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Back to the future

Filed under: — gavin @ 30 April 2008 - (Español)

A few weeks ago I was at a meeting in Cambridge that discussed how (or whether) paleo-climate information can reduce the known uncertainties in future climate simulations.

The uncertainties in the impacts of rising greenhouse gases on multiple systems are significant: the potential impact on ENSO or the overturning circulation in the North Atlantic, probable feedbacks on atmospheric composition (CO2, CH4, N2O, aerosols), the predictability of decadal climate change, global climate sensitivity itself, and perhaps most importantly, what will happen to ice sheets and regional rainfall in a warming climate.

The reason why paleo-climate information may be key in these cases is because all of these climate components have changed in the past. If we can understand why and how those changes occurred then, that might inform our projections of changes in the future. Unfortunately, the simplest use of the record – just going back to a point that had similar conditions to what we expect for the future – doesn’t work very well because there are no good analogs for the perturbations we are making. The world has never before seen such a rapid rise in greenhouse gases with the present-day configuration of the continents and with large amounts of polar ice. So more sophisticated approaches must be developed and this meeting was devoted to examining them.

The first point that can be made is a simple one. If something happened in the past, that means it’s possible! Thus evidence for past climate changes in ENSO, ice sheets and the carbon cycle (for instance) demonstrate quite clearly that these systems are indeed sensitive to external changes. Therefore, assuming that they can’t change in the future would be foolish. This is basic, but not really useful in a practical sense.

All future projections rely on models of some sort. Dominant in the climate issue are the large scale ocean-atmosphere GCMs that were discussed extensively in the latest IPCC report, but other kinds of simpler or more specialised or more conceptual models can also be used. The reason those other models are still useful is that the GCMs are not complete. That is, they do not contain all the possible interactions that we know from the paleo record and modern observations can occur. This is a second point – interactions seen in the record, say between carbon dioxide levels or dust amounts and Milankovitch forcing imply that there are mechanisms that connect them. Those mechanisms may be only imperfectly known, but the paleo-record does highlight the need to quantify these mechanisms for models to be more complete.

The third point, and possibly the most important, is that the paleo-record is useful for model evaluation. All episodes in climate history (in principle) should allow us to quantify how good the models are and how appropriate are our hypotheses for climate change in the past. It’s vital to note the connection though – models embody much data and assumptions about how climate works, but for their climate to change you need a hypothesis – like a change in the Earth’s orbit, or volcanic activity, or solar changes etc. Comparing model simulations to observational data is then a test of the two factors together. Even if the hypothesis is that a change is due to intrinsic variability, a simulation of a model to look for the magnitude of intrinsic changes (possibly due to multiple steady states or similar) is still a test both of the model and the hypothesis. If the test fails, it shows that one or other elements (or both) must be lacking or that the data may be incomplete or mis-interpreted. If it passes, then we a have a self-consistent explanation of the observed change that may, however, not be unique (but it’s a good start!).

But what is the relevance of these tests? What can a successful model of the impacts of a change in the North Atlantic overturning circulation or a shift in the Earth’s orbit really do for future projections? This is where most of the attention is being directed. The key unknown is whether the skill of a model on a paleo-climate question is correlated to the magnitude of change in a scenario. If there is no correlation – i.e. the projections of the models that do well on the paleo-climate test span the same range as the models that did badly, then nothing much has been gained. If however, one could show that the models that did best, for instance at mid-Holocene rainfall changes, systematically gave a different projection, for instance, of greater changes in the Indian Monsoon under increasing GHGs, then we would have reason to weight the different model projections to come up with a revised assessment. Similarly, if an ice sheet model can’t match the rapid melt seen during the deglaciation, then its credibility in projecting future melt rates would/should be lessened.

Unfortunately apart from a few coordinated experiments for the last glacial period and the mid-Holocene (i.e. PMIP) with models that don’t necessarily overlap with those in the AR4 archive, this database of model results and tests just doesn’t exist. Of course, individual models have looked at many various paleo-climate events ranging from the Little Ice Age to the Cretaceous, but this serves mainly as an advance scouting party to determine the lay of the land rather than a full road map. Thus we are faced with two problems – we do not yet know which paleo-climate events are likely to be most useful (though everyone has their ideas), and we do not have the databases that allow you to match the paleo simulations with the future projections.

In looking at the paleo record for useful model tests, there are two classes of problems: what happened at a specific time, or what the response is to a specific forcing or event. The first requires a full description of the different forcings at one time, the second a collection of data over many time periods associated with one forcing. An example of the first approach would be the last glacial maximum where the changes in orbit, greenhouse gases, dust, ice sheets and vegetation (at least) all need to be included. The second class is typified by looking for the response to volcanoes by lumping together all the years after big eruptions. Similar approaches could be developed in the first class for the mid-Pliocene, the 8.2 kyr event, the Eemian (last inter-glacial), early Holocene, the deglaciation, the early Eocene, the PETM, the Little Ice Age etc. and for the second class, orbital forcing, solar forcing, Dansgaard-Oeschger events, Heinrich events etc.

But there is still one element lacking. For most of these cases, our knowledge of changes at these times is fragmentary, spread over dozens to hundreds of papers and subject to multiple interpretations. In short, it’s a mess. The missing element is the work required to pull all of that together and produce a synthesis that can be easily compared to the models. That this synthesis is only rarely done underlines the difficulties involved. To be sure there are good examples – CLIMAP (and its recent update, MARGO) for the LGM ocean temperatures, the vegetation and precipitation databases for the mid-Holocene at PMIP, the spatially resolved temperature patterns over the last few hundred years from multiple proxies, etc. Each of these have been used very successfully in model-data comparisons and have been hugely influential inside and outside the paleo-community.

It may seem odd that this kind of study is not undertaken more often, but there are reasons. Most fundamentally it is because the tools and techniques required for doing good synthesis work are not the same as those for making measurements or for developing models. It could in fact be described as a new kind of science (though in essence it is not new at all) requiring, perhaps, a new kind of scientist. One who is at ease in dealing with the disparate sources of paleo-data and aware of the problems, and yet conscious of what is needed (and why) by modellers. Or additionally modellers who understand what the proxy data depends on and who can build that into the models themselves making for more direct model-data comparisons.

Should the paleo-community therefore increase the emphasis on synthesis and allocate more funds and positions accordingly? This is often a contentious issue since whenever people discuss the need for work to be done to integrate existing information, some will question whether the primacy of new data gathering is being threatened. This meeting was no exception. However, I am convinced that this debate isn’t the zero sum game implied by the argument. On the contrary, synthesising the information from a highly technical field and making it useful for others outside is a fundamental part of increasing respect for the field as a whole and actually increases the size of the pot available in the long term. Yet the lack of appropriately skilled people who can gain the respect of the data gatherers and deliver the ‘value added’ products to the modellers remains a serious obstacle.

Despite the problems and the undoubted challenges in bringing paleo-data/model comparisons up to a new level, it was heartening to see these issues tackled head on. The desire to turn throwaway lines in grant applications into real science was actually quite inspiring – so much so that I should probably stop writing blog posts and get on with it.

The above condensed version of the meeting is heavily influenced by conversations and talks there, particularly with Peter Huybers, Paul Valdes, Eric Wolff and Sandy Harrison among others.


523 Responses to “Back to the future”

  1. 451
    Rod B says:

    CobblyWorlds (448): Certainly it doesn’t pay to get real precise, maybe like trying to estimate ocean turbulence’s effect on solar reflection, though it wouldn’t hurt to at least address it. It just seems that the level that I am questioning (even though within even that scientific estimates have to be made) is pretty rudimentary and doesn’t take a lot of effort. The scientists who simply blab their gut-feel to convince people the loss of ice is imminent (and bad) do themselves a disservice. There is no need for them to make themselves look stupid when it is so easy to do it otherwise, especially if the outcome might change little.

    I’m not addressing nor refuting (yet, and maybe not ever – unless and until I get smarter) all of the other impacts and effects on Arctic ice/snow.

  2. 452

    #446 Rod B:

    Problem: I could swear I have seen (but not read) studies that assessed just such transitions. Could you have these transitions and maintain momentum within a system of (many) molecules? Or is this idea just goofy.

    You could, but it would involve a collision with a second molecule. It can be understood classically: just imagine a molecule rapidly spinning or resonating like a church bell, and you throw another molecule against it. Can you see it come back with a greater speed than it went in with? At the expense of the rotational/vibational energy of the other molecule. Or coming out with less energy than going in, having spun up or started vibrating the other molecule?

    Any snooker experience?

  3. 453
    Rod B says:

    Barton (449), With my 8 years in the computer industry (coupled with a zillion yrs. In telecom) I successfully avoided knowing much about coding. If pressed I could put out a DOS Batch pgm, but that’s about it. None-the-less I pretty much followed it, and while I would have a hard time accepting a 0.3K worldwide temperature increase for an ice/snow mass that receded from 60 to 61 degrees without getting into the specifics, you are doing, correctly or not, what I’ve been saying should’ve been done all along – a rigorous assessment rather than what appears to be pure handwaving. (To be fair, I don’t know that the others didn’t use rigor – they just didn’t present it as such.)

    Looks neat. I’ll try to wade through it and see if I have any scientific or mathematical questions or disagreements.

  4. 454
    Rod B says:

    Martin, Ray, et al: In a long ago time I played just a little pool: rotation, 8-ball, 9-ball. a little less billiards, and hardly any snooker… though I have been snookered a time or two.

    Without prejudice to some esoteric low probabilities as Ray implied, for general purposes: ignoring angular momentum can it be said that a molecule’s momentum exists only by virtue of its translation velocity, and the only way that momentum can change is through a collision with another molecule. It can not change by shifting momentum within itself because it has no comparable momentum to share. Then it follows that neither can energy be transferred from rotation/vibration DoFs to translation (nor the reverse) because if it did the molecule’s velocity (and temperture… couldn’t resist ;-) ) would have to increase by virtue of 1/2mv^2, but it can not increase by virtue of conservation of momentum. True?

    The one fly in the ointment, as Martin hinted, is the ignoring of angular momentum. Is that a valid assumption? Or can a body (or self contained system) convert some of its angular momentum to tangential linear momentum and maintain momentum equilibrium? Is the angular momentum of say a molecule sufficient to matter (I have no idea what a normal molecular’s w (omega) is)? I would guess that coming from vibration the molecular momentum averages to zero — though would this depend on the mode of vibration and the precise moment of any transition?

  5. 455

    #454 Rod B:

    … True?

    False.

    Energy is a scalar, momentum is a vector, and they have different formulas: momentum is (mv_x, mv_y, mv_z), its magnitude mv. The momentum of a particle pair can be zero with both flying fast in opposite directions and kinetic energy being large.

    Think baseball: when the swinging batsman hits the ball, he gets pushed backward by the ball — and would be pushed over if not actively equilibrating. The ball, and the whole system, increases its linear kinetic energy from the batsman’s muscles, but total momentum doesn’t change.

  6. 456
    CobblyWorlds says:

    #450 Phil Felton,

    The initial fissuring happened previously in April, I did a video to illustrate a point I was making in another discussion.
    http://www.youtube.com/watch?v=l8vi–s1T8E
    It’s from day 90 to 106, and it implied to me that the whole of that dense white mass was “lifting off” the Canadian Archipelago.

    Looking at timelapses of QuikScat this “lifting” happens when the AO is in negative mode, though not all the time that it is. I think it’s from wind forcing off the Archipelago into the Arctic Ocean. But aside from IR/Visible HRPT satellite (cloud movement) and some spot reporting of winds off the net, I can’t find something like a traditional synoptic polar-centred pressure map. What I can find is way over my non-metereologist’s head.

    Fissures in this region have happened in previous years. For example there’s a dark line visible across the Archipelago on QuikScat around day 100 of 2006. Stressing is to be expected there, it’s the boundary between fast and ocean ice. The latter being mobile and more affected by tides. However the recent open water there is very early, and not at all typical for mid May from what I can see. QuikScat only goes back to 2006, but Cryosphere Today has shown the Polnya off Banks Island and that goes back for decades. So despite similar past behaviour, what is heppening now really looks more extreme and worthy of note.

    The Banks Island Polnya has frozen over in today’s QuikSCAT, but as we’re in May that ice will offer little resistance to melt. Even if the Polnya closes again from ocean currents or wind, the new ice over it will offer some resistance to compression as it ridges and compacts. So there will still be a region of new ice in there, and it will still be susceptible to melting.

    Above I said:

    I’m watching the coast off the Canadian Archipelago like a hawk right now. Because I suspect a prominent role for mechnical failure in the final stages of the perennial ice.

    Here is what is concerning me.

    With the ongoing melt to continue throughout June, July, August, and much of September. There’s now a spot that’s likely to be open water, in such a position that it could cause a propagation of melting to open water along the coast of the Archipelago. Not to mention inroads of open water into the perennial mass.

    And on the other side, the Pole could be ice free this year(Serreze) because it’s covered with thick first year ice over 1.2 metres thick (National Ice Service). First year ice melts at a lower temperature than perennial ice because it contains salt and brine (perennial doesn’t). In a regime where the temperature’s moving up from well below zero the lower melting point fist year ice preferentially melts before the perennial ice (think of road-gritting “in reverse” in terms of temperature).

    So it looks possible to me that a substantial chunk of the perennial mass could float “loose” in the Arctic Ocean. That’s why I bought up the gif of the Beaufort ice fracture as an example of the sort of thing that we could see, if such a thing were to happen then you could have an Arctic Ocean full of bergs, and effectively no ice cap. I am aware that there’s a great physical mass of ice, and that consequently (due to inertia) the movement of an entire ice pack drifing in the ocean would only be a small fraction of any wind speeds. So it would not necessarily “float out” as such. But smaller bergs are more mobile and rapid propagation of cracking as seen in the Beaufort ice sheet could turn much of what is presently behaving like a coherent mass off the Canadian Archipelago into a mass of bergs, each one (more or less) surrounded by waters warming in the sun (direct or through clouds).

    QuikSCAT:
    http://manati.orbit.nesdis.noaa.gov/cgi-bin/qscat_ice.pl
    Environment Canada’s Satellite page (scroll down to “HRPT NOAA (Polar Orbiting)”:
    http://www.weatheroffice.gc.ca/satellite/index_e.html
    National Ice Centre:
    http://www.natice.noaa.gov/products/arctic/index.htm
    And for the casual browsers here:
    Cryosphere Today:
    http://arctic.atmos.uiuc.edu/cryosphere/
    NSIDC’s “Official Blog” on this summer’s Arctic Melt:
    http://nsidc.org/arcticseaicenews/index.html

    #451, Rod B,

    As far as I can see the scientists are being conservative in the face of the evidence, as they should. There’s a real process happening here (as indeed with AGW), groundless overstatement will be exposed by reality, so such overstatement is pointless.

  7. 457
    Ray Ladbury says:

    Rod, Molecular collisions can become arbitrarily complicated, but when you are dealing with a large assemblage of molecules, what matters is how often they occur. Imagine that we come up with a game of billiards where the balls are connected via springs of varying rigidity. When ball 1 collides with ball 2, ball 2 may just take off, dragging any ball(s) it is attached to behind it. Or it may start oscillating wrt the balls it is attached to, or if we hit it at a glancing angle, we might start it rotating with its attached balls around their common center of mass.
    If we want another level of complication, we can simulate (sort of) emission of IR photons by imagining that one of the balls will spontaneously shoot off a smaller ball with a certain energy and momentum (remember photons have momentum h/wavelength). Maybe this will help you visualize things a bit. Quantum mechanics complicates things a bit, but not irreparably. At the very least, I think I’ve designed an interesting video game.

  8. 458
    Rod B says:

    Martin (455), but if you transfer scalar energy from, say, vibration to translation that increased translation energy has to be manifested by increased center-of-mass velocity (larger 1/2mv^2); then how can you get increased center-of-mass velocity but not increase the molecule’s vectored momentum? I see only two ways: 1) the energy can not be transferred because of the dilemma, or 2) the molecule somehow also transfers internal vibration (or rotation/angular) momentum to its translation momentum, though I’m having a tough time getting my arms around that.

    I understand a passive ball striking a passive bat (or another ball), but it seems a batter swinging the bat will increase the momentum of 1st the bat, then the ball by virtue of force = time derivative of momentum. True? False?

  9. 459
    CobblyWorlds says:

    PS to my above post:

    BBC’s David Schukman:

    Scientists travelling with troops found major new fractures in giant ice shelves in Canada’s far north. David Shukman reports.

    http://news.bbc.co.uk/1/hi/sci/tech/7418041.stm

  10. 460
    Rod B says:

    Ray (457), Thanks. I understand the conversion of translation momentum to angular momentum ala imparting spin to the struck billiard ball (or the standard: measuring the velocity of a bullet) but in your good example it is two separate bodies interacting. However the question is transferring momentum within a single body (molecule). Also the molecular analysis is further complicated, as you imply, by the quantum nature of the various “stores” of energy and momentum in the molecule. Some stores being near frozen out at various times, or some stores picking up (or losing) energy or momentum is only discrete quanta — none of which applies to billiards.

  11. 461
    Rod B says:

    ps interesting insight, Ray (which now seems totally obvious): the vibration DoFs must have a net positive momentum (though I don’t readily see how) if, from time to time, a photon is emitted.

  12. 462

    Rod B #458

    but if you transfer scalar energy from, say, vibration to translation that increased translation energy has to be manifested by increased center-of-mass velocity (larger 1/2mv^2); then how can you get increased center-of-mass velocity but not increase the molecule’s vectored momentum? I see only two ways: 1) the energy can not be transferred because of the dilemma, or 2) the molecule somehow also transfers internal vibration (or rotation/angular) momentum to its translation momentum, though I’m having a tough time getting my arms around that.

    (This is getting a bit of a private classical mechanics lesson, I hope there are other interested readers. Don’t take this badly, but I have a bit of a problem getting my head around your stated intent of meaningfully critiquing elements of climate science esp. physical modelling :-) )

    1) is not the case. 2) is closer, though angular momentum is not momentum. It is a whole separate quantity also to be conserved.

    Consider the following situation. A and B are two balls connected by a rod, rotating in the xy plane around the center point of the rod. At a certain point in time, ball A moves in the x direction with velocity v, while ball B moves in the -x direction with velocity -v. Total momentum of AB is mv – mv = 0, where m is the mass of one ball. AB has no net linear movement.

    No ball C comes from the right, velicity -v, and hits ball A. It bounces off with velocity +v (moving right), kinetic energy unchanged. Ball A comes off vith velocity -v (moving left, just like B). In fact the whole AB now moves left with velocity -v, kinetic energy 1/2*(2m)*(-v)^2 = mv^2. Its (linear) kinetic energy has increased. It no longer rotates though, that’s where the energy came from.

    Please check that total (linear) momentum both before and after collision is -mv.

  13. 463
    Arch Stanton says:

    Martin (462) also Ray and Rod;

    I for one am lurking along and following with interest.

    I doubt I am the only one.

    Thank you all for your time and patience with the private lesson in classical mechanics. ;-)

  14. 464

    #462 correction: in the last sentence, the momentum should be the vector (-mv, 0, 0), i.e., only the x component nonzero.

    Thank you for your kind words Arch.

  15. 465
    Rod B says:

    Martin (462), et al, My rationale for all of this questioning is understanding (some of) the heat and temperature transfer mechanism ala climate warming at the atomic/molecular level. I’m not sure if it really is all that important to anybody else (and if a few complain that I’m taking up valuable space, I’ll desist), but, what I find interesting is the actual atomic-level mechanism of heating the atmosphere is (or doesn’t seem to be…) not fully understood.

    To make matters worse, I find myself jumping from pillar to post as the onion is peeled back (I think mixed metaphors are greatly over criticized.), which must make it sound even more helter skelter. This new (to me) business of momentum is a case in point. To repeat: If a molecule can transfer energy between DoFs (the interest is in translation to/from one of the other two) how does the momentum also get transferred? Or must it transfer along with energy (still being intra-molecular, not inter)?

    How is vibrational momentum evidenced? I can understand rotation and translation, but, with my limited vision, can’t see vibrational momentum. Simply, if the two (for simplicity) atoms are oscillating in opposite directions along the atomic bond, it would seem the net instantaneous momentum must be zero. If oscillating in congruent direction, instantaneous momentum ought to be sinusoidal, with an average of zero, and what momentum gets transferred depends on the exact instant of transfer. Bending poses similar scenarios, but with goofy-seeming complications.

    When a CO2 molecule absorbs a photon and its energy is picked up in a vibration DoF, where is the (granted mucho smaller) momentum picked up, and, if in a vibration, how do it do dat? The same question in reverse: when CO2 emits a photon where does it get its momentum?

    Or is it impossible for molecules to transfer intra-molecular energy and/or momentum to/from translation? (I don’t have the same difficulty understanding inter-molecular (collision) energy and momentum transfer.) Somewhat related (and again…), if a CO2 molecule at 250K absorbs a photon into a vibration DoF, does that make it out of kilter re the quantum distribution of energy among DoFs and, as I asked earlier, does it want to equipartition things out with some sort of intra-molecular transfer? Or is a single molecule being out of equipartition kilter make any difference ala the quantum mechanical distribution over a large number of molecules – given it will return to normal soon enough through collision or emission?

  16. 466
    Ray Ladbury says:

    RodB, OK, think of the billiard balls on springs again. First, we can have motion where the oxygen on either end of the carbon oscillates longitudinally along the axis passing through all 3 molecules. We can also have a situation where the oxygen molecules oscillate back and forth transverse to that axis (2 such motions). So we have our oxygen atoms actually moving physically–they have real momentum that is minimum at the extremes of their range of motion and maximum when they reach what would be their equilibrium picture if they weren’t moving.
    Now let’s say we have a single billiard ball (call it Ar)sitting there minding its own business. Classically, the CO2 molecule could strike one of the oxygen atoms so that it transfers just enough momentum to stop the motion of the O2 mulecules with the Ar atom picking up the extra momentum. Quantum mechanically, it has to transfer the full quantum of motion–and given the large number of collisions during the relatively long vibrational life of the CO2 mode, such a transfer is likely.

    As to radiative transfer of momentum–when a molecule absorbs or emits a photon, you change the momentum of the molecule to compensate. The same thing happens in radioactive decay when you give off, say, and alpha particle. And again, the concept of equilibrium only applies to ensembles of molecules.

    Finally, all of this is understood in extremely gory detail–it’s how laser jocks put bread on the table. If my explication is imperfect, it’s because it ain’t my day job. I’m delving back over 20 years when I had to learn this stuff. I recommend the following site for visualization aids:

    http://www.ems.psu.edu/~bannon/moledyn.html

    I think it requires Quicktime.

  17. 467

    Rod, Ray’s link looks good.

    Some direct responses:

    When a CO2 molecule absorbs a photon and its energy is picked up in a vibration DoF, where is the (granted mucho smaller) momentum picked up, and, if in a vibration, how do it do dat?

    Like a bullet hitting a target. Target+stuck bullet recoil, now containing the same momentum as bullet before impact. Kinetic energy after impact is (much) less than before; the difference is dissipated in the target, causing damage (as bullets are supposed to do).

    Vibration or rotation of a molecule as a whole (not looking at individual atoms) contains no momentum, just energy. Only the linear motion of the whole molecule contains momentum.

    [And just to keep you out of trouble: angular momentum has nothing to do with momentum. Totally different quantity, separately conserved.]

    The same question in reverse: when CO2 emits a photon where does it get its momentuum?

    Like a bullet fired from a gun, causing the gun to recoil.

    Before firing, momentum of bullet+gun is zero; after firing, this is still the case, momenta of bullet and recoiling gun being equal and opposite.

    The kinetic energy of bullet + gun has increased from zero to a large value, most of it in the lighter, faster bullet (remember 1/2 mv^2). This energy comes from the explosive in the cartridge; for a molecule, from, e.g., a decaying vibrational state.

    if a CO2 molecule at 250K absorbs a photon into a vibration DoF, does that make it out of kilter re the quantum distribution of energy among DoFs and, as I asked earlier, does it want to equipartition things out with some sort of intra-molecular transfer? Or is a single molecule being out of equipartition kilter make any difference ala the quantum mechanical distribution over a large number of molecules – given it will return to normal soon enough through collision or emission?

    Yes it will be “out of kilter” in that way, but for a single molecule it means thermodynamically — nothing. Yes, it tries to equipartition the energy, both inter- as well as intra-molecularly.

    It does make a difference if it happens to a lot of molecules within a very short time span (compared to molecular collisions / excited state decays), like when firing at them with a nanosecond pulse laser. Not relevant for the (lower) Earth atmosphere.

  18. 468

    Martin,

    You wrote:

    Yes it will be “out of kilter” in that way, but for a single molecule it means thermodynamically — nothing. Yes, it tries to equipartition the energy, both inter- as well as intra-molecularly.

    It does make a difference if it happens to a lot of molecules within a very short time span (compared to molecular collisions / excited state decays), like when firing at them with a nanosecond pulse laser. Not relevant for the (lower) Earth atmosphere.

    In the lower atmosphere there are a lot of CO2 molecules being blasted with thermal radiation from the surface of the earth. They will lose their absorbed vibrational energy to the translational energy of the air molecules through collisions which preserve momentum but transfer energy.

    In other words, the surface air will be radiatively heated, just as food is heated in a microwave oven.

    Cheers, Alastair.

  19. 469
    Rod B says:

    Thanks for the help. I’ll try to summarize what I still believe and/or what I’ve learned and concluded.

    The transfer (or absorption) of a photon’s momentum is still a little loose, but doesn’t matter. It can’t manifest itself in vibration since it seems there is zero net momentum within vibrational movements. Plus the concept of momentum in one “body” be transferred in an non-elastic collision to just a part of another sounds odd. It might be converted to angular momentum and show up in molecular rotation, though I have no idea if the numbers make sense. (I still think linear momentum can be converted to angular momentum as in a bullet fired into a stationary rotatable absorbent mass off center.) Predominantly (always?) it gets picked up by the molecule’s linear momentum (translation), increasing the translation velocity, its energy, and temperature. When a molecule emits a photon its translation momentum (and energy) decreases. However, the photon’s momentum is so miniscule that none of this really matters one way or another. Which is the same conclusion I arrived at by default when contemplating (and getting a headache) a photon’s angular momentum — who cares?!?

    For all practical purposes the energy from a photon at 15microns is absorbed in a CO2 molecule’s vibration,. This energy is more than twice the kinetic energy of a molecule’s translation at a nominal 300K, and does not affect the temperature, other than from the likely trivial momentum transfer just mentioned. By equipartition quantum probabilities it will strongly want to relax This relaxation is predominately transferring energy to another atmospheric molecule’s translation via collision, at other than very low pressure (density). This does raise the atmosphere’s temperature. It can, though unlikely, transfer to its own translation; however this becomes trivial since the molecule will still quickly relax its new found translation energy again via collision. It can also re-emit a photon instead, the probability of which increases as pressure decreases (among other factors). Conversely, a CO2 can sometimes collide with another molecule, pick up some kinetic energy and immediately relaxing via photon emission, transiting through vibration, and provide a net cooling of the atmosphere

    Some of my calculations as a reference (and a check if anyone is so inclined):
    The energy of a 15micron photon is 1.325×10^-20 joules; its momentum is 4.4×10^-29 [you'd think by now physicists would have come up with momentum units; I suggest OOMPHS!]
    The kinetic energy of a CO2 molecule at 300K is 6.214×10^-21joules (3742 joules for a mol); its momentum is 3.0×10^-23; its velocity (or the average velocity for a mol) is 412.5 m/sec.
    One photon’s energy going into one molecule’s translation will raise its temperature from 300K to 630K; the mol’s avg. temp would increase to 300.038K.

    The standard relaxation process has a complex formula of probabilities. Of interest is the vibration to translation transfers usually require a large number of collisions before occurring. 10,000 to 100,000 is often quoted, but that is usually at high temperatures (500-1000K) where the molecule is more “comfortable” with its vibration. At atmospheric 200-300K (and normal pressure) it is less : 100-10,000 I would guess (haven’t done the cumbersome math), and more likely to make the transfer. The lower rotation energy takes 5-100 collisions to make a translation transfer and is highly likely — H2O to N2 or O2, e.g. Going the other direction — translation to vibration to emission — I would think (no math again) at low temp and pressure, e.g. stratosphere, somewhere around the 10,000-100,000 collisions range is more the normal and not as likely to occur.

    I haven’t sorted through the ramifications of all this yet. Maybe there is none. But it’s a small piece of the puzzle: a clearer understanding of the molecular process of energy transfers and temperature changes. I’m sure I have left something out, but I’ve already worn out my welcome. (Plus the other stuff like blackbody radiation….to do ;-) )

  20. 470
    Ray Ladbury says:

    Rod B., Your biggest problem is still conceptual–you still persist in equating temperature with kinetic energy of molecules–even of a single molecule. This will inevitably lead to contradictions and confusion. I’ve pointed out, for instance, that in a laser, the medium in which the population inversion occurs will have a negative temperature. The conceptual difficulties increase the more one departs from equilibrium or moves toward quantum mechanics.
    In addition, relaxation of the CO2 molecule has the same (long) lifetime regardless of how it is excited. And when it decays collisionally, the most likely outcome will be that the other molecule (probably N2) and the CO2 will both increase in kinetic energy.

    You’re getting there. The interpretation of thermodynamics is inherently statistical–and there are a lot of subtleties that even most physicists don’t appreciate.

  21. 471

    Alastair,

    In the lower atmosphere there are a lot of CO2 molecules being blasted with thermal radiation from the surface of the earth

    “Blasted” definitely conveys the wrong impression — “slowly roasted” more like it. Those molecules leasurely equipartition their received energy, LTE is never in danger. Same in a microwave oven.

  22. 472

    Rod,

    you’ve been working hard!

    > the mol’s avg. temp would increase to 300.038K

    Not reasonable… you just demonstrated that you know Avogadro’s number, and it’s huge :-)

    Otherwise numbers seem OK.

    > It can, though unlikely, transfer to its own translation;

    But what about momentum consevation? It would need help from another molecule.

    > (I still think linear momentum can be converted to angular
    > momentum as in a bullet fired into a stationary rotatable
    > absorbent mass off center

    What happens there is that the incoming bullet has angular momentum relative to the rotation axis (how AM is always defined!) by virtue of being off-center; this gets transferred to the absorbent. As does, separately, the linear momentum, producing the overall recoil.

    Otherwise seems OK. Remember with interaction probabilities that they are equal in opposite directions. This is why their precise values are immaterial under LTE.

  23. 473
    Rod B says:

    Thanks guys. Ray (470), we will always disagree on the temperature-molecule thing. I contend the translation kinetic energy is the only thing that manifests (real) temperature. The other “containers” have a construct called “characteristic temperature” which is useful, handy, easily understood, yet still a construct. And until you convince me that a single molecule can not have 1/2 mv^2 kinetic energy, I will contend it has (real) temperature, even if unmeasurable. Maybe we should just quietly maintain our respective beliefs ;-) .

    I agree that the relaxation of vibration energy through collision can (and does) distribute translation energy to both the CO2 and the N2 (e.g.) molecules.

    Martin (471 & 472): one minor clarification: the relaxation of CO2s vibration energy is not near as leisurely as H2O’s rotation energy.

    Avogadro’s number IS the difference between mol stuff and molecular stuff. If one CO2 molecule in a mol of CO2 at 300K absorbs one 15 micron photon and relaxes, the mol will then be 300.038K. ???

    “…But what about momentum conservation? It would need help from another molecule…”

    I hadn’t thought of that, but it makes sense ala Ray’s collision above.

    The linear moving bullet has angular momentum by virtue of the rotating blob it is about to hit but presently has no cognizance of its existence? What if the aim is bad and it misses? Does it still have the AM? Does the AM change as the bullet flies toward the blob by virtue of changing tangential velocity and “radius”? I’m having a hard time getting wrapped around this. Are you sure??

  24. 474

    Rod:

    You use one operational definition of temperature, which is not wrong but gets cumbersome as you dig deeper into statistical mechanics. Those “characteristic temperatures” are real temps too, as theory will tell you once you study it. If it barks like a dog… Same with the “temperature” of one molecule. You are inventing your own terminology there, differing from accepted usage. Not quite illegal, but expect confusion down the road.

    Avogadro’s number IS the difference between mol stuff and molecular stuff. If one CO2 molecule in a mol of CO2 at 300K absorbs one 15 micron photon and relaxes, the mol will then be 300.038K. ???

    Show me the computation… pretty sure it’s wrong. More like
    300K + 330K/(6.022×10^23), which is extremely close to 300K.

    The linear moving bullet has angular momentum by virtue of the rotating blob it is about to hit but presently has no cognizance of its existence?

    Huh? It has AM relative to the axis of rotation of the target, which we choose (for convenience) as the origin of reference to describe AM on. You always have to choose such an origin and use it consistently, and then AM will be conserved.

    What if the aim is bad and it misses? Does it still have the AM?

    Yes… it just continues with its AM (relative to the chosen reference origin) intact.

    Does the AM change as the bullet flies toward the blob by virtue of changing tangential velocity and “radius”?

    No, those changes precisely cancel each other for linear uniform motion (as you may easily prove for yourself). This even remains true in a central (relative to the reference origin) force field… you get Kepler’s Law of Areas, a geometric way to state conservation of AM. [Moving from a molecule to a bullet to a planet... welcome to the universality of physical law.]

    I’m having a hard time getting wrapped around this. Are you sure??

    Oh yes. These are not matters of opinion :-)

  25. 475
    Ray Ladbury says:

    Rod B., If you insist on equating temperature and translational kinetic energy, you are going to run into some pretty interesting philosophical problems. For instance, you definition of temperature will become temperature dependent. We’ve already talked about the problems that arise wrt a temperature inversion–where adding energy to the system makes the temperature more negative. You also wind up having some issues as temperature rises and more and more degrees of freedom become unfrozen. The idea of a single molecule having a temperature is even more problematic.

    As to angular momentum, the definition is the cross product of the momentum and position vectors (p and r, or L=p x r). Any system in which the elements have momenta perpendicular to lines joining the elements will have angular momentum.

  26. 476

    Re 475.

    Ray,

    Rod’s equating of temperature with translational kinetic energy is its normal use. It is the temperature measured by a (mercury or gas) thermometer. Because gaseous molecules have other forms of internal energy: rotational, vibrational and electronic then they also have three other temperatures: Tr, Tv, and Te.

    The way that the rotational temperature Tr is determined is explained here.

    Because of the Equipartition Theorem we would expect all four temperatures to be the same. However, in the Earth’s atmosphere the temperature is far too low for any atoms to become electronically excited hence Te is zero. Tv is also less than T, but Tr will be equal to T.

    Since LTE only applies when Kirchhoff’s Law of blackbody radiation Ilamda= k Blambda(T)is obeyed, then LTE does not exist in the Earth’s atmosphere see I = B(T) See Eric Weisstein’s World of Physics.

    HTH (but it probably won’t :-)

    Cheers, Alastair.

  27. 477
    Hank Roberts says:

    If I may veer back onto the topic for a moment, away from the replay:

    This is an interesting scenario, I wonder if the modelers here have anything to add to it? The “Cited by” links suggest it’s had a bit of attention, but most of them are paywalled.

    A change in the slow geochemical sink for CO2 can, and did, cause a rise in CO2, followed by extreme warming. That was the fastest event we know of before the present excursion we’re causing now. It was far slower than what’s happening now as we burn fossil carbon:

    http://dx.doi.org/10.1016/S0031-0182(03)00667-9
    Palaeogeography, Palaeoclimatology, Palaeoecology, Volume 203, Issues 3-4, 15 February 2004, Pages 207-237

    Causes and consequences of extreme Permo-Triassic warming to globally equable climate and relation to the Permo-Triassic extinction and recovery

    David L. Kidder Corresponding Author
    and Thomas R. Worsley

    ——excerpt from abstract, breaks added for readability——-

    Permian waning of the low-latitude Alleghenian/Variscan/Hercynian orogenesis led to a long collisional orogeny gap that cut down the availability of chemically weatherable fresh silicate rock resulting in a high-CO2 atmosphere and global warming. The correspondingly reduced delivery of nutrients to the biosphere caused further increases in CO2 and warming.

    Melting of polar ice curtailed sinking of O2- and nutrient-rich cold brines while pole-to-equator thermal gradients weakened. Wind shear and associated wind-driven upwelling lessened, further diminishing productivity and carbon burial. As the Earth warmed, dry climates expanded to mid-latitudes, causing latitudinal expansion of the Ferrel circulation cell at the expense of the polar cell.

    Increased coastal evaporation generated O2- and nutrient-deficient warm saline bottom water (WSBW) and delivered it to a weakly circulating deep ocean. Warm, deep currents delivered ever more heat to high latitudes until polar sinking of cold water was replaced by upwelling WSBW. With the loss of polar sinking, the ocean was rapidly filled with WSBW that became increasingly anoxic and finally euxinic by the end of the Permian. Rapid incursion of WSBW could have produced [approx.] 20 m of thermal expansion of the oceans, generating the well-documented marine transgression that flooded embayments in dry, hot Pangaean mid-latitudes. The flooding further increased WSBW production and anoxia, and brought that anoxic water onto the shelves.

    Release of CO2 from the Siberian traps and methane from clathrates below the warming ocean bottom sharply enhanced the already strong greenhouse. Increasingly frequent and powerful cyclonic storms mined upwelling high-latitude heat and released it to the atmosphere. That heat, trapped by overlying clouds of its own making, suggests complete breakdown of the dry polar cell.

    Resulting rapid and intense polar warming caused or contributed to extinction of the remaining latest Permian coal forests that could not migrate any farther poleward because of light limitations. Loss of water stored by the forests led to aquifer drainage, adding another [approximately] 5 m to the transgression. Non-peat-forming vegetation survived at the newly moist poles.

    Climate feedback from the coal-forest extinction further intensified warmth, contributing to delayed biotic recovery that generally did not begin until mid-Triassic, but appears to have resumed first at high latitudes late in the Early Triassic.

    Current quantitative models fail to generate high-latitude warmth and so do not produce the chain of events we outline in this paper…..

  28. 478
    Rod B says:

    Martin (474), I think we have a semantic difference (I thought we had resolved this already, but…). When I say “real temperature” I actually mean temperature as we sense it: hot water, cold ice, measurable (theoretically) with a thermometer. Rotational and vibration energy has “characteristic temperature”, commonly shortened to just “temperature” which is not “real” by the above definition, but is real in terms of a usable concept or construct for analyzing things and making accurate calculations. Take a mole of H2O vapor at 300K as measured by my thermometer. Now allow me to magically raise every molecule’s rotation energy to its first quantum level without any relaxation. The temperature of the mole is now…. tada…300K. Even though there is energy, oft-called heat, sometimes called kinetic, in the rotation of the molecule’s that one can define a “temperature” to that energy level that makes it easier to think about and to make some calculations and therefore be ‘real’, but it ain’t “warm”!

    Microwaves (the boxy type, not the wave type — don’t want to raise the latter ’cause it makes Ray mad :-) ) impart energy to the rotation levels of water molecules. But the food does not increase its “temperature” until the H2O relaxes and transfers rotation energy to translation/kinetic energy via molecular collisions.

    Ray, as more degrees of freedom free up as the temperature rises, more and more energy gets added that goes into those degrees of freedom and does not increase the “real temperature” as defined above.

    I second Alastair (476) with teeny clarification. The Tv and Tr are not only different numbers (can be), they are different physical entities, of a sort.

  29. 479
    Martin Vermeer says:

    Alastair,

    No it doesn’t :-) Your quoted abstract clearly refers to a non-LTE situation. Under LTE, all these temps are equal (and T_e is not zero but undefined/meaningless for low temps).

    As for Eric Weisstein, I noticed too that he uses a different definition of LTE that does include the radiation field. So do some astrophysicists. Babel :-(

    If you google for “local thermodynamic equilibrium” you’ll find many references agreeing with the definition I gave (which I find more fruitful).

  30. 480
    Ray Ladbury says:

    Alastair says, “Since LTE only applies when Kirchhoff’s Law of blackbody radiation Ilamda= k Blambda(T)is obeyed, then LTE does not exist in the Earth’s atmosphere”

    Kirchoff’s law is an idealization–it never applies perfectly in any real physical system. However, given the long lifetime of the CO2 vibrational state, energy transfer from the radiation to kinetic energy is pretty efficient–so it’s a pretty good approximation. What matters is the physical temperature–the derivative of energy wrt entropy. Stick with that and you won’t go wrong. The reason I object to equating temperature and translational kinetic energy is because it leads to absurd ideas like assigning temperature to individual molecules.

  31. 481

    Martin,

    I was only using the abstract to show that talking about vibrational and rotational temperatures was legitimate. I don’t think the abstract had any other relavanc to the current discussion, but then I haven’t even read it :-)

    LTE is an astrophysical term. The concept was first proposed by Karl Schwarzschild in 1906 for the interior of the sun, which does emit as a blackbody. It was his brother-in-law Robert Emden, best known for his book on stars “Gaskuglen”, who proposed that it could be applied to the earth’s atmosphere. AFAIK, it was the astrophysicist A. E. Milne who first named it, when describing gas nebulae. The great astrophysicist Chandrasekhar seems to have assumed it also applied to terrestrial atmospheres, and his prestige has decided the matter.

    So not only is it a term invented and used by astrophysicists, it seems to have been adopted by the climate modelers who were in awe of them.

    As you suggested I have goggled yet again for “local thermodynamic equilibrium” with the same result as before. Most results are for non-LTE. The rest, apart from Wikipeadia, are for its astrophysical use.

    One is interesting which is a paper by H Roos entitled “On the Problem of Defining Local Thermodynamic Equilibrium”. I will try to read it yet again, but perhaps in the mean time you could provide your definition of LTE?

    Cheers, Alastair.

  32. 482

    Alastair writes:

    Since LTE only applies when Kirchhoff’s Law of blackbody radiation Ilamda= k Blambda(T)is obeyed, then LTE does not exist in the Earth’s atmosphere see I = B(T) See Eric Weisstein’s World of Physics.

    Every authority in atmospheric studies says local thermal equilibrium applies to the troposphere and even the stratosphere, and that Kirchhoff’s law applies as well. Every computer model of the atmosphere uses it and gets the right answers.

    Readers should note that Alastair has his own unique view of radiation physics.

  33. 483
    Martin Vermeer says:

    Alastair:

    > but perhaps in the mean time you could provide your definition of LTE?

    The Wikipedia one. Energy in molecules equipartitioned over degrees of freedom, Maxwell-Bolzmann statistics, Saha law, etc. Radiation field being irrelevant to the definition. Every point in the distribution of matter has a single well defined temperature.

    Rod:

    > Now allow me to magically raise every molecule’s rotation
    > energy to its first quantum level without any relaxation.

    …meaning non-LTE. “Magic” is the operative word. What you’re doing is stepping off the cliff and pondering gravity for a moment before starting the journey down, like the cartoon figure. Doesn’t happen in the real Earth atmosphere (below 100 km or so). Or in the Solar atmosphere below the corona.

    For LTE, all those different temperatures are the same. That’s one definition of LTE I have seen (in the Roos paper?): a state in which every point (or small environment) has a single well defined number called temperature, which it will share with a thermometer if you stick one in at that point.

    The “thermometer” can be pretty much anything, including natural processes that are already there: observing doppler broadening of spectral lines will give you the kinetic temperature, observing the vibration spectrum the Tv, and so on. And under LTE they are all the same.

    The distinction you make between these different temperatures is completely artificial. They are all just as real. The origin of defining temperature in terms of the kinetic temperature is historical: the gas thermometer was the first operational way to produce an accurately linear temperature scale. Now we have many more. And as I earlier tried to point out, many solids don’t even have a well defined kinetic temperature.

  34. 484
    Rod B says:

    Ray, Martin, I’m wearing out my welcome with this momentum stuff but its racking my brain. While I continue my search, a couple of Qs: 1) If the bullet has angular momentum of r X P referenced to the axis of the spherical absorbent blob, it would seem the angular momentum of the bullet must vary with time as r and the angle between r and P vary. But there is no noticeable force/torque on the bullet which means momentum is not being conserved. You can define the system as the bullet, the bullet and blob, or the bullet, blob, and gun and get the same results. How is this ostensible non-conversation of angular momentum explained?

    2) Why not this: The bullet’s PB (=mv) goes from zero to something from the force of the gunpowder ala F = d(mv)/dt. (Gun gets same negative P by virtue of equal and opposite reaction.) When struck the blob applies an inertial force on the bullet, changing its linear momentum back to zero. The equal and opposite reactive force applies a torque of FR X r on the blob, which accelerates the blob in a rotation, giving it angular momentum L
    where L = d(PB)/dt X r.

    Maybe I’m dense, but to quote Red Skelton, “…it just don’t look right to me.”

  35. 485

    Re #477 Where Hank quotes the abstact to a paper which concludes:

    Current quantitative models fail to generate high-latitude warmth and so do not produce the chain of events we outline in this paper.

    I find it hard to reconcile that with Barton’s assertion in #482:

    Every computer model of the atmosphere uses it[LTE]and gets the right answers.

    Just because my views are unique does not make them wrong, and just because all computer models use LTE does not make them right. It is not as if all computer modelers have come to that conclusion independently. Those who are now involved in that code believe they need only copy what was written before.

    In the abstract for Santer et al. 2005, of which Gavin was a co-author, they conclude:

    “These results suggest that either different physical mechanisms control amplification processes on monthly and decadal timescales, and models fail to capture such behavior, or (more plausibly) that residual errors in several observational datasets used here affect their representation of long-term trends.”

    Despite the passage of three years and the publication of yet more scientific papers trying to find “residual errors in” all the relevant “observational datasets” , “The Tropical Lapse Rate Quandary” still exists.

    It seems that the climate modelers are just like any other computer programmers, and will never admit that there is a bug in their program :-(

    Returning to Hank’s problem, it should be possible to get the models to give the polar amplification needed to reproduce the paleoclimate data if the absurd theory based on LTE that greenhouse warming operates by affecting the lapse rate high in the troposphere. The greenhouse effect is due to the absorption of infrared radiation from the surface of the earth and it is the air there which warms. This raises the latitude and altitude of the snow line. It is the change in albedo from loss of snow and ice cover which drives the polar amplification. That would explain the Phillipona Dilemma see: A Busy Week for Water Vapour

    Cheers,Alastair.

    [Response: Alastair, please no more nonsense about LTE. Your notions about LTE are flatly incompatible with direct observations of the outgoing infrared spectrum of Earth and Mars, which confirm Kirchoff's law to a high degree of accuracy. That should convince you, even if the basic physics isn't good enough(see e.g. Chamberlain and Hunten's textbook on atmospheric physics, which has a particularly good explanation of why calculations based on Einstein transition coefficients give the same answer as straight thermodynamic equilibrium, even when the full photon spectrum is not blackbody) . As for polar amplification in paleoclimate data, please note that the revision of tropical temperatures eliminates a lot of the problem. The PETM warmth from the new polar core could put some of the problem back, but even if that relatively new data holds up, there are lots more plausible mechanisms than abandoning LTE -- things to do with cloud feedbacks, changes in aerosol concentration, hurricane-induced ocean mixing, and so forth. --raypierre]

  36. 486
    Ray Ladbury says:

    Rod–the cross product depends only on the component of the r that is perpendicular to p–in magnitude it is |r||p|sin(theta), where theta is the angle between r and p. L remains constant throughout the trajectory of the bullet and, indeed after collision. And you’ve almost got the collision part right–indeed, the bullet and blob are accelerated by the same force acting over the same distance, but the angular momentum of the system is conserved. Now think about the special case where the bullet passes through the center of mass of the system–there is no perpendicular component to r, so angular momentum is zero throughout–and although the blob will accelerate, it will not rotate wrt its center of mass.

    Hey, at least this is science–it’s better than talking about corporate ethics.

  37. 487
    tamino says:

    Re: #484 (Rod B)

    There’s no conflict in the definition of angular momentum. As the bullet moves, the radius changes according to r = r(closest) / sin(angle), where “r(closest)” is the radius at closest approach and “angle” is the angle between the present direction and the direction in which the bullet is travelling. The vector cross product in “r x p” gives |r| times |p| times sin(angle). So the magnitude of the result is r(closest) / sin(angle) times |p| times sin(angle) = r(closest) times |p|, which is constant through time.

  38. 488
    Martin Vermeer says:

    Rod:

    1) If the
    bullet has angular momentum of r X P referenced to the axis of the
    spherical absorbent blob, it would seem the angular momentum of the
    bullet must vary with time as r and the angle between r and P vary.

    No, Rod. It remains strictly constant. The variations in r and the sine of the angle cancel precisely.

    Look at it this way: the size of the bullet’s angular momentum is

    |r x P| = m |r x v| = m |r| |v| sin (angle r,v)

    perpendicular to the (r,v) plane. In a fixed unit of time dt this is

    |r x P| dt = m dt |v| |r| sin (angle r,v) = m |dr| |r| sin (angle r,dr) = 2 m * area,

    where “area” is the surface area swept out by the line connecting origin and bullet over the amount of time dt, and dr = v dt the travelled distance in that time. We have

    area = 1/2 |dr| |r| sin(angle r,dr).

    Now look at this surface area. Make the distance dr travelled by the bullet the base of the triangle, and the shortest distance between the origin and the bullet path, i.e.,

    |r| sin(angle r,dr),

    its height. [I'm trying to draw an ASCII picture here...] this is a constant. Also |dr| is a constant. It follows that “area”, and thus |r x P| dt, and thus |r x P|, is a constant. This is Kepler’s second law and conservation of angular momentum. At least, of its size. Its direction is also constant as it is the direction perpendicular to both r and v, as you will agree from geometric intuition.

    Phew.

    L = d(P[B])/dt X r.

    If you want L to be angular momentum, it should be

    L = P(B) x r and thus dL/dt = d(P(B)/dt x r = m dv(B)/dt x r.

    If you call the torque N = F(R) x r, you have

    dL/dt = N.

    (And linear and angular momentum are still separately conserved…)

    Hope this gets you a little further. I miss a blackboard.

  39. 489
    Joel Shore says:

    Alastair (#485): You say, “It seems that the climate modelers are just like any other computer programmers, and will never admit that there is a bug in their program.” First of all, I hardly think that 3 years is enough time to resolve all of the problems that exist with the data…and progress has been made on that front. As for admitting a bug, I think it is probably pretty challenging to come up with an intelligent way to change the climate models so that they do a better job of agreeing with the data on multidecadal timescales while not messing up the agreement on the shorter timescales (where we know the data is actually more trustworthy!) I’m not saying that no reasonable mechanism can be found that would do this but I haven’t seen any proposals along this line.

    And, by the way, as someone who does computational modeling for a living (not in the climate science field), I would say that contrary to your assertion about never admitting a bug, I do admit bugs…and when the experiments and models disagree, I try to give my best assessment of which is more likely to be incorrect. Furthermore, I can tell you that, if anything, previous history has shown that I have erred more often on the side of having too little confidence rather than too much confidence in my modeling. And, in the case of the tropical tropospheric amplification, I would say that there are some very good reasons to believe that most of the problem is likely with the data and not the models.

  40. 490
    Rod B says:

    Ray, tamino, Martin: Well, that’s a helluva note! I was responding to Ray with a curt calculation of the L bullet at two different locations as it speeds toward the blob. Problem: L came out exactly the same at all positions. I got to get my calculator fixed (a wonderful old HP with reverse Polish stuff, btw…).

    This blows my whole fundemental concept of angular momentum. It’s still on topic (I think) though pretty shaky, so I’ll be brief. All of this tells me that angular momentum as a numerical quantity of something, associated with a mass (or virtual mass) is virtual, pseudo. For the same individual unique bullet at the same instant I can calculate many (an infinite number!) angular momentums: it will have angular momentum relative to any fixed mass with an axis. One L for my blob; a different L for the bottle on a stick 10 meters distant from my blob; a different L for the globe sitting in the library; a different L for the moon, etc. Each of those Ls, themselves, will be conserved. Is this accurate?

    [Actually this might help my long-held secret doubts about angular momentum: "a mass moving in a straight line will want to continue moving along that line" is an intuitively easy concept; "a mass moving around in a circle will want to continue moving around the circle" ain't near that easy!]

  41. 491
    Rod B says:

    ps, still one tiny missing piece. I didn’t follow Martin’s stuff on how linear momentum is conserved. Is it that since linear momentum converts to a torque (F = d(mv)/dt) when stopped by the blob, but not into angular momentum… though the angular momentum of the bullet gets applied to mass of both the bullet and the blob at impact…

  42. 492
    Rod B says:

    Martin (483), I’m not sure if you agree or disagree. I claim that only the 1/2mv^2 translation energy has “real” temperature, and that the term temperature applied to a molecule’s molecular rotation or vibration energy, or E-M radiation (oops!!) is a construct that’s very useful and understandable but its not the same essence as translation temperature. These might also be “real” in the sense that “there they are! right there on the paper of my calculations!”, but not “real” as I’m using it — the old thermometer measured stuff. (btw, how would you stick your thermometer probe into the rotation of a molecule, or the vibration of the atomic bonds?) You said

    “The distinction you make between these different temperatures is completely artificial. They are all just as real. The origin of defining temperature in terms of the kinetic temperature is historical: the gas thermometer was the first operational way to produce an accurately linear temperature scale. Now we have many more….”

    which I think you meant to disagree with me but what you said in essence is that we used to have kinetic energy = temperature but then we went and came up with more “temperatures” — pretty much describes inventing a construct to me.

    In any event, you had realism problems with my entire mole of CO2 having excited rotation and no relaxation — despite relaxation being a quantum mechanical probability function. How about if the mole is completely excited for just an instant (and I say at 300.000000K). You pick the timing of molecules relaxing (let’s say all through collision for simplicity, but you can divvy it up between emission and collision), tell me how many are relaxed at any instant and I will calculate the “real” temperature of the mole. The energy that still resides in rotation (or that has been emitted) WILL NOT enter into the calculation of the “real” temperature.

    I thought we agreed once, ala the Wikipedia reference. I suspect you might just be hung up with the semantics (though maybe I’m wrong here). We call “A” temperature and we call “B” temperature, but A and B might be totally different. “Blind” can be something that covers a window; “blind” can be lack of eyesight. Would you insist that lack of eyesight is exactly the same as the thing on the window?

  43. 493
    Martin Vermeer says:

    Rod #492

    (btw, how would you stick your
    thermometer probe into the rotation of a molecule, or the vibration of
    the atomic bonds?)

    Simple. A “thermometer” is anything outputting a reading allowing you to derive a temperature value (cf. Einstein: time is what you read off a clock). And you don’t stick a thermometer into a single molecule but always into a macroscopic amount of molecules (as Ray argued too).

    The thermometer probe here is the molecules itself emitting radiation from transitions between vibration/rotation levels. Every intensity ratio between two spectral lines is one “thermometer”.

    So it doesn’t have to be a liquid-in-glass thingy. The principle of thermometry is, as you say “sticking it in” so temperature can be shared between sample and probe… but that can mean many things, like sample and probe being the same molecules. And the “reading” can travel even over interstellar distances.

    About LTE (in my, not Alastair’s meaning), it is really valid almost everywhere, except in extreme situations like the Solar corona. So there is a single well-defined temperature at every point you care to measure it, shared by linear molecular motion, vibrations, rotations, electronic excitations (and ionization where appropriate).

    The historical role of kinetic temperature is due to metrological considerations (the art of measuring). The gas thermometer based on the ideal gas law was the first thermometer known on theoretical grounds to measure correctly and linearly. You can then calibrate other thermometers (mercury…) to measure on the same scale, creating a traceability chain. Astrophysical measurements of temperature are similarly theoretically based and not part of the same traceability chain. If that’s what you mean by “artificial”, then yes, I suppose you have a point. But in science, most “thermometers” do not contain mercury, and you wouldn’t recognize them if you saw them… like also a mass spectrograph measuring oxygen isotope ratios in an ice core sample ;-)

    “Blind” is a homonym. Not a valid metaphor.

    Sorry if this is a bit rambling; preparing to travel.

  44. 494
    Martin Vermeer says:

    How about if the mole is completely
    excited for just an instant (and I say at 300.000000K). You pick the
    timing of molecules relaxing (let’s say all through collision for
    simplicity, but you can divvy it up between emission and collision),
    tell me how many are relaxed at any instant and I will calculate the
    “real” temperature of the mole. The energy that still resides in
    rotation (or that has been emitted) WILL NOT enter into the calculation
    of the “real” temperature.

    What you are calculating, called the “real” temperature, is apparently the kinetic temperature. OK. Highly nonstandard, but not as such wrong — for gases. It will get you into trouble down the line in your study though.

    What happens is that the moment the mole is excited, it moves from LTE into (serious!) non-LTE. In my temperature language, at that moment it ceases to have a well defined temperature. It has measurable kinetic, rotation and vibration temperatures, all different; and if you wish to call anything artificial here, I would these three temperatures including the kinetic temperature.

    Then, when all molecules relax, the sample is restored to LTE and has a real temperature again.

    That’s my take on it.

  45. 495

    Raypierre,

    Thank you for your response and the recommendation of Chamberlain and Hunten’s textbook. I have ordered it. Hopefully it will answer at least some of the questions I have.

    You responded “Alastair, please no more nonsense about LTE.”

    Should that not have been “Alastair, please no more about that nonsense LTE.”? :-)

    It is quite clear from the replies to my recent posts that LTE can be defined in various ways, and that defenders of the paradigm that is used to explain greenhouse gas induced climate change, have no clear idea of what is meant by it.

    So you will hear no more from me at present. It is obvious that I am not going to have any success in explaining that the mechanism of the greenhouse theory of climate change, which originated in the nineteenth century with Arrhenius, is wrong. Where that mechanism does give the correct results, they are used to ‘prove’ that it is correct, and when it gives the wrong results they are brushed aside or explained away. That is the way old invalid paradigms are defended.

    I realise that since you too are a believer in the old paradigm it is impossible to convince you that it is wrong. But if I do not reply to your disparaging comment, then it will be construed by others that I have conceded defeat. This is not correct, so I will briefly outline my arguments before resting my case.

    Kirchhoff’s Law in its simplest form is just restatement of the Law of the Conservation of Energy. In other words, in thermal equilibrium the radiation absorbed must equal radiation emitted, since they are the only energy flows. In the troposphere not only do we lack thermal equilibrium, we also have an additional flow of energy from latent heat. Therefore, a balance in absorption and emission is impossible there.

    It is only in the upper atmosphere above the stratopause that radiation is the sole energy transfer mechanism, so there Kirchhoff’s Law must be obeyed. Since this is where the radiation to space originates it is unsurprising that the spectra of outgoing longwave radiation measured by satellites agrees with the calculations of the current paradigm. This does not mean that similar calculations in the troposphere are correct!

    In the troposphere the lapse rate does not match the calculations as is shown by the MSU and radiosonde data. This is because the heating effect there is by the absorption of radiation by the surface air, not by surface heating from Kichhovian back emission of greenhouse gases as proposed by the current paradigm.

    The point to realise is that the match with the OLR does not prove that the current model is correct. The lack of match with the tropical lapse rate, and many other difficulties with the models, does prove that they are wrong.

    This error in the models has important consequences. They have been unable to predict the rapid melting of the Arctic sea ice. They are unable to explain rapid climate change, and so are unable to predict the rapid warming that will happen when the Arctic sea ice disappears.

    (That event could even happen this year. The Arctic sea ice extent is already less than it was at this time last year. See: NSIDC Arctic Sea Ice Extent. That year the ice hit a record minimum extent in September. It now seems extremely likely that there will be another record low in three months time.)

    So you see I really ought to be acquiring more evidence that my ideas are correct, rather than arguing with the upholders of the conventional wisdom here. They will never be convinced :-(

    Cheers, Alastair.

  46. 496
    Ray Ladbury says:

    Rod, Of course if you switch origins, you wind up with a different angular momenmtum. Likewise, if you change the velocity of your origin, you get different momenta and kinetic energies, too. The momentum, energy and angular momentum are not properties of the bullet, but of the bullet wrt the coordinate system in which they are calculated. This in no way decreases their utility, since I know how to transorm the momentum/energy/angular momentum for any coordinate system I need them in. Moreover, while you have different values in different reference frames, there are clearly some frames that make the physics easier.
    As to the issue with temperature–you need to be careful. What do you mean by “excited to a temperature of 300 million K”. If the gas is to be at equilibrium at that temperature, then equipartition will apply. OTOH, if you mean that somehow Maxwell’s demon has gotten busy with his tennis racket and accelerated the linear momentum of each molecule to the right distribution of velocities for a gas at 300 million K, then you are not in equilibrium, and there will be an exchange of energy flowing into the modes that were not excited (rotation, vibration, electrical excitation, ionization…). And eventually, you will come to equilibrium at some lower temperature. Until it is in equilibrium, the system will not even really have a well defined temperature. We really only know how to treat systems that are either in equilibrium or near equilibrium. There are a few systems where the energetics are sufficiently simple that we can treat them far from equilibrium (e.g. the population inversion in a lasing medium), but the thermodynamics yields some really odd results, such as negative temperature. Trust me on this–there’s a reason why physicists do things the way they do.
    The thing is that you definition of temperature purely in terms of kinetic degrees of freedon doesn’t buy you anything. For instance, what happens when you think of the temperature of a solid? Here, the motions of atoms about their lattice positions are restricted, so most of the energy is in vibrational modes. Do you contend that a thermometer brought into contact wouldn’t register a temperature?

  47. 497
    Rod B says:

    Martin , (sigh!), maybe we will just always disagree; but one more shot:
    To sum it up, I was claiming “temperature” a homonym, not using “blind” as a metaphor. My “real temperature”…wait, I’m going to change its name since the other “temperatures”, as I said, are also “real” (homonyms) in some sense. I’m calling it “sensory temperature” — what you measure with the standard thermometer; or if you touch it you get a sensation commonly known as “temperature” by the public at large though maybe not (all) physicists. The other characteristics of certain molecular phenomena and radiation are not the same at all; physicists just call them “temperature”, a homonym, because its easier, handy, shorter, aids the understanding of the phenomena, and is understood by physicists within the context.

    A far out but hopefully insightful (as opposed to inciteful) example. I have a 2000kg vehicle going 45MPH (20.115m/sec). If I want a descriptive parameter that would help me understand what’s going on physically, I could first calculate its “real” kinetic energy (as a unit body), which is 404,613.2 joules. Now I’m going to define a parameter analogous to temperature (the sensory kind) calculated as if it was a relaxed mole of gas: kinetic energy = 3/2kT. T = 1.955×10^18 degrees. I’m going to call this number the car’s “temperature” because the calculation is somewhat similar to other calculations of temperature, and also, well…, the units are degrees Kelvin. (btw, when the car is stopped its “T” is zero.) If this is helpful to me as, say, a crew chief of a race car because it gives me some measure that combines speed and mass, I can ask my timer, “what’s the temperature?” because it’s understood by the crew within its context (and easier and shorter than asking, “what’s the equivalent of what the temperature of a mole of gas at STP would be assuming the car has three degrees of freedom (though hopefully using only one at any instant!) at its current kinetic energy?” By the time I get the question out, the car has completed another lap!) The timer says “about 1-1/2×10^18 degrees.”, which in our vernacular is probably shortened to “1-1/2 – 18.” I can use this information. I know what it means. I also know what it doesn’t mean. It’s not THE (sensory) temperature, which is probably around 80 degrees Celsius or so (and which my driver is very happy about), even though I call it “temperature”.

    You can measure a molecule’s emission frequency and associate it with a temperature related to Planck’s function, and then refer to the energy that generated the photon as having “temperature”. Everybody in the circle knows what’s being talked about. But, 1) the emission is NOT ala blackbody/Planck, and 2) does NOT have (sensory) temperature. If you could somehow stick a regular thermometer, not just in the molecule, but in the area within the molecule just where the atoms are vibrating back and forth, it would read zero. You can stick a spectroscope (which you can even call a “temperature probe”) and in fact measure the frequency of the emission and relate that to an equivalent blackbody temperature, and call it temperature, and get a lot of understanding and figure stuff out about molecular energy levels, emission, equipartion, etc. But it is not really temperature, which this by itself should be a no-op and have no effect what-so-ever on your endeavors — until one starts to think of it as sensory temperature because it’s called “temperature”

    I haven’t thought the following through in detail, but I think LTE and Equipartition are two separate and different qualities. When a molecule gets excited either vibrationally or rotationally the molecule is, for a while, out of whack, equipartionally. But it is not off ala LTE; in fact I don’t think LTE applies at all to an individual molecule (and especially intramolecular). The temperature of individual molecules in a mole — call it kinetic energy to assuage Ray — vary all over the place yet the mole can be in LTE with its next door neighbor. Last, to repeat (sorry), since H2O relaxation is in part dependent on quantum mechanics probability, it is certainly mathematically possible to have all molecules in a mole excited and make my thought experiment reasonable.

    Lest we forget ala RC, all this means is that when infrared radiation is absorbed by CO2 or H2O into their vibration or rotation modes, the (sensory) temperature of the atmosphere does NOT change — until the absorbed energy is transferred to another molecule’s translation energy through a collision.

    The real problem we both might be facing is the law of science that says, “the longer the explanation, the less likely it’s true.” ;-)

  48. 498
    Rod B says:

    Ray (496), if you define a fixed frame or coordinate system then linear momentum can be explicitly and inherently associated with the bullet, pure and simple. The bullet has a fixed mass and a fixed velocity in the fixed coordinate system and has a definite momentum equal to mv. I understand the case of different or moving frames, and have no problem with it. But, if my blob, bottle, globe and the moon are in the same fixed coordinate system, my bullet, it seems, has different angular momentums which, different from the linear momentum, are not explicit or inherent but varies depending on what it is related to. I don’t think I disagree or misunderstand this; it’s just a new foreign concept that I have to get used to. Are we on the same page here?

    re the temperature stuff, you missed the (obscure) decimal point. My base case was a mole at a normal 300.(point)00000000K (or so) to emphasize the precision. In any case I think your first point agrees with my point that internal modes do not exhibit (sensory — see my reply to Martin) temperature, which is why the (sensory) temperature drops as the molecules start to spread the massive abnormal energy around into vibration and rotation, in part.

    Don’t get me wrong. When I say physicists invented a construct and called things “temperature” that weren’t really temperature, I’m not being critical in the least. It makes eminent sense and I’m all in favor of it and support it for the reasons mentioned. They’re just not real (sensory) temperature. But I do not care if the term is used; I’m not insisting, nay even suggesting, nay would even want, that translation kinetic energy have exclusive rights to the word.

    It seems that the vibration of molecules in a crystal lattice solid is not at all the same as vibrating atoms within a molecule, (other than there’s a whole lotta vibratin’ goin’ on).

    Please let me know if you wish to revise and amend given my misleading “300″ degrees.

  49. 499
    Joel Shore says:

    Re #495: Alastair says “So you see I really ought to be acquiring more evidence that my ideas are correct, rather than arguing with the upholders of the conventional wisdom here.”

    Yes…You really ought to. And, you also ought to remember that probably 99.9% of the people who believe they are overturning an “old invalid paradigm” have simply come to a wrong conclusion. So, you need to constantly be asking yourself, “Am I really so sure that I am in that special 0.1%.” Or…As Clint Eastwood would put it, “So, you gotta ask yourself kid, do you feel lucky?” ;)

  50. 500
    Ray Ladbury says:

    Rod, I think you must be making an error–the magnitude of the angular momentum is |r||p|sin(theta), where theta is the angle between them. This will not change during the bullet’s flight, and the angular momentum of the system is conserved even after the collision. You can’t keep the coordinate system fixed for the linear momentum and allow it to change for angular.
    Rod, until the modes start spreading energy around to acheive equipartition, the temperature of the system is not well defined because it is not in equilibrium. The temperature will seem to change even though energy is not changing in the system–you’re starting in a low-entropy state and temperature = partial of E wrt S. No, the vibration is exactly the same whether it’s in a solid a liquid or a gas–just the allowed modes are different. There is absolutely no advantage in the way you are thinking about this. When you say “sensory temperature,” what you are actually feeling is the flow of energy–which depends on temperature difference.


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