If the Second Law of Thermodynamics is always true, then what about big isolated nebulas in intergalaxial space? Doesn't a collapsing neblua go from disorder to a relatively ordered star system?

-FredrikM

Originally posted by FredrikM
If the Second Law of Thermodynamics is always true, then what about big isolated nebulas in intergalaxial space? Doesn't a collapsing neblua go from disorder to a relatively ordered star system?

-FredrikM

you know, i don t know the answer to this one. but i think it s not such an easy problem. read this page (http://www.math.ucr.edu/home/baez/entropy.html) for more on this question.

sorry i don t know the answer.

I would say that as a Nebula collapses to become a galaxy with individual stars and planets, a tremendous amount of heat and light is released. When you calculate the entropy, including the heat, the total entropy of the system will increase with the collapse.

Originally posted by James R
I would say that as a Nebula collapses to become a galaxy with individual stars and planets, a tremendous amount of heat and light is released. When you calculate the entropy, including the heat, the total entropy of the system will increase with the collapse.

according to that page, the entropy of the nebula decreases. but i guess James is correct. we are forced to conclude that there is enough entropy in the radiation produced from the collapse, that the second law is saved.

No, No, No

Entropy is not a measure of disorder, but is related to the abilty of a system to do work.

The gravitational potential energy of the nebulae is greatest when the gas is dispersive, and less when the gas forms a star. In both instance the rest mass energy of the particles are an additive constant and thus may be ignored.

A dispersive configuration only maximises the energy when the potential is a hard sphere potential. For attractive forces, the tendancy towards configurations with the lowest potential energy drives the system towards a state of higher entropy.

Entropy is one of the most misinterpreted quantities in physics. It is not a measure of disorder.

Originally posted by ryans
No, No, No

Entropy is not a measure of disorder, but is related to the abilty of a system to do work.

...

Entropy is one of the most misinterpreted quantities in physics. It is not a measure of disorder.

are you sure? i have always learned that entropy is the logarithm of the number of microstates (which is related to the notion of disorder). i have never seen a definition of entropy that says anything about the ability to do work.

can you please provide the definition of entropy you have in mind? in what way is it related to the ability to do work?

dS=dQ/T

S= entropy
Q=internal energy
T= temperature

That's the definition. You don't need to consider discrete states here. In Thermodynamics, work is the ability to transfer heat. This is how the heat death of the universe is related entropy. If the universe is at a uniform temperature, then heat can't flow, work cannot be done, at which point entropy is maximised.



i have always learned that entropy is the logarithm of the number of microstates

That is for the quantum canonical ensemble, which is prescribed as constant number of particles, constant temperature and constant volume, hardly applicable.

If you simply take dQ denote the work performed by the particles (they compress) then dQ is positive and hence so is dS

In Thermodynamics, work is the ability to transfer heat

Sorry edit above. Work is the change in the internal energy of a system.

For say the universe, if it can be considered a closed system where conservation of energy stricly holds, once entropy is maximised, work is required to be performed on the system by an external source to change the canonical quantities of the universe. If the universe is everything, then there is no external source.

Originally posted by ryans

That is for the quantum canonical ensemble, which is prescribed as constant number of particles, constant temperature and constant volume, hardly applicable.

If you simply take dQ denote the work performed by the particles (they compress) then dQ is positive and hence so is dS

ahh... you are talking thermodynamics, i am talking statistical mechanics. i believe that the two definitions coincide in the thermodynamic limit.

in statistical mechanics, entropy is the logarithm of the number of states, which is a direct measure of the disorder of the system. since the two definitions coincide, your definition is also a measure of the disorder of this system. why do you think this notion is not applicable? the statistical definition is not restricted to cases with constant temperature and constant number of particles.

also, in thermodynamics, dQ is not internal energy, it is heat transfer. it is related to internal energy of course, by the first law of thermodynamics: dU = dQ - dW. where dW is work done by the system.

which is a direct measure of the disorder of the system. since the two definitions coincide, your definition is also a measure of the disorder of this system.

Then how come at the triple point of water, or the liquid solid interface, the entropy of each phase is the same, even though we have the apparently ordered crystal coexisting next to an amorphous liquid/gas. Because work is required to be done on the system by an outside source to further change the configuarion of the system. Without further work done on or by the system, this arrangement shall not change. The model you apply is for closed system where pressure and temperature are constant and single phases are present. For this arrangement then yes, you may consider entropy as a form of disorder. But for other systems. this is not the definion prescribed 2 and in fact leads to paradoxes, which are easily demonstratable if you are so inclined to know.

For example imagine the simple arrangement where we have 2 atoms both in the first excited state. Each atom can emit and absorb a photon promoting it or lowering it to the E2 or E0. How to you calculate how the system will evolve i.e. which configuration represents equilibrium

A B
E0 E2
E1 E1
E2 E0

Originally posted by ryans

For example imagine the simple arrangement where we have 2 atoms both in the first excited state. Each atom can emit and absorb a photon promoting it or lowering it to the E2 or E0. How to you calculate how the system will evolve i.e. which configuration represents equilibrium

A B
E0 E2
E1 E1
E2 E0

for a system with only 3 microstates, the law of large numbers doesn t apply, and neither does the second law of thermodynamics. all three states are equally likely.

But see, you have already misinterpreted the statistical mechanics formulation of entropy, as the value of the entropy applies to the whole system and not individual states.
Entropy is given as

S=-k[SUM]Plog(P)

where P is the probabilty of the microstate.
Thus the entropy of the above state is

S=-k(1/3log(1/3)+1/3log(1/3)+1/3log(1/3)
= -klog(1/3)
=klog3

Individual states don't have values of entropy, and this is where the notion of entropy is misinterpreted. It is a property of a whole system, not individual states of the system.

Originally posted by ryans
But see, you have already misinterpreted the statistical mechanics formulation of entropy, as the value of the entropy applies to the whole system and not individual states.
Entropy is given as

S=-k[SUM]Plog(P)

where P is the probabilty of the microstate.
Thus the entropy of the above state is

S=-k(1/3log(1/3)+1/3log(1/3)+1/3log(1/3)
= -klog(1/3)
=klog3

Individual states don't have values of entropy, and this is where the notion of entropy is misinterpreted. It is a property of a whole system, not individual states of the system.

it is macrostates that have entropy. but in fact, i don t think i misinterpreted anything. i just claimed that the whole ball of wax does not apply to this system. remember, the thermodynamic limit is where the number of particles goes to infinity. 2 is simply not close enough to infinity for me to even talk about entropy in this case.

For work do be done in the above system, either the photon has to escape, or a photon is admitted, but in which case the system is no longer closed. The canonical ensemble allows only changes in the Entropy, Pressure and chemical potential of the system. For the system described above, entropy is constant.

Incidentally if the particle are indistinguishable, the entropy of the system is changed, with only 2 states accessible

S=-k(2/3log(2/3)+1/3log(1/3)
=-k/3log4/27

i just claimed that the whole ball of wax does not apply to this system. remember

But it applies. What is the difference between 2 microstates and 1000. This is how statistical mechanics is built up. Its misapplication of the rules which leads to the incorrect answers. It goes the same with HUP. How often have you seen that misapplied?

Originally posted by ryans
It goes the same with HUP. How often have you seen that misapplied?

a lot

ryans:

You have made several statements which seem to conflict with my understanding of entropy. I hope you can help me with this.

<i>dS=dQ/T

S= entropy
Q=internal energy
T= temperature

That's the definition.</i>

My understanding is that Q is the heat, not the internal energy.

The first law of thermodynamics says dU = dQ - dW, where U is the internal energy, Q is the heat into the system, and W is the work done by the system.

Notice that the change in the internal energy depends on both the work done and the heat transfer.

<i>If you simply take dQ denote the work performed by the particles (they compress) then dQ is positive and hence so is dS</i>

Consider just the particles in the collapsing nebula. Then, I would say that dU is negative and dW is positive. dQ is negative, because heat is released in the process. I think that under these circumstances, dS is probably negative too, so the entropy of the particles alone increases.

Now include the radiated heat in the system. dU is negative due to the change in gravitational potential. dQ, by definition, is zero. dW is positive. I'm not sure about the entropy change in this case. It is either zero or positive.

<i>For example imagine the simple arrangement where we have 2 atoms both in the first excited state. Each atom can emit and absorb a photon promoting it or lowering it to the E2 or E0. How to you calculate how the system will evolve i.e. which configuration represents equilibrium

A B
E0 E2
E1 E1
E2 E0</i>

Since all states have equal energy, I would say that all are equally likely. Therefore, there is no equilibrium solution. In an ensemble of atoms like this, I would expect to find each combination equally, taking the atoms in pairs. I would also imagine that the configurations have equal entropy. All this assumes, of course, that the atoms are in equilibrium with a radiation field.

<i>For work do be done in the above system, either the photon has to escape, or a photon is admitted, but in which case the system is no longer closed.</i>

Yes, in which case we have a whole new ball game, in which states such as E0 E0 are allowed.

<i>The canonical ensemble allows only changes in the Entropy, Pressure and chemical potential of the system. For the system described above, entropy is constant.</i>

I agree.

ryans/James R.,

Notice that the change in the internal energy depends on both the work done and the heat transfer.

While I clearly prefer to not agree with ryans :D, I think I must in this case.

However, I also think it is more interesting than just the ultimate entrophy question.

The overall scenario seems to suggest that gravity is a form of energy (i.e. - does work), and I still suspect under proper control (localized conditions) has the ability to convert its potential into useful energy for our consumption.

We just have to figure out how. :D

James
My understanding is that Q is the heat, not the internal energy.

Same thing. The amount of heat (NOT THE TEMPERATURE) something holds is its internal energy.

Consider just the particles in the collapsing nebula. Then, I would say that dU is negative and dW is positive. dQ is negative, because heat is released in the process. I think that under these circumstances, dS is probably negative too, so the entropy of the particles alone increases.

No. What about the gain in kinetic energy of the particles. You cannot treat the system as an ideal gas anymore, specifically, the frequency of higher order collisions (3, 4 or more bodies colliding instantaneously) increases, highly non-ideal. You cannot even use boltzmanns relationship between average kinetic energy and temperature, as axcitation of particles also involves rotational energy.



Since all states have equal energy, I would say that all are equally likely. Therefore, there is no equilibrium solution. In an ensemble of atoms like this, I would expect to find each combination equally, taking the atoms in pairs. I would also imagine that the configurations have equal entropy.


.Entropy cannot be attributed to individual configurations, but is a property of the whole system. To calculate entropy you must take an ensemble of all configurations, work out the probability of each and use the formula I states earlier

Conservation of energy. What goes up must come down

Originally posted by ryans
James


Same thing. The amount of heat (NOT THE TEMPERATURE) something holds is its internal energy.


heat and internal energy are not the same thing. if they were, the first law of thermodynamics would state that no system can ever do work.

let me remind you the form of the first law of thermodynamics

dU = dQ - dW.

if you think that Q and U are the same thing, you have badly misunderstood beginning thermodynamics. if Q and U are the same thing, the first law of thermodynamics becomes

dW = 0.

thus no system can ever do work.

Seems as though I'm telling people they're sloppy with their language, then I go and say that. You are totally right.

ryans:

Oops. I made a mistake. The part you quoted above should have read:

Consider just the particles in the collapsing nebula. Then, I would say that dU is negative and dW is positive. dQ is negative, because heat is released in the process. I think that under these circumstances, dS is probably negative too, so the entropy of the particles alone <b>decreases</b>.

You replied:

<i>No. What about the gain in kinetic energy of the particles.</i>

The change in kinetic energy is due to the work done. Since dW is positive, there is an increase in kinetic energy.

<i>You cannot treat the system as an ideal gas anymore...</i>

I didn't mention ideal gases. The laws of thermodynamics aren't restricted to ideal gases.

<i>...specifically, the frequency of higher order collisions (3, 4 or more bodies colliding instantaneously) increases, highly non-ideal. You cannot even use boltzmanns relationship between average kinetic energy and temperature, as axcitation of particles also involves rotational energy.</i>

I agree with all that.

<i>Entropy cannot be attributed to individual configurations, but is a property of the whole system. To calculate entropy you must take an ensemble of all configurations, work out the probability of each and use the formula I states earlier </i>

I disagree. Entropy is a state variable related directly to a particular configuration of a system. In that sense, it is similar to things like volume, number of particles etc.

.I disagree. Entropy is a state variable related directly to a particular configuration of a system. In that sense, it is similar to things like volume, number of particles etc.

No, that is simply wrong. In thermodynamics, you cannot measure, or even in fact calculate the entropy of a system. Reasons.

In thermodynamics, energy changes continously hence dS=dQ/T
Only changes in entropy are important, and this goes only one way, up, and its definition is the ability to do work. Look it up in a scientific dictionary

If you want to treat individual configurations, you must use statistical mechanics. Entropy does not nessesarily have to increase on microscopic time scales, but averaging over all ensembles (Hence the term ensemble average) gives an increase in entropy of system not in equilibrium.

YOU CANNOT CALCULATE THE ENTROPY OF ONE CONFIGURATION. You must average over all ensemble, and formulate the required canonical partition function. Your above statement is simply incorrect, and I will argue that with more vigour than trying to convince Mac relativity is correct.

ryans,

YOU CANNOT CALCULATE THE ENTROPY OF ONE CONFIGURATION. You must average over all ensemble, and formulate the required canonical partition function. Your above statement is simply incorrect, and I will argue that with more vigour than trying to convince Mac relativity is correct.

Aww come on. :D

Originally posted by ryans

In thermodynamics, energy changes continously hence dS=dQ/T
Only changes in entropy are important, and this goes only one way, up, and its definition is the ability to do work. Look it up in a scientific dictionary


i am sorry, but how can you say that the definition of entropy is the ability to do work? it doesn t even have the right units for that. and the formula you gave (dQ/T) doesn t mention work, only temperature and heat, and we just agreed that work and heat are not the same thing.

i have my thermodynamics book on my lap right now (schroeder), and its definition of entropy is the statistical definition: entropy of a macrostate is the log of the number of microstates available.

no changes in energy required. no constant temperature assumed, or fixed particle number.

lemme look a little further:

then Schroeder says that the formula dS = dQ/T is only valid for quasistatic processes, and is not true for, say, the free expansion of a gas. in this case, the best we can say is dS > dQ/T. so i don t see how this formula can be sensibly take as the definition of entropy.

i am just quoting out of my textbook here.

Ryans, please have a read of http://farside.ph.utexas.edu/teaching/sm1/lectures/node42.html to see where Lethe is coming from.

Now back to the OP. The second law of thermodynamics says entropy increases in a closed system. A Nebula is hardly a closed system. It either collapses due to gravity, hence energy enters the system, or it is shocked by something, again energy enters the system.

The classic trick question is, "Why does a flower conreadict the 2nd law". Answer, it's not a closed system. Same for Nebula.

Originally posted by thed

Now back to the OP. The second law of thermodynamics says entropy increases in a closed system. A Nebula is hardly a closed system.
this is, of course, the only reasonable conclusion. however...

It either collapses due to gravity, hence energy enters the system,
graviational collapse is not energy entering the system. it is the exchange of gravitational potential energy for kinetic or thermal energy. it can happen in a closed system, which i think is the whole source of the confusion.

or it is shocked by something, again energy enters the system.
gravitational collapse is usually spontaneous. it happens when the gravitational pressure excedes the radiation pressure.

radiation is probably the key for realizing that this is not a closed system. so the nebula collapses, until its radiation pressure grows to be equal to its gravitational pressure, at which point, it is turning into a star, and losing lots and lots of energy due to radiation, so it is not a closed system.

Lethe
entropy of a macrostate is the log of the number of microstates available.

Wrong
It is the sum over the log of all states time the probabilty of the state. SUM

You have to sum over all possible states.LISTEN TO ME

SINGLE CONFIGURATIONS DO NOT HAVE ENTROPY.

Where's cris, he knows this shit.

i am sorry, but how can you say that the definition of entropy is the ability to do work?

Oxford dictionary of science

"Entropy-A measure of the unavailabilty of a system's energy to do work; In a closed system an increase in energy is iaccompanied by a decrease in energy available"

Originally posted by ryans
Lethe


Wrong
It is the sum over the log of all states time the probabilty of the state. SUM

simply saying "Wrong" is not a very strong argument.

You have to sum over all possible states.LISTEN TO ME

SINGLE CONFIGURATIONS DO NOT HAVE ENTROPY.

a macrostate is not a single configuration. there are many possible microstates available that are all the same macrostate.

a macrostate has an entropy. i am looking at this statement in my textbook as we speak. do you want me to transcribe my book for you?



Where's cris, he knows this shit.
yes, where is Crisp, he could probably settle this more authoritavely than i. maybe i should PM him? he hasn t been to sciforums in almost 2 weeks, it seems.



Oxford dictionary of science

"Entropy-A measure of the unavailabilty of a system's energy to do work; In a closed system an increase in energy is iaccompanied by a decrease in energy available"
i will take a thermodynamics and statistical mechanics textbook definition over a dictionary, any day of the week.

and what the fuck is with that definition? an increase in energy is accompanied by a decrease in energy? what the fuck does that mean?

you can keep your oxford dictionary of science, thanks.

ryans:

<i>No, that is simply wrong. In thermodynamics, you cannot measure, or even in fact calculate the entropy of a system.</i>

Hmm. I'm not sure about measuring, but I'm pretty sure you can calculate it. I'm with lethe as regards the statistical mechanics definition of entropy (which can be shown to give all the correct macroscopic features).

<i>Only changes in entropy are important, and this goes only one way, up, and its definition is the ability to do work. Look it up in a scientific dictionary</i>

The ability to do work is a definition of <b>energy</b>, not entropy.

<i>If you want to treat individual configurations, you must use statistical mechanics. Entropy does not nessesarily have to increase on microscopic time scales, but averaging over all ensembles (Hence the term ensemble average) gives an increase in entropy of system not in equilibrium.</i>

An ensemble average is not an average over many ensembles. It is an average over the properties of the individual particles in a single ensemble. Isn't it?

<i>YOU CANNOT CALCULATE THE ENTROPY OF ONE CONFIGURATION.</i>

I think you can. If this goes on for much longer, I might even have to resort to a textbook or something to back myself up. :)

and what the fuck is with that definition? an increase in energy is accompanied by a decrease in energy? what the fuck does that mean?

Sorry, I was getting a bit excited, and it's the only reference I hav on me at the moment. It should read

Entropy-A measure of the unavailabilty of a system's energy to do work; In a closed system an increase in entropy is accompanied by a decrease in energy available"

You two smut are confusing ensembles.

Check this out

here (http://bessie.che.uc.edu/tlb/teach/grad/722w03/tbnotes/node4.html)

<i>"Entropy-A measure of the unavailabilty of a system's energy to do work; In a closed system an increase in energy is iaccompanied by a decrease in energy available"</i>

A measure of the <b>un</b>availability of a system's energy to do work is very different from a measure of the system's ability to do work (which is its internal energy).

JAMES

An ensemble average is not an average over many ensembles. It is an average over the properties of the individual particles in a single ensemble. Isn't it?

Definately not.

1) ergodic hypothesis

In short
Time averages may be replaced by ensemble averages

Originally posted by ryans

Check this out

here (http://bessie.che.uc.edu/tlb/teach/grad/722w03/tbnotes/node4.html)

umm... equation 1.30 of that link says that entropy of a system with energy E is the log of the number of states in that energy.

this is what i have been saying all along.

Originally posted by ryans
Oxford dictionary of science

"Entropy-A measure of the unavailabilty of a system's energy to do work; In a closed system an increase in entropy is iaccompanied by a decrease in energy available"

Entropy is a notion used both in thermodynamics and statistical physics.
The definition that you give from the dictionary is apparentely the thermodynamic one.

Since Statistical physics intend to give the results of thermodynamics (i.e thermodynamics should be a consequence of statistical physics), the definition given in statistical physics is the correct one, while the definition of entropy in thermodynamics is a consequence.
Historically, thyermodynamics existed before statistical physics, so all the thermodynamical notions allready existed, and had their definitions, butthese are not necessarilly the basic ones.

Entropy is a state variable.

see here (http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/ThermLaw2/ThermalProcesses.html).

Originally posted by ryans
JAMES



Definately not.

1) ergodic hypothesis

In short
Time averages may be replaced by ensemble averages

"may be replaced with" does not mean the same thing as "is the same thing as". ensemble averages means average over the microscopic states, as james claims.

Entropy is a state variable.

So, you still have to average over ensembles. That's why it is called statistical mechanics, you look at all possible states of the system, not just the one the system is in at the moment.

A measure of the unavailability of a system's energy to do work is very different from a measure of the system's ability to do work (which is its internal energy).

Pedantic, and you kno it. The system can have all the internal energy in the world, but if the energy of the medium surrounding the system is at the same energy, then the system cannot do work(e.g. expand). Thus only changes in internal energy matter

Lethe, it's the same as renormalisation, you set your zero where you want it becasue the actual fact is that we can only measure changes in a system.

Let's take a very simple (artificial) system. It is a box which contains two distinguishable particles, A and B. Each particle can exist in either the left side of the box or the right.

There are four possible configurations of the system:

|AB| |
| |AB|
|A|B|
|B|A|

These are microstates.

Now, define three macrostates:

State 1: Two particles in left side
State 2: Two particles in right side
State 3: One particle on each side

Now, the entropy of a state is defined (leaving out a multiplicative constant) to be:

S = log N

where N is the number of microstates corresponding to each macrostate.

The entropies of the three macrostates are:

State 1: zero
State 2: zero
State 3: log 2

Notice that entropy is a state variable. Each possible (macro)state of a system has a certain value of entropy. The system can only exist in one state at a time, so the system has a particular value of entropy at any particular time.

The equipartition theorem assumes that all microstates are equally probable, so at any time the probabilities of the system being found in the three states are:

State 1: 1/4
State 2: 1/4
State 3: 1/2

If we start the system in, say, state 1, and let it evolve randomly, so it can occupy any microstate, then, on average, we will be more likely to find it in a state with higher entropy.

Now, ryans, my question is: what does all this have to do with averaging over ensembles,and with entropy not applying to a single system?

Originally posted by James R

The equipartition theorem assumes that all microstates are equally probable, so at any time the probabilities of the system being found in the three states are:


not to nitpick, but i am under the impression that the statement that all microstates are equally likely is a postulate, not a theorem...

is that so?

You have done that wrong, because by this conjecture. For distinguishable particles there is 4 distinct configurations.
YOU HAVE USED ENSEMBLES.

The first configuration is the first ensemble, the next the second etc

Your macrostates are ensembles, your microstates are configurations.

2. The energy of this sytem does not change, and thus neither does the energy.

Using this model, how do you digress what the result is of the addition of a third compartment. How does the system evolve

Hi all,

I'm working on some things to say in this discussion, but I need to think it over first. The main problem is that the confusion about "entropy = disorder", and I am thinking on how I can explain it best in this collapsing nebula scenario.

I'll be back soon ;)

Bye!

Crisp

The main problem is that the confusion about "entropy = disorder"

That's it god damn it! that's the crux of the misconception

Originally posted by lethe
graviational collapse is not energy entering the system. it is the exchange of gravitational potential energy for kinetic or thermal energy. it can happen in a closed system, which i think is the whole source of the confusion.

Ooops, very sloppy language on my part. OK, it was fundamentaly wrong. Yes, you are right.

gravitational collapse is usually spontaneous. it happens when the gravitational pressure excedes the radiation pressure.

Indeed, but this spontaneity occurs mostly in the early universe, IIRC. After that the major cause of collapse becomes supernova shockwaves. Unless I am seriously forgetting my galactic structure models.

Brainssssss.

My entropy meter just pegged. Can I have a new one please.

Hi all,

Some remarks are in order I think.

For the nebula collapse: I have absolutely no idea what the final state of a collapsing nebula will be, but I am convinced that it will be an equilibrium state and that the entropy is maximal. The problem with the "entropy is disorder" picture is that gravity invalidates the idea of "the maximum entropy state is the most disordered one". One can prove that particles that interact only through gravity evolve towards an equilibrium state where matter is clustered (and this evolution maximes the entropy). To understand why this gravitational interaction breaks this "entropy = disorder" picture, you need to look at the statistical mechanical definition of entropy.

Before delving into these details, the main conclusion is that for the nebula collapse, no violation of the second law needs to happen, basically because of a misinterpretation of the
concept of entropy.

First some remarks concerning the definition of entropy:

<UL>
<LI>Ryans, your definition of entropy:

S(<FONT FACE="Symbol">m</FONT>) = - <FONT FACE="Symbol">å</FONT>_m <FONT FACE="Symbol">m</FONT>(m) ln <FONT FACE="Symbol">m</FONT>(m)

is not always the "physical" entropy. Your specific definition is called the Shannon entropy, and it is quite often used in information theory and statistics (this particular entropy
reflects of the amount of information contained in the measure <FONT FACE="Symbol">m</FONT> ). It is only equal to the thermodynamic entropy when in equilibrium.

For non-equilibrium systems, the Shannon entropy is constant in time under Hamiltonian evolutions (Liouville theorem), and hence does not reproduce a second law of thermodynamics. I personally prefer not to use it for that reason, but there are some (good) counter arguments aswel.

Also note that the Shannon entropy is an entropy defined for measures, and not for (macro)states!
</LI>
<LI>Next, the Boltzmann entropy

S(M) = k ln |M|

where |M| is the phasespace Liouville volume of the macrostate M (i.e. the number of microstates corresponding to the macrostate M) is another kind of entropy, which was proposed by Boltzmann when he first started formulating statistical mechanics. This definition is once again only equal to the thermodynamic entropy when in equilibrium.

The Boltzmann entropy almost immediatelly reproduces the second law of thermodynamics for Hamiltonian evolutions. The reason behind is that after a while, the system is simply more
lickely to be found in a macrostate that has many, many, many corresponding microstates (because such a macrostate takes up almost the entire phasespace volume for real systems).

This entropy, in contrast to the Shannon entropy, is defined on macrostates, and, once you have chosen what the possible macrostates of your system are, you can calculate all the entropies, and the macrostate with the highest energy will be the one that your system evolves to.
</LI>
</UL>

My personal feeling is that the Shannon entropy is used more in information theory and in (some) non-equilibrium statistical mechanics, while the Boltzmann entropy is used more to determine equilibrium states (as being the one with the largest number of corresponding microstates). The caveat is ofcourse that you have to take into account constraints that are imposed on the evolution. Perhaps this can be explained well when looking at the nebula collapse example:

First of all, this is a highly non-equilibrium system and hence there is no point in talking about it in the context of equilibrium entropies. You need to pick a better entropy that works also outside of equilibrium and the sad truth is that there is no consensus yet on which entropy to choose. But that does not prevent us to say something sensible about what happens to
the entropy there.

Given a non-equilibrium starting position, we first take into account the constraints that are imposed, such as the conservation of energy (!!!), conservation of angular momentum, .... This usually filters out a small subspace of the phasespace of the system on which the evolution can take place. In this subspace, you need to look for the macrostate that has the highest Boltzmann entropy in order to find the (final) equilibrium state. Ofcourse it is by looking at this restricted subspace that the non-intuitive things can happen.

Don't know what else to add at the moment... ;)

Bye!

Crisp

Originally posted by Crisp
::snip very good research::
Don't know what else to add at the moment... ;)


Let me add The Entropy of Nebulae by J. Baez (http://math.ucr.edu/home/baez/entropy.html) and The Virial Theorem (http://math.ucr.edu/home/baez/virial.html) on which it is based. Also read links there from.

My brain may be completely addled but at least I can remember a few salient points to search on :).

Originally posted by thed

Indeed, but this spontaneity occurs mostly in the early universe, IIRC. After that the major cause of collapse becomes supernova shockwaves. Unless I am seriously forgetting my galactic structure models.


could be you re right. i don t really know much about stellar evolution and collapse.

i thought how it works is that the star runs out of fuel, the radiation pressure drops too low, and the star collapses.

but that might not apply to nebulae at all, which have no radiation pressure to speak of.

i don t know anything about this shockwave stuff, so i can t really claim to be right or wrong either way.

Originally posted by Crisp
Hi all,

Some remarks are in order I think.

For the nebula collapse: I have absolutely no idea what the final state of a collapsing nebula will be, but I am convinced that it will be an equilibrium state and that the entropy is maximal.

i would hope so

The problem with the "entropy is disorder" picture is that gravity invalidates the idea of "the maximum entropy state is the most disordered one". One can prove that particles that interact only through gravity evolve towards an equilibrium state where matter is clustered (and this evolution maximes the entropy).
well, if you look at the link to Baez page that i posted above, and thed posted below, you will see him do some calculations (admittedly handwavy) in which he shows that the entropy has lessened after gravitational collapse. which is, of course, the whole question behind the original post on this thread, before we went wildly off topic arguing what the definition of entropy is.

i don t really know the resolution to that apparent decrease in entropy, except that i suppose there must be lots of entropy in the radiation that is carried off after the collapse reaches equilibrium.

To understand why this gravitational interaction breaks this "entropy = disorder" picture, you need to look at the statistical mechanical definition of entropy.

Before delving into these details, the main conclusion is that for the nebula collapse, no violation of the second law needs to happen, basically because of a misinterpretation of the
concept of entropy.

i don t see where the misinterpretations of entropy have been. calculation of the entropy of a nebula as an ideal gas, which Baez uses on his page, is not valid because of this misinterpretation?

namely, the formula he uses is:
S = kN [(3/2) ln T + ln (V/N)]

is this one of those formulae that is only valid at equilibrium? surely the nebula is at equilibrium at the beginning and end of its collapse?


First some remarks concerning the definition of entropy:

<UL>
<LI>Shannon entropy....
</LI>
<LI>Next, the Boltzmann entropy</LI>
</UL>

how does the thermodynamic definition dS=dQ/T compare with these other two?


This entropy, in contrast to the Shannon entropy, is defined on macrostates, and, once you have chosen what the possible macrostates of your system are, you can calculate all the entropies, and the macrostate with the highest energy will be the one that your system evolves to.
its nice to hear you say that, since that was one of the major points of disagreement between me and ryans.

lethe,

i don t really know the resolution to that apparent decrease in entropy, except that i suppose there must be lots of entropy in the radiation that is carried off after the collapse reaches equilibrium.

To inject my unqualified thoughts about your statement.

Are you not in a way suggesting the "Undetected Graviton" as a radiative component that could account for the change.

Escaping gravitons could be projected radially outward increasing the expansion push in lieu of "Dark Energy.

Originally posted by MacM
lethe,



To inject my unqualified thoughts about your statement.

Are you not in a way suggesting the "Undetected Graviton" as a radiative component that could account for the change.

i am not at all suggesting that. i just mean normal radiation. while there would probably be a component of the radiation that is gravitational, probably it would be only one part in a billion billion billion. the rest would be electromagnetic radiation.

your alternative theories of gravity are not really relavant here, nor is even einstein s theory of relativity.

Escaping gravitons could be projected radially outward increasing the expansion push in lieu of "Dark Energy.

ugh... it seems you still don t know what dark energy is.

Lethe
its nice to hear you say that, since that was one of the major points of disagreement between me and ryans.

Seems we were both right and both wrong in a round about way.

which is, of course, the whole question behind the original post on this thread, before we went wildly off topic arguing what the definition of entropy is.

On the contrary, I believe it forms the basis of the question.

Crisp stayed pleasingly neutral in his expoundment, something which he should be commended for. Truely a man of science.

Originally posted by ryans

Seems we were both right and both wrong in a round about way.

since i never said anything like "YOU ARE WRONG" in all capitals, like i was screaming, rather, i only stated my impressions, or quoted from textbooks, it is hard to claim that i was actually wrong, only that i was misinformed.

you on the other hand, argued vehemently things that were completely false, like that heat and internal energy are the same thing, or that entropy is not a state variable. both of these statements are false, and expose a misunderstanding of basic thermodynamics.

can you please show me with a verbatim quote exactly which statements of mine you feel were wrong? i like to know when i am wrong and be corrected.


Crisp stayed pleasingly neutral in his expoundment, something which he should be commended for. Truely a man of science.

saying that i think that something you said was wrong doesn t mean i am not neutral. when i see you post physics on this message board that i think is wrong, everytime, i will be compelled to point out that i think it is wrong.

everytime.

that doesn t mean that i am not neutral, it just means that my biggest priority is seeing correct physics.

lethe,

your alternative theories of gravity are not really relavant here, nor is even einstein s theory of relativity.

ANS:What ever that means. None of the above have anything to do with my view. You might start by reviewing "Strings" and "Brane" theories for graviton propagation.


quote:
--------------------------------------------------------------------------------
Escaping gravitons could be projected radially outward increasing the expansion push in lieu of "Dark Energy.
--------------------------------------------------------------------------------



ugh... it seems you still don t know what dark energy is.


ANS: May I suggest that if you do that you should write it down and have it notarized, publish it and go accept your Nobel.

Originally posted by MacM
lethe,



ANS:What ever that means. None of the above have anything to do with my view. You might start by reviewing "Strings" and "Brane" theories for graviton propagation.

what on earth does string theory have to do with nebula gravitational collapse?


quote:
--------------------------------------------------------------------------------
Escaping gravitons could be projected radially outward increasing the expansion push in lieu of "Dark Energy.
--------------------------------------------------------------------------------



ugh... it seems you still don t know what dark energy is.[/b]


ANS: May I suggest that if you do that you should write it down and have it notarized, publish it and go accept your Nobel. [/B]
no, no you may not suggest anything to me regarding my physics career.

thank you.

Originally posted by lethe
well, if you look at the link to Baez page that i posted above, and thed posted below, you will see him do some calculations (admittedly handwavy) in which he shows that the entropy has lessened after gravitational collapse. which is, of course, the whole question behind the original post on this thread, before we went wildly off topic arguing what the definition of entropy is.

i don t really know the resolution to that apparent decrease in entropy, except that i suppose there must be lots of entropy in the radiation that is carried off after the collapse reaches equilibrium.


I do not believe Baez's statement if he claims that the entropy decreases. I'm gonna check the calculations today (didn't have time yesterday when the link was posted) and I'll check back to you.


i don t see where the misinterpretations of entropy have been. calculation of the entropy of a nebula as an ideal gas, which Baez uses on his page, is not valid because of this misinterpretation?


I'll get back to you on this one. I am sure that I read somewhere that a free gas, gravitational interaction has an entropic increase when it clusters (i.e. collapses). You don't have to do any wierd things like taking escaping photons into account. But I stayed deliberately vague about it in my original post because I did not remember the exact details :).


namely, the formula he uses is:
S = kN [(3/2) ln T + ln (V/N)]

is this one of those formulae that is only valid at equilibrium? surely the nebula is at equilibrium at the beginning and end of its collapse?


No. At the end it will be in equilibrium (in the long run), but in the beginning it could start out in an highly non-equilibrium situation. To come back to the particles-in-a-box with gravitational interaction: if the entropy is maximal for clustered particles, then a uniform distribution of the particles in the box has a (much?) lower entropy. Hence it is a non-equilibrium situation since equilibrium by definition (for not externally manipulated systems) is defined as the (macro)state of maximal entropy.

Sidenote: this changes when you start doing work and manipulate the system. Then you can get thermodynamic situations where you go from one equilibrium position to another, but keep in mind that such a system is open (external work is performed on it) and then things become much more difficult from a statistical mechanical point of view.


how does the thermodynamic definition dS=dQ/T compare with these other two?

Their values coincide in equilibrium. For example, filling out the canonical equilibrium distribution exp(-<FONT FACE="Symbol">b</FONT>H) in the Shannon entropy, you get:

S
= - <FONT FACE="Symbol">å</FONT>_m exp(-<FONT FACE="Symbol">b</FONT>H(m)) (-<FONT FACE="Symbol">b</FONT>H(m))
= (-<FONT FACE="Symbol">b</FONT> <FONT FACE="Symbol">å</FONT>_m H(m) exp(-<FONT FACE="Symbol">b</FONT>H(m))
= <FONT FACE="Symbol">b</FONT>&lt;E&gt;_c

where I did not write the partition function and used the notation <A>_c for the canonical expectation value. Then very hand-wavingly one could say that for infinitesimal changes we get something like:

dS = <FONT FACE="Symbol">b</FONT>&lt;d E&gt;_c

Identifying the canonical expectation value of the energy with dU in the first law of thermodynamics (dU = dQ + dW) you find back the relation you specified when no work is performed on the system. But work is ofcourse not included in the Gibbs distribution for the canonical ensemble: it is assumed to describe equilibrium when no work is being performed.

You can do similar calculations for the Boltzmann entropy, but they are slightly harder because you would have to use the microcanonical ensemble to start counting. It was actually one of Boltzmann's major contributions to science to prove that this indeed yields the thermodynamic entropy. Another way of looking at it is even simply the following: the value of the Boltzmann constant is defined such that S = k ln |M| coincides with the thermodynamic entropy ;).


Bye!

Crisp

Seems we have a major spat here about entropy and disorder.

Way I was taught this was as follows. If you have a sealed box with a gas in it then the system is ordered if all the molecules exist in, say, one half the box only. For this to happen, typically work is done and entropy decreases. When the molecules are evenly distributed, in equilibrium at low energy states, the system is disordered and entropy is at a maximum.

Ergo, a nebula starts at high entropy, or in a disordered state, collapses and ends up in a ordered or lower entropy state.

This is counterintuitive perhaps.

The whole point of entropy increases leads to the heat death of the Universe, when all stars die and all you have is homogeneous gas in the lowest energy states you have maximal entropy. Anything else is a lower entropy state.

Originally posted by thed
Seems we have a major spat here about entropy and disorder.

Way I was taught this was as follows. If you have a sealed box with a gas in it then the system is ordered if all the molecules exist in, say, one half the box only. For this to happen, typically work is done and entropy decreases. When the molecules are evenly distributed, in equilibrium at low energy states, the system is disordered and entropy is at a maximum.


I believe that is only because there is a tendency for the gas to 'fill' the container completely. In the nebula collapse, there is no such tendency but to collapse to a lower gravitational energy.

Originally posted by lethe
could be you re right. i don t really know much about stellar evolution and collapse.

i thought how it works is that the star runs out of fuel, the radiation pressure drops too low, and the star collapses.

but that might not apply to nebulae at all, which have no radiation pressure to speak of.

Fair enough. A lot is known about the end points of stellar evolution, what mass stars end as what stellar remnant. As well as what happens in the middle Admittedly I studied this a while ago but no one appeared to fill in the details about stellar birth. I suspect more observational evidence has been found since, E.G. Proplyds, that can be used to build more detailed models. You are also getting in to the realms of the ISM here, that was always a notoriously difficult field to model properly. My copy of Stars I, Bowers and Deeming and other texts on Stellar Models hardly mention the birth process before the so called Zero Age Main Sequence.

This may be worth raising as a new topic in the Astro forums as well as over on Physicsforums.com. I am sure someone can add a bit more detail.

i don t know anything about this shockwave stuff, so i can t really claim to be right or wrong either way.

Basic overview is that you have a large cloud of Gas/dust in a HI or HII region in equilibrium. Nearby star goes supernova, shock wave goes out and impacts the cloud triggering collapse. New stars are born, some of them go supernova causing collapse in another region. The original supernova is now a nebula suitable for later star formation. As you can see, the details rapidly get complicated as these nebula are not perfect gases, can be highly ionised, the processes are dynamic you get involved in plasma physics a lot.

Way back when, they where dabbling with crude models of this in supercomputers to see if this explained why Spiral Galaxies had the structure they did. In short, they did plus you get stable 'bars' in the galaxies. Look up 'stochastic models of spiral galaxies' if you are intersted. Also look up 'Lin-Shu density wave model' as an alternative.

I think why Baez was hand waving is that the models are still fairly crude. We simply don't observe enough collapsing nebula to build models.

Originally posted by thed
If you have a sealed box with a gas in it then the system is ordered if all the molecules exist in, say, one half the box only. For this to happen, typically work is done and entropy decreases. When the molecules are evenly distributed, in equilibrium at low energy states, the system is disordered and entropy is at a maximum.

This is perfectly ok for non-interacting particles.

Ergo, a nebula starts at high entropy, or in a disordered state, collapses and ends up in a ordered or lower entropy state.

No, here you are assuming that entropy does not depend on the interaction, while it most certainly does. If gravitational interacting particles behave just like free particles, then there would be no way to distinguish between the two from an entropic viewpoint, and this is obviously not the case. Just one way to see this (and a way that pops into my mind at the moment) is the following: the free energy is the logarithm of the (canonic) partition function, and hence depends on the Hamiltonian. From this free energy you can calculate the entropy through a derivate, hence also dependant on the Hamiltonian and the interactions.

This is counterintuitive perhaps.

... and a large violation of the second law ;).


The whole point of entropy increases leads to the heat death of the Universe, when all stars die and all you have is homogeneous gas in the lowest energy states you have maximal entropy. Anything else is a lower entropy state.

Once again incorrect for the same reason mentioned above. This is not true for gravitational interactions.

Bye!

Crisp

Originally posted by thed
I think why Baez was hand waving is that the models are still fairly crude. We simply don't observe enough collapsing nebula to build models.

True, this prevents us from knowing what happens exactly. We do know enough from general principles to establish what happens roughly, and this is the clustering of matter through gravity, which generates rotating planets because of the conservation of angular momentum. To predict the exact planet and stellar formation is a whole different story ofcourse ;).

But I am not sure if the original question was about these exact details, but rather about the apparent violation of the second law of thermodynamics.

Bye!

Crisp

Originally posted by Crisp
This is perfectly ok for non-interacting particles.

No, here you are assuming that entropy does not depend on the interaction, while it most certainly does.

OK, help me out here. I'll admit I've learnt more about Entropy in the last 2 hours[1] than I realised existed. Must have fell asleep during that part. But I can't find anything about how gravity (interacting particles) changes the entropy of the system.

I can't be bothered to go digging through my copy of Rief(sp) to find details. It's buried in the top of the attic somewhere.

since i never said anything like "YOU ARE WRONG" in all capitals, like i was screaming, rather, i only stated my impressions, or quoted from textbooks, it is hard to claim that i was actually wrong, only that i was misinformed.

You haven't even gpt your Ph.D yet and you are already a pompus arse. There is nothing wrong with saying you are incorrect, or else one day you'll be made a fool.



you on the other hand, argued vehemently things that were completely false, like that heat and internal energy are the same thing, or that entropy is not a state variable. both of these statements are false, and expose a misunderstanding of basic thermodynamics.

Why turn to personal attacks Lethe. Show me the quote where I states that entropy is not a state variable? I get passionate when I debate physical processes but never revert to personal attacks against those who are constructively adding to the arguement.

You can do the math and plug numbers into your equations, but what else?

When was the last time I made a personal comment about you? You think you're such a hot shot, But you wouldn't know the first thing about anything someone hasn't written about.

So lay off.

I had a respect for you, but seems I have hit a soft spot when exposing areas of physics you know shit about!

Originally posted by ryans
You haven't even gpt your Ph.D yet and you are already a pompus arse.
you don t know me. don t assume anything about me.

There is nothing wrong with saying you are incorrect, or else one day you'll be made a fool.
i will let the onlookers decide who looks the fool here. my only desire is to have correct physics. so this doesn t interest me.



Why turn to personal attacks Lethe.
have i said anything personal about you, ever? please show me the quote where i ever talk about anything but your physics.


Show me the quote where I states that entropy is not a state variable?

OK, here:

Originally posted by ryans

Originally posted by James R
Entropy is a state variable.

So, you still have to average over ensembles. That's why it is called statistical mechanics, you look at all possible states of the system, not just the one the system is in at the moment.


remember, a state variable is one that depends only on the state of a system.



I get passionate when I debate physical processes but never revert to personal attacks against those who are constructively adding to the arguement.
i would like you to show me where i made a personal attack. all i can recall doing was saying that your physics is wrong.

on the other hand, you have assumed things about my level of education, threatened that i will look foolish, and claimed that i cannot think independently, but rather can only plug numbers.

now who is getting personal?


When was the last time I made a personal comment about you? You think you're such a hot shot, But you wouldn't know the first thing about anything someone hasn't written about.
see above.


So lay off.

I had a respect for you, but seems I have hit a soft spot when exposing areas of physics you know shit about!

i still would like you to point out any statements that i have posted in this thread that were false. really. i want to be corrected, and so far, you have not convinced me that anything i said was false. you have already made more than one statement that you have admitted was false (heat and energy are the same thing, for example), now show me which statement that i made that was false.

remember, my top priority in having physics conversations is correct physics. you can tell me that i am wrong all you want, and call me a fool, but please, show me what is wrong.

You can do the math and plug numbers into your equations, but what else?

you accuse me of being too closed-minded to know any physics except what i have read. but do you know me? how do you know what physics i know?

Hi thed,

Originally posted by thed
OK, help me out here. I'll admit I've learnt more about Entropy in the last 2 hours[1] than I realised existed. Must have fell asleep during that part. But I can't find anything about how gravity (interacting particles) changes the entropy of the system.

I can't be bothered to go digging through my copy of Rief(sp) to find details. It's buried in the top of the attic somewhere.

Well, first is to know that the simplest way to calculate (an) entropy is to count the number of microstates that correspond to one macrostate. This is the Boltzmann prescription for entropy, and it coincides with the thermodynamic entropy in equilibrium.


Example:

Take N Ising spins &sigma;<SUB>i</SUB> on a lattice, with &sigma;<SUB>i</SUB> in {-1,+1}. The phasespace, where all possible configurations live, is given by

&Gamma;<SUB>i</SUB> = {-1,+1}<SUP>N</SUP>

This &Gamma; contains all the 2<SUP>N</SUP> microstates of the system. In order to define macrostates, we have to choose macroscopic variables so we can divide the microstates into different groups. For spins, the interesting macrostate is the magnetisation

M = </SUP>&sum;<SUB>i=1</SUB><SUP>N</SUP> &sigma;<SUB>i</SUB>

where we should also divide by N to keep M finite in thermodynamic limits etc, but this weights heavy on the notation here. This divides the 2<SUP>N</SUP> into (2N+1) macrostates being {-N, -N+1, ... , 0 , ... N-1 , N}.

Now let's talk dynamics and equilibrium for that dynamics:


1. No interaction

If there is no interaction between spins, then it is quite easy to find the equilibrium macrostate. We define a dynamics that independently and randomly performs spinflips at sites. Then the entire phasespace is accessible through this dynamics (from any starting configuration in &Gamma; , you can reach any other). The system can reach any microstate, and hence also any macrostate. Therefore we must evaluate the entropy of all macrostates and look which is the highest. This is obviously the macrostate for

M = 0

since the number of microstates that correspond to a macrostate of magnetisation M<SUB>0</SUB> is given by Binomial( N choose (N+M<SUB>0</SUB>)/2 ) . This is simply the number of ways you can arrange (N+M<SUB>0</SUB>)/2 up-spins (with value +1) over N spins in total. Note that (N+M<SUB>0</SUB>)/2 up spins implies (N-M<SUB>0</SUB>)/2 down spins, to give a magnetisation of:

</SUP>&sum;<SUB>i=1</SUB><SUP>N</SUP> &sigma;<SUB>i</SUB> = (+1)*(N+M<SUB>0</SUB>)/2 + (-1)*(N-M<SUB>0</SUB>)/2 = M<SUB>0</SUB>

as it should.

Hence, for no interaction, only spinflips the macroscopic equilibrium value of the magnetisation is M = 0. This also corresponds to the intuition of M = 0 being the state which has the highest "disorder": as many spins are pointing up as there are pointing down, which is completely the opposite of the highly ordered magnetisation state M = +N , where all spins are nicely pointing up.

2. With interactions

From the moment we include spin interactions, the above is no longer true. For example, if we include Ising-like spin interactions and a magnetic field that tries to align the spins up, then it is clear that M = 0 will not be the equilibrium value of the magnetisation anymore, but rather a positive magnetisation.

It should be put in formulas, but the idea is now that a large part of phasespace is no longer accessible for the dynamics; the field drives you towards the part of phasespace where M > 0 and you stay there. I am staying deliberately vague here because it is a very tricky subject; you should write down spin-flip hamiltonians, take into account external fields, temperature effects to account for random flips (stochasticity)... and this becomes very difficult to handle quite fast.




For gravitational interaction, something similar happens: if the particles are uniformely spread, then a lot of the system energy is put in the "positional" part of phasespace (R^3N of the total phasespace R^6N), and only a small amount of kinetic energy is available. Hence: there are a lot of positional configurations compatible with that macrostate, but only a few momentum configurations.

When the particles get closer together (cluster) then gravitational energy is freed and more configurations in momentumspace become available. The number of "positional" configurations stays roughly the same, but the number of momentum configurations grows rapidly.

The net effect is more or less that there are more microstates/configurations compatible with the macroscopic clustered state, but only when you look at the constant-energy surface in phasespace.

Hope this more or less clarifies things, I am being deliberately vague because I never read the articles where they once calculated this gravitational interaction, but I remember having a discussion about this once with my supervisor. I asked him for references two days ago but he couldn't remember them by heart either, but I suppose that a decent book on (non-equilibrium?) statistical mechanics should contain something like that.

Bye!

Crisp

ryan wrote
You haven't even gpt your Ph.D yet and you are already a pompus arse.

I don't know anything about Lethe, to whom this was directed, but am I correct in interpreting this as saying that getting a Ph.D. MAKES one a pompous arse?

Getting a Ph.D might be a matter of time for lethe. Even then, i doubt, whether he would be pompus. I could say myself & few others learned quite some stuff from lethe's posts. The combination of advanced physics with maths and with attitude to teach is a killer combination indeed.

Ryans also could do well if he drops his cryptographic language for the sake of would-not-be PH.Ds (atleast in near future) who are 99% here.

Originally posted by everneo
I could say myself & few others learned quite some stuff from lethe's posts. The combination of advanced physics with maths and with attitude to teach is a killer combination indeed.


thanks for the support. i really enjoy teaching physics, but only if i think there is someone listening, trying to learn. so it is quite encouraging to hear you say stuff like that.

Originally posted by ryans
I had a respect for you, but seems I have hit a soft spot when exposing areas of physics you know shit about!

you don t have to respect me, but i wish you would be more modest. there is no reason to assault peoples physics knowledge. everyone here, including you (and me), has limitations to his knowledge, and you would do well to keep that in mind before claiming that you have a correct understanding of thermodynamics, and that everyone else does not.

doing that can make one appear a pompous arse.

Getting back on topic:


Originally posted by Crisp
I do not believe Baez's statement if he claims that the entropy decreases. I'm gonna check the calculations today (didn't have time yesterday when the link was posted) and I'll check back to you.

I'll get back to you on this one. I am sure that I read somewhere that a free gas, gravitational interaction has an entropic increase when it clusters (i.e. collapses). You don't have to do any wierd things like taking escaping photons into account. But I stayed deliberately vague about it in my original post because I did not remember the exact details :).


OK, today i went searching the archives of spr, and found the thread where this topic was discussed. here are the points, as i understand them.

1. gravitational systems are never in equilibrium, and always "boil off" some particles.

2. for large, cold gas clouds, we can neglect this process, but then the system is not virialized, meaning it does not obey the virial theorem T=-U/2, which is only true for a closed system on a time scale long enough to average the energies, which we cannot do, since we are neglecting "boiling off" which makes the system not closed.

3. when the system is colder than the virial temperature, the heating process due to gravitational collapse makes the entropy increase so that the total entropy of the (closed) system increases.

4. at some point during the collapse, the system heats up enough to virialize. from this point on, the increase in entropy due to increase in temperature due to collapse does not offset the decrease in entropy due to decrease of volume: the entropy of the gas cloud decreases

5. the second law is not violated, however, even for the virial system. once the temperature is high enough that the virial theorem is satisfied, we can no longer neglect the "boiling off" process. the system is no longer closed, and radiates photons, as well as particles. the entropy of the cloud + the entropy of the radiated particles still increases. the second law is saved.

find the discussion here (http://groups.google.com/groups?threadm=8ok45v%248g9%241%40Urvile.MSUS.EDU)

what do you think? that thread is quite long, and it was hard to glean the correct concepts. Baez mentions that there are mistakes on the entropy website, that have not been corrected, but i didn t quite figure out what they were. i also didn t verify any of the calculations, but might be interested in trying it. i don t know too much thermodynamics, and i don t know the virial theorem very well, so i hope my summary above sounds correct.

I have the impression that it is a very difficult and desperate attempt to try to save the second law :)

I've been looking for some more references on the entropy vs. gravity debate. I also asked my PhD supervisor who pointed me out to the "salvation of the second law" in that scenario three years ago, but he could not come up with the reference either.

An interesting read is already:

<A HREF="http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/961223a.html">NASA: Ask an astronomer</A> debate on entropy vs. gravity.

I think that Penrose's "The Emperor's new mind" holds an explanation of gravity and entropy, but I have to look it up first. And by searching the net, I came up with:

"Chance in Physics: Foundations and Perspectives, edited by Jean Bricmont, Detlef Durr, Maria C. Galavotti, Giancarlo Ghirardi, Francesco Petruccione, and Nino Zanghi, Lecture Notes in Physics 574", published by (Springer-Verlag)

which should hold an article by Michael Kiessling on "the arrow of time". That article should also discuss this issue.

I'll look them both up, all this virial theorem stuff is simply too much magic, I don't buy it :).

Bye!

Crisp

well, if one of those references that you mentioned addresses this issue explicitly, please mention it, i d love to check it out...

Well, at least I was already unsuccesful in finding the Emperor's new mind at work... it turns out it has been returned to the university library so I have to go there :) ... I'll do this when I have some spare time.

The other reference could be a bit harder to find, appearantly I cannot find it in any scientific library in my country. Luckily I know two of the editors, perhaps they can help me out.

Bye!

Crisp

Originally posted by Crisp
Well, at least I was already unsuccesful in finding the Emperor's new mind at worki

Errr, Crisp. The Emperors New Mind is a book written by R. Penrose in respone to critiques against his stance on complexity Vs. simplicity wrt AI. I've got a copy (largely unread I'll admit) in my loft. The overall thrust of the text is to do with AI and not entropy, IIRC.

YMMV.

Thed,

You should read more in it then. I first got confronted with the book when someone (higher in rank) shoved it under my nose saying "Penrose writes here something about the arrow of time, I want to understand it". I also didn't completely read it ofcourse, but I doubt it is as limited in content as you think. I should look it up to be certain, but did not yet have time.

Bye!

Crisp