entropy and gravitation

[Phillip Helbig (undress to reply)]

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.

[Eric Flesch]

On 30 May 2017 Phillip Helbig wrote:
>it seems strange that the entropy should change at some value of G.

For a while I thought thermodynamics was the most marvellous science,
but now I think that "entropy" is just a fudge to fill in the gap
after the enthalpy is measured. So who ever proved that "disorder" is
a full explanation of so-called entropy? I had the same thoughts as
Phil about the effects of scale on that interpretation (as well as how
it is different in space than on a planet), but casting it in the
light of the value of G is a new one.

[Jos Bergervoet]

On 5/30/2017 6:55 AM, 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

But if everything is in one big black hole, and the black hole
would need only mass and angular momentum and charge to describe
it, then that would be extremely low entropy (and essentially we
would have back the "ordinary" behavior you described first).

So the difference is only in the entropy that is in the "soft
supertranslation hair" (if that is the correct theory..)

If the oxygen in one corner of the room would also have this
extra entropy that black holes seem to have (for whatever reason),
then the cases would be the same.

Provided of course that before black hole formation occurs
the normal behavior (lumpy distribution has lower entropy) is
respected by gravity as it is by other forces. So the question
is: would there still be a reason, in cases without black holes,
to expect that gravity is different?

--
Jos

[Phillip Helbig (undress to reply)]

In article <592d593d$0$789$***@news.xs4all.nl>, Jos Bergervoet
<***@xs4all.nl> writes:

> But if everything is in one big black hole, and the black hole
> would need only mass and angular momentum and charge to describe
> it, then that would be extremely low entropy (and essentially we
> would have back the "ordinary" behavior you described first).

The first clause doesn't really make sense, since if "everything"
(presumably meaning all matter in the universe) were "in one big black
hole", this would have to be something different than what is normally
thought of as a black hole, e.g. a static solution in a background of
Minkowski space.

[Gerry Quinn]

In article <ogiti8$1k0t$***@news.kjsl.com>,
***@asclothestro.multivax.de says...
>
> 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.

The smooth distribution always has the same entropy. Start with the
smooth distribution and no gravity, and increase the gravitational
constant. Now high entropy states start to become available that were
not available withouy gravity.

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.

The clumpiness versus dispersion 'paradox' isn't such a paradox either.
When gravitational clumping takes place, energy is released which will
eventually become widely dispersed as low-grade thermal energy. This
ultimately applies even if black holes are formed, although the thermal
energy will for a long time be locked up in black holes.

- Gerry Quinn

---
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[Phillip Helbig (undress to reply)]

In article <***@news.eternal-september.org>,
Gerry Quinn <***@bindweed.com> writes:

> The smooth distribution always has the same entropy. Start with the
> smooth distribution and no gravity, and increase the gravitational
> constant. Now high entropy states start to become available that were
> not available withouy gravity.

Sounds plausible.

> 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.

[Gerry Quinn]

In article <ogptl5$s26$***@news.kjsl.com>, ***@asclothestro.multivax.de
says...
>
> In article <***@news.eternal-september.org>,
> Gerry Quinn <***@bindweed.com> writes:
>
> > The smooth distribution always has the same entropy. Start with the
> > smooth distribution and no gravity, and increase the gravitational
> > constant. Now high entropy states start to become available that were
> > not available withouy gravity.
>
> Sounds plausible.
>
> > 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?

Consider a smooth universe full of hydrogen, with non-zero density and
no gravity. This universe is clumpy too, it's just that the clumps are
mostly H2.

You could make the same paradox by imagining a universe full of H atoms,
and slowly turning on atomic interactions.

- Gerry Quinn

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[Phillip Helbig (undress to reply)]

In article <***@news.eternal-september.org>,
Gerry Quinn <***@bindweed.com> 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.)

[Roland Franzius]

Am 02.06.2017 um 12:07 schrieb Phillip Helbig (undress to reply):
> In article <***@news.eternal-september.org>,
> Gerry Quinn <***@bindweed.com> 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

[Martin Brown]

On 02/06/2017 11:07, Phillip Helbig (undress to reply) wrote:
> In article <***@news.eternal-september.org>,
> Gerry Quinn <***@bindweed.com> 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

[Tom Roberts]

On 6/1/17 6/1/17 3:36 PM, Phillip Helbig (undress to reply) wrote:
> In article <***@news.eternal-september.org>,
> Gerry Quinn <***@bindweed.com> 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
G>0. 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

[jacobnavia]

Le 30/05/2017 à 06:55, Phillip Helbig (undress to reply) a écrit :
> 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.

What about time?

An aribtrarily small G will take an almost infinite time to manifest
itself. Weaker gravity will EVENTUALLY get matter clumpy but if gravity
is weak, it will take MUCH more time for gravity effects to manifest
themselves.

An arbitrarily small gravity will take arbitrarily long time to have any
effect.

Does the contradiction disappear if we take time into the picture?

[Poutnik]

Dne 02/06/2017 v 09:04 jacobnavia napsal(a):

>
> What about time?
>
> An aribtrarily small G will take an almost infinite time to manifest
> itself. Weaker gravity will EVENTUALLY get matter clumpy but if gravity
> is weak, it will take MUCH more time for gravity effects to manifest
> themselves.
>
> An arbitrarily small gravity will take arbitrarily long time to have any
> effect.
>
> Does the contradiction disappear if we take time into the picture?

Thermodynamics generally does not care,
what time it takes for a system
to get into the preferred final state.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.

[jacobnavia]

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.

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?

[Poutnik]

Dne 02/06/2017 v 23:43 jacobnavia napsal(a):
> 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.
>
> 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.

[Martin Brown]

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

[Steven Carlip]

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

[Steve Willner]

In article <ogv6ba$dtp$***@dont-email.me>,
Steven Carlip <***@physics.ucdavis.edu> writes:
> 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.

More generally, I think one has to consider velocities as well as
positions. The initial state with random positions _but all zero
velocities_ (in co-moving coordinates) has low entropy. In
equilibrium, the velocities would follow a Maxwell distribution. With
gravity, you can raise the entropy of velocities at the expense of
introducing clumpiness, which lowers entropy of the positions. As
Martin (I think) indicated, the tradeoff between the two depends on
temperature. I'm not at all sure I have all the details right, but
this looks like at least one way to think about the problem.

For Martin in another message, the baryon census of the universe is a
subject of active research. Until recently, about half the baryons
were unaccounted for, but it now seems they are located in very hot
gas associated with galaxy clusters. The fraction of baryons in
stars increases with cosmic time but is only of order 10% now. One
recent paper, which apparently still finds baryons to be "missing" is
http://adsabs.harvard.edu/abs/2016A%26A...592A..12E
I have not researched this question in any detail, but the
Introduction of the above paper has lots of relevant references.

For another poster, 'entropy' is a defined physical quantity, not
some general synonym for disorder.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone [phone] ***@cfa.harvard.edu
Cambridge, MA 02138 USA

[Steve Willner]

[[Mod. note -- This article was originally submitted on 2017-07-15
(about a week ago), but I mistakenly misfiled it and have only just
(re)discovered it. I apologise to the author and to readers for the
mixup & long delay.
-- jt]]

In article <ogv6ba$dtp$***@dont-email.me>,
Steven Carlip <***@physics.ucdavis.edu> writes:
> 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.

More generally, I think one has to consider velocities as well as
positions. The initial state with random positions _but all zero
velocities_ (in co-moving coordinates) has low entropy. In
equilibrium, the velocities would follow a Maxwell distribution. With
gravity, you can raise the entropy of velocities at the expense of
introducing clumpiness, which lowers entropy of the positions. As
Martin (I think) indicated, the tradeoff between the two depends on
temperature. I'm not at all sure I have all the details right, but
this looks like at least one way to think about the problem.

For Martin in another message, the baryon census of the universe is a
subject of active research. Until recently, about half the baryons
were unaccounted for, but it now seems they are located in very hot
gas associated with galaxy clusters. The fraction of baryons in
stars increases with cosmic time but is only of order 10% now. One
recent paper, which apparently still finds baryons to be "missing" is
http://adsabs.harvard.edu/abs/2016A%26A...592A..12E
I have not researched this question in any detail, but the
Introduction of the above paper has lots of relevant references.

For another poster, 'entropy' is a defined physical quantity, not
some general synonym for disorder.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone [phone] ***@cfa.harvard.edu
Cambridge, MA 02138 USA

[Gregor Scholten]

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

***@asclothestro.multivax.de (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.

[Richard D. Saam]

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

[Phillip Helbig (undress to reply)]

> 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.

[Martin Brown]

On 08/06/2017 20:26, Phillip Helbig (undress to reply) wrote:
>> 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.

That suggests an interesting question.

Is it possible to compute either by simulation or from observations what
percentage of ordinary matter is tightly bound together (either
gravitationally or electromagnetically) as a function of length scale
(or mass).

There is clearly everything from ionised hydrogen, neutral hydrogen
(which must be a fair chunk in itself) dust grains and upto ~300Msun.
Does it obey some power law or are the preferred mass/length scales?

--
Regards,
Martin Brown

[Gerry Quinn]

In article <ohdl81$150i$***@gioia.aioe.org>,
'''newspam'''@nezumi.demon.co.uk says...
>
> On 08/06/2017 20:26, Phillip Helbig (undress to reply) wrote:
> >> 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.
>
> That suggests an interesting question.
>
> Is it possible to compute either by simulation or from observations what
> percentage of ordinary matter is tightly bound together (either
> gravitationally or electromagnetically) as a function of length scale
> (or mass).
>
> There is clearly everything from ionised hydrogen, neutral hydrogen
> (which must be a fair chunk in itself) dust grains and upto ~300Msun.
> Does it obey some power law or are the preferred mass/length scales?

I would imagine that there has been a lot of work done in this area in
conjunction with studies of dark matter. Obviously the density and
distribution of baryonic dark matter (i.e. ordinary matter that's not in
stars) is a basic starting point for this research.

In fact, googling 'baryonic dark matter distribution' gives links which
will probably be in the ballpark of what you are interested in.

- Gerry Quinn

---
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

[Nicolaas Vroom]

On Sunday, 11 June 2017 21:46:50 UTC+2, Gerry Quinn wrote:

> I would imagine that there has been a lot of work done in this area in
> conjunction with studies of dark matter. Obviously the density and
> distribution of baryonic dark matter (i.e. ordinary matter that's not in
> stars) is a basic starting point for this research.
>
> In fact, googling 'baryonic dark matter distribution' gives links which
> will probably be in the ballpark of what you are interested in.
>

Baryonic dark matter? Is there something I'am missing?

When you go directly to the link:
https://en.wikipedia.org/wiki/Dark_matter#Composition_of_dark_matter:_Baryonic_vs._nonbaryonic
Or https://en.wikipedia.org/wiki/Dark_matter and select paragraph 4
You will read:
"Dark matter can refer to any substance which interacts predominantly via
gravity with visible matter (e.g. stars and planets). Hence in principle
it need not be composed of a new type of fundamental particle but could,
at least IN PART, be made up of standard baryonic matter, such as protons
or electrons."

What is the current main stream opinion about "in part"?
IMO darkmatter is (was?) always considered as non-baryonic as compared to
normal matter which is considered as baryonic.

The problem is that the name dark matter is linked to the human sense: see.
visible versus invisible. And as such it is a very unlucky name.
A much better way is to make a distinction solely between baryonic and
non baryonic matter.

The problem is that in order to explain a galaxy rotation curve you can
assume a certain amount baryonic matter which density is so low that
it becomes invisible. The question is here what is this limit and how
much baryonic matter is involved.

Nicolaas Vroom.

[Steve Willner]

In article <66415215-eeff-4f5e-9430-***@googlegroups.com>,
Nicolaas Vroom <***@pandora.be> writes:
> Baryonic dark matter? Is there something I'am missing?
>
> When you go directly to the link:
> https://en.wikipedia.org/wiki/Dark_matter#Composition_of_dark_matter:_Baryonic_vs._nonbaryonic
> Or https://en.wikipedia.org/wiki/Dark_matter and select paragraph 4
> You will read:
> "Dark matter can refer to any substance which interacts predominantly via
> gravity with visible matter (e.g. stars and planets). Hence in principle
> it need not be composed of a new type of fundamental particle but could,
> at least IN PART, be made up of standard baryonic matter, such as protons
> or electrons."

As the OP has noticed, there is confusion in the terminology. Many
people use "dark matter" to refer only to non-baryonic dark matter,
but others use the term more generally to include baryonic matter in
any form that's not yet detected. There have been many suggestions
for removing the ambiguity, but none of them has caught on. Until
something does, readers have to understand the context in which the
term is used. Careful authors will define which way they are using
the term.

> What is the current main stream opinion about "in part"?

The Concordance Cosmology puts the total of baryonic matter at about
5% of the critical density. That comes most precisely from the CMB
fluctuations, but it's consistent with Big Bang Nucleosynthesis.
About half of that value is well accounted for (stars and gas plus
some other odds and ends). Until the last couple of years, the other
half has been "missing," but recent observations have found very hot
gas associated with galaxy clusters. As far as I can tell, the
weight of opinion is that this gas accounts for all the missing
baryonic matter, but I don't think it's 100% established as yet.

In contrast, non-baryonic dark matter accounts for around 26% of the
critical density according to the Concordance Cosmology.

Web searches should produce more reliable sources than Wikipedia.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone [phone] ***@cfa.harvard.edu
Cambridge, MA 02138 USA

[Phillip Helbig (undress to reply)]

In article <66415215-eeff-4f5e-9430-***@googlegroups.com>,
Nicolaas Vroom <***@pandora.be> writes:

> Baryonic dark matter? Is there something I'am missing?

Maybe. Today, "dark matter" is often used as an abbreviation for
"non-baryonic matter apart from neutrinos, the identity of which is
unknown".

Technically, electrons are also non-baryonic matter, but a) they are not
dark and b) their mass is negligible in cosmology.

Also, keep in mind that "dark" really means "does not interact
electromagnetically". Not only does it not glow, neither does it
reflect and is transparent.

Of course, not all baryonic matter emits light so, in this sense, it is
baryonic dark matter.

We have a good idea from the CMB and big-bang nucleosynthesis how many
baryons there are. Known baryonic matter in stars, gas etc is actually
substantially less.

> When you go directly to the link:
> https://en.wikipedia.org/wiki/Dark_matter#Composition_of_dark_matter:_Baryonic_vs._nonbaryonic
> Or https://en.wikipedia.org/wiki/Dark_matter and select paragraph 4
> You will read:
> "Dark matter can refer to any substance which interacts predominantly via
> gravity with visible matter (e.g. stars and planets). Hence in principle
> it need not be composed of a new type of fundamental particle but could,
> at least IN PART, be made up of standard baryonic matter, such as protons
> or electrons."

Yes, that's one definition.

> What is the current main stream opinion about "in part"?

This means that we know that most dark matter cannot be baryonic,
because cosmological observations imply that there is much more than is
compatible with the CMB and big-bang nucleosynthesis.

> IMO darkmatter is (was?) always considered as non-baryonic as compared to
> normal matter which is considered as baryonic.

It depends on the definition.

Again, this is an area where different notation can be confusing but
learning the different schemes gives some insight into the matter.

> The problem is that the name dark matter is linked to the human sense: see.
> visible versus invisible. And as such it is a very unlucky name.
> A much better way is to make a distinction solely between baryonic and
> non baryonic matter.

Not necessarily; it depends on the question one is asking.

> The problem is that in order to explain a galaxy rotation curve you can
> assume a certain amount baryonic matter which density is so low that
> it becomes invisible. The question is here what is this limit and how
> much baryonic matter is involved.

The question here is whether the dark matter needed to explain galaxy
rotation curves could a) be baryonic and b) escape detection by other
means.

[Phillip Helbig (undress to reply)]

In article <ohdl81$150i$***@gioia.aioe.org>, Martin Brown
<'''newspam'''@nezumi.demon.co.uk> writes:

> > 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.
>
> That suggests an interesting question.
>
> Is it possible to compute either by simulation or from observations what
> percentage of ordinary matter is tightly bound together (either
> gravitationally or electromagnetically) as a function of length scale
> (or mass).

At larger scales, dark matter is important, but we don't know what it
is. In particular, we don't know whether it is self-interacting (other
than via gravity) and even if it isn't, it might not be in the form of
isolated particles (though that is what many people assume); the was a
paper by Bernard Carr and co-authors recently which pointed out that
there is still a mass range where it could be in primordial black holes.

At smaller scales, the last I heard, the IMF (initial mass function) for
stars was not computable from first principles. From observations, we
have a pretty good idea what it is locally, but it was probably
different at high redshift.

With certain assumptions, the Press-Schechter formalism allows one to
calculate a mass function, and, not surprisingly (but a good consistency
test and sanity check), this also comes out of simulations with the same
assumptions.

[Gregor Scholten]

"Richard D. Saam" <***@att.net> 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?

As you can read here:

https://en.wikipedia.org/wiki/Jeans_instability#Jeans.27_length

Jeans' length depends on T^(1/2) for constant mass density and constant G.
So, for high temperatures, the length is very big, allowing only for big
clouds to collaps, e.g. a proto-galactic cloud to form a galaxy, whereas
for low temperatures, also smaller clouds can collaps, e.g. a proto-stellar
cloud to a star.

[Richard D. Saam]

On 6/10/17 1:54 AM, Gregor Scholten wrote:
> "Richard D. Saam" <***@att.net> wrote:
>> 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?
>
> As you can read here:
>
> https://en.wikipedia.org/wiki/Jeans_instability#Jeans.27_length
>
> Jeans' length depends on T^(1/2) for constant mass density and constant G.
> So, for high temperatures, the length is very big, allowing only for big
> clouds to collaps, e.g. a proto-galactic cloud to form a galaxy, whereas
> for low temperatures, also smaller clouds can collaps, e.g. a proto-stellar
> cloud to a star.
>
>
What is the origin of these different
proto-galactic or proto-stellar cloud formation temperatures
in the context of the accepted
ubiquitous present CMBR 2.7 K temperature observation
that can be redshifted to any proto-galactic or proto-stellar cloud era
by (1+z)?

[Gregor Scholten]

"Richard D. Saam" <***@att.net> wrote:

>> As you can read here:
>>
>> https://en.wikipedia.org/wiki/Jeans_instability#Jeans.27_length
>>
>> Jeans' length depends on T^(1/2) for constant mass density and
>> constant G. So, for high temperatures, the length is very big,
>> allowing only for big clouds to collaps, e.g. a proto-galactic cloud
>> to form a galaxy, whereas for low temperatures, also smaller clouds
>> can collaps, e.g. a proto-stellar cloud to a star.
>>
>>
> What is the origin of these different
> proto-galactic or proto-stellar cloud formation temperatures
> in the context of the accepted
> ubiquitous present CMBR 2.7 K temperature observation
> that can be redshifted to any proto-galactic or proto-stellar cloud
> era by (1+z)?

In today's universe, the average temperature is 2.7 K, which is low
enough for proto-stellar clouds to collaps. In the early universe,
the average temperature was higher, allowing only for proto-galactic
clouds to collaps. That's why star formation started some time
later than galaxy formation.