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entropy and gravitation



 
 
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  #11  
Old June 2nd 17, 12:30 PM posted to sci.physics.research,sci.astro.research
Roland Franzius
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Posts: 2
Default 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


  #12  
Old June 2nd 17, 01:47 PM posted to sci.physics.research,sci.astro.research
Tom Roberts
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Posts: 344
Default 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  
Old June 2nd 17, 01:47 PM posted to sci.physics.research,sci.astro.research
Martin Brown[_2_]
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Posts: 4
Default 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


  #14  
Old June 2nd 17, 10:43 PM posted to sci.physics.research,sci.astro.research
jacobnavia
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Posts: 105
Default 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?



  #15  
Old June 3rd 17, 08:07 AM posted to sci.physics.research,sci.astro.research
Poutnik[_5_]
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Posts: 68
Default 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.


  #16  
Old June 4th 17, 07:47 AM posted to sci.physics.research,sci.astro.research
Steven Carlip
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Posts: 9
Default 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

  #17  
Old June 7th 17, 08:08 AM posted to sci.physics.research,sci.astro.research
Gregor Scholten[_2_]
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Posts: 3
Default entropy and gravitation

[I already tried to send this on Saturday, June 3. This is the second
attempt]

(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.


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.

Imagine a van der Waals gas in a bottle: above the boiling point, the
gas phase with smooth distribution of the atoms is preferred, below the
boiling point, the liquid phase with lumpy distribution is preferred.
The value of the boiling point itself depends on the strength of the
attractive forces betweens the atoms. The stronger these forces are, the
higher is the boiling point.

So, for a small gravitational constant G, the universe has to be very
cold for lumpy distributions (galaxies, stars, planets) being preferred,
i.e. having higher entropy. For a higher value of G, the temperature can
be higher. In the limit G - 0, the critical temperature is running to T
= 0, too, making the behaviour being the same as in a universe without
gravity.

An example: in the 1980's, many physicists believed that neutrinos would
be a good candidate for Dark Matter. Since their masses are very small,
they would be "hot" Dark Matter, i.e. even for low temperatures around 3
K, their average velocities are very high, in the range of the speed of
light. Computer simulations showed then that such hot Dark Matter
couldn't yield the structures we observe in the universe. Hot Dark
Matter would allow for super-clusters and voids to form, but not for
galaxies or even stars. So, hot Dark Matter would be still above the
"boiling point", even at 3 K. That's why physicists search for "cold"
Dark Matter instead today.


  #18  
Old June 8th 17, 04:07 PM posted to sci.physics.research,sci.astro.research
Richard D. Saam
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Posts: 240
Default 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



  #19  
Old June 8th 17, 04:07 PM posted to sci.physics.research,sci.astro.research
Martin Brown[_3_]
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Posts: 189
Default 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


  #20  
Old June 8th 17, 08:26 PM posted to sci.physics.research,sci.astro.research
Phillip Helbig
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Posts: 38
Default 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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