hopeful”.

  - Spencer’s argument, in a several-page article of his critiquing the mainstream view: “I believe that all the feedbacks will ultimately cancel out and stabilize the Earth’s temperature.” That was the whole argument!: “I believe“.

  - Good discussions of the enhanced greenhouse effect are hard to find. But I have located one that is reasonably succinct, although still mathematical in nature: below Figure 3.6, page 111 of http://geosci.uchicago.edu/~rt.....teVol1.pdf

  To try to summarize it in a nutshell:
  a) In the troposphere, the temperature profile is essentially set by convection. This implies that the temperature will fall roughly linearly with altitude (starting from ground temperature).

  b) As infrared radiation in the 15-micron range passes from the surface of the Earth up through the troposphere, it interacts with the CO2. As this “beam” travels through cooler and cooler gas, its intensity changes to reflect the temperature it is passing through, although with a kind of “delay”: The intensity reflects the blackbody radiation at that frequency for the temperature that is farther back by optical-depth = 1.

  c) So when the “beam” eventually breaks free of gas, its intensity reflects the temperature corresponding to the point optical-depth = 1 back from the edge of the atmosphere: the photosphere for 15-micron IR.

  d) So the total amount of CO2 in the atmosphere determines the location of the 15-micron photosphere.

  e) Therefore, if the amount of CO2 in the atmosphere increases, the location of the photosphere (where optical-depth = 1, as measured inward from space) moves upward in altitude, and thus the temperature goes down.

  f) Since the temperature of the new location of the photosphere is lower, the intensity of the 15-micron radiation is also lower.

  g) Thus, there is less 15-micron radiation leaving the Earth. But the amount of incoming solar radiation remains unchanged: This is the radiative imbalance that is called the enhanced greenhouse effect.

  h) This imbalance will remain until the ground temperature increases, and thus the temperature profile all the way up to the top of the troposphere incrementally increases, until the new temperature at the position of the new photosphere is the same as the old temperature of the old photosphere (before the additional CO2).

  i) The reduction in outgoing 15-micron radiation from the photosphere is why the stratosphere, just above it, cools: It’s getting less radiation from below.

  This description is considerably more complicated than the high-school version of the story, that “CO2 absorbs IR”; you can see why it’s a story seldomtold. Even so, I’ve left out a number of points, which I think are not necessary for the basic idea: However, a careful thinker might find points that need further amplification or qualification, and which might even require a considerable expansion of the scope of discussion.

  So if we have to discuss this further, I don’t promise not to bring up other considerations.

• 102 Sadun Kal // Jul 28, 2008 at 9:02 pm

  #98
  I’m just inclined to think that since the vegetation effects the CO2 traffic and since warming seems to increase vegetation… Well you know what I mean… But maybe they’re already included in the models, I don’t know…

  2)I don’t know anything at all about stratospheric cooling. But if it’s that significant then it should be thoroughly discussed. I didn’t hear Al Gore talking about it. If warming turns out to be mostly natural and not much to worry about and if the problem would turn out to be only this cooling then the strategy to deal with the problem would change dramatically.

  (By the way I don’t fully agree with what you said inside the brackets. I don’t think it’s that simple: Politics play a bigger role in my opininon and many people are easily seduced by short-term benefits of their actions.)

  About Dyson’s views, I’m not picking anything, just talking about probabilities. If you disagree with my conclusion then say so.
  My trust in any scientist is limited, I cited him because he was relevant to our dialog. And note that Feynman would also be a “dilettante”… also Einstein, Sagan whoever…

  #99

  I know.
  It was to present a somewhat balanced view. Here’s another one:
  http://www.whoi.edu/page.do?pi.....amp;ct=162

  I don’t know enough about oceans or physics to determine who’s right and I’m not that interested to read all that stuff anyway. I just thought that somebody else would benefit from those sources if they want to examine the case. If I were actively contributing to the climate science I would definetely want to know for sure before trusting those “experts” or someone at RealClimate.
  But what I intuitively can say is that it’s hard to believe massive volcanoes under the arctic have less effect on the ice than the greenhouse gases. And it’s also important to note that this story wasn’t seriously discussed in the media…

• 103 Raven // Jul 28, 2008 at 10:15 pm

  Neal,

  The IPCC scientific claims involve a number of steps and getting any one of them wrong can invalidate their conclusions even if the other steps were done correctly. For example, the calculations that show that increasing CO2 will cause an increase in radiative forcing is based on well understood physics and most likely correct. However, the calculations that translate an increase in radiative forcing due to CO2 into an increase in surface temperature do not appear to have any theoretical derivation and are largely guesstimates based on analysis of paleo records and climate model outputs.

  The lack of any theoretical derivation for CO2 sensitivity is why the IPCC has resorted to attribution studies where they use models to demonstrate that the recent warming can only be explained if the CO2 is the only driver and that their sensitivity guesstimates are correct. However, this is an extremely weak argument because it can be refuted by showing that other mechanisms could explain some of the warming over the last 30 years. That is why the work of Roy Spencer on clouds and Peilke Sr. on land use changes is so significant because they both demonstrate that other plausible mechanisms exist and have been operating in the last few decades. If any portion of the warming can be attributed to these mechanisms it will have the effect of reducing the net CO2 sensitivity of the climate system.

  The work of Leif Svalgaard has also cast doubt on the IPCC’s CO2 sensitivity guesstimates because the IPCC uses the sun to explain changes in temperature before human emitted CO2 became a factor. However, Leif’s latest research has shown that the sun’s TSI levels have not varied much at all over the last 1000 years or so which means that the sun is either affecting the climate in ways other than TSI or that it is hypersensitive to changes in TSI. It is hard to predict what effect this change will have on the CO2 sensitivity guesstimates, however, Leif is apparently quite disappointed that the AGW supporters refuse to revisit their previous conclusions (probably because they realize it would open up a can of worms wrt CO2 sensitivity).

  I realize that many supporters of AGW have tried to reverse the burden of proof and insist that it is up to the skeptics to conclusively demonstrate that the IPCC guesstimates are wrong. I disagree with that position for two reasons: First, the temperature trend over the last ten years is inconsistent with the IPCC’s CO2 sensitivity guesstimates even if one takes into account weather noise. Second, the cost of acting on GHGs is huge so we can’t afford to invest trillions on a guesstimates that could be wrong.

  Of course, uncertainity does not mean policy makers do nothing – it just means that policies much be adopted that would make sense even if CO2 sensitivity turns out to be lower than what the IPCC has claimed.

• 104 Neal J. King // Jul 28, 2008 at 11:29 pm

  #101, sadunkal:

  - Dyson: My point is that if you cite him as an authority, you are assuming he is authoritative. But then you have to treat him as an authority, not pick & choose. For reasons I have already discussed, I don’t regard him as an authority.

  - Comparison of Feynman with Dyson: There’s a big difference. Feynman played at a lot of different things (Mayan mathematics, biology, hexaflexagons) but when he was doing physics, he was not a dilettante, he was a polymath: He became a master of the area he was focusing on: quantum electrodynamics, superfluidity, general relativity, quantum gravity, the weak interaction. He was freaking amazing. By contrast, Dyson’s work has been scattered around different areas of science, and the depth of his contribution has been limited (by comparison to Feynman’s) except for quantum electrodynamics, where he really contributed in a big way. He knew Feynman well, and he would not begin to compare himself to him (and I know that from a personal conversation with Dyson).

  - But even the third article you post is entirely negative on the idea that underwater volcanic eruptions could be responsible for the melting of the ice-caps. When all the experts seem to be saying no, why should you be surprised that the media is not pushing “yes”?

• 105 Mike N // Jul 28, 2008 at 11:32 pm

  Neal (100), Regarding your “i)”, are you sure this is the case? I’d been under the impression that reduced 03 (less UV absorption) was considered to be the likely dominant effect in lower stratospheric cooling, and the dominant effect likely causing cooling in the upper stratosphere as being either increased H2O or CO2 (as well as other trace GHG’s)?

  Do you know of any freely available, recent (Clough and Iacano ‘95 LBL type) research online? I’m not sure, but I seem to remember C&I95 held a static strat. H2O concentration, while I *think* H2O concentrations in the strat. have risen since then, which should obviously affect the cooling rate.

  Anyway, sorry about being off topic

• 106 Neal J. King // Jul 28, 2008 at 11:58 pm

  #102, Raven:

  - Using palaeo-proxy records is still based on empirical facts, and so still tied to something measurable. It would be awfully nice to have a “first principles” full-on estimate - but it’s a lot more than what Lindzen or Spencer have for their hopeful canceling-out.

  - I’m not familiar with Svalgaard’s work. What I recall from looking at TSI studies a year or so ago was that, although TSI was shown to be linked to temperature hundreds and thousands of years ago, that linkage had essentially vanished in the last 100 years.

  - I don’t know where you get the impression that the IPCC report assumes or concludes that CO2 is the only driver for climate change. That sounds extremely far off the mark.

  - Didn’t you mention the PDO as a multi-decade oscillation? Given phenomena such as this, expecting to extract any kind of trend over 10 years is hopeless.

  - What we have is uncertainty, and the situation could be worse, as well as better, than expected. Lomborg (who does not favor action for AGW) nonetheless did an estimate of the cost of fixing the CO2 problem, at a few trillion$ over the rest of the century. He evaluated it as putting back the world’s development (global product) by one year by the year 2100: one year out of 92.

• 107 Jan Pompe // Jul 29, 2008 at 12:21 am

  Neal,

  “Net positive feedback does not mean that the thermometer will forever be pegged at max. ”

  Tell me Neal where do I make a claim that it does?

  Strawman arguments being made of straw don’t carry much weight.

  ” It would indeed be remarkable if all of these were to somehow cancel out.”

  Given the paleoclimate record it would be remarkable if they didn’t.

  “It is essentially postulated to have the magnitude necessary to tamp down GW. ”

  Was it that or was it an attempt to explain the general stability of the climate system over the pas million years. Again it’s straw.

  “That was the whole argument!: “I believe“.”

  Yes he allows himself to be persuaded by the evidence. Funny that, I like to do the same, how about you?

  Really Neal I am quite familiar with the story seldom told. As it turns out the story seldom told leads to a theoretical temperature discontinuity at the surface which is in fact not observed so the story seldom told must have at least one error. Again I allow myself to be persuaded by the evidence, how about you?

• 108 Neal J. King // Jul 29, 2008 at 12:55 am

  #104, Mike N.:

  I’m pretty sure that the reduction of outgoing IR due to increased CO2 contributes to stratospheric cooling, but I don’t claim that that is the only reason that it is cooling: I think I’ve also heard that ozone depletion is a bigger effect. It comes up in this context because Dyson refers to this point in his video.

  When you refer to increased H2O and CO2 and other GHGs, these should all have the similar effect:
  adding more gas
  => pushes up the photosphere (for the relevant wavelengths)
  => cools the surface of the relevant photosphere
  => reduces the outgoing radiation from troposphere
  => reduces the radiation heating the stratosphere from below

  The in-progress textbook by Pierrehumbert is the only thing I know that’s free:
  http://geosci.uchicago.edu/~rtp1/ClimateBook/ClimateVol1.pdf

  I got my general understanding of the radiative-transfer picture from John Houghton’s The Physics of Atmospheres: disappointingly, even Weart’s site at AIP (The Discovery of Global Warming) doesn’t even try as hard as I did above to explain the story.

  Something that does puzzle and irritate me: Nobody seems to consider it at all controversial that a doubling of CO2 will lead to a radiative imbalance of 3.7 Watts/m^2. However, the only calculation that anyone cites for this is from about 1979 (Ramanathan, Lian and Cess), and it doesn’t crank the formulas and numbers in front of you. I have a good idea of how I’d go about calculating this number, and people have pointed me to on-line programs that could, in principle, provide the absorption coefficients as a function of frequency and temperature, but I’m not a grad student, and I really don’t have time to master the interface and do it this way. I find it no end irritating that someone doesn’t sit down and do a walk-through of the whole calculation, just so I know that I’ve seen it done fully and am not missing something. But I’ve never seen a modern reference. Maybe they use it as a term project, the way they used a supernova simulation when I was taking a stellar structure class decades ago; so nobody wants to write it up and make it too easy for lazier grad students…

  The RLC paper can be found here:
  http://www-ramanathan.ucsd.edu/publications/Ram%20Lian%20and%20Cess%20JGR%201979.pdf

  There’s also a paper from 1977 (Augustsson and Ramanathan):
  http://ams.allenpress.com/arch.....-3-448.pdf

• 109 sadunkal // Jul 29, 2008 at 3:06 am

  #103, Neal,
  I must admit that I’m impressed by your ability to answer all the comments from multiple sources.

  Anyway… I agree that there’s a big difference between Dyson and Feynman, but technically you wouldn’t be able to call Feynman an expert either. At least he could be attacked as not being an expert in the field if necessary, wouldn’t probably stop him though…

  “…why should you be surprised that the media is not pushing “yes”? ”

  I never implied such a thing. I said that it hasn’t been discussed and I don’t think that’s normal considering how significant its role can be. By the way:

  “Science is the belief in the ignorance of the experts”.
  Richard Feynman

• 110 Neal J. King // Jul 29, 2008 at 3:31 am

  sadunkal,

  Feynman might not have been an expert in a field before he started, but he had such a quick, deep and penetrating mind that he very rapidly got the heart of any problem he turned his attention to. He was able to almost immediately arrive at the frontier of what was already known, and quite often able to take a step, or two or three, beyond it. It was great fun watching him tackle a new problem: He would go at it with everything, try one approach after another after another. You could learn more from watching Feynman fail to solve a problem than from another professor that actually knew how to solve it.

  Dyson is a very creative guy, and has contributed a lot in many areas. But he doesn’t compare; and that’s not an insult.

  wrt the volcanic non-influence on arctic ice: There are probably very few experts on undersea volcanoes. If ALL of them say, “There’s no connection,” it’s pretty hard to make a story out of it. You need at least one person with some credibility to take a contrary view, in order to say there’s a controversy.

• 111 Raven // Jul 29, 2008 at 4:10 am

  Neal,

  Are you speaking for yourself or quoting someone?

  “Something that does puzzle and irritate me: Nobody seems to consider it at all controversial that a doubling of CO2 will lead to a radiative imbalance of 3.7 Watts/m^2.”

• 112 Neal J. King // Jul 29, 2008 at 7:18 am

  Raven,

  Yes, that’s me. I’ve never seen anyone professional (not Lindzen, not Spencer) or non-professional (of course it’s not clear they’d know enough to raise the issue) cast any doubt on the 3.7 Watts/m^2 radiative imbalance . I have talked to people about how to calculate it, and I know how I’d go about it. Pierrehumbert’s book, that I mentioned before, sketches it out. But I would like to see the whole thing done explicitly, not just the way it’s written up for a journal article, but the way you would do a homework set: step by step.

  The fact that I’ve never seen it written up that way doesn’t necessarily mean anything: If you look at Jackson’s big fat book on Classical Electrodynamics, there are loads of problems in each chapter, and most of those problems take pages and pages of calculations to do (I’ve done quite a few of them). Journal articles are never written up with that kind of detail: Probably one in 10 equations would show, at best.

  And in this case, quite a lot of the computation would be numerical, because it depends explicitly and implicitly on the absorption coefficient as a function of frequency and temperature.

• 113 Mike N // Jul 31, 2008 at 10:02 pm

  #108, Neal,

  Your first paragraph, agreed. The second, agreed, but I think I wasn’t clear enough with my question. Let me throw out some make believe numbers. Suppose the stratosphere received 20% of it’s energy from absorption of IR from below, 80% from UV absorption and 0% from the ignorosphere. So, reduced IR from below will cause cooling, likewise less ozone - reduced UV absorption, both reducing power input. The stratosphere must cool radiatively, and H2O and CO2 are the major players affecting the cooling rate. In the earlier line by line calculations I’ve seen, doubled CO2 increased the cooling rate of the upper stratosphere quite considerably, but I believe they held the stratospheric H2O at a fixed concentration. Additional H2O in the upper stratosphere will also radiatively increase the cooling rate, I’m just wondering how big of a role the recent H2O increases have had.

  Anyway, thanks for your responses, and it must have been really something to be able to correspond with Feynman!

  Mike

  P.S. I think Raven in #111 is referring to a few Climateaudit posts where Steve also tries to find the calcs for the 3.7W/m^2.

• 114 Peter Gallagher // Aug 2, 2008 at 1:14 am

  For those of us who tend to stop reading when the sentence contains an exponent sign, may I refer to the elegant one-sentence description of M’s idea by ex-Astronaut Walter Cunningham here: http://launchmagonline.com/index.php/Viewpoint/In-Science-Ignorance-is-not-Bliss.html

  ” When former NASA mathematician Ferenc Miskolczi pointed out that “greenhouse warming” may be mathematically impossible, NASA would not allow him to publish his work. Miskolczi dared to question the simplifying assumption in the warming model that the atmosphere is infinitely thick. He pointed out that when you use the correct thickness—about 65 miles—the greenhouse effect disappears! Ergo: no AGW. Miskolczi resigned in disgust and published his proof in the peerreviewed Hungarian journal Weather.”

  Peter Gallaghers last blog post..Doha defeated by contrary goals, rear-view mandate

• 115 Neal J. King // Aug 3, 2008 at 12:41 am

  #113, Mike N:

  - I would think that additional water-vapor in the upper stratosphere would reduce the cooling rate. But it seems to be a moot point: There doesn’t seem to be much water vapor above 10 km (http://www.spectralcalc.com/atmosphere_browser/atmosphere.php) , but the stratosphere lies between 8 and 50 km.

  - I didn’t correspond with Feynman, I attended his Monday evening Q&As at CalTech for a year, and had a lot of fun. A few times, I ran into him at the cafeteria, and he invited me to join him for lunch.

  - In #111, Raven is quoting me in #108: I guess he was having trouble conceiving that someone who basically accepts the mainstream view of atmospheric physics would have any complaint about the landmark publications on this topic. wrt the discussion in ClimateAudit: Yes, I participated in part of that discussion. I agree that it’s harder to dig out than it should be, but I don’t credit CA’s general paranoia that it’s all a grand conspiracy. I think it’s laziness.

  - In this connection, I remember Feynman complaining about the usual explanation of airplane-wing lift as due to some line integral of flow around the cross-section of the wing. He didn’t say it was wrong (it’s not really wrong) but he did say that it didn’t give a clear physical understanding. “We haven’t done a very good job of explaining it. What’s going on is that the air is being deflected downward by the wing, and the force applied to the air causes an upward force applied to the wing. It’s kind of obvious. “

• 116 Neal J. King // Aug 3, 2008 at 12:51 am

  Peter Gallagher, #114:

  The issue is not what Miskolczi claims: That is clear enough.

  The issue is what Miskolczi proves.

  And the point is that there are a number of people who have pointed out issues that are not just quibbles, but conceptual problems in validating his argument. A “proof with holes” is not a proof.

  And the support of a 76-year-old ex-astronaut who has since focused more on business than science (Harvard Business School 1974) does not really add any more credibility to the argument.

• 117 Jan Pompe // Aug 3, 2008 at 3:57 am

  Neal

  Here is an abstract of a paper that finds U = 5P/2 which is a bit different from your U=3P/2. You can’t both be right. It’s even more improbable the three of you are right.

  So who is?

• 118 Neal J. King // Aug 3, 2008 at 9:28 am

  Jan Pompe, #117:

  “Here is an abstract of a paper that finds U = 5P/2 which is a bit different from your U=3P/2.”

  Where is here?

  What is P? I have been talking about total kinetic energy, total potential energy.

  “You can’t both be right… So who is?” I expect that I am . But the ratio depends on how potential energy is defined. Before getting into who is right, one has to determine whether we are talking about the same thing.

• 119 Mike N // Aug 3, 2008 at 9:38 am

  Re: 115, Neal, thanks again for your response and the link. I found my copy of I&C 1995, and their estimate for H2O cooling contribution is quite small, so I was *way* off wondering if it was competing with CO2. Here are their estimates:

  10-500cm^-1 ~6% (mainly H20)
  500-820cm^-1 ~72% (mainly CO2)
  820-980cm^-1 ~0?
  980-1180cm^-1 ~18%(mainly O3)
  1180-3000cm^-1 ~4% (weak H2O, CO2, O3 bands)

• 120 Jan Pompe // Aug 3, 2008 at 2:11 pm

  “Where is here?”

  The link didn’t post for some reason so google
  “atmosphere virial theorem pacheco”

  It will get you there the abstract in in NASA Physics abstract service.

  U is total internal energy P total gravitational potential energy. I would have thought that you could work this out from U=3P/2. The difference is probable due to different handling of the kinetic energy as well as a different agenda those writers wanted to show that U/P = c_v/R = 5/2. C-v is the isochoric heat capacity of air and R the gas constant.

• 121 Neal J. King // Aug 3, 2008 at 3:46 pm

  Jan Pompe,

  I can get to a brief abstract that doesn’t help much. But I don’t find a link to the actual paper: maybe if one has institutional privileges. This is as far as I get:
  http://adsabs.harvard.edu/abs/2003NCimC..26..311P

  It is not my custom to spend a lot of time guessing about other people’s notation. If someone wants to be understood, it’s usually not hard to be clear. If one doesn’t, why should I waste my time?

• 122 Jan Pompe // Aug 3, 2008 at 4:41 pm

  Neal

  I though it was clear - never mind.

  I got it from an online library in the UK.

  I’ll try the link
  try this

• 123 Neal J. King // Aug 3, 2008 at 4:43 pm

  This paper is not worth 8.35 GBP to me.

• 124 Sadun Kal // Aug 3, 2008 at 5:10 pm

  Maybe Jan Pompe can post the relevant parts here, I don’t think anyone would sue anybody because of that.

• 125 Jan Pompe // Aug 4, 2008 at 3:08 am

  Sadun

  “Maybe Jan Pompe can post the relevant parts here”

  Not today I have much more to do of a more pragmatic and necessary nature this medium doesn’t really lend it self to it.

  However the main distinction is that in order to get his ratio he employs the equipartition principle and allows for 5 degrees of freedom by including the 2 rotational degrees of freedom and then allowing the KE only 1 so then 1/5 of the total internal energy balances the potential energy. Is it valid? I don’t think so but will think on it some more yet.

• 126 Neal J. King // Aug 4, 2008 at 7:53 am

  Jan Pompe,

  1) What formula is he using for the potential-energy function? U(r) = -a/r , or U(z) = mgz ?

  2) Rotational KE is not part of the virial theorm, so it shouldn’t be counted as part of a (total KE)/(total PE) ratio. Of course, rotational KE is part of total internal energy of the gas, and the equipartition theorem should apply. (I use it in my derivation of the (3/2) factor.)

• 127 Neal J. King // Aug 4, 2008 at 11:07 pm

  Jan Pompe, #125:

  Here’s a guess:

  If Pacheco et al. are using
  gravitational potential = mgz, then I would get
  (total KE) = (3/2) (total PE)

  But if there are 2 rotational degrees of freedom in addition to 3 translational degrees of freedom, then:
  (internal energy/molecule) = (5/2)kT
  (KE/molecule) = (3/2)kT

  So (internal energy)/(KE) = (5/3)

  and hence
  (total internal energy)/(total potential energy)
  = (TIE)/(TPE)
  = ( (TIE)/(TKE) )*((TKE)/(TPE))
  = (5/3)*(3/2)
  = (5/2)

  which would agree with Pacheco et al.

  So if that’s what they’re doing, we’re both right.

  Still leaves Miskolczi out in the cold…

• 128 Jan Pompe // Aug 5, 2008 at 2:04 am

  of course you are and 5 = 3 a good result in anyone’s book I’m sure.

• 129 Neal J. King // Aug 5, 2008 at 2:13 am

  Jan Pompe,

  I believe you miss my point.

  I showed
  (total KE)/(total PE) = (3/2)

  From what I can tell of your rendition of the abstract, Pacheco et al. may well have shown
  (internal energy)/(total PE) = (5/2), assuming diatomic molecules.

  Since from the equipartition theorem,
  (total internal energy)/(total KE) = 5/3 for diatomic molecules, then putting these two results together implies:
  (total internal energy)/(total PE)
  = ((total internal energy)/(total KE)) * ((total KE)/(total PE))
  = (5/3)*(3/2) = (5/2)

  which is what Pacheco claims.

  But I haven’t seen the paper, so (as I said before) I can only guess about what they are exactly trying to do. And if they are really using the 1/r potential, I probably don’t agree. It depends, among other things, on whether they’re assuming a finite earth, etc.

  But, as I said, I’m not really interested enough to spend 8 pounds on this. YOU raised the question, you deal with it.

• 130 Jan Pompe // Aug 5, 2008 at 3:56 am

  Your are right about one thing I did miss it.

• 131 Neal J. King // Aug 5, 2008 at 9:35 am

  Jan Pompe, #130:

  That’s a pretty graceless way to admit that you were wrong.

  That’s about three times in this thread, I believe.

• 132 admin // Aug 5, 2008 at 10:05 am

  Guys, Askimet checked a couple of your posts as spam. I can’t see any reason for it. Sorry. Believe me, I am making progress with spam.

• 133 Franko // Aug 5, 2008 at 10:42 am

  Interesting how slippery and complicated to get a handle on temperature. Only so mutch, from torturing the classical laws.

  Orbital kinematics and quantum mechanics, almost enough if you neglect the raindrops.

  Attack from a programming simulation effort, to get generalized results. Imaginary, large and simple property objects as gas molecules. Photons as delayed action at a distance. Scale, and re-calculate for smaller and smaller molecular size. Enough petaflops for insight, but still insufficient for accuracy ?

• 134 Jan Pompe // Aug 5, 2008 at 1:31 pm

  “That’s a pretty graceless way to admit that you were wrong”

  Yep was in a hurry I’m going to just as graceless now have to be back at work in 7 hours for a 16 hour shift.

  Problem I still have is that I can see no reason to eliminate 4 degrees of freedom of the total internal energy to get the result he does.

• 135 Neal J. King // Aug 6, 2008 at 9:59 am

  By the way, my calculations on the VT (and other questions posed to Miskolczi) can be found here:
  http://landshape.org/stats/wp-.....ions-3.pdf

  I have re-sent them to Miskolczi, who should be back home by now.

• 136 Jan Pompe // Aug 6, 2008 at 1:46 pm

  “I have re-sent them to Miskolczi, who should be back home by now.”

  It’s the beginning of August could be a few more days yet IIRC ~15th.

• 137 Ken Gregory // Aug 7, 2008 at 5:35 pm

  Neal, I read your questions to Miskolczi. I also was not able to follow all the steps in the math.

  In section 3, you ask how equation 7 is determined from energy conservation. I too had emailed both Miskolczi and Zagoni requesting an explanation of how equation 7 was determined. Zagoni gave a reply which just repeated what the paper says, which was unsatisfactory.

  The equation 1 through 6 apply to all atmospheres without restrictions. Equation 7 is used to derive equation 8, with applies only to a saturated atmosphere. Equation 8, and therefore equation 7 both requires that there is an infinite source of greenhouse gases in the form of water vapor available to the atmosphere, so is restricted to only Earth like atmospheres.

  Equation 9 is the general form of the equation. The paper says this applies to all atmospheres.

  Unfortunately, the paper did not make the restrictions on equation 7 clear. In deriving the equations, I don’t know if Miskolczi first derived the saturated case of equation 8, then generalized it to equation 9, or the other way around. I think it would make more sense to first present the derivation of the general case equation 9, then apply the specific restrictions to determine equation 8.

  In section 4, you note that reducing the strength of the greenhouse effect causes equation 8 to fail because Su = OLR in an atmosphere without greenhouse gases. But equation 8 requires a saturated greenhouse atmosphere, so your comment does not apply to equation 8.

  A second form of equation 9 (3 lines below equation 9), which I’ll call equation 9A says:

  3/5 + 2Ta/5 = OLR/Su - general equation.

  In the Earth type atmosphere, Ta = 1/6.

  If you put this into equation 9A, you get

  3/5 + 2/(6X5) = 2/3 = OLR/Su

  This is equation 8, which is shown as

  Su = 3 OLR/2 equation 8 Earth type

  If you put Ta = 1 into equation 9A, which is the no greenhouse gas case, you get

  3/5 +2/5 = 1=OLR/Su

  If you assume Ed/10 is much less than OLR, so put Ed = 0 in equation 9, and using Ta = St/Su, you get equation 10, which is the thin atmosphere, or Mars case:

  2/3 + Ta/3 = OLR/Su equation 10 Mars type

• 138 Neal J. King // Aug 7, 2008 at 7:29 pm

  Jan Pompe,

  Thanks for providing the paper by Pacheco & Sanudo, that you mentioned in #117.

  I’ve looked it over, and the situation is pretty much what I had guessed in #127 & #129. In fact, the general approach taken in P&S and in my calculation (#135) are very similar; although I put a lot of work in to show that the results of the spherical-Earth case go over into the flat-Earth case in the appropriate limit. P&S don’t do that. And they cleverly show that you can do the flat-Earth case using hydrostatic equilibrium, and I didn’t do that. But we pretty much cover the same ground.

  P&S do not suppress 4 degrees of freedom:

  a) In the first half of section 1 of P&S, they show that, in the flat-Earth case (they call it the “planar atmosphere”), that the total gravitational potential energy (taken to = 0 at altitude 0) is equal to the integral over altitude of the pressure. Therefore,
  [total PE] = integral p(z)dz
  (Eqns. (3) & (4).)

  Since the perfect-gas law tells us that:
  p = n*kT
  where n = number density of molecules,
  the integral of p(z)dz is equal to
  the integral of n(z)*kT dz = [total (kT)]

  This total kT is the total over all molecules in the gas.
  Since the KE of a molecule = (3/2)kT by the equipartition theorem (because of 3 degrees of freedom),
  [total KE] = (3/2)[total kT]
  = (3/2) integral of p(z)dz
  = (3/2) [total PE]

  (This is eqn(3) in my calculation - derived from the virial calculation.)

  When calculating the internal energy, with 5 degrees of freedom,
  [total U] = (5/2)[total kT]
  = (5/2)[total PE]

  For diatomic molecules, c_V = (5/2)R; for monatomic molecules, c_V = (3/2)R.

  So all degrees of freedom are properly taken into account.

  However, I do see that in eqn. (1) and (2) they forget the factor of the molecular mass; however these two omissions cancel out.

  What I would consider the “main course” of the calculation - the virial sum - is done quite similarly in P&S and in my Table 3. For the spherical-Earth case, compare their eqn(9) to my “Virial Equation” in row 11 of the 3rd column of Table 1. For the flat-Earth case, compare their eqn(23) to my “Conclusion” at the bottom of the 1st column. (They do the calculation with a cylindrical column of air, whereas I used a square-based column. I just didn’t want to introduce another factor of pi where it wasn’t really needed.)

  So I would say that the paper by Pacheco & Sanudo and the calculation by me are completely in-line (give or take a few typos or easily correctable errors).

• 139 Neal J. King // Aug 7, 2008 at 7:51 pm

  #137, Ken Gregory:

  - What is it about Equations (8) and (7) that make them apply only to a saturated atmosphere? Unless I see a saturation-related rupture in the logic, I would conclude that these equations either apply to both saturated and non-saturated atmospheres, or to neither saturated nor non-saturated atmospheres.

  - wrt my Section 4: Again, if my comment does not apply to equation (8), please explain at what specific point the logical step does not follow - and why it does follow for a saturated atmosphere.

  - You say, “In the Earth type atmosphere, Ta = 1/6.” But at the bottom of page 7, M says: “For the Earth obviously the T_A ~ 0 condition apply…”

  What is missing for me here is the logical coherence. “Extraordinary claims require extraordinary evidence.” Logical coherence is merely the bare minimum; but I don’t even see that, at this point.

• 140 Jan Pompe // Aug 8, 2008 at 12:31 am

  Neal

  “Thanks for providing the paper by Pacheco & Sanudo, that you mentioned in #117.”

  You are more than welcome. I’ll look at the rest later.

• 141 Ken Gregory // Aug 8, 2008 at 1:40 am

  Neal, yes, we are missing “logical coherence”.

  Miskolczi will have to answer your question of why equation 7 applies only to saturated atmosphere. It is disappointing that he hasn’t been more communicative.

  The statement “For the Earth obviously the Ta ~ 0 condition apply…” mean that for optimal conversion of Fo to OLR, Ta = 0, but the Earth can not have optimal conversion because Ta does not equal 0, but equals 1/6.

  The paper is confusing, but I think it is worth pursuing because the predictions match observations.

  I sent the following question to Miskolczi, and did not receive a reply. Can you answer it?

  Page 7 of the paper says:

  In Eq. (9) the Sv=St/2 - Ed/10 virial term will force the hydrostatic equilibrium while maintaining the radiative balance.

  And the virial theorem applied to the atmosphere says Su = 2Eu

  How do we get the virial term Sv = St/2 - Ed/10 from the virial theorem?

• 142 Neal J. King // Aug 8, 2008 at 2:05 am

  #141, Ken Gregory:

  - I cannot even figure out how M gets to eqn.(7). In my own comments, I’ve already expressed puzzlement at eqn.(8). (9) is said to follow from (8).

  - Neither I, nor any of the folks above, have been able to explain how M uses the virial theorem to deduce ANYTHING. The way I’ve asked it is really simple: “What are the equations you are using to relate X to Y?” where X is some bulk quantity like total potential energy or total kinetic energy (which appear in the VT) and Y is a flux of some sort. The most that he says is stuff like, “The surface air temperature t_A is linked to the total gravitational potential energy through the surface pressure and air density. The temperature, pressure, and air density obey the gas law, therefore, in terms of radiative flux S_A = sigma*t_A^4 represents also the total gravitational potential energy.” Linked how? Represents how? None of M’s defenders have been able to suggest any specific meaning beyond these words.

• 143 Jan Pompe // Aug 8, 2008 at 2:33 am

  Neal,

  “that the total gravitational potential energy (taken to = 0 at altitude 0)”

  Yes I saw that but is the graviational potential really zero at the surface for this purpose?

  r ~ 6,38o,ooo m gravitational potential energy is given by P = -G mM/r just to remind you here r is the distance between the centres of mass.

  If you want to say that the gravitational potential is zero at the surface you would also have to say the weight is zero and this is simply untrue.

  BTW it was that line that caused me to suppose that you analyses would be similar.

• 144 Neal J. King // Aug 8, 2008 at 2:49 am

  #143, Jan Pompe:

  - In principle, you can set the point of 0 gravitational potential where you like, because the force here is due to the pull from the big Earth, not to the little molecules attracting each other, so you can add any constant you like.

  - In particular, in this section, P&S explicitly state: “It is important to realize that in this calculation of U and P, we have adopted as the state of minimum energy, or level of reference, that where T(z) = 0 for the whole column, and consequently all the gas molecules are deposited at ground level, that is U_o = P_o = 0.”

  - It is also clear in this section that they are doing the flat-Earth calculation, so your reference to the (1/r)-potential is premature. The flat-Earth system is a perfectly valid system: You could actually imagine a huge flat planar sheet, and if you stayed away from the edges and didn’t go too far up, it would look like that. And for lots of situations, it’s perfectly fine to model one’s experience on the surface of the Earth that way. But you have to know the limits of applicability.

• 145 Jan Pompe // Aug 8, 2008 at 4:58 am

  “- In principle, you can set the point of 0 gravitational potential where you like, because the force here is due to the pull from the big Earth, not to the little molecules attracting each other, so you can add any constant you like.”

  This is true but it also has a problem if you set o at a point on the surface you can use it to determine the kinetic energy an object falling to the surface when it reaches the surface and as such it is useful but it is useless in telling us what the gravitational potential which gives the downward pressure exerted on the surface but that body at rest.

  “But you have to know the limits of applicability.”

  Indeed and you might start with KE = PE =O at the surface. The force due to gravity does not disappear at the surface and it is this that also constrains all degrees of freedom except the tangential and that kinetic energy is also set to zero.

  All in all not a good start in the context of the virial theorem which balances [total KE> and in a spherical system.

  Now if you want to invoke the equipartition principle and take into account all thermal energy or total internal energy then you indeed have 5 degrees of freedom and the kinietic component is 3/5(total internal energy) so then we have without the convoluted argument:

  [total ke] = -5/6[total PE]

  and in the light of what Eli has to say in 32 one has to wonder how much of a role rotational energy actually plays and whether we are justified in assigning 40% of the internal energy to it. Just a thought but I’m fairly certain that the rotational components won’t be contributing much to E_U or OLR.

  “The flat-Earth system is a perfectly valid system: You could actually imagine a huge flat planar sheet, and if you stayed away from the edges and didn’t go too far up”

  Sure good thinking you won’t fall of the edge that way I do believe those sailors who didn’t sail with Chris Columbus would have agreed with you. It also avoids problems with singularities when r = 0.

  I do agree with you that for lots of situations it’s perfectly valid - like how many bones you might break if you jump from a high place onto a solid surface it does not apply to a situation where we a dealing with a gravitationally bound multi-particle system which is keeping it’s distance from the surface.

• 146 Neal J. King // Aug 8, 2008 at 10:23 pm

  #145, Jan Pompe:

  - “…but it is useless in telling us what the gravitational potential which gives the downward pressure exerted on the surface but that body at rest.”: Sorry, but I absolutely don’t understand what you are trying to say here.

  - “All in all not a good start in the context of the virial theorem which balances [total KE> and in a spherical system.": The factor of (5/2) does not apply to a spherical system. It only applies in the planar/flat-Earth geometry, in which the absolute value of the potential function has no natural value.

  - "[total ke] = -5/6[total PE]“: I guess you get this by trying to fit the 5 degrees of freedom into the “classical” result of the VT. But this is wrong for two reasons:
  1) the KE that fits into the VT only includes translational KE, with 3 degrees of freedom.
  2) in the planar/flat-Earth setting, the “classical” result that
  [total translational KE] = -(1/2)[total PE]
  does not apply: that result applies for a gas cloud held together by mutual gravitation, or by a gas cloud held together by attraction to a point-mass, not by a planet with appreciable size; in which case you get eqn.(9).

  To understand these points, you need to understand how the “classical” result was derived. Both P&S and I explain this in some detail.

  - “in the light of what Eli has to say in 32 one has to wonder how much of a role rotational energy actually plays and whether we are justified in assigning 40% of the internal energy to it.”: Nothing Eli says has any implications for the 40%. As long as c_V/R = (5/2), the rotational KE is contributing fully to the internal energy of the gas. If you look at
  http://en.wikipedia.org/wiki/Heat_capacity_ratio
  you can see that for room-temperature conditions, c_P/c_V = 1.4 (implying 5 degrees of freedom). At very low temperatures, the rotational degrees of freedom will stop contributing, due to quantum effects.

  - “it does not apply to a situation where we a dealing with a gravitationally bound multi-particle system which is keeping it’s distance from the surface.” But the planar model is exactly what P&S are solving to get the factor of (5/2), in section 3 and the first half of section 1. This factor does not apply to the spherical-atmosphere case, for which one gets eqn.(9).

  I’ve tried to clarify your questions, but frankly, the clearest way I can explain what is going on I have already expressed in my own write-up, here:
  http://landshape.org/stats/wp-.....ions-3.pdf

  You may not be particularly interested in what I have to say, but I guarantee you that if you understand what I have done in Sections 1 , and Section 1.1 with the Flat-Earth column of Table 1, I can explain to you what P&S are doing in 5 minutes, because the physics is the same. If you even just read Section 1, it will at least be clear that they are doing two separate calculations with two separate results - something that may not always be clear in their article.

• 147 Jan Pompe // Aug 9, 2008 at 1:00 am

  Neal

  “You may not be particularly interested in what I have to say, but I guarantee you that if you understand what I have done in Sections 1 , and Section 1.1 with the Flat-Earth column of Table 1, I can explain to you what P&S are doing in 5 minutes, because the physics is the same.”

  Yes I know the physics and the main mistake is the same. I’ve been trying to get you to see it for yourself but I don’t think that is going to happen now.

  It’s here
  I find that:
  [KE] = -1/2[U_s] - 2*pi* r^3 * P_o

  but from kinetic theory

  P_o = 2/3(N/V)([1/2]mv^2)

  your equation then becomes

  [KE] = -1/2[U_s] - (2*pi*r^3) * 2/3 (N/V) *[KE]

  you and P & S are counting the kinetic energy twice. The pressure exerted on the surface is due to the kinetic energy of the gas at the surface. Then you add the pressure applied by the surface it’s the same force double counted.

  Now while it’s intuitive that solid dense core should make a difference before we can decide just what difference it makes we really need to work out what the quanititive difference between particles bouncing elastically off a surface and particles sinking below (now imaginary) the surface to be replaced by particles or the same energy from beneath under the fundamental assumption that

  dG/dt = 2[T] - U = 0 (see eq 1.5.2 of that online book you referrenced)

  over long periods of time for stable systems.

  I do concede that I expressed the consequences of Eli’s remark badly rotational energy does indeed make up part of the total internal energy it don’t think it contributes much to radiative flux. I’m perfectly happy to be corrected on this.

• 148 Neal J. King // Aug 9, 2008 at 9:41 am

  #147, Jan Pompe:

  “but from kinetic theory”:
  The equation
  [KE] = -1/2[U_s] - 2*pi* r^3 * P_o

  is an equation obtained by integration over the entire gas, the entire atmosphere.

  You counterpose to this your equation:
  P_o = 2/3(N/V)([1/2]mv^2)

  I guess you get this from:
  (1/2)mv^2 = (3/2)kT

  so that
  P_o would = (2/3)(N/V)*(3/2)kT
  =(N/V)kT
  which looks like the perfect-gas law, which is

  P = nkT

  But it’s not: the perfect-gas law (PGL)
  relates the pressure, density and temperature variables at one point. It applies to a bulk (like a definite volume of gas) only when that bulk can be approximated well as being homogeneous: when the pressure, density and temperature can be considered as substantially constant throughout. When that it not the case, you cannot set number density (n) equal to total number (N) divided by total volume (V). Likewise, you cannot set the pressure variable (which varies throughout the atmosphere) equal to the pressure at the surface of the Earth (P_o): It’s not constant.

  You can of course apply the PGL at the Earth’s surface itself, and then you get:
  P_o = n_o * kT_o

  But n_o = local molecular number density, it is certainly not = N/V; at least not if you mean by N the total number of molecules in the atmosphere and by V the total volume of the atmosphere.

  So my equation from the line “Virial Equation” is not in contradiction with the PGL or kinetic theory.

  “we really need to work out what the quanititive difference between particles bouncing elastically off a surface and particles sinking below (now imaginary) the surface to be replaced by particles or the same energy from beneath“:
  The VT is applicable to a definite set of molecules that are confined to a bounded region of space: If you do an imaginary switch of molecules, you can’t apply the time-average process that gives the result of the VT. In order to use the VT, you have to take explicitly into account the fact that the molecules are kept out of the planet’s bulk. (In fact, in this particular point I was inspired by a derivation of the PGL from the VT that I saw years ago, which applied the VT to a fixed box of gas (in the absence of gravity): You have to take explicit acknowledgment of the walls of the container and the pressure applied. If you don’t, the set of molecules under consideration is not spatially bounded, and the VT cannot be applied.

• 149 Jan Pompe // Aug 9, 2008 at 11:16 am

  “But n_o = local molecular number density, it is certainly not = N/V; at least not if you mean by N the total number of molecules in the atmosphere and by V the total volume of the atmosphere.”

  Now you are being silly doesn’t really matter if it’s a ball resting near/on the surface or a very thin layer over the whole surface the pressure felt by the gas comes from the gas itself hence you are double-counting the force.energy.

  You might also claim that it’s the weight then it comes from the gravitational potential and you would be double-counting that.

  “If you do an imaginary switch of molecules, you can’t apply the time-average process that gives the result of the VT.”

  You are missing the point is it deliberate?

  The basis of he virial theorem is long term average or over whole periods of an oscillation in short there is no quantitative difference before and after.

  “In order to use the VT, you have to take explicitly into account the fact that the molecules are kept out of the planet’s bulk.”

  Not if there is no quantitative difference between molecules bouncing elastically off the surface and as in the above case exchange with particles of the same energy or momentum. In both cases total momentum is conserved as is total potential.

  We are not dealing with a gravitation free box of gas but a gravitation bound system and we need to be very careful about double-counting the forces. In the box it will be quite easy because it is easy to see the pressure exerted on the gas by the walls is due to the momentum of the gas particles and the pressure of the walls won’t be added to that (no double-counting).

• 150 Neal J. King // Aug 9, 2008 at 11:44 am

  #149, Jan Pompe:

  No silliness: the PGL is an equation of state that relates local variables:
  - local pressure p(r)
  - local number density n(r)
  - local temperature T(r)

  In particular, the pressure is NOT constant throughout the gas: P_o is just p(r=R_earth), which is the highest pressure throughout the gas. With increasing altitude, pressure drops towards zero.

  The VT is proven by doing the virial sum over a definite set of molecules and averaging over time. You cannot exchange molecules at the border and hope it does the same: It doesn’t.

  I am not double-counting anything: I am adding up the r_i.F_i term, molecule by molecule. Gravity applies to all of them (giving rise to my term, “Virial_grav”); the pressure applies only to the molecules that are interacting with the ground, in the process of bouncing (giving rise to my term, “Virial_bottom”). I do not add pressure as a separate constraint or force separate from what is included in the calculation of the virial.

  You are trying to pretend that the walls/floors are not there, and that you can model the system by an exchange trick. My point is that the virial calculation can be done straightforwardly, and gives rise to what I calculated. There is no need to play equivalence tricks to avoid the difficulty, because there is no difficulty; and from the perspective of having done the calculation, it’s not hard to see that these tricks DO NOT give rise to an equivalent result.

• 151 Jan Pompe // Aug 9, 2008 at 12:32 pm

  - local pressure p(r)
  - local number density n(r)
  - local temperature T(r)

  In particular, the pressure is NOT constant throughout the gas: P_o is just p(r=R_earth), which is the highest pressure throughout the gas. With increasing altitude, pressure drops towards zero.

  It’s no reason to ‘find’ that the surface is providing a pressure on the atmosphere which is due to the KE of the atmosphere and add that back in.

  “My point is that the virial calculation can be done straightforwardly, and gives rise to what I calculated.”

  Double-counting forces is an error it cannot considered straightforward. The particles obey Newton’s classical laws and I know that the third law says that to ever action there is an equal and opposite reaction but if there is a single force applied there is only one force it does not double because of it and the object will move accordingly even the the pushing object will feel the same force.

  The error starts with at the surface with no kinetic energy i.e. [KE] = [PE]=0. As I’ve pointed out before the gravitational potential is still GmM/R which is the work done bringing the particle to distance R from the centre and the force it gives rise (just divide by R) to you might know as the weight these particles don’t become weightless when they lose KE. At what KE do you suppose we would get lift off?

• 152 Neal J. King // Aug 9, 2008 at 4:30 pm

  #151, Jan Pompe:

  It’s no reason to ‘find’ that the surface is providing a pressure on the atmosphere which is due to the KE of the atmosphere and add that back in.
  True: That’s why I’m not doing it. The P_0 term in my equation comes from the r.F terms for the molecules that are bouncing off the ground.

  Double-counting forces is an error
  True: That’s why I’m not doing it.

  The error starts with at the surface with no kinetic energy i.e. [KE] = [PE]=0.
  That point relates to the planar/flat-Earth calculation - which has absolutely nothing to do with the spherical-Earth problem that we were talking about wrt 4*pi*R_earth^3*P_0. These are completely different calculations.

  We would probably make more progress if you would work through my actual argument, instead of picking different parts (that actually don’t relate to each other) and trying to find issues. I invite you to study my Section 1 and identify problems that you have with that. This is a kind of overview, but it should still give you an idea of what I am trying to show (and what I am not trying to show). You might well have issues with the overview, even before the detailed math.

• 153 Jan Pompe // Aug 9, 2008 at 11:56 pm

  Neal

  The P_0 term in my equation comes from the r.F terms for the molecules that are bouncing off the ground.

  so they magically gain potential energy by not changing r good one Neal.

  “True: That’s why I’m not doing it.”
  but you just told me you do

  Here:
  “The P_0 term in my equation comes from the r.F terms for the molecules that are bouncing off the ground.?”

  Particles bouncing is just a short term case of

  dG/dt = [KE] + 1/2[OE] = 0 counting them before and after the bounce is simply double counting.

• 154 Neal J. King // Aug 10, 2008 at 1:20 am

  #153, Jan Pompe:

  No:

  - The VT says:
  [KE] = - (1/2)[r.F]

  as time averaged.

  - [r.F] is the sum over all molecules of the position of the molecule dotted into the total force acting on that molecule. I call [-r.F] “the virial”; I think that is standard terminology.

  - All of the molecules are affected by gravity, so it is straightforward to represent gravity’s contribution to the virial. This is shown on the “Virial_grav” line of the 4th column of the table. The result is also described as “(1/2)[-U_s], because it is the sum of (1/2)GMm_i/r_i
  (i = index for molecules).

  - At the instant of doing the sum, some of the molecules are in the process of colliding with the ground. These are experiencing a non-gravitational force. The effect of all theses forces is to reflect the molecules; the reaction force is the pressure applied to the ground; so the magnitude of the collision force sum is the opposite of the pressure: an upward force per unit area, of magnitude exactly equal to the P(r=R) (which is the downward force the atmosphere is applying to the ground). So the virial contribution from the ground bounce is what I denote Virial_bottom; it is equal to -(1/2)(4piR^3)*P(r=R), and is simply a term in the virial. It only comes from the molecules interacting with the ground, because molecules in the “interior” of the atmosphere are not colliding with the ground. And there is no double counting: the sum is taken at one instant, and since it takes a finite amount of time for a molecule to change its momentum, at that instant there will be a finite number of molecules that are experiencing that turn-around and therefore the upward force. In principle, you could calculate how many molecules are in that collision mode at one instant, and their average area density with the ground, and how big the force is; but Newton’s 3rd law tells us that the upward force per unit area experienced by the gas is equal and opposite to the downward force applied by the atmosphere onto the ground, aka “the pressure”.

  If this is still not clear, I can give another example: a derivation of the perfect-gas law. But I don’t want to spend a lot of time on that unless you need further discussion.

  - The Virial Equation sets [KE] to the sum of the two contributions to the virial, as shown in the Table, resulting in:
  [KE] = -(1/2)[U_s] - (2piR^3)*P(r=R)

• 155 Neal J. King // Aug 10, 2008 at 2:03 am

  #154, continued:

  Here’s another way of thinking about it:

  - Model the ground as force field which is zero everywhere except between z = d and z = 0, where it is upward with force f.

  - In doing the virial calculation, the gravitational force affects all molecules, but the ground force affects ONLY those molecules that are located, at this instant, between z = 0 and z = d (I call these “bouncing molecules”). All of those experience force f, and are located at r = R + d/2, to within (+/-) (d/2). So each one of the r.F terms has magnitude about:
  -(1/2) (R + d/2)*f.

  - When you turn the sum into an integral, the integral is taken over the area of the Earth: If nc is the area-density of bouncing molecules, the force for differential area da is:
  (1/2)f*nc*da

  and the differential contribution to the virial is:
  d(Virial_bottom) = - (1/2)(R+d/2)*f*nc*da

  But by Newton’s 3rd law,
  f*nc*da = P(r=R)*da

  so:
  d(Virial_bottom) = -(1/2)(R+d/2)*P(r=R)*da
  ~-(1/2)R*P(r=R)*da

  Completing the integral:
  Virial_bottom ~ -(1/2)R*P(r=R)*(4pi*R^2)
  = -(2pi*R^3)*P(r=R)

• 156 Jan Pompe // Aug 10, 2008 at 3:30 am

  Neal,

  “- The VT says:
  [KE] = - (1/2)[r.F]

  as time averaged.”

  and
  “I call [-r.F] “the virial”; I think that is standard terminology.”

  No the virial is G = Sum[p_i . r_i]

  where p = momentum

  My apologies I did not realise that you were harbouring such a basic misconception.

  “These are experiencing a non-gravitational force. ”

  These are experiencing the equal and opposite force that the surface ‘feels’ exerted by the particles due to gravity it changes the direction of motion it is NOT and extra force nor is it an extra source of energy. You are double counting the force and as a consequence the energy.
  Precisely that can be expressed as

  dG

• 157 Neal J. King // Aug 10, 2008 at 2:19 pm

  #156, Jan Pompe:

  a) “No the virial is G = Sum[p_i . r_i]

  where p = momentum

  My apologies I did not realise that you were harbouring such a basic misconception.”
  According to eqn. (1.2.6) of “The Virial Theorem in Stellar Astrophysics” ( http://ads.harvard.edu/books/1978vtsa.book/ ): The last term on the right is known as the Virial of Clausius and it happens to be the Sum(i)[f_i.r_i]

  I’m really getting tired of your pointless rudeness. It’s particularly pointless when you’re wrong, as in this case. But it’s generally pointless anyway.

  b) These are experiencing the equal and opposite force that the surface ‘feels’ exerted by the particles due to gravity it changes the direction of motion it is NOT and extra force nor is it an extra source of energy. You are double counting the force and as a consequence the energy
  No, the f in the virial is the actual force felt by each particle. Each particle feels a gravitational pull towards the center of the Earth. Only the particles within a small distance of the ground feel an upward force: this is what keeps the particles from going through the floor. Think of the floor as adding an additional potential field:
  Un(r) = 0, for r > R + d
  Un(r) = (V_o/d)*(R+d-r), for R < r < R+d
  Un(r) = V_o, for r < R

  V_o is a positive constant that is large enough that particles won’t get over this barrier: which is what a floor is. If there were no such barrier, the particles would go through the floor - which is not the problem that is being addressed.

  So if you evaluate the contribution to the virial coming from the particles which are in the range
  R < r < R+d , you will get something that is proportional to the radius R and to the area 4*pi*R^2. It becomes obvious that the magnitude of the force per unit area is the same as the pressure because the pressure (pushing down from the atmosphere) is just the reaction “felt” by the floor to the force felt by the particles.

• 158 Neal J. King // Aug 10, 2008 at 2:23 pm

  #157 continued:

  If you don’t like this approach, please explain how you would model the inpenetrability of the floor in the context of a virial calculation.

• 159 Jan Pompe // Aug 10, 2008 at 2:38 pm

  the virial of Clausius is part of the derivative of the virial

  G=sum[p_i.r_i] which is the first term of 1.2.2

  I’m not wrong I’m getting tired of you pointless twisting and turning around the fact that you are double-counting the kinetic energy that book warns about it several times I notice.

• 160 Neal J. King // Aug 10, 2008 at 2:53 pm

  Jan Pompe:

  - Look at eqn.(1.2.6), and the line directly below it.
  Just look at it. You are wrong.

  - I’m not even counting kinetic energy: I’m calculating the virial. As I suggested above, if you have another way of doing the calculation, please expound.

  In the meantime, consider the fact that my methods and results are essentially identical with P&S, who got got their paper published in a peer-reviewed international journal. Probably not a first-tier journal (Italy is not a big physics country these days) but still pretty respectable. Where is your point of view represented in the published literature?

• 161 Jan Pompe // Aug 10, 2008 at 3:04 pm

  “- Look at eqn.(1.2.6), and the line directly below it.
  Just look at it. You are wrong.”

  I did before I posted the last answer that ‘G’ in dG/dt is the the virial it’s defined in eq’n 1.2.2 it might not be named there but it’s the virial all right I just checked it is named in wikipedia (it’s a good dictionary). You will find that the kinetic energy is there in the derivative there is no need to add it again disguised as a pressure term.

  You owe me an apology

• 162 Neal J. King // Aug 10, 2008 at 3:31 pm

  So, for you, a wikipedia article is more authoritative than a 101-page textbook that has been published since 1978, by a professor at Case Western Reserve?

  You have got to be kidding me. I don’t owe you a thing.

  And beside the point of terminology, are you actually interested in understanding the physics, or just trying to score points? If you’re interested in understanding the physics, meet my challenge: Do the straightforward virial calculation.

  If you’re just trying to score points, I have other things to do. I’ve already worked my way through the problem, and I’m pleased that P&S agree with me: It’s nice to have some confirmation.

• 163 Pat Cassen // Aug 10, 2008 at 3:32 pm

  Jan – I have been following your discussion with Neal on the subject of the application of the virial theorem to the atmosphere because I could not fathom what Miskolczi was doing in this regard. As you are both getting a little testy with each other, I will offer my perspective, FWIW.

  Perhaps part of the difficulty you are having with Neal’s analysis is due to the fact that the virial theorem is seldom applied to systems with solid boundaries. Such boundaries (the ground, in this case) are capable of exerting forces on the particles of the system which must be included in the f(dot)r terms. Collins does not treat these kinds of systems, and I have found no other examples in standard texts. (Indeed, it is bizarre that Miskolczi attempted to apply the virial theorem, because atmospheric structure is adequately described by the equations of motion without having to deal with the complications introduced in the virial integrals.)

  I think that Neal has done a good job of working through the details of a proper treatmnet in this unorthodox application. He is not “double-counting”. (Collins’ warning applies to double-counting the forces between particles within the system.) At this point, I suggest that you do your own derivation, and test it with specific atmospheric structures (isothermal, polytropic) as Neal has done. See what you get. Does it make sense?

• 164 Jan Pompe // Aug 10, 2008 at 8:40 pm

  Neal,

  “So, for you, a wikipedia article is more authoritative than a 101-page textbook that has been published since 1978, by a professor at Case Western Reserve?”

  No I don’t and they don’t disagree.

  Pat,

  I don’t think it’s particularly unorthodox I’m pretty sure Miskolczi took that approach because of it’s simplicity not to make it more difficult. I’m also pretty sure the solid surface isn’t preventing the further collapse of the atmospheric gases and is quite passive in the whole business.

  OK I will do my own derivation but for now I must go to work.

• 165 Pat Cassen // Aug 10, 2008 at 10:44 pm

  Jan says “I’m also pretty sure the solid surface isn’t preventing the further collapse of the atmospheric gases and is quite passive in the whole business.”

  Gee. Do the thought experiment: Consider the Earth with radius Re and mass Me, surrounded by its atmosphere of mass Ma << Me. Now shrink the Earth to radius r < Re, keeping Me and Ma fixed. The atmosphere proceeds to fill the space between r and Re, and beyond Re it adjusts to the new equilibrium, despite the fact that the gravitational potential beyond Re is unchanged. (The effect of the changed surface location is felt through the new pressure gradient.) The pressure at the surface is now greater than “one atmosphere” (or we have to define a new “one atmosphere”). No, the surface is not passive.

• 166 Neal J. King // Aug 10, 2008 at 11:54 pm

  Pat Cassen,

  My approach to the problem was inspired by a derivation of the perfect-gas law I saw long ago, I believe in Marion’s book on classical dynamics:

  - Assume a box with dimensions X, Y, Z filled with gas.

  - Neglect gravity, so everything is held together by the box.

  - The VT says:
  2[KE]= -[r.f]

  - The virial [r.f] gets contributions only from the walls, because gravity is ignored and inter-molecular collisions cancel out (equal & opposite forces acting at the same point, when one sets the size of the molecules to 0).

  - The sum of the virial can be broken into three separate sums: one for the virial contribution from the two walls perpendicular to the x-axis, one for the walls perpendicular to the y-axis, and one for the z-axis.

  - For the x-axis: the virial contribution turns into surface integrals over the two walls. A little thought reveals that the sum over the two walls is: X*(P)*(YZ) = (XYZ)*P = V*P

  - For the y-axis, you get: Y*(P)*(ZX) = V*P

  - For the z-axis: Likewise V*P

  - So the virial sums to 3PV, and hence:
  2[KE] = 3PV

  - Since from the equipartition theorem, the average KE per molecule is (3/2)kT, 2KE = 3kT, and 2[KE] = 3[kT] = 3NkT

  - Thus 3NkT = 3PV, or PV = NkT. Which is the perfect-gas law.

  So this is not the first time that people have applied the VT to systems bounded by walls. I think you can also derive the van der Waals gas law with a modification of this approach, too.

• 167 Pat Cassen // Aug 11, 2008 at 2:37 am

  Re: #166
  Very neat. Thanks.

• 168 Jan Pompe // Aug 11, 2008 at 7:15 am

  Pat,

  Do the thought experiment: Consider the Earth with radius Re and mass Me, surrounded by its atmosphere of mass Ma << Me. Now shrink the Earth to radius r < Re, keeping Me and Ma fixed.

  Yes Pat I’ve done this already, and do agree that if you shrink the radius that the pressure at the surface will be greater. That’s good because you at least will be able to see what I’m getting at.

  [KE] = -(1/2)[U_s] - (2piR^3)*P(r=R)

  Now unless I miss my guess by a wide margin the magnitude of the pressure term increases with radius increase and decreases with radius decrease and did we just agree that as the radius gets smaller the pressure must increase?

  So it does look as though it’s back to the drawing board for either you and I or for Neal.

  I don’t agree though that this shows that the hard surface is in any way active the pressure is still derived from the kinetic energy needed to balance the increase in gravitational potential. At best we can say it sets an upper limit in the gravitational potential of the gas and consequently the kinetic energy then in turn on the pressure.

  I don’t think that the way to express that limiting effect is by adding an additional kinetic energy (pressure) term.

• 169 Neal J. King // Aug 11, 2008 at 7:42 am

  - As R decreases, both [KE] and (1/2)[-U_s] will increase, because the molecules will be spending more time closer to the source of the gravitational pull. I’ll see if I can find a way to do an independent (non-virial) estimate of P(r=R), but my expectation is that as R decreases, the quantity (R^3)P(r=R) will decrease monotonically: Certainly when R = 0, that term will vanish, because at that point I fully expect to get
  [KE] = -(1/2)[U_s] , because at that point there would be no boundary term.

  In other words: I agree that P(r=R) will increase as R decreases, but it since R^3 decreases as R decreases, I fully expect that the product of the two will decrease. Otherwise, P(r=R) would have to have a stronger dependence than 1/R^3.

  - As pointed out many times, the pressure is not an “added kinetic energy”. The pressure comes in as a boundary term - precisely analogous to the case in #166, which shows the boundary term in very pure form (because there is no bulk force to take into account). As shown, it leads directly to the perfect-gas law.

• 170 Jan Pompe // Aug 11, 2008 at 7:53 am

  Neal,

  “Certainly when R = 0, that term will vanish, because at that point I fully expect to get
  [KE] = -(1/2)[U_s] , because at that point there would be no boundary term.”

  At R = 0 [U_s] = GMm/R explodes.

• 171 Neal J. King // Aug 11, 2008 at 11:19 am

  #170:

  Yes, [KE] will be going through the roof as well.

  That’s how stars work: the pressure (proportional to the kinetic-energy density) builds up near the center high enough that the nuclear collisions are high enough in energy to overcome their electrical repulsion and smash into each other and do the nuclear fusion trip.

  If the KE intensity did not build up high enough, none of us would be here.

• 172 Neal J. King // Aug 11, 2008 at 1:39 pm

  Something else I just noticed about the formula:
  [KE] = -(1/2)[U_s] - (2piR^3)*P(r=R)

  For the fixed Earth radius R, and total amount of gas having mass M_gas, let the temperature of the gas be set to zero: T= 0.

  Then [KE] = 0, and all the gas molecules are sitting on the Earth’s surface, at r = R.

  So
  [-U_s] = -(-GM*M_gas)/R) = GM*M_gas/R

  P(r=R) will be the pressure required to support the gas:
  P(r=R) = (GM*M_gas/R^2)/(4pi*R^2)
  = (GM*M_gas/(4pi))/R^4

  so
  (4pi*R^3)*P(r=R) = GM*M_gas/R

  Sticking all this together:
  0 = [KE] = [-U_s] - (4pi*R^3)*P(r=R)
  = (GM*M_gas/R)*(1 - 1) = 0

  It works: In the low-temperature limit, it’s clearly true.

  I’m liking this result better all the time.

• 173 Jan Pompe // Aug 11, 2008 at 1:42 pm

  Neal,

  “Yes, [KE] will be going through the roof as well.

  That’s how stars work”

  Yes I know however there is no term in the that will cause it to explode i.e. no singularity so it’s the only term that remains.

  I think the point here is if there is one that {KE} need not indeed for the bulk of the body of gravitationally gas is not = 1/2{PE} for an atmosphere or body of particles remain bounded by gravity that {KE} <= 1/2{PE}.

  If the {KE} > {PE}/2 it becomes unbounded i.e. it boils off.

  If the force due to the kinetic energy of a body or particle is greater than the force holding it in it departs. No?

  Then if the holding force is gravity then the force du kinetic energy must be less than or equal to the inward force due to gravity.

  centrifugal force <= gravitational force
  mv^2/R <= -GmM/R^2
  or
  mv^2 <= -GmM/R
  dividing through by 2
  1/2mv^2 <= -GmM/2R

  so a gravitational field cannot hold a body that has a greater kinetic energy than 1/2 of the bodies potential energy. I really don’t see how particles or even a collection of particles can have three times that kinetic energy can remain bound to that planet by gravity.

• 174 Jan Pompe // Aug 11, 2008 at 2:14 pm

  Neal,

  “Then [KE] = 0, and all the gas molecules are sitting on the Earth’s surface, at r = R.”

  Ouch!!! and we are each of us buried under 10 kilotons of solidified/liquified gas. What is that pressure in mega-pascals?

  I’m sorry Neal it just doesn’t work, but even though I don’t really have time for it I am enjoying it.

  “I’m liking this result better all the time.”

  So am I Neal so am I.

• 175 Neal J. King // Aug 11, 2008 at 5:56 pm

  #173:

  I’m not quite sure what you’re getting at, but I’ll try to respond:

  - As long as the KE of a particle is less than the magnitude of the binding PE, that particle will be bound. A gravitational field can CERTAINLY hold a particle, the KE of which exceeds (1/2)(-PE).

  - The factor of (1/2) applies to the time-average bulk energies (KE and PE) of the whole collection of particles. That factor is specific to gravity (based on its 1/r dependence), other factors apply to other power-law dependencies.

  - You’re thinking about the case of a circular orbit. If you look at an elliptical orbit, at its perihelion it can have much more KE than -PE/2; and at the aphelion, the opposite can be the case. All that’s needed for binding is that the total energy, KE + PE , is negative.

• 176 Neal J. King // Aug 11, 2008 at 6:02 pm

  #174:

  What I’m pointing out is that the equation is doing the right things in the limiting case of KE = 0: the particles are resting on the surface, the planetary surface is supporting them, and the result of the VT calculation is met.

  Pressure is the normal force felt by the surface which is additional to what would have been there if there was no gas. In this case, all the molecules would be resting in the (R, R+d) zone, experiencing the upward push of V_o/d, the pressure is the area density of molecules on the surface * V_o/d.

  So the formula is doing exactly what it should be doing.

• 177 Neal J. King // Aug 11, 2008 at 7:39 pm

  #173, further thought:

  There is no upper limit to temperature. So there’s nothing to stop the temperature from getting as hot as it needs to “keep pace” with the PE at 1/R.

• 178 Jan Pompe // Aug 11, 2008 at 8:23 pm

  Neal,

  “I’m not quite sure what you’re getting at, but I’ll try to respond:”

  Newton’s second law. It’s sometimes easy to lose track of the simple stuff that underlies it all.

  “A gravitational field can CERTAINLY hold a particle, the KE of which exceeds (1/2)(-PE).”

  No it can’t because as I have just shown for the second time in this thread that when the force holding a body in circular motion is equal to the outward centrifugal force i.e. it cannot move out the kinetic energy is equal to half the potential. This holds for a single particle around a massive body and as the virial theorem shows for a many particle gravitationally bound system system.

  “- The factor of (1/2) applies to the time-average bulk energies (KE and PE) of the whole collection of particles”

  Or as I have just shown a single body orbiting a more massive one like a moon around a planet or bullroarer whirring around an indigenous Austalian’s head.

  “All that’s needed for binding is that the total energy, KE + PE , is negative.”

  That would be KE + PE/2 is negative and if KE > -PE/2 the body is no longer held but starts to accelerate out because the outward bound force is greater than the holding force. If KE = 3PE/2 the outward force will be 3 times greater than the holding force.

• 179 Neal J. King // Aug 11, 2008 at 8:37 pm

  #178:

  You are over-focusing on circular orbits. For an elliptical orbit, there are times when the KE is significantly larger than the -PE: that’s why the particle moves out.

  For someone being so snotty about Newton’s second law, you are wildly off-base on classical mechanics. I suggest you spend some time learning the basics. One possible starting point:
  - Circular orbits: http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravPotEnergy.htm
  - Elliptical orbits:
  http://galileo.phys.virginia.e.....Orbits.htm

  Hint: As long as the orbit is elliptical (embracing circular), the particle is bound.

• 180 Neal J. King // Aug 11, 2008 at 8:39 pm

  #179: Correction:

  “For an elliptical orbit, there are times when the KE is significantly larger than the -PE: that’s why the particle moves out.”
  => “For an elliptical orbit, there are times when the KE is significantly larger than the -PE/2: that’s why the particle moves out.”

• 181 Jan Pompe // Aug 12, 2008 at 7:35 pm

  Neal;

  “You are over-focusing on circular orbits. For an elliptical orbit, there are times when the KE is significantly larger than the -PE: that’s why the particle moves out.”

  Of course it does but have you forgotten already that the viral theorem is only concerned about averages or complete cycles of oscillations or orbits? The averages of {KE} < {PE}/2 over a complete orbit will hold it the circular orbit is just a trivial case where the averages = the instantaneous. Now for an elliptical orbit if the average {KE} > {PE}/2 the planet will leave the orbit.

• 182 Pat Cassen // Aug 12, 2008 at 9:19 pm

  Jan -
  “Now for an elliptical orbit if the average {KE} > {PE}/2 the planet will leave the orbit.”

  A particle (planet) in an elliptical orbit is in a bound orbit, by definition.

  This discussion of orbits is getting pretty far from the virial theorem and even further from Miscolczi’s mangling of it.

• 183 Neal J. King // Aug 12, 2008 at 9:21 pm

  I kind of see what you’re getting at, but there are still problems:

  1) It is still a fact that an object of mass m is gravitationally bound if its total energy,
  E = KE + PE
  = mv^2/2 - GMm/r

  is ≥ 0.

  2) However, since KE and PE vary with time, whilst in an undisturbed orbit E does not,
  E = avg-E = avg-KE + avg-PE

  So if we apply the VT to the system of two particles, one little particle m and one big heavy particle M, we get
  E = [E] = [KE] + [PE]
  = -(1/2)[PE] + [PE]
  = (1/2)[PE]

  So, over the course of the orbit,
  [KE] = - E = -(1/2)[PE] if the orbit is bound (which means 0 ≥ E) ; if E > 0, the VT doesn’t apply, because the system is unbounded.

  3) So you can talk about what happens in the case that [KE] > (-1/2)[PE], but it doesn’t make a lot of sense to attribute an unbounded orbit to this fact: the calculation of the average depends on the trajectory, the trajectory does not depend on the average over the trajectory. You can select particular sets of trajectories to focus on, based on their avg-KE/avg-PE ratios, but the trajectories have caused the ratio of averages, the ratio of averages have not caused the trajectories.

  4) And an elliptical orbit has the shape and extent of an ellipse: It is bounded.

  5) And a planet can never “leave its orbit” unless it’s bumped: Its orbit is its trajectory, which is “whatever it does”.

  So are you making progress with the virial calculation?

• 184 Neal J. King // Aug 12, 2008 at 9:26 pm

  #183, correction on item 1):

  “is ≥ 0.” should be “is ≤ 0.”

  Pat Cassen’s posting was done while I was drafting #183. I am in total agreement with him.

• 185 Neal J. King // Aug 12, 2008 at 9:49 pm

  #174:

  Going back to this question:
  “Ouch!!! and we are each of us buried under 10 kilotons of solidified/liquified gas. What is that pressure in mega-pascals?”

  The answer is, “Very slightly more than 0.1 MPa”: in other words, just slightly more than atmospheric pressure right now. Since the atmosphere only goes up a short distance compared to the radius of the Earth, we can ignore the geometrical effects and say that the pressure exerted on the gas from the ground neutralizes the pull from the Earth (this also ignores the variation of the gravitational attraction with distance). So today’s pressure already supports the weight of the atmosphere, and on a T=0 Earth, it will still do that.

• 186 Jan Pompe // Aug 13, 2008 at 10:01 am

  Pat,

  “A particle (planet) in an elliptical orbit is in a bound orbit, by definition.”

  Remarkable I would never have guessed. Then if you somehow cause {KE} > {PE}/2 it’s no longer an elliptical orbit and the planet is off to a new home. I do believe this little factoid is put to good use by the folks sending probes to the outer planets.

  What about an answer to my question in #168?

• 187 Neal J. King // Aug 13, 2008 at 11:35 am

  wrt #168: As R approaches 0, both [KE] and [-PE] can become quite large (in the case that T is not = 0), so I don’t see a problem. And in the case that T = 0, #172 & #185 show that the formula describes the situation perfectly: the gas molecules just sit on the ground, and the pressure supports them against gravity.

  So overall, I don’t see any contradiction.

  What result are you proposing instead?

  wrt #186: As I discussed in #183, item 3), talking about “causing [KE] to be > [PE]/2″ is like putting the cart before the horse. The [] terms are averages over time, and are properties of the trajectory as a whole.

  If a probe is in a bound orbit, its energy = E = KE + PE is < 0.

  If you want it to go into a bound orbit that goes farther out, you give it more KE, such that its new energy is still negative, but less negative than before.

  If you want it to go into an unbound orbit, you give it enough KE for the total energy to go ≥ 0: the extra energy kick is going to be
  d(KE) ≥ - (KE_now + PE_now) = - E_now

  As suggested by the terminology, the relevant required energy-kick is determined by the current values of KE, PE and thus E; not by reference to some average over time: You can only determine the average over time after you determine the trajectory.

• 188 Jan Pompe // Aug 13, 2008 at 1:30 pm

  Neal,

  “The answer is, “Very slightly more than 0.1 MPa”: in other words, just slightly more than atmospheric pressure right now.”

  Oh a mass of ~99,800 Kg resting on every square metre is .1MP ? Actually its .0978MP (to 2 sig figures) with a current T !=0 pressure of .1013 MP it is slightly less.

  “we can ignore the geometrical effects and say that the pressure exerted on the gas from the ground neutralizes the pull from the Earth”

  It only neutralises the downward pull due to gravity it doesn’t neutralise the upward pull due to KE and that is where it matters. The gas has no upward pull for the gravity the neutralise and that is the relevant metric to the virial theorem.

• 189 Neal J. King // Aug 13, 2008 at 2:36 pm

  #188:

  - I’m not sure where you’re deriving your atmospheric mass from, but I’m getting my information from the equation of hydrostatic equilibrium:
  dP/dr = - (mass-density)g

  where g = local acceleration of gravity.

  Integrate from bottom of the atmosphere (r = R) to the top (about 100 km up). Since g does not change very much over 100 km (when R = 6400 km), as an excellent approximation we can set g equal to the constant = 9.8 m/s^2. So then we get:
  0 - P(r=R) ~ - Integral (mass-density)*g
  = - g*Integral (mass-density)

  or:
  P(r=R) ~ g * Integral (mass-density)

  This is true right now, with the current atmosphere.

  When we change the situation to T = 0, the height of the atmosphere becomes 0, so the approximation becomes exact. Since at finite temperature, the atmosphere had some thickness, so at higher altitudes the gravitational acceleration is slightly less, hence the T=0 case will result in a slightly higher pressure than 1 atmosphere. Not by much: dg/dr = -2GM/r^3, so the variation in g will be about
  dg/dr*height = -(2GM/R^3)*height
  = - (2*height/R)*g

  So
  delta-g/g ~ -2*100/6400 = 1/32 or about 3%.

  So the result of freezing the atmosphere will result in the total atmospheric pressure increasing by less than 3%.

  That was my point.

  And, yes there is an upward force on the gas, otherwise it would only feel gravity and pass through the ground towards the center: In freshman physics it’s generally a “normal” force, and in this case it is the reaction-partner to what is detected as the atmospheric pressure.

• 190 Pat Cassen // Aug 13, 2008 at 5:20 pm

  Jan – “…I would never have guessed.”

  Sorry, I didn’t intend to be condescending, but I couldn’t make sense of your “Now for an elliptical orbit if the average {KE} > {PE}/2 the planet will leave the orbit.” I see now what you were trying to say, but still find it irrelevant.

  The answer to your question in #168 is Yes, we agree that if you shrink the radius that the pressure at the surface will be greater. (This does not contradict anything that Neal J. King has said.) The point of the thought experiment was that changing only the location of the surface would change atmospheric structure, even where the gravitational potential remained constant, so the surface was NOT “passive”. I apologize if I am misunderstanding what you mean by “passive”.

  The discussion has drifted to simple orbits and the hydrostatic equation, the former understood since Newton and the latter at least since Euler. Are there further mysteries to be discovered here? As Pascal’s imaginary dog said, I think not.

  I came by here with the hope of finding out what Miskolczi (did I get the spelling right this time?) was doing with the virial theorem. Neal J. King has, with patience, explained quite clearly what he should have been doing. NJK derived the proper equation, which agrees with the P&S paper, and has shown that it behaves properly in various examples. You seem to have a problem with his analysis; I don’t.

  I still don’t know what Miskolczi was doing, but it may not be worth trying to figure out what’s going on in the mind of someone who can write down so many wrong equations.

• 191 Pat Cassen // Aug 13, 2008 at 5:23 pm

  Oops. Misplaced an italic indicator in the above. Sorry for all that unwarranted emphasis.

  No Preview; how do I correct that kind of thing?

• 192 sadunkal // Aug 13, 2008 at 5:51 pm

  At this stage, I’d like to point out that Miskolczi was working at NASA until he resigned. Assuming that he wrote down many wrong equations, it’s interesting that NASA hires scientists as incompetent as he ‘is’. Maybe questioning the competence of NASA scientists should become a more important part of the debate…

  But before I make such an assumption about Miskolczi, I’d like to see his response to Neal’s review. Did he respond yet?

  sadunkals last blog post..HIV/AIDS Hypothesis - Medical Industrial Complex?

• 193 Neal J. King // Aug 13, 2008 at 6:04 pm

  #192, sadunkal:

  When I first sent him the list, he answered right away that:
  - He would be out of town until August, and planned to respond in that period;
  - That some clarification could be found in the presentation by Zagoni and on blogs.

  I replied that I hadn’t found Zagoni and he blogs helpful in pinning down the logic I was inquiring about; and that I looked forward to reading his response when he got back home.

  Several days ago, I re-sent the letter, but haven’t heard anything. It’s possible that his trip was extended. I make no assumptions.

• 194 Pat Cassen // Aug 13, 2008 at 6:26 pm

  Re: sadunkal, #192

  Miskolczi’s resignation letter, which is posted at http://hps.elte.hu/zagoni/Proo.....theory.htm
  refers to AS&M as his employers, not NASA.

  AS&M is a NASA contractor, which provides technical services. So Miskolczi apparently worked at a NASA center, but was employed by AS&M. This is conjecture; I have no first-hand knowledge of the situation.

  By the way, the same website informs us that his paper was rejected by reviewers at three prominent journals, and was withdrawn from a fourth by NASA. Such an action would be entirely within NASA’s (uncontroversial) operating procedures if a NASA supervisor made the judgment that, after the paper had received a fair evaluation, it was found to be scientifically unsound.

• 195 Neal J. King // Aug 13, 2008 at 6:53 pm

  Pat Cassen:

  - Yes, it seems clear that Miskolczi was employed by AS&M, not by NASA directly. I knew someone who also did something like that through a different company: She was principally hired to work on programs for projects directed by others, but I think she was also allowed to take some time for her own research; however, my impression was that that was low-priority, from the viewpoint of her management. To get the full story, we would have to contact someone who knew him from his time at NASA.

  - What further surprises me is that there does not seem to be any sign of an internal review: When I worked for Bellcore years ago, you were allowed and even encouraged to submit papers for publication - but the paper had to survive an internal review by two folks outside of your immediate business area, before it could be sent outside. I would think that NASA would have a similar policy as well (perhaps without the demand that the internal reviewers be outside your area, since Bellcore is a commercially related organization and NASA is a scientifically oriented organization).

  - Preview? The only preview I’m aware of on this site is foresight and careful checking…

  - And, yes, I think the formula holds up pretty well in special cases; and the calculation, while using peculiar boundaries, is fairly straightforward. P&S certainly didn’t seem to have had any trouble with peer review: It was approved within 5 months of reception.

• 196 Jan Pompe // Aug 13, 2008 at 8:17 pm

  Neal,

  And, yes there is an upward force on the gas, otherwise it would only feel gravity and pass through the ground towards the center: In freshman physics it’s generally a “normal” force, and in this case it is the reaction-partner to what is detected as the atmospheric pressure.

  In freshman physics it would also fall straight through to an equal distance from the centre on the other side if there was nothing to stop it. Your extra pressure term is unnecessary and your walls in 166 are imaginary.

• 197 Neal J. King // Aug 13, 2008 at 8:28 pm

  #196:

  In freshman physics (and real physics), the normal force keeps the gas from going through the floor; and that is the case of interest.

  Go ahead and do your own calculation. We’re all interested to see what you come back with.

• 198 sadunkal // Aug 13, 2008 at 8:31 pm

  #194, Pat Cassen:

  Oh, alright… I shift the direction of my suggestion from NASA to AS&M Inc. -whatever that is- and their cooperations with NASA then…
  And I’d still be interested in getting the details and the full story about all this, either from Miskolczi or anyone else, doesn’t matter…

  sadunkals last blog post..HIV/AIDS Hypothesis - Medical Industrial Complex?

• 199 Neal J. King // Aug 13, 2008 at 9:06 pm

  #198, sadunkal:

  - AS&M: But if the case of my friend (who had a Ph.D in astrophysics) is typical, at least some of these contractors are hired to provide technical services (like programming) not scientific insight. They could be allowed to publish, without being encouraged (by NASA management) to publish. Of course, I can’t say what was expected specifically of Miskolczi.

  - As far as getting the details & full story: So would we all…

• 200 Nick Stokes // Aug 14, 2008 at 4:13 am

  Neal, I’ve worked for a NASA-like organisation (CSIRO). All papers published using your CSIRO affiliation required internal review before being sent out. You could publish under your own name without citing affiliation (unless you used proprietary information etc) but you wouldn’t get credit for it. Of course, they could insist that you didn’t do it on company time. I suspect that NASA didn’t say he couldn’t publish - just that he couldn’t cite his NASA affiliation.

• 201 Neal J. King // Aug 14, 2008 at 6:05 am

  #200, Nick Stokes:

  In this case, it seems that his NASA supervisor withdrew the paper from consideration at one journal after the paper had been given negative judgments at two or three others.

• 202 Nick Stokes // Aug 14, 2008 at 6:15 am

  #201
  Neal, the NASA supervisor couldn’t “withdraw” the paper - that’s between the author and editor. He could have directed the author not to use NASA affiliation. Or if he wanted to get really heavy, he could have demanded that the journal not cite that affiliation.

• 203 Neal J. King // Aug 14, 2008 at 6:33 am

  #201, Nick Stokes:

  I’m just reporting what Zagoni’s site said, in their section on the results of the peer reviews. Look at Section 7, “Endnote” of hps.elte.hu/zagoni/Proofs_of_the_Miskolczi_theory.htm :
  “Applied Optics: Withdrawn by NASA.

  Dr. Miskolczi’s reaction to this latter event was fast and adequate; here is the copy of his letter of resignation. ”

  Unfortunately, right now the site appears to be unreachable; but perhaps later you can read the letter for yourself. That’s all I’m going on.

• 204 Sadun Kal // Aug 14, 2008 at 2:10 pm

  Although I tend to agree with #201, I guess NASA is powerful enough to literally prevent papers from getting published without giving an explanation. I believe that they’re perfectly capable of asking/pressuring a journal to not to publish a certain paper. We’re not living in a perfect world, sometimes the rules don’t matter so much.

  A paper examining modern science can be found here at WikiChecks, you can review it if you wish:
  http://landshape.org/stats/sci.....h-cartels/

  A quote from the abstract:

  “…Science is now altogether different from the traditional disinterested search, by self-motivated individuals, to understand the world. What national and international organizations publicly proclaim as scientific information is not safeguarded by the traditional process of peer review. Society needs new arrangements to ensure that public information about matters of science will be trustworthy…”

• 205 Neal J. King // Aug 16, 2008 at 5:15 pm

  In the meantime, the discussion on the Virial-Theorem calculation is going on at:
  http://www.climateaudit.org/ph.....9474#p9474
  and thereabouts.

  We seem to be getting closer to the same sheet of music.

• 206 Neal J. King // Aug 18, 2008 at 12:21 am

  Nick Stokes:

  I have been trying to explain to Jan Pompe (on the ClimateAudit discussion thread, at http://www.climateaudit.org/ph.....9509#p9509) how to derive the result of the Virial Theorem in the case of a solid planet, but it hasn’t worked.

  So tomorrow I will post a completely independent proof that is based on the equations of hydrostatic equilibrium.

  I believe both proofs are perfectly valid, but the one based on hydrostatic equilibrium is harder to get confused about.

• 207 Jan Pompe // Aug 18, 2008 at 8:21 am

  how to derive the result of the Virial Theorem in the case of a solid planet, but it hasn’t worked.

  Of course it hasn’t because what you have been doing is incorrect.

  First of all the virial theorem
  dG/dt = 2{KE} + {PE}

  is the derivative (rate of change) of the summed product of the momenta and their distance form the CoG of an ensemble of particles (the Virial). This in turn is the time derivative of the moment of inertia of that ensemble (google for description) I = Sum_i{m_i.r_i^2} which does not describe or provide the correct I for a planet with an atmosphere I = Sum_i{m_i.r_i^2} + 2MR^2/5. The time derivative (rate of change due changes in mass distribution) of this is once again the virial because since the second term cannot change it’s mass distribution it is a constant so it disappears. Thus we are left with the very same expression for the rate of change of the virial.

  Of course the virial does not apply where 2{KE}<{PE} so in the attempt to generalise the virial theorem to cover this contingency he fiddles with the derivative and changes it to:
  G = Sum_i{p_i . r_i} (using product rule)
  dG/dt = 2{KE} - {PE} + C 1
  (where C is the “time averaged” Normal upward force at the surface constrained so it can’t change IOW a constant.

  If what he has done is OK then we should be able to integrate the rate of change over time and return to the original description (this time I’m not going to correct the sign).

  we can expand 1 leaving out subscripts:
  dG/dt = Sum{mv^2} - Sum{dp/dt.r} + C

  integrating from t=O to t
  G = int [ Sum{mv^2} - Sum{dp/dt.r} + C] dt
  = Sum{mv.r} - Sum{p.r}
  = Sum{p.r} - sum{p.r} + Ct
  =Ct

  We have an expression here that has lost completely the structure of the original problem and grows with limit (but has no soil to grow on).

  OF course correcting the sign on the second term leads to
  G=Sum{p.r} + Ct which also grows without limit.

  Now for the virial to apply the lower atmosphere must have enough energy to balance the gravitational force at the surface for it to do this a particle needs to have a speed of around 8000m/s at 288K a Nitrogen molecule has a speed of around 500m/s. Under these circumstances I’d suggest using hydrostatic equilibrium rather than the orbital equilibrium of the virial theorem.

• 208 Neal J. King // Aug 18, 2008 at 8:29 am

  #207, Jan Pompe:

  No, the reason that it “didn’t work” was because you refused to do the calculation. Instead you insist on integrating something that was just differentiated, forgetting that there was an average in-between. You can’t find out anything new by doing that.

  OK, here is the proof of the same result, without recourse to the VT. It was inspired by the P&S proof of the (3/2) factor for the planar/flat-Earth case; in fact, I’m a little surprised they didn’t do this as well.

  Equation of hydrostatic equilibrium:
  dP/dr = - (mass_density)*(local_acceleration of gravity)
  = - n*m(GM/r^2)
  = - (GMm)n/r^2
  = - an/r^2

  eqn(1): dP/dr = - an/r^2

  Multiplying by r:

  -an/r = r(dP/dr)
  = (d/dr)(rP) – P

  So:
  eqn(2): P – an/r = (d/dr)(rP)

  Multiplying by (4pi*r^2):

  eqn(3): (4pi*r^2)*P - (4pi*r^2)(an/r) = (4pi*r^2)(d/dr)(rP)

  Because of spherical symmetry, when we integrate the LHS of eqn(2) over r, we are actually getting the volume integral over all angles, from the inner radius to the outer radius:

  Integral (4pi*r^2)*P dr – Integral (4pi*r^2)*(an/r) dr
  = Integral P (d^3)x + Integral (-an/r) (d^3)x

  Note that since the gravitational potential is –a/r, the second term is actually:
  Integral (n*U(r.)) (d^3)x = [U_s]

  So we get from the above and eqn(3):
  eqn(4): Integral P (d^3)x +[U_s] = Integral (4pi*r^2)(d/dr)(rP) dr

  The RHS of eqn(4) can be written as:
  Integral (4pi) [(d/dr)(r^2 * (rP)) – (rP)*(d/dr)(r^2)] dr
  = (4pi)(r^3)*P – (4pi) Integral 2(r^2)*P dr
  = (4pi)(r^3)*P – 2*Integral (4pi*r^2)*P dr

  So we get:
  eqn(5): Integral P (d^3)x +[U_s] = (4pi)(r^3)*P – 2*Integral (4pi*r^2)*P dr

  But notice that
  2*Integral (4pi*r^2)*P dr

  is just 2*Integral P (d^3)x, so eqn(5) becomes

  eqn(6): Integral P (d^3)x +[U_s] = (4pi)(r^3)*P – 2* Integral P (d^3)x

  or:
  eqn(7): 3*Integral P (d^3)x + [U_s] = (4pi)(r^3)*P

  Now the end is in sight: Just as shown before in both calculations,

  Due to the perfect-gas law, P = nkT, so

  Integral P (d^3)x = Integral nkT (d^3)x

  Due to the equipartition theorem, KE per molecule = (3/2)kT, so
  Integral nkT (d^3)x = (2/3)[KE]

  Therefore,
  eqn(8): Integral P (d^3)x = (2/3)[KE]

  Combining (7) and (8) give:
  eqn(9): 2[KE] + [U_s] = (4pi)(r^3)*P

  where the RHS has to be evaluated at the two extreme values of r.

  When r = large, P = 0, so the upper value does not contribute.

  When r = R, P = P(r=R), so the lower value gives:
  -(4pi)R^3 P(r=R)

  We end up with:
  eqn(10): 2[KE] + [U_s] + (4pi)R^3*P(r=R) = 0

  which is the same result we obtained through the direct calculation of the VT.

• 209 Jan Pompe // Aug 18, 2008 at 8:48 am

  “Neal”

  “nstead you insist on integrating something that was just differentiated, forgetting that there was an average in-between. You can’t find out anything new by doing that.”

  You can find out something very important that way: whether you did it right and since we get nonsense going back the other way it’s obvious you did something very wrong.

  It is the simplest of sanity tests and it failed.

  I really can’t be bothered with the rest of the post I’m sure you got the hydrostatic equilibrium calculations right. As I’ve tried to point out before the virial theorem applies to multi-particle systems that are bounded solely by gravity (no upward forces) they could be collapsing as they shed energy and get hotter. Do you think that describes earths atmosphere?

  I don’t I think earth has already collapsed (all the evidence I can see points to it) now the atmosphere at least where we live has a positive heat capacity not a negative one as I’ve just described.

• 210 Neal J. King // Aug 18, 2008 at 9:23 am

  #209, Jan Pompe:

  I’ve proven the exact same formula by two independent means. The fact that you cannot follow the reasoning on one or both of these arguments does not disturb me, as I have examined your complaints and found them groundless. As someone once said, “I am obliged to find you an explanation; I am not</i< obliged to find you an understanding.”

  I would have no hesitation about submitting the VT derivation to an expert on the area, and I’m confident it would be accepted without a second thought. My confidence is reinforced by the fact that P&S got this same argument published in an internationally respected peer-reviewed journal of physics in 2003, in a short and very easy-to-understand article, with the minimal period of review (a few months): in other words, the result is not seen as controversial.

  Unless you have actually new material on this matter, I do not plan to entertain further discussion with you on this point. If others want to comment, I will respond to them. But I have been over the same ground with you many times, and I find your objections without further interest.

• 211 Jan Pompe // Aug 18, 2008 at 9:40 am

  Neal

  “I’ve proven the exact same formula by two independent means.”

  But failed the simplest of sanity checks of a type which it must pass to be considered valid.

  give it up Neal whatever it is that you’ve verified by independent means does not describe an invariant atmosphere around a rigid core.

  If I can’t follow your reasoning have ever considered the possibility that it is because it’s nonsense?

  Any way I’m done at this point I don’t think you have anything worthwhile more to offer.

  Perhaps a course in first year calculus will help.

• 212 Pat Cassen // Aug 18, 2008 at 3:08 pm

  Jan, Neal -
  I have been following your discussion and, although I agree that it should be ended, I am sorry that you did not come to agreement.

  I have gone over most (but not all) of Neal’s analysis and am satisfied that it is correct. I cannot understand Jan’s objections; they make no sense to me.

• 213 Nick Stokes // Aug 18, 2008 at 9:25 pm

  What Pat said. I would have followed it even more closely, except that I keep remembering that Miskolczi has not shown how his statements about IR fluxes could be connected to any energy relation that could come out of the virial theorem.

• 214 Jan Pompe // Aug 18, 2008 at 9:41 pm

  Pat

  “I cannot understand Jan’s objections; they make no sense to me.”

  Perhaps you can take your own advice and do your own calculation from a good description of the moment of inertia of a planet + it’s atmosphere. See if you come up with something different.

• 215 Neal J. King // Aug 18, 2008 at 9:44 pm

  #212, Pat Cassen:

  - One analysis just applies the VT with boundaries.
  - The other calculation is basically hydrostatic equilibrium, perfect-gas law, and a clever choice of integrations by parts.

  #213, Nick Stokes:

  Yes, my still-outstanding question to him on that point is, Even assuming the “classic” statement of the VT, I do not understand the relationship you describe in item (g) of Section 2 connecting [PE] and temperature. What are the actual equations that define this relationship?

  Likewise, in Section 3.1 you describe a relationship between [KE] and EU. But [KE] is the total kinetic energy, and EU is an energy flux: How do you identify a flux with a bulk quantity? Again, what are the actual equations that define this relationship?I hope he gets back to me on these questions.

• 216 Neal J. King // Aug 18, 2008 at 9:46 pm

  OK, quotations don’t work. I’ll try again on the response to Nick Stokes:

  Yes, my still-outstanding question to him on that point is,

  “Even assuming the ‘classic’ statement of the VT, I do not understand the relationship you describe in item (g) of Section 2 connecting [PE] and temperature. What are the actual equations that define this relationship?

  Likewise, in Section 3.1 you describe a relationship between [KE] and EU. But [KE] is the total kinetic energy, and EU is an energy flux: How do you identify a flux with a bulk quantity? Again, what are the actual equations that define this relationship?”

  I hope he gets back to me on these questions.

• 217 Jan Pompe // Aug 19, 2008 at 1:26 am

  Pat,

  According to Neal #215

  “- One analysis just applies the VT with boundaries.
  - The other calculation is basically hydrostatic equilibrium, perfect-gas law, and a clever choice of integrations by parts.”

  They are both hydrostatic each one just has a different shaped container.

• 218 Neal J. King // Aug 19, 2008 at 7:14 am

  #217, Jan Pompe:

  No: Only one equation is based upon the equation of hydrostatic equilibrium.

  The other is an application of the Virial Theorem to a situation in which there is:
  a) One central attracting mass and
  b) An excluded spherical volume.

  The physical configuration described by these two is the same, but the physical principles applied are different. However, since the results involve the same physical quantities (average KE, average PE), the results have to be the same: Physics is self-consistent.

• 219 Jan Pompe // Aug 19, 2008 at 9:28 am

  Neal,

  “The other is an application of the Virial Theorem to a situation in which there is:
  a) One central attracting mass and
  b) An excluded spherical volume.”

  Glad you agree it’s a different shaped container. There is conceptually no difference here between this and the physically constrained on 4 sides column of air open at the top on your flat earth model just the shape.

  I agree physics is self consistent.

• 220 Neal J. King // Aug 19, 2008 at 10:38 am

  #219, Jan Pompe:

  No: it’s not a differently shaped container. If you follow the derivation carefully, you see that the hydrostatic equilibrium calculation:
  - assumes a spherical geometry
  - assumes the gravitational force is due to a central attracting mass
  - has a particular pressure at r=R
  So the physical configuration is the same as used in the VT derivation.

  However, the result from the hydrostatic equation is more general, because it gives the result for any given R, whether or not there is a floor at that value. What that means is that you can calculate the KE and PE averages for a portion of the atmosphere (all r > X) and relate it to the pressure at r = X. But in particular, it applies when there IS a floor at r = R, which is the case that is applicable to the VT (because you need an actual floor to keep the molecules bounded within a region).

• 221 Neal J. King // Aug 19, 2008 at 10:42 am

  #220, cont’d:

  And there are no other walls than the floor, in this case. They are not needed in a spherical geometry, because the surface of sphere is a closed surface.

• 222 Jan Pompe // Aug 20, 2008 at 2:17 am

  Neal,

  “They are not needed in a spherical geometry, because the surface of sphere is a closed surface.”

  Exactly, I think you are beginning to learn.

• 223 Neal J. King // Aug 20, 2008 at 12:55 pm

  #222, Jan Pompe:

  - Vertical walls and a floor are needed in the 1-D case.
  - Floors are needed in the 3-D spherical case.

  This applies to the VT calculations.

  For hydrostatic calculations, it’s not necessary that the system be bounded.

  The only thing I expect to learn from you is whether you’ll ever face the music and admit that I’m right. To pin it down very precisely: I gave two formulas for the [KE]-[PE] relationship in the flat-Earth and spherical Earth in July, and since then have produced alternative derivations for the same formulas via the equations of hydrostatic equilibrium. I stand by those formulae.

• 224 Neal J. King // Aug 20, 2008 at 4:51 pm

  I have revised my letter to Miskolczi to fix a typo and to add a reference to Pachecho & Sanudo.

  http://landshape.org/stats/wp-.....ions-4.pdf

  I hope he gets back to me on it: It’s way past 1 August by now.

• 225 Jan Pompe // Aug 21, 2008 at 2:09 am

  Neal,

  “- Vertical walls and a floor are needed in the 1-D case.
  - Floors are needed in the 3-D spherical case.”

  That’s right floor at the bottom open at the top limited area :- as I said same story different shape.

  “This applies to the VT calculations.”

  That’s right too.

  You do get some things right. I have absolutely no trouble saying so.

• 226 Neal J. King // Aug 21, 2008 at 9:39 am

  #225, Jan Pompe:

  The real issue is the equations; upon which pliny and I are in complete agreement.

  Pat Cassen would be fully appraised of this also if he were to look at pliny’s cleaned-up version at:
  http://www.climateaudit.org/ph.....;start=790
  posted by pliny on Mon Aug 18, 2008 9:01 pm.

• 227 Jan Pompe // Aug 21, 2008 at 10:01 am

  Neal

  “The real issue is the equations; upon which pliny and I are in complete agreement.”

  I agree the equations are the real issue and I’m sure you gain great comfort that Nick is in agreement with you Nicks a good mathematician but the equations don’t describe the system in question very well. In fact I would say when you changed

  2 + = 0

  to

  2 - = -4*pi*R^3P(R)

  or was it

  2 + = -4*pi*R^3P(R)

  you started heading into the long tall grass.

  That’s not the only problem of course do you want a list?

• 228 Neal J. King // Aug 21, 2008 at 10:15 am

  #227, Jan Pompe:

  Not really: I’ve already satisfied myself that you haven’t been able to follow the argument. I see that hasn’t changed, and I don’t find it amusing to try to re-educate you.

• 229 Sadun Kal // Aug 21, 2008 at 10:41 am

  I think it’s a shame that it came to this… There are obviously problems with more than just the equations. Maybe both of you should take a step back and take a deep breath before you keep going. You have a common goal after all as far as I can see; to figure out what Miskolczi had in mind and if he was correct.

  Maybe simply wait for him to respond -or even pressure him to respond… Because your discussion isn’t productive at all currently.

• 230 Jan Pompe // Aug 21, 2008 at 10:57 am

  Neal,

  “Not really: I’ve already satisfied myself that you haven’t been able to follow the argument.”

  Perhaps it’s the sheer novelty of your procedure that I can’t get my head around. I’ve never seen anyone add a constant of integration to the function before doing the integration before.

• 231 Jan Pompe // Aug 21, 2008 at 11:08 am

  Sadun Kal

  “You have a common goal after all as far as I can see; to figure out what Miskolczi had in mind and if he was correct.”

  On this issue I’ve not changed my views since I wrote this in March Neal has given me no reason to. Ferenc pointed it out as a reasonable explanation as to what he had in mind. I’m satisfied that I do have an idea of what that is. I don’t believe that Neal is the slightest bit interested in what he had in mind.

  http://www.climateaudit.org/ph.....t=20#p2939

• 232 Neal J. King // Aug 21, 2008 at 11:22 am

  Sadun Kal:

  I have gone around the bush numerous times with Jan Pompe, and he never really understands. Nick Stokes and Pat Cassen have also looked at my argument, and both seem to be fine with it.

  wrt “not caring about what Miskolczi thinks”: No doubt, that explains why I have sent Miskolczi a letter with my questions. He responded with an indication he would get back to me in August.

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