sal:
Thank you much for the response.

I confess to some surprise. I've seen lots of conflict between you and
PMB over definitions and semantics, but this is the first case I've seen
(or, anyway, the first I've recognized) where there was a clear
disagreement over facts.

I didn't bother to read pmb's responses, so all I can say in that
regard is that it doesn't surprise me.

Bilge wrote:
sal:

Yes, that's exactly what I had in mind. Qualitative analysis here, just
to check intuition.

At that point we can remove the power source
so that the "extra" energy is due to the motion of the centrifuge
itself. In newtonian mechanics, the source of the gravitational field
would be the mass of the centrifuge. In general relativity, the
source of the gravitational field is the stress energy tensor which
includes the momentum and energy of the masses in a system as well as
the forces which hold the centrifuge together.

It's the "forces which hold the centrifuge together" where I'm confused
at this point, I think...

When the centrifuge spins, something has to prevent the molecules in
the centrifuge from coming apart. As the angular velocity increases,
more force is required to prevent the centrifuge from disintegrating.


2) OK, now something less obvious: I have a large balloon filled
with low-pressure gas, and I weigh it (in a vacuum, of course).
Now, I compress the gas into a small tank. I let it cool off so
it's the same temperature it was in the balloon. I weigh it again.
It weighs more, due to the pressure terms in the stress-energy
tensor ... right? Or does it? (I added "free energy" when I
compressed it, which presumably is where the extra mass/energy --
if any -- came from.)

Yes. The stress energy tensor is of the form:

T^ab = (\rho + P)U^a U^b + P g^ab

where \rho is the density, P is the pressure, U^a and U^b are four
velocities and g^ab is the metric tensor. Since the initial and final
temperatures are the same and the velocities are prortional to kT,
the initial and final stress-energy tensors differ only in the second
terms. The change in energy is then due to the additional force
required to confine the gas to a smaller volume.

Ah hmm. This is what I think I've read, and it's sure how the stress
energy tensor looks. But it seems to lead to something unpalatable,
which implies that there's something more going on than I've grokked
as yet.

Specifically, this appears to violate conservation of energy.

How so? By your own construction, you added energy to compress the gas.
So, the process isn't adiabatic.


If I put a compressor, battery, tank, and gas supply in a sealed
laboratory, and put the whole thing on a spring, then when the
compressor runs and the gas is compressed into the tank, if the
weight (mass/energy) increases then the spring will be compressed.

Why is that? If the laboratory is sealed, then all of the energy
that compresses the gas has to come from inside the laboratory, so
whatever energy is transferred to the gas, comes from something else
in the lab. That's a different question than before where you were
only talking about the gas and adding and subtracting energy.

I can use the motion of the laboratory as the spring is compressed
to do work. If I then let the gas out of the tank again, the lab
gets lighter, the spring relaxes a bit, and I can use the upward
motion of the laboratory to do more work. The total mass/energy
of the universe has increased.

Again, why would that happen? The energy you release from the gas,
can't go anywhere but inside the laboratory.


That seems wrong -- it seems like there must be a balancing effect as well.

Now, no mass/energy entered or left the sealed laboratory. However, the
amount of Gibbs free energy in the lab decreased as the battery's charge
was drained off and was converted to heat.

The heat goes into the kinetic energy of the molecules inside the lab.

Is _that_ the solution to the conundrum?

The reason I told you to decide what happens at each step was to
avoid doing just what you've done. You've strung together a lot
of processes without analyzing (or even carefully specifying)
how each process takes place.

3) I have a tank of gas with a membrane down the middle. All the
gas is squeezed into one half of the tank. I rupture the membrane.
The gas expands to fill the tank. The pressure in the tank drops;
the temperature doesn't change, since the gas did no work while
expanding. Does the tank get lighter? (Note that no energy -- or
anything else -- flowed between the tank and the outside world.)

Yes, it did. Before rupturing the membrane, there were stresses in
the membrane that confined the gas to one region in the tank. By
allowing the pressure to equalize on both sides, you removed the
force acting to confine the gas.

Hmmm.

In this case it seems that we have a flat violation of mass/energy
conservation.

Why is that? You obviously believe there is stored energy in a
compressed (or stretched) spring. That's exactly what your membrane
is.

But it's a one-time thing so you can't build a perpetual
motion machine out of it: there's no way to put the gas _back_ in the
smaller volume without doing work on it. Is that a fair statement?

Why does that matter?

In other words, you can destroy free energy, and that reduces the
overall mass/energy of the universe. But you can't create more free
energy, so when it's gone, it's gone -- there's no way to run it in a
cycle. Right? Close? Totally off-base?

I think you first need to specify each step of the "experiment"
well enough to answer the question "what happened to the energy".

4) [snip] I made this last example too complex -- didn't realize the
implications of all that was going on.

I have one additional question to add to the list (one which I asked of
PMB earlier). Since the answers to the others were surprising, perhaps
this will be, too...

We have two particles at relative rest, separated by distance L.
They're held at rest relative to each other by a massless rigid rod, or

First of all, a massless, rigid rod violates relativity, so using
you shouldn't be surprised if you get a non-sensical answer when
using such a construct.

they're only momentarily at rest, or they're both suspended from strings
-- I don't care which. I measure the distant gravitational field of the
pair.

Now, I move them together to distance L/2, and assure that they're again
at relative rest (stick them to a shorter rod, or do whatever else is
necessary).

The distant gravitational field of the pair of particles should
_decrease_, right? The reason is that the total energy of the system
decreased, due to the loss of potential energy -- to bring the particles
to rest a second time energy had to be removed from the system. Yes?

OK, this isn't really a relativity question, since potential energy
is not really a relativistic idea. In newtonian mechanics, lowering
the potential energy means making the value a _larger_negative_number.
For example, if m is a mass in the gravitational field of M, then,
U = -GMm/r and the kinetic energy of the mass m is T = (1/2)mv^2, then
the total energy of is (1/2)mv^2 - GMm/r. That gives:

(1/2)mv^2 = GMm/r

For an escape velocity of 11,118 m/s. Now suppose the masses are closer.
The postential energy os smaller, because it's more negative, not because
the magnitude of the number is smaller. If r wrer half the earth radius,
then the escape velocity is 15,800 m/s.

First cousin to these issues is a much messier question: Does the
distant gravitational field of a star increase, decrease, or stay the
same when it collapses into a black hole? But that's a question I'm
happy to leave for next year...

If all of the mass falls into the black hole, then the distant
gravitational field doesn't change.

Tom Roberts wrote in message ...
Pmb wrote:
"sal" wrote in message
...
Ah hmm. This is what I think I've read, and it's sure how the stress
energy tensor looks. But it seems to lead to something unpalatable,
which implies that there's something more going on than I've grokked as

yet.

Caution: What bilge gave you was the energy-momentum tensor of the *gas*
only. Cooling the balloon means that the proper mass density rho decreased.
Since the balloon is a rest then you can ignore the container walls itself
since the stress in the walls does not contribute to the the energy density
when the walls are at rest. Just add the weight of the mass density of the
balloon to the mass density of the gas. The proper mass of the whole deal is
defined as the integral of T^00 over the whole object.

Further caution: The original question was phrased in terms of "gets
heavier", which implies "weight", not (proper) mass. While the stress in
the walls does not contribute to the (proper) mass of the whole deal, it
does contribute to its "weight".

Not as he phrased it. The balloon is at rest and therefore the stress
does not contribute. The weight is W = M_o*g where M_o is the proper
mass of the body

But if you claim otherwise let's see a calculation

sal:

But the field equations take the form G^ab = 8*pi*T^ab, which means the
pressure terms are carrying across directly to the Ricci tensor and the
metric. It's beyond my as yet feeble grasp of this stuff to say exactly
what the effect of boosting the pressure would be on the curvature and
hence on the gravitational field as we measure it, but it sure seems
like there should be _some_ effect. And that seems to be Bilge's point
-- it's the pressure terms which are affecting the field directly, not
just the proper mass.

Bingo.


(FWIW, that's certainly what Schutz said as well, in English, but all
English statements about things like tensors tend to be a little
suspect. From "A first course in general relativity", p196: "But again,
if T^00 alone were the source, one would have to specify a frame in
which T^00 was evaluated. An invariant theory can avoid introducing
preferred coordinate systems by using the _whole_ of the stress-energy
tensor T as the source of the gravitational field". I'm as yet still
unsure just how to interpret this statement, which was, of course, the
point in the questions I've been asking.)


It appears that you interpreted it correctly above.

Tom Roberts wrote in message ...
[I have not had time to look at the detailed calculations, so I snip them.]

Pmb wrote:
"sal" wrote
(FWIW, that's certainly what Schutz said as well, in English, but all
English statements about things like tensors tend to be a little
suspect. From "A first course in general relativity", p196: "But again,
if T^00 alone were the source, one would have to specify a frame in
which T^00 was evaluated. An invariant theory can avoid introducing
preferred coordinate systems by using the _whole_ of the stress-energy
tensor T as the source of the gravitational field". I'm as yet still
unsure just how to interpret this statement, which was, of course, the
point in the questions I've been asking.)

GR is an "invariant theory" in Schutz's sense here. That is, the
Einstein field equation is simply:

G = 8pi T [G is the Einstein curvature tensor, T is
the energy-momentum tensor; here I use your
units, but I normally use units in which
the 8pi is inside T, not external as here.]

There is no coordinate system in that at all, much less any "preferred"
one. Clearly the "whole" of T is indeed involved.

The mathematical object which plays the role of source in Einstein's
field equations is the energy-momentum tensor, T, since mass (density)
= T^00/c^2 (no - not proper mass, mass-energy) in one frame is
momentum in another frame etc. Same in EM where the mathematical
object which plays the role of source in Maxwell's field equations is
the 4-current, J, since charge (density) = J^0/c^2 in one frame is
current in another other. The physical quantity which is the source of
an EM-field is charge which is the time component of J. The physical
quantity which is the source of the G-field is mass and mass density
is the time-time component of T.

The question is with regard to weight. Sal was refering to weighing
the an object in the rest frame of the object. Its assumed the
gravitational field is simple. The most general answer is too detailed
to get into here since details of both source and object must be known
and that can get very very messy.

The weight of an object at rest is always related to proper mass and
nothing else. This holds true since the 4-momentum of a object is
given as P = MU where P = total 4-momentum and U = 4-velocity of
object and M = proper mass of object = integral over T^00. The weight
of the object is found in the usual way. Accelerate the object. In
rezt frame of the object the weight will be W = Mg.

However life is not that easy since the act of accelerating the object
can actually change the object and therefore change the
energy-momentum tensor.

In the simple case of a moving particle the weight will depend on the
velocity of the particle relative the the observer. You do know what
that means right?


It would be wise to look at EM and see what it means for J^u = (rho*c, j) to
be the source of the EM field where rho is charge density and j is current
density. What you'll see is that charge in one frame is current in another.

Yes, but a better analogy is the fields of EM -- rank-2 tensors are more
complicated than the 4-vector current. One can always find a
locally-inertial frame in which J is pure rho (...

That is not generally true. That holds true only in special cases.
I.e. cases where there is no relative motion of the charges.



Example:
Wrt its rest frame the field of an isolated charge is pure E, and wrt a
frame in which it is moving the field is a mixture of E and B -- there
is no frame in which it is purely B. If one has a frame in which the
field is pure B, there is no frame in which it is pure E. This is all
wrapped up in the fact that the E and B fields are really components of
a 2-form....

Back to the actual analogy! If the charge is at rest in the EM field
then the force on the charge is purely electric and the current
associated with the charge is absent in this frame. Therefore there is
no interaction in the rest frame with the magnetic field - only the
electric field of the source and only the time component of the
4-current of the object we are interested in which has a force on it.
I.e. Force = qE

Bilge wrote:

SAL wrote:
2) OK, now something less obvious: I have a large balloon
filled with low-pressure gas, and I weigh it (in a vacuum, of
course). Now, I compress the gas into a small tank. I let it
cool off so it's the same temperature it was in the balloon. I
weigh it again. It weighs more, due to the pressure terms in
the stress-energy tensor ... right? Or does it? (I added
"free energy" when I compressed it, which presumably is where
the extra mass/energy -- if any -- came from.)

Bilge:
Yes. The stress energy tensor is of the form:

T^ab = (\rho + P)U^a U^b + P g^ab

where \rho is the density, P is the pressure, U^a and U^b are
four velocities and g^ab is the metric tensor. Since the initial
and final temperatures are the same and the velocities are
prortional to kT, the initial and final stress-energy tensors
differ only in the second terms. The change in energy is then due
to the additional force required to confine the gas to a smaller
volume.

SAL:
Ah hmm. This is what I think I've read, and it's sure how the
stress energy tensor looks. But it seems to lead to something
unpalatable, which implies that there's something more going on
than I've grokked as yet.

Specifically, this appears to violate conservation of energy.

Bilge:
How so? By your own construction, you added energy to compress the
gas. So, the process isn't adiabatic.

SAL:
If I put a compressor, battery, tank, and gas supply in a sealed
laboratory, and put the whole thing on a spring, then when the
compressor runs and the gas is compressed into the tank, if the
weight (mass/energy) increases then the spring will be compressed.

Bilge:
Why is that? If the laboratory is sealed, then all of the energy that
compresses the gas has to come from inside the laboratory, so
whatever energy is transferred to the gas, comes from something else
in the lab. That's a different question than before where you were
only talking about the gas and adding and subtracting energy.

That's right, it is. In the first case the system was "open"; I'm
attempting to close it so I can understand where the energy is coming from.


I can use the motion of the laboratory as the spring is compressed
to do work. If I then let the gas out of the tank again, the lab
gets lighter, the spring relaxes a bit, and I can use the upward
motion of the laboratory to do more work. The total mass/energy of
the universe has increased.

Again, why would that happen? The energy you release from the gas,
can't go anywhere but inside the laboratory.

Ah. This is the point I'm trying to understand -- if we account for all
the energy, rather than just part of it, does everything balance? A
closed system (a sealed laboratory) is just an attempt to force an
accounting.



That seems wrong -- it seems like there must be a balancing effect
as well.

Now, no mass/energy entered or left the sealed laboratory.
However, the amount of Gibbs free energy in the lab decreased as
the battery's charge was drained off and was converted to heat.

The heat goes into the kinetic energy of the molecules inside the
lab.

Yes, that's right. All mass-energy is retained inside the lab.
However, the overall "free energy" decreased -- it's not conserved and
some was used up running the compressor.


Is _that_ the solution to the conundrum?

The reason I told you to decide what happens at each step was to
avoid doing just what you've done. You've strung together a lot of
processes without analyzing (or even carefully specifying) how each
process takes place.

Yes, you are correct; that's exactly what I've done. Does that actually
make it impossible to draw any conclusions about what happens? Consider:

If this were a problem in classical mechanics, we could close the system
and then say, "Energy and mass must both be conserved within the closed
system!" and we could draw conclusions from that, without knowing
exactly what happened inside the box.

If this were a problem in thermodynamics we could, again, draw
conclusions from the fact that the system was closed, without analyzing
everything which happens inside the system in detail.

But this is a problem in GR and I don't know the rules. So, I'm asking
you: If we close the system, then must the total mass/energy of the
system be conserved? Alternatively, can we state, in advance of
analyzing every process in the closed system, that the spring-balance
weight of the system as a whole must remain constant?

(Come to think of it the system isn't really closed, after all -- energy
can still leak out as gravity waves. But at any rate, nothing's going in.)



3) I have a tank of gas with a membrane down the middle. All
the gas is squeezed into one half of the tank. I rupture the
membrane. The gas expands to fill the tank. The pressure in
the tank drops; the temperature doesn't change, since the gas
did no work while expanding. Does the tank get lighter? (Note
that no energy -- or anything else -- flowed between the tank
and the outside world.)

Yes, it did. Before rupturing the membrane, there were stresses
in the membrane that confined the gas to one region in the tank.
By allowing the pressure to equalize on both sides, you removed
the force acting to confine the gas.

Hmmm.

In this case it seems that we have a flat violation of mass/energy
conservation.

Why is that? You obviously believe there is stored energy in a
compressed (or stretched) spring. That's exactly what your membrane
is.

But !

I've got a sealed tank sitting on a spring balance. Nothing goes into
it, nothing comes out of it. The membrane in the tank pops; still,
nothing goes in or comes out. Yet, the thank becomes lighter -- the
needle on the balance moves.

In the universe as a whole, _something_ just decreased.



But it's a one-time thing so you can't build a perpetual motion
machine out of it: there's no way to put the gas _back_ in the
smaller volume without doing work on it. Is that a fair statement?

Why does that matter?

In other words, you can destroy free energy, and that reduces the
overall mass/energy of the universe. But you can't create more
free energy, so when it's gone, it's gone -- there's no way to run
it in a cycle. Right? Close? Totally off-base?

I think you first need to specify each step of the "experiment" well
enough to answer the question "what happened to the energy".

Perhaps I didn't phrase this properly. In elementary thermodynamics a
half-filled tank of gas with a barrier in it is a standard "gedanken"
item. If the gas is allowed to expand into the whole tank _without_
doing any work, the temperature of the gas doesn't change, and no
(classical) energy enters or leaves the system. That's all I'm trying
to portray.

But when the gas expands, we lose the potential to do work with the gas,
so something does go away. As far as I know, what goes away is "Gibbs
free energy" -- which is exactly the capacity to do work.


I have one additional question to add to the list (one which I
asked of PMB earlier). Since the answers to the others were
surprising, perhaps this will be, too...

We have two particles at relative rest, separated by distance L.
They're held at rest relative to each other by a massless rigid
rod, or

First of all, a massless, rigid rod violates relativity, so using you
shouldn't be surprised if you get a non-sensical answer when using
such a construct.

OK, if the rod must not be massless and totally rigid, make it very
light and give it a very high spring rate. Then it can enter into the
analysis. Does that affect the overall conclusion? (I'm not looking
for a contradiction -- I'm just trying to understand what happens!)

For that matter, I'm also happy to assume the masses are momentarily at
rest. In that case they'll be accelerating, so they'll be emitting
gravitational radiation but I don't think that affects the overall
conclusion.


they're only momentarily at rest, or they're both suspended from
strings -- I don't care which. I measure the distant gravitational
field of the pair.

Now, I move them together to distance L/2, and assure that they're
again at relative rest (stick them to a shorter rod, or do whatever
else is necessary).

The distant gravitational field of the pair of particles should
_decrease_, right? The reason is that the total energy of the
system decreased, due to the loss of potential energy -- to bring
the particles to rest a second time energy had to be removed from
the system. Yes?

OK, this isn't really a relativity question, since potential energy
is not really a relativistic idea.

I beg your pardon!

It certainly is a relativity question! I'm describing a situation and
asking what relativity predicts would happen to the gravitational field.
I used some phrases from classical mechanics in an effort to describe
the situation clearly and to try to see if I understand the principles
involved, but I'm not asking about classical physics here. I know
perfectly well that the far field of two masses in Newtonian mechanics
doesn't vary as the masses move together or move apart. My question is
what GR predicts happens.

If the description I gave is inadequate to answer the question, then I'd
like to clean up the description and try again. So, what else is needed
to come to a conclusion?



In newtonian mechanics, lowering
the potential energy means making the value a
_larger_negative_number.

Yes, I know that; it's not what the question concerned. The question
concerned GR. I have read that in GR the "potential energy" also
contributes to the gravitational field. So, I'm asking you -- in this
case, does the far field of two masses increase, decrease, or stay the
same as they move closer together, and their classical potential energy
drops?

All my talk of rigid rods was just an attempt to make it clear that as
they move closer together, I'm pulling kinetic energy out of the system,
so in classical terms _just_ the potential energy changed.




First cousin to these issues is a much messier question: Does the
distant gravitational field of a star increase, decrease, or stay
the same when it collapses into a black hole? But that's a
question I'm happy to leave for next year...

If all of the mass falls into the black hole, then the distant
gravitational field doesn't change.

Well! This is certainly consistent with a decrease in the far field in
the much simpler case I tried to describe above.

--
To email me directly, take out nospam and put back foobox.

I think I finally got it. See comments following "I think I'm
beginning to catch on" below for what I think I got...


Pmb wrote:
"sal" wrote

Bilge said:

Yes. The stress energy tensor is of the form:

T^ab = (\rho + P)U^a U^b + P g^ab

where \rho is the density, P is the pressure, U^a and U^b are
four velocities and g^ab is the metric tensor. Since the initial
and final temperatures are the same and the velocities are
proportional to kT, the initial and final stress-energy tensors
differ only in the second terms. The change in energy is then due
to the additional force required to confine the gas to a smaller
volume.

SAL:
This pretty clearly means Bilge believes the weight changes even
when temperature is held fixed.

PMB:
bilge gave the wrong expression for this tensor.

I think Bilge uses a Lorentz metric signature of -1,1,1,1. Using
that, his expression looks right. In the MCRF, U^a*U^b is 1 if
a==b==0 and 0 otherwise, and P*g^ab == diag(-P,P,P,P), so T reduces to
diag(\rho,P,P,P), which, at any rate, agrees with Schutz p. 107 eq
(4.36) ("proof by quoting wise men"). Schutz also uses -1,1,1,1 for
the signature, BTW.

You've mentioned that you prefer a signature of 1,-1,-1,-1 for the
metric, and in that case I think you need to flip the sign of the
second term for the stress-energy tensor.


What you said was the temperature remains the same. That is
true. The pressure remains the same.

Howzat? Typo? The gas occupies a smaller volume at the same
temperature, so by PV = nRT the pressure must be higher.

Or have I failed to understand what "pressure" means? I believe
pressure at a surface is defined as momentum transfer across the
surface, which agrees with the classical notion of force per unit
area. If the gas is not expanding then the momentum transfered from
one "cell" to an adjacent "cell" is exactly balanced by the transfer
back, but the pressure terms in the stress energy tensor are still
nonzero in that case, and still correspond to the classical notion of
"pressure". Right...?


What changed is the proper mass density, what [Bilge] calls "rho."

That changed, too, of course.


When the gas cooled rho decreased.

But not back to its original value, of course, since the density is
higher due to the smaller volume. Rather, if we halve the volume, we
should end up with rho' = 2 * rho (assuming temperature is fixed).


The stress in the walls does not contribute the the energy density
in the rest frame of the gas. If you changed from one frame to a
moving frame then this would be a different story though. Stress in
one frame contributes to mass-energy in another frame

Just as well, since I left the walls of the container out of my
original "image" of the problem. Hmm.


to which PMB said:

Caution: What bilge gave you was the energy-momentum tensor of
the *gas* only. Cooling the balloon means that the proper mass
density rho decreased. Since the balloon is a rest then you can
ignore the container walls itself since the stress in the walls
does not contribute to the the energy density when the walls are
at rest. Just add the weight of the mass density of the balloon
to the mass density of the gas. The proper mass of the whole deal
is defined as the integral of T^00 over the whole object.

As above, the stress energy tensor, as Bilge gave it, was this:

T^ab = (\rho + P)U^a U^b + P g^ab


That is an incorrect expression. It should be

T^ab = (\rho + P)U^a U^b - P g^ab

I think you're both right. I think the difference here is just that
Bilge uses -1,1,1,1 for the metric signature, and you're using
1,-1,-1,-1.


See Eq. (1.58) in
http://assets.cambridge.org/05214227...21422701WS.pdf

Yes; that paper also uses a signature of 1,-1,-1,-1 for the metric,
and shows the same form you do for the stress-energy tensor.



Now, in the MCRF for the gas, T^00 == \rho. So, if we integrate
T^00 over the volume of the gas we'll get something that looks a
lot like nRT.

Why? Please provide a proof or an explanation

As above, with signature -1,1,1,1,

T^00 = (rho + P)(U^0 * U^0) + P * g^00

In the MCRF, g^00 == -1 and U^0 == 1, so this comes out to

T^00 = rho

rho is the particle density times the energy per particle (or
molecule), right? The energy per particle is very much like the value
captured by "RT" in the gas equation (tho I'm not sure they're totally
identical). The integral over a volume of rho will give the total
number of particles in that volume times the energy per particle,
which is what's expressed by "nRT".

If we halve the volume without changing the temperature, we don't
affect the "nRT" term, and "PV" remains fixed. By the same token, the
density of particles will double, but the energy per particle won't
change. So, rho will double, the volume will halve, and the integral
over the volume of rho -- or of T^00 -- will remain fixed, just as
nRT did.

SAL:
But the field equations take the form G^ab = 8*pi*T^ab, which means
the pressure terms are carrying across directly to the Ricci tensor
and the metric. [...]

PMB:

Consider the weak field approximation. It can be shown that

del^2 Phi = 4*pi*G(rho_o + p/c^2)

The effective active gravitational mass is rho_eff = rho_o + p/c^2.
So pressure does act as as source of gravity.

Ah! I think I'm beginning to catch on. Let us reason together.

In elementary terms, we integrate del^2 Phi over a volume to get the
field pointing out through the surface of the volume. Classically Phi
would depend only on rho, but in GR it depends on rho _and_ on p.

Now, we squeeze the gas into half the volume, holding T constant.
'rho' must double. BUT 'p' doubled too. So, the integrand, which is
something times (rho + p/c^2), will also exactly double. When we
integrate that over the new smaller volume, we'll get the same value
for the flux of the field that we got to start with.

So, the pressure plays a role: it causes the field to be stronger than
it would have been classically. But changing the pressure and hence
the volume (assuming we do it without adding or removing energy)
doesn't change the field, either classically or in GR.

Is that correct?


Suppose there is a spherical shell which contains a gas. Then this
is the equation for the field. Suppose that its a gas of photons and
all of a sudden the shell absorbs the photons. Then if rho_eff
remains the same the gravitational field won't change. This can
happen if rho_ increase to rho_eff. I've never investigate this but
it seems logical. Not sure though so don't quote me on that! :-)

OK!


And that seems to be Bilge's point -- it's the pressure terms which
are affecting the field directly, not just the proper mass.

Are we talking about the gravitational field or are we talking about
the mass?

Gravitational field. Back at the beginning, I specified "spring
balance weight" on Earth's surface, which, unless I'm totally
confused, is a measure of the gravitational field of the object being
weighed.

I have also mentioned "far field" or "distant field" in a few places.
All I had in mind with that phrase was that I'm not concerned with the
exact shape of the field -- I want to move far enough away so we can
assume the field is spherically symmetric without affecting the
qualitative results.

I'm currently pretty unclear on what "mass" is as distinct from "thing
that makes gravity", and that's no doubt part of the confusion here.
I thought they were identical until I wandered into the pressure term
swamp.


It would be wise to look at EM and see what it means for J^u =
(rho*c, j) to be the source of the EM field where rho is charge
density and j is current density. What you'll see is that charge in
one frame is current in anotgher. Thinking about that will give you
insight into GR

But there's nothing equivalent to a magnetic field in GR, is
there? The 'j' term gives rise to a force which depends on velocity
in a way that we never see with gravity ... AFAIK.


Pmb



--
To email me directly, take out nospam and put back foobox.

"sal" wrote in message
...
I think I finally got it. See comments following "I think I'm
beginning to catch on" below for what I think I got...


Pmb wrote:
"sal" wrote

Bilge said:

Yes. The stress energy tensor is of the form:

T^ab = (\rho + P)U^a U^b + P g^ab

where \rho is the density, P is the pressure, U^a and U^b are
four velocities and g^ab is the metric tensor. Since the initial
and final temperatures are the same and the velocities are
proportional to kT, the initial and final stress-energy tensors
differ only in the second terms. The change in energy is then due
to the additional force required to confine the gas to a smaller
volume.

SAL:
This pretty clearly means Bilge believes the weight changes even
when temperature is held fixed.

PMB:
bilge gave the wrong expression for this tensor.

I think Bilge uses a Lorentz metric signature of -1,1,1,1.

Could be. I'm used to another one. It occured to me later that he may have
used such a metric and hence my retraction


What you said was the temperature remains the same. That is
true. The pressure remains the same.

Howzat? Typo? The gas occupies a smaller volume at the same
temperature, so by PV = nRT the pressure must be higher.

Recall how you stated the problem

I let it cool off so it's the same temperature it was in the balloon. I
weigh it again.

The only way to let it cool is to let heat leave - thus it looses energy.

Recall what I said in my response - "Cooling the balloon means that the
proper mass density rho decreased."

Or have I failed to understand what "pressure" means?

Nah. I think you neglected to take into account the fact that cooling the
body requires a decrease in thermal energy and thus a decrease in mass.

Uh oh! .... Damn it! Sorry. Yes. You're right. The pressure changed as a
result of cooling. Wha the hell was I thinking!

Sorry. Now I see how obvious it is since

Initial state: P_1*V_1 = nRT_1

Compress gas. Requires work. Increases internal energy and raises T

Final state: P_2*V_2 = nRT_2

Divide first eq by second eq

P_1*V_1 nRT_1 T_1
--------- = ------- = -----
P_2*V_2 nRT_2 T_2

Cool gas so that T_1 = T_2. Mass density decreases.

P_1*V_1 = P_2*V_2

P_2 = (V_1/V_2)*P_1

V_1 V_2 -- Pressure increases.

Sorry. I was working on other things last night and I guess I was confused
on this. I guess I was thinking pressure remained constant rather than
temperature remained constant.

Sorry for the confusion sal.

I believe
pressure at a surface is defined as momentum transfer across the
surface, which agrees with the classical notion of force per unit
area.

That is all worked out in detail here
http://www.geocities.com/physics_wor...ass_tensor.htm

If the gas is not expanding then the momentum transfered from
one "cell" to an adjacent "cell" is exactly balanced by the transfer
back, but the pressure terms in the stress energy tensor are still
nonzero in that case, and still correspond to the classical notion of
"pressure". Right...?

The gas does no work so there is no change in internal energy. The final
state is related to the initial state as

P_1*V_1 T_1
--------- = -----
P_2*V_2 T_2

When the gas cooled rho decreased.

But not back to its original value, ...

I don't know. Is that true? I didn't calculate/think about that.

In the MCRF, g^00 == -1 and U^0 == 1, so this comes out to

T^00 = rho

rho is the particle density times the energy per particle (or
molecule), right?

Yes.



Consider the weak field approximation. It can be shown that

del^2 Phi = 4*pi*G(rho_o + p/c^2)

The effective active gravitational mass is rho_eff = rho_o + p/c^2.
So pressure does act as as source of gravity.

Ah! I think I'm beginning to catch on. Let us reason together.

In elementary terms, we integrate del^2 Phi over a volume to get the
field pointing out through the surface of the volume. Classically Phi
would depend only on rho, but in GR it depends on rho _and_ on p.

Now, we squeeze the gas into half the volume, holding T constant.
'rho' must double. BUT 'p' doubled too. So, the integrand, which is
something times (rho + p/c^2), will also exactly double. When we
integrate that over the new smaller volume, we'll get the same value
for the flux of the field that we got to start with.

That is my guess. I spoke to a GRist once about this and he disagreed with
me. But I think that what you said was right. Too bad we can't actually do
an experiment and test it huh? :-)

I'm sure there is some literature around which explains that what you've
said is right since I can't imagine how it could be wrong.

So, the pressure plays a role: it causes the field to be stronger than
it would have been classically. But changing the pressure and hence
the volume (assuming we do it without adding or removing energy)
doesn't change the field, either classically or in GR.

Is that correct?

Seems right to me. However I have concerns about it all since active
gravitational mass should equal passive gravitational mass and thus rho
should have added to passive gravitational mass density.


But there's nothing equivalent to a magnetic field in GR, is
there? The 'j' term gives rise to a force which depends on velocity
in a way that we never see with gravity ... AFAIK.

It's a well known fact that the gravitational force in GR is velocity
dependant. See
http://www.geocities.com/physics_wor...grav_force.htm

Pmb

Gauge wrote:
The weight of an object at rest is always related to proper mass and
nothing else. This holds true since the 4-momentum of a object is
given as P = MU where P = total 4-momentum and U = 4-velocity of
object and M = proper mass of object = integral over T^00. The weight
of the object is found in the usual way. Accelerate the object. In
rezt frame of the object the weight will be W = Mg.

All true when the object in question can be considered a "test
particle". But in this discussion one cannot do that -- the questions
specifically relate to the other terms of the stress-energy tensor and
how they affect gravitational interactions. Note these other terms are
down typically by a factor of v/c or (v/c)^2 from the (0,0) component of
T, so neglecting them for a test particle is quite justified (here v is
some typical relative velocity of the stuff in question).

Note I am ignoring the fact that the e-m tensor is a field on
spacetime, and is therefore not really associated with any
object(s). When the world can be separated into objects the
e-m tensor can be written as a sum over them, and I'm really
discussing the object's individual contribution.


However life is not that easy since the act of accelerating the object
can actually change the object and therefore change the
energy-momentum tensor.

Hmmm. Acceleration need not change the object, but can still change its
e-m tensor (and usually does):

If the object is unchanged, then its e-m tensor projected
onto its rest frame will be unchanged. But after acceleration
its rest frame has been changed, and therefore so has the
basis; since the components are unchanged the tensor itself
must have changed.

Exercise: describe a situation for which an accelerated
object's e-m tensor does not change. Hint: determine
carefully what "acceleration" means here; beware of puns!

[N.b. this is an excellent example of where confusing a
tensor with its components would be a disaster.]


Yes, but a better analogy is the fields of EM -- rank-2 tensors are more
complicated than the 4-vector current. One can always find a
locally-inertial frame in which J is pure rho (...
That is not generally true. That holds true only in special cases.
I.e. cases where there is no relative motion of the charges.

It is generally true for any given point (remember J is a 4-vector field
on spacetime). This is basically the same statement as that one can
always find a locally-inertial frame in which a given timelike particle
is at rest. Remember J must be a timelike 4-vector.

If there are multiple charges headed in different directions
at a given point, J is simply their (4-vector) sum -- E&M is
linear. There is still a frame in which J is purely rho at
this point.

Also please note the difference: J is a 4-vector field on spacetime, but
the 4-velocity of a particle is not -- that is the tangent 4-vector to
the particle's trajectory, not a field of any sort.


Tom Roberts