From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Entropy Date: 15 Aug 2000 16:56:18 GMT Newsgroups: sci.physics.research Summary: [missing] In article <8n3hfs$bvl$1@nnrp1.deja.com>, wrote: >John Baez wrote: >> We assume the molecules >> interact only gravitationally, so by the virial theorem, the kinetic >> energy K and potential energy P are related by >> >> K = -P/2 >Ain't the virial theorem only valid for thermal equilibrium?? Since collections of gravitating point particles can never be in thermal equilibrium (as I explained in another recent post on this thread), it would be rather sad if the virial theorem only applied in this case! Luckily, the conditions for the virial theorem are weaker than that. Basically we need to assume that: 1) the time averages of K and P exist and 2) the positions and velocities of the particles are bounded as a function of time. We then get K = -P/2 as an equation between *time averages*. The proof of the virial theorem is in Goldstein's _Classical Mechanics_, and I urge everyone to look at it, because it's quite simple and pretty. A good example of a system that's not in equilibrium where the virial theorem still applies is the 2-body problem, in the case of an elliptical orbit. Clearly conditions 1) and 2) hold here. But the catch is that if we have a large collection of point particles interacting gravitationally, particles are likely to "boil off" in the long run, so condition 2) will not hold. However, in the "medium run", we can try to ignore "boiling off". Instead of taking time averages over an infinite stretch of time, we can do so over a finite stretch of time. And then we can hope that these finite time averages roughly satisfy K = -P/2. Apparently it works pretty well. As you see, this stuff gets subtler and subtler the more you think about it! I'm not really sure why the virial theorem gives pretty good results, despite the "boiling off" phenomenon. I guess the boiling off tends to be rather slow. ============================================================================== From: ted@rosencrantz.stcloudstate.edu Subject: Re: Entropy Date: 15 Aug 2000 18:32:43 GMT Newsgroups: sci.physics.research In article <8nbsni$ljd$1@Urvile.MSUS.EDU>, John Baez wrote: >Luckily, the conditions for the virial theorem are weaker than >that. Basically we need to assume that: > >1) the time averages of K and P exist > >and > >2) the positions and velocities of the particles are bounded as a >function of time. > >We then get K = -P/2 as an equation between *time averages*. That doesn't seem useful for the problem we were considering, which probably means that I was considering a slightly different problem from you. We're considering a cloud that's collapsing under its own weight. In that case, we want to know about K and P at particular times, not about time averages. Both K and P are going to be changing as time passes, so a statement about time averages isn't terribly useful. I guess you want to assume that the time over which you have to average to determine those time-averages (to a good approximation) is short compared to the time scale on which things are shrinking, so that the virial theorem can be taken to apply instantaneously rather than just in the time average. Is that right? That's OK if you're considering a cloud that's virialized and is shrinking *very* slowly by evaporation (as, for instance, globular clusters do). But that misses a big part of the story of entropy change in gravitational collapse. The big cloud of gas that shrinks to become a star, for instance, starts off very far from virialized: K << |P|. You can certainly imagine such a cloud shrinking and heating up without much evaporation at all, and it's worth thinking about what happens to the entropy in that case. So I'm going to try to adapt your original calculation to the case where the initial conditions are much colder than virial. Say you've got a spherical cloud with N particles and radius R. The potential energy is P ~ -N^2 / R. The total energy is E = K + P. I want to assume that the cloud collapses without evaporation, and without any other way for energy to escape. So R gets smaller, |P| gets larger, and K gets larger by the same amount, leaving E constant. Specifically, if the cloud's radius changes by a small (negative) amount dR, then dP / P = -dR / R. Conservation of energy tells us that dK = -dP, so dK / K = -dP / K = -(P/K) dP/P = (P/K) dR/R. The temperature is proportional to K, so dT/T = (P/K) dR/R. The entropy is S = kN [(3/2) ln T + ln (V/N)], so the change in entropy as the cloud shrinks a bit is dS = kN [(3/2) dT/T + dV/V] = kN [(3/2)(P/K) + 3] dR/R. Now, if the cloud is shrinking, then dR < 0, so for dS to be positive as the cloud shrinks the term in the square bracket must be negative. That means that P/2K + 1 < 0, or K < -P/2. So a cloud can shrink without violating the second law if and only if it's cooler than its virial temperature. Summary: If you start with a large, cold cloud with K < -P/2, it can (and typically does) shrink fairly rapidly without much evaporation. During this phase, S increases (as of course it must), and the cloud heats up. Eventually, the cloud reaches a "final" radius such that K = -P/2 (the virial radius). From this time on, further change occurs only by evaporation, which is very slow. Change is so slow that you can apply the virial theorem to relate K and P at any particular time. To account properly for the entropy in this case, you have to consider both the entropy of the bound system (which decreases) and the entropy of the particles that are escaping (which provide enough extra entropy that the total increases). Does that sound right? -Ted ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Entropy Date: 15 Aug 2000 20:29:48 GMT Newsgroups: sci.physics.research In article <8nc2cb$n58$1@Urvile.MSUS.EDU>, wrote: >In article <8nbsni$ljd$1@Urvile.MSUS.EDU>, >John Baez sketched a precise statement >of the virial theorem: >>Basically we need to assume that: >> >>1) the time averages of K and P exist >> >>and >> >>2) the positions and velocities of the particles are bounded as a >>function of time. >> >>We then get K = -P/2 as an equation between *time averages*. >That doesn't seem useful for the problem we were considering, which >probably means that I was considering a slightly different problem >from you. We're considering a cloud that's collapsing under its own >weight. In that case, we want to know about K and P at particular >times, not about time averages. Both K and P are going to be changing >as time passes, so a statement about time averages isn't terribly >useful. > >I guess you want to assume that the time over which you have to >average to determine those time-averages (to a good approximation) is >short compared to the time scale on which things are shrinking, so >that the virial theorem can be taken to apply instantaneously rather >than just in the time average. Is that right? Right! You have to crack eggs to make an omelette. The full-fledged dynamical description of a rapidly collapsing initially cold gas cloud was too complicated for me to understand. So instead, I considered the situation you mention below: a "fully virialized" ball of gravitationally interacting particles that's shrinking very slowly by emitting energy in the form of "boiling off" and/or radiation. As you note, this might be a good approximation for the motion of stars in a globular cluster. This allows us to apply the virial theorem in an *approximate way*, averaging over a time period that's long enough for all the particles to wiggle all over the place, yet short enough that not much shrinkage occurs. By the way, I hope it's clear by now that ALL really interesting applications of the virial theorem are only approximate, because as soon as you have enough particles that you can't *prove* one won't fly off, you can't justify condition 2) of the virial theorem. A good example is the solar system: people have been debating the stability of the solar system for a long time. Will a planet ever be flung off? These days the experts say that yes, several will. If this is true, the virial theorem doesn't really apply. Nonetheless, I bet it gives pretty good predictions if we time-average over a period that's long compared to the year of Jupiter but short compared to the time when planets get flung off. In short, it's a dirty business - real physics, not that effete mathematically rigorous stuff! My calculation (available at http://math.ucr.edu/home/baez/entropy.html) also made some other sleazy assumptions, e.g. that the particles are distributed homogeneously in a sphere in position space, and distributed in a Maxwellian way in momentum space. Again, the main reason for doing this was to make life easy, NOT because it's actually correct. At best, we'll get some insight and then go about slowly improving the calculation by eliminating the worst unrealistic assumptions one at a time - like you did below. (By the way, Ted, I know you know most of this stuff about the need to make simplifying assumptions to do calculations involving complex phenomena. I'm not hectoring *you* about this: I'm just taking advantage of this opportunity to the hector the crowd.) >That's OK if you're considering a cloud that's virialized and is >shrinking *very* slowly by evaporation (as, for instance, globular >clusters do). But that misses a big part of the story of entropy >change in gravitational collapse. The big cloud of gas that shrinks >to become a star, for instance, starts off very far from virialized: >K << |P|. You can certainly imagine such a cloud shrinking and >heating up without much evaporation at all, and it's worth thinking >about what happens to the entropy in that case. So I'm going to try >to adapt your original calculation to the case where the initial >conditions are much colder than virial. Cool! By the way, maybe I should tell the masses what "virialized" means. The word "virial" always reminds me of "virile", so I wouldn't be surprised if Joe Six-Pack out there thinks a "fully virialized" gas cloud is one that works out at the gym regularly. Basically, a cloud of gravitationally interacting particles is "virialized" if K ~ -P/2. If K is much smaller, the particles are moving slowly, and the force of gravity will pull them together, making the cloud collapse. If K is much bigger, the particles are moving fast, and they will fly out in all directions. >Say you've got a spherical cloud with N particles and radius R. >The potential energy is > >P ~ -N^2 / R. > >The total energy is E = K + P. I want to assume that the cloud >collapses without evaporation, and without any other way for energy to >escape. So R gets smaller, |P| gets larger, and K gets larger by the >same amount, leaving E constant. Specifically, if the cloud's radius >changes by a small (negative) amount dR, then > >dP / P = -dR / R. > >Conservation of energy tells us that dK = -dP, so > >dK / K = -dP / K = -(P/K) dP/P = (P/K) dR/R. > >The temperature is proportional to K, so > >dT/T = (P/K) dR/R. > >The entropy is > >S = kN [(3/2) ln T + ln (V/N)], > >so the change in entropy as the cloud shrinks a bit is > >dS = kN [(3/2) dT/T + dV/V] > = kN [(3/2)(P/K) + 3] dR/R. > >Now, if the cloud is shrinking, then dR < 0, so for dS to be positive >as the cloud shrinks the term in the square bracket must be negative. >That means that > >P/2K + 1 < 0, > >or > >K < -P/2. > >So a cloud can shrink without violating the second law if and only >if it's cooler than its virial temperature. That's neat! But now I'm confused. I thought my calculation showed that right *at* the virial temperature, the cloud did lose entropy as it shrunk. Yours seems to say it won't change at all: dS = 0 when K = -P/2. This seems weird, since my calculation is supposed to be a special case of yours. I'll have to think about this. >Summary: If you start with a large, cold cloud with K < -P/2, it can >(and typically does) shrink fairly rapidly without much evaporation. >During this phase, S increases (as of course it must), and the cloud >heats up. Eventually, the cloud reaches a "final" radius such that >K = -P/2 (the virial radius). From this time on, further change >occurs only by evaporation, which is very slow. Change is so slow >that you can apply the virial theorem to relate K and P at any >particular time. To account properly for the entropy in this case, >you have to consider both the entropy of the bound system (which >decreases) and the entropy of the particles that are escaping (which >provide enough extra entropy that the total increases). > >Does that sound right? It certainly SOUNDS right. It's an intuitively appealing story! But I'll have to compare your calculation to mine to ease my above-mentioned worry. ============================================================================== From: ted@rosencrantz.stcloudstate.edu Subject: Re: Entropy Date: 15 Aug 2000 22:14:31 GMT Newsgroups: sci.physics.research In article <8nc5vb$ohn$1@Urvile.MSUS.EDU>, John Baez wrote: >In article <8nc2cb$n58$1@Urvile.MSUS.EDU>, > wrote: >>So a cloud can shrink without violating the second law if and only >>if it's cooler than its virial temperature. >That's neat! But now I'm confused. I thought my calculation showed >that right *at* the virial temperature, the cloud did lose entropy as >it shrunk. Yours seems to say it won't change at all: dS = 0 when K = >-P/2. This seems weird, since my calculation is supposed to be a >special case of yours. The same worry occurred to me shortly after I sent off my post, but I think I know what the answer is. Your calculation is not a special case of mine; the two are based on different assumptions. I'm assuming the cloud shrinks with energy being conserved, while you're assuming it shrinks with the virial condition remaining satisfied. So I'm assuming dK = -dP and you're assuming dK = -dP/2. So my cloud will heat up twice as much as yours as the radius shrinks a bit. That's why my cloud has constant entropy (to first order in dR) if it tries to shrink at the virial radius, while yours decreases in entropy. -Ted ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Entropy Date: 16 Aug 2000 22:01:24 GMT Newsgroups: sci.physics.research John Baez wrote, in reply to Ted Bunn: >That's neat! But now I'm confused. I thought my calculation showed >that right *at* the virial temperature, the cloud did lose entropy as >it shrunk. Yours seems to say it won't change at all: dS = 0 when K = >-P/2. This seems weird, since my calculation is supposed to be a >special case of yours. I'll have to think about this. Oh... duh! The assumptions are different: you were requiring that the kinetic plus potential energy stay constant, whereas I was letting energy leak off in the form of radiation or "boiling off" or something. The calculations really apply to different regimes: yours applies to the "fast regime" when the cloud is getting virialized but (you assume) acting like a closed system, whereas mine applies to the "slow regime" when it's already virialized and is slowly losing energy by radiation or boiling off. It all makes sense now. Great! What's really cool about your calculation is that you arrive at the conclusion of the virial theorm (K = -P/2) by a very different route than the usual one. You used thermodynamics, while the usual argument doesn't even mention entropy or temperature or anything like that - just some primitive Newtonian mechanics. ============================================================================== From: ted@rosencrantz.stcloudstate.edu Subject: Re: What is the virial theorem? Date: 20 Aug 2000 06:10:43 GMT Newsgroups: sci.physics.research In article <8nih9o$1hp$1@cantuc.canterbury.ac.nz>, Bill Taylor wrote: >It's come up a lot here recently. I had a look in the library, and saw >references to virial coefficients, and virial tensors, but few to the theorem. >And none of the above was at all helpful... no explanations at all. Have you tried classical mechanics textbooks? As I think John B. mentioned in a recent post, there's a proof of the theorem in Goldstein's textbook. The statement and proof of the theorem are on pp. 82-85 of the 2nd Edition. The theorem says the following: = -(1/2) . Here r_i is the position of the ith particle in the system, F_i is the force on that particle, T is the total kinetic energy, and <...> denotes a time-average. The theorem applies if the motion is periodic (in which case the average is over a period), or if all positions and velocities are bounded, in which case it applies in the limit as the time over which the average is taken tends to infinity. If the particles are subjected to a central force F = A r^n, then this becomes = -(1/2) = ((n+1)/2) , where V is the potential energy. For the particular case of gravity, n = -2, and we have = -(1/2) . The virial theorem is extremely important in astronomy. Very often, a bunch of particles will collapse to form a gravitationally bound system. If that system is in some sense in equilibrium (i.e., the coarse-grained distribution of particles is independent of time), then the virial theorem implies that T = -(1/2) V. This is a terrific thing, because it lets you find the masses of bound systems. In fact, it's really the reason we think that dark matter exists. To be specific, suppose you measure the speeds of a bunch of visible objects in your system, and infer T. Then the virial theorem tells you V. If you find out that the potential well is deeper than what you'd get by adding up the contributions from the masses of everything you see, you know there's dark matter. People do this for spiral galaxies, elliptical galaxies, and galaxy clusters, getting strong evidence for dark matter in all cases. By the way, most descriptions of the evidence for the existence of dark matter in spiral galaxies phrase the argument in terms of Kepler's third law, or equivalently Newton's second law applied to circular motion, G M m / r^2 = m v^2 / r, instead of the virial theorem. But this is really just a special case of the virial theorem, as you can check for yourself if you're so inclined. (The proof is actually immediate from the equation above.) If you want to generalize to, say, galaxy clusters, in which things don't move in nice circular orbits, Kepler's law doesn't apply, and you need to go straight to the virial theorem. As far as the recent thread is concerned, John was imagining a gravitationally bound cloud of particles that was in quasi-equilibrium, so that the coarse-grained distribution was changing only extremely slowly (by evaporation). In that case, the virial theorem could be taken to apply, at least approximately, without the time-averages. That gives a relation between the temperature (proportional to T) and the radius (related to V) of the system, which is what you need to infer the energy. My own analysis considered the preamble to the case John was considering. I imagined a cloud that had not yet reached quasi-equilibrium. So at first the virial theorem did not apply (instantaneously). Such a cloud collapses (if T < -V/2) or expands (if T > -V/2) until it reaches equilibrium (T = -V/2), or "virializes." -Ted ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: What is the virial theorem? Date: 23 Aug 2000 22:53:45 GMT Newsgroups: sci.physics.research Bill Taylor wrote: > [....] will someone please tell us in simple terms, what it says, > what significance it has, and how it impacted on the recent thread > it appeared in so often. Ted Bunn and others have answered this, and you can get more details from Goldstein's book - the bible of classical mechanics. But just for the heck of it, I've made a little webpage on the virial theorem, including the proof. Here it is: http://math.ucr.edu/home/baez/virial.html I've borrowed some of Ted's description for this post, so he should let me know if he objects. (Naturally I credited him.) My webpage has a link to another one that's eventually supposed to be a decent explanation of the thermodynamics of gravitating systems. However, right now that one is a bit rough and does not take into account all the stuff we've just learned by discussing the subject here on s.p.r..