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#11
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entropy and gravitation
Am 02.06.2017 um 12:07 schrieb Phillip Helbig (undress to reply):
In article , Gerry Quinn writes: To put it another way, the 'clumpy' states in the non-gravitational universe have lower entropy than the smooth state, but the clumpy states in the gravitational universe have higher entropy than the smooth state. Imagine a clumpy universe with no gravity. It has low entropy (lower than the smooth universe). Now G starts increasing from zero to, say, its current value (at which point the clumpy universe has a higher entropy than the smooth universe). At some value of G, the clumpy universe must have the same entropy as the smooth universe (which you say has the same entropy with or without gravity). So for this value of G, the entropy is independent of the clumpiness. Someone has made an error somewhere. Why should it not be independent of the clumpiness? Because it's not. A room full of air with the same density everywhere has higher entropy than a room with all of the air squeezed into one corner. (In the case where gravity can be neglected. When gravity plays a role, then the clumpier distribution has higher entropy.) This kind of comparison needs a gas, a process that is adiabatic for one leg and isothermal for the other leg of a reversible path in state spece and therefor at least one thermal bath. Because all such things do not exist in the universe of lets say a gas of galaxies or photons or hydrongen and helium all kinds of modelling of entropy along the classical examples of gas in a variable volume and and two temperatur baths at hand are highly doubted in the community. Finally, the two volumes of a system at two times are the 3d-boundaries of a 4-volume, bottom and ceiling orthogonal to the direction of time. With a nonstationary 3-geometry in the rest system volume changing has no thermodynamic effect because all particles and fields follow their unitary or canonically free time evolution in a given Riemann space. That does not change the von Neumann entropy because of Liouvilles theorem of constancy of any 6-volume element of spce and momentum. Finally for interacting system of fermionic particles and fields at temperatures below the Fermi temperature, a state with lumpy matter and a small fraction of free gas over its surface is the state of maximal entropy. Interacting matter evenly distibuted in a given volume that it does not fully occupiy as a condensed body is highly improbable. -- Roland Franzius |
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entropy and gravitation
On 6/1/17 6/1/17 3:36 PM, Phillip Helbig (undress to reply) wrote:
In article , Gerry Quinn writes: [...] Someone has made an error somewhere. Yea. I believe it is in your entire approach: I don't think that entropy is an absolute quantity as you implicitly assume. It does make sense to compare entropy between states of a given universe, but not between states of different universes. After all, entropy is an extensive quantity, and in considering different universes you cannot possibly ensure they have the same values of all other extensive quantities. In "changing the value of G" you really have a different universe for each value. You cannot change the laws of physics in a universe, you must consider an ensemble of universes. For instance, consider two universes, one with G=0 and one with G0. Give them the exact same initial conditions, in that the initial geometries are the same, as are all field distributions and derivatives. Since the Lagrangians are different, the evolutions of the fields and geometries will be different, the total energies will be different, etc. -- such extensive [#] differences surely invalidate any comparison of their entropies. This presumes it is possible to give them the same initial conditions. It is not obvious that this is so.... Certainly GR has strong constraints that the initial conditions must meet; different values of G could yield different constraints. [#] This is a pun: "extensive" as in type of property, and "extensive" as in vast or widespread. Tom Roberts |
#13
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entropy and gravitation
On 02/06/2017 11:07, Phillip Helbig (undress to reply) wrote:
In article , Gerry Quinn writes: To put it another way, the 'clumpy' states in the non-gravitational universe have lower entropy than the smooth state, but the clumpy states in the gravitational universe have higher entropy than the smooth state. Imagine a clumpy universe with no gravity. It has low entropy (lower than the smooth universe). Now G starts increasing from zero to, say, its current value (at which point the clumpy universe has a higher entropy than the smooth universe). At some value of G, the clumpy universe must have the same entropy as the smooth universe (which you say has the same entropy with or without gravity). So for this value of G, the entropy is independent of the clumpiness. Someone has made an error somewhere. It is a failure of intuition rather than of physics. The apparent paradox is because a self gravitating clump of material gets hotter as shrinks under the influence of its own gravity. Adding gravity makes the smooth uniform matter distribution metastable wrt perturbations. Why should it not be independent of the clumpiness? Because it's not. A room full of air with the same density everywhere has higher entropy than a room with all of the air squeezed into one corner. (In the case where gravity can be neglected. When gravity plays a role, then the clumpier distribution has higher entropy.) The difference is that once gravity gets involved there is potential energy available to be released when a clump of matter collapses under the influence of mutual gravitational attraction (gravity is always and attractive force). The shrinking material heats up as it is compressed. The original uniform maximum entropy state is not the lowest energy state for the system and so it is vulnerable to collapse if density fluctuations arise sufficient to allow self gravitating clumps. It would behave like a short lived star collapsing in on itself and then getting smaller and hotter as a result without any nuclear fusion to hold it up for longer. Martin Rees describes this far better than I can on p116 of Just 6 Numbers in the section about Gravity and Entropy. You now have a significant temperature difference between your new gravitational star and the background which can be used to do work. Originally it was Lord kelvin that did the lifetime computation of a star powered only by gravitational collapse as a means of discrediting the very long geological timescales needed for Darwinian evolution. -- Regards, Martin Brown |
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entropy and gravitation
Le 02/06/2017 Ã* 09:16, Poutnik a écrit :
Thermodynamics generally does not care, what time it takes for a system to get into the preferred final state. "Final state" implies time. Atoms A and B in gas state are inserted into a container in proportion 1:1. The final state is an almost perfect distribution of a mixture of both gases. We do not expect the gases to appear separated after some time. That is the accepted final state of a smooth distribution for two gases in a container at room temperature say. But is it the final state? Surely not, since if not given any external energy, the final state of the mixture could be a separated mixture of frozen A and B at almost absolute zero. Let's suppose that when freezing, gases A and B do not mix easily. Time is always there in all physics. The concept of "final state" implies time, you see? |
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entropy and gravitation
Dne 02/06/2017 v 23:43 jacobnavia napsal(a):
Le 02/06/2017 Ã* 09:16, Poutnik a écrit : Thermodynamics generally does not care, what time it takes for a system to get into the preferred final state. "Final state" implies time. Atoms A and B in gas state are inserted into a container in proportion 1:1. The final state is an almost perfect distribution of a mixture of both gases. We do not expect the gases to appear separated after some time. That is the accepted final state of a smooth distribution for two gases in a container at room temperature say. But is it the final state? Surely not, since if not given any external energy, the final state of the mixture could be a separated mixture of frozen A and B at almost absolute zero. Let's suppose that when freezing, gases A and B do not mix easily. Time is always there in all physics. The concept of "final state" implies time, you see? "does not care what time" does not mean "time is not implied" As thermodynamics is just one side of the coin. The other is kinetics. Some states are thermodynamically stable, some are kinetically stable, some both. -- Poutnik ( The Pilgrim, Der Wanderer ) A wise man guards words he says, as they say about him more, than he says about the subject. |
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entropy and gravitation
On 5/29/17 9:55 PM, Phillip Helbig (undress to reply) wrote:
A smooth distribution corresponds to high entropy and a lumpy one to low entropy if gravity is not involved. For example, air in a room has high entropy, but all the oxygen in one part and all the nitrogen in another part would correspond to low entropy. If gravity is involved, however, things are reversed: a lumpy distribution (e.g. everything in black holes) has a high entropy and a smooth distribution (e.g. the early universe) has a low entropy. Let's imagine the early universe---a smooth, low-entropy distribution---and imagine gravity becoming weaker and weaker (by changing the gravitational constant). Can we make G arbitrarily small and the smooth distribution will still have low entropy? This seems strange: an ARBITRARILY SMALL G makes a smooth distribution have a low entropy. On the other hand, it seems strange that the entropy should change at some value of G. I think the mistake here is thinking about "smooth" and "lumpy" as a binary choice. What G affects is *how* lumpy the maximum entropy system is. Suppose first that there are no forces except gravity. As soon as you turn on G, a smooth system becomes unstable -- the Jeans length is zero. Thermodynamically, the gravitational potential energy can become arbitrarily negative, and at fixed energy it's entropically favorable for the system to collapse a little more, lowering the gravitational energy, and kick out a particle with extra kinetic energy. The classic analysis of this is Lynden-Bell and Wood, "The Gravo-Thermal Catastrophe in Isothermal Spheres and the Onset of Red-Giant Structure for Stellar Systems," MNRAS 138 (1968) 495. Now suppose there are other forces that are not purely attractive. Dynamically, the Jeans length is now finite, and this determines the typical size of lumps. If you turn up G, the Jeans length decreases, and you get more, smaller lumps. Thermodynamically, you can still increase entropy by collapsing and kicking out particles with high kinetic energy, but this process is now limited, since the collapse will eventually be stopped by other forces. Bigger G allows more collapse before this equilibrium is reached, and more lumpiness. I don't know of anywhere this has been worked out, but I suspect that if you found a measure of the amount of lumpiness in the maximum entropy state you'd find that it varies smoothly with G. (There's probably some nice way to use the Jeans length for this.) Steve Carlip |
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entropy and gravitation
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#18
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entropy and gravitation
On 6/7/17 2:08 AM, Gregor Scholten wrote:
The solution is that it is a matter of temperature. That a lumpy distribution has higher entropy than a smooth distribution as soon as gravity is involved is only true for low temperatures. For high temperatures, the smooth distribution still has the higher entropy. That's why the universe has to be cold enough before galaxies and stars can form. In as much as galaxy and star planetary system size distributions are different, are two different formation temperatures required within the concept of Jeans' length? Richard D Saam |
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entropy and gravitation
On 02/06/2017 22:43, jacobnavia wrote:
Le 02/06/2017 à 09:16, Poutnik a écrit : Thermodynamics generally does not care, what time it takes for a system to get into the preferred final state. "Final state" implies time. No it doesn't apart from perhaps having to wait an eternity to get there. Metastable system states may persist almost forever if the activation energy to escape from a local minima is too large. Atoms A and B in gas state are inserted into a container in proportion 1:1. The final state is an almost perfect distribution of a mixture of both gases. We do not expect the gases to appear separated after some time. That is the accepted final state of a smooth distribution for two gases in a container at room temperature say. Give or take a random fluctuations yes. If there are N atoms in the volume and you choose to split the space down the middle with an imaginary line then the probability of a split N-n, n across that line is given by the terms in the binomial expansion with 2^N states in total. The most common states being determined by their degeneracy factor - essentially one derivation of entropy. N!/((N-n)!n!) Formally it is obviously maximised when n=N/2 although you would be very surprised if you didn't see variation. If N is small enough then it isn't so long to wait to catch all of them in one half by chance. But if N is Avagadro's number - well you do the maths. But is it the final state? Surely not, since if not given any external energy, the final state of the mixture could be a separated mixture of frozen A and B at almost absolute zero. Let's suppose that when freezing, gases A and B do not mix easily. Perhaps more apposite to the original question one of the methods of separating U235 from U238 relies on making UF6 gas and putting it through a cascade of centrifuges to impose a large potential gradient on the maximum entropy distribution in the spinning centrifuges. Time is always there in all physics. The concept of "final state" implies time, you see? The concept of "final state" implies that ultimately it has a lower energy than all other possible states. Although something may still be long term metastable despite a lower energy state being available if there is an activation energy needed to get there that isn't available. Common window glass for example. -- Regards, Martin Brown |
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entropy and gravitation
In as much as galaxy and star planetary system size distributions
are different, are two different formation temperatures required within the concept of Jeans' length? The Jeans length is important for star formation, but the stuff which forms (rocky) planets is only a small fraction of a larger cloud which collapsed (as described by Jeans) to form a star. There doesn't seem to be a lower limit on the size of "planets". There is an obvious upper limit for (gaseous) planets---stars. The sizes of planets are determined more by accretion, where gravitation is only one factor. |
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