Equation of state in gravity vs microphysics


by Jip
Tags: equation, gravity, microphysics, state
Jip
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#1
Dec26-12, 05:47 AM
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Hi,
I have a very simple question. Consider a free scalar field in the realm of GR. Then its stress energy tensor, in a Roberton-Walker Universe, is the one of a perfect fluid with pressure=denity, hence an equation of state : w=p/rho=1

However this scalar model is an archetype of maslesss (spin 0) particles. Now if you consider massless particles in a box, and do your thermo/statistical physics homework on it, you find that, as any relativistic (masless) particles, the pressure must be 1/3 of the density, just as it is the case for photons. Hence here we find w=p/rho=1/3 for such a scalar field in a box.

Where's the discrepancy coming from?

Thanks for comments!
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bcrowell
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Dec26-12, 09:58 AM
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Possibly useful:
http://www.physicsforums.com/showthread.php?t=134682
http://faculty.washington.edu/mrdepi...rk_Energy2.pdf

Quote Quote by Jip View Post
I have a very simple question. Consider a free scalar field in the realm of GR. Then its stress energy tensor, in a Roberton-Walker Universe, is the one of a perfect fluid with pressure=denity, hence an equation of state : w=p/rho=1
I'm not sure this is right, in general. The cosmological constant can be considered as a free scalar field, and it has w=-1, not 1. (A constant is a solution of the m=0 Klein–Gordon equation.) In general (see the pdf link above), a scalar field that's not spatially varying has
[itex]\rho = (1/2)\dot{\phi}^2+V(\phi)[/itex]
[itex]p = (1/2)\dot{\phi}^2-V(\phi)[/itex]
In the case of a cosmological constant, the time derivatives are zero. In general, you could get any [itex]-1 \le w \le 1[/itex].

As a side issue, why do you say, "its stress energy tensor, in a Roberton-Walker Universe" -- I don't think it matters what the cosmology is, does it? Or is this because you're invoking the assumption that the field doesn't vary spatially?

Quote Quote by Jip View Post
However this scalar model is an archetype of maslesss (spin 0) particles. Now if you consider massless particles in a box, and do your thermo/statistical physics homework on it, you find that, as any relativistic (masless) particles, the pressure must be 1/3 of the density, just as it is the case for photons. Hence here we find w=p/rho=1/3 for such a scalar field in a box.
So this may or may not be headed down the right road to solve your problem, but I think we can get an issue with photons that's the same as the one I referred to above for a scalar field. You can have electromagnetic fields for which [itex]w\ne 1/3[/itex]. For example, a uniform, static electric field E in the x direction has [itex]T_{00}=(1/2)E^2[/itex], [itex]T_{11}=-(1/2)E^2[/itex], which isn't consistent with w=1/3. [Edit: fixed a mistake here]

So for example it seems to me that a cosmological constant is able to evade our expectation of w=1/3 for massless particles in a box because it's not a state in thermal equilibrium. Maybe the thermal equilibrium state of a massless scalar field does have w=1/3.
Jip
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#3
Dec26-12, 10:35 AM
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Well, some comment on your post.
1. I assume cosmological background in order to set the spatial derivatives to zero, indeed.
2. Then you get the two equations you gave for rho and p of the scalar field. By the way, you jut prove my claim (it is well known): take V(phi)=0, and see how your formulas give indeed p=rho, hence w=1
3. Yes probably the answer lies in the thermal equilibrium assumption, but I don't know precisely how!
4. Can you elaborate of the electric field example you give? How do you compute the pressure here, and find p=-rho?

5. Indeed my question is more general : what is the link (for any kind of field) between the w computed in GR with the w computed from "microphysics", e.g partition function Z => p, U => p, rho=U/V => w

I'll look at the links you provided, thanks

Jip
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#4
Dec26-12, 10:48 AM
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Equation of state in gravity vs microphysics


Ok I just saw quickly the references you gave me. I want to stress that my question is not about the cosmological constant or Dark Energy modeling, etc, but about the thermodynamical (or satistical physics) interpretation of the pressure and density as defined in GR through the perfect stress energy tensor derived from the action.

I think that we all agree (?) that this way of defining presure and density is a priori very different from the definition coming from partition function and so on. And I gave one particular example that it could lead to different results for the equation of state.

Now maybe I'm just completely wrong here. If someone can help :D
bcrowell
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Dec26-12, 11:40 AM
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Quote Quote by Jip View Post
2. Then you get the two equations you gave for rho and p of the scalar field. By the way, you jut prove my claim (it is well known): take V(phi)=0, and see how your formulas give indeed p=rho, hence w=1
It doesn't prove your claim, it disproves it. It gives a counterexample to your claim that w=1. Giving any number of examples doesn't prove a claim. Giving one counterexample disproves it.

Quote Quote by Jip View Post
4. Can you elaborate of the electric field example you give? How do you compute the pressure here, and find p=-rho?
Sorry, I messed that up. In that example we have [itex]T=diag(E^2/2,-E^2/2,+E^2/2,+E^2/2)[/itex], where the field is in the x direction. See http://en.wikipedia.org/wiki/Electro...3energy_tensor .But the point is that it doesn't have the form [itex](\rho,p,p,p)[/itex] with w=1/3.

Quote Quote by Jip View Post
3. Yes probably the answer lies in the thermal equilibrium assumption, but I don't know precisely how!
The factor of 1/3 is thermodynamic in origin. The logic is that we assume thermal equilibrium, this leads to equipartition, and therefore the energy is partitioned equally in the 3 spatial degrees of freedom.

A uniform electric field doesn't represent a thermal-equilibrium state of the electromagnetic field, so the above logic fails, and we don't get a stress-energy tensor of the form [itex](\rho,p,p,p)[/itex] with w=1/3. Similarly, a uniform scalar field isn't a state of thermal equilibrium.

Quote Quote by Jip View Post
5. Indeed my question is more general : what is the link (for any kind of field) between the w computed in GR with the w computed from "microphysics", e.g partition function Z => p, U => p, rho=U/V => w
They're the same. GR is locally the same as SR.

Quote Quote by Jip View Post
Ok I just saw quickly the references you gave me. I want to stress that my question is not about the cosmological constant or Dark Energy modeling, etc, but about the thermodynamical (or satistical physics) interpretation of the pressure and density as defined in GR through the perfect stress energy tensor derived from the action.
The point is that the cosmological constant is a counterexample to your claim that w has certain values for a scalar field.

Quote Quote by Jip View Post
I think that we all agree (?) that this way of defining presure and density is a priori very different from the definition coming from partition function and so on.
No, I don't agree. There aren't two definitions. There is a definition, and then there is a particular thermodynamic approximation for calculating the thing that's been defined, under the assumption of thermal equilibrium.
atyy
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Dec26-12, 12:30 PM
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Carroll seems to get the 1/3 in Eq 8.27 of http://ned.ipac.caltech.edu/level5/M.../Carroll8.html .

Perhaps a matter of definitions?
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Dec26-12, 12:59 PM
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Quote Quote by atyy View Post
Carroll seems to get the 1/3 in Eq 8.27 of http://ned.ipac.caltech.edu/level5/M.../Carroll8.html .

Perhaps a matter of definitions?
He gets the 1/3 for electromagnetic fields under the assumption of a perfect fluid, which means a fluid that's isotropic in its rest frame. There is no frame in which a uniform electric field is isotropic. More fundamentally, I don't think it's valid to talk about a fluid and an equation of state unless you have some kind of system that is in thermodynamic equilibrium.

A real scalar field has two degrees of freedom, [itex]\phi[/itex] and [itex]\dot{\phi}[/itex]. If these are in equilibrium, then I think the expectation value of the two corresponding energies should be the same, [itex]\langle\dot{\phi}^2/2\rangle=\langle V(\phi)\rangle[/itex]. Presumably this leads to a traceless stress-energy tensor if the field is massless? If so, then zero trace along with isotropy implies w=1/3. But I think what's going on here is that the interesting examples, such as inflation or a cosmological constant, are not in equilibrium.
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Quote Quote by bcrowell View Post
He gets the 1/3 for electromagnetic fields under the assumption of a perfect fluid, which means a fluid that's isotropic in its rest frame. There is no frame in which a uniform electric field is isotropic. More fundamentally, I don't think it's valid to talk about a fluid and an equation of state unless you have some kind of system that is in thermodynamic equilibrium.
Actually, he seems to define a perfect fluid with arbitrary w (Eq 8.21). He sets w=1/3 in Eq 8.27, which he says holds for radiation by comparing Eq 8.25, 8.26 and 8.15, 8.19. For other perfect fluids, he gets other values of w (Eq 8.31,8.48-8.51). He uses w=1/3 for radiation-filled open, flat and closed universes in Eq 8.52-8.54. I think the idea is that although pressure is hard to define in a general non-equilibrium case, as long as the expansion is "slow" then the quantities make approximate physical sense. After all, we can apply thermodynamics in everyday life where things are in "equilibrium" over our finite time of observation, even though we know nothing is in true equilibrium over an infinite time of observation (since the universe is expanding).
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Dec26-12, 02:31 PM
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Quote Quote by atyy View Post
Actually, he seems to define a perfect fluid with arbitrary w (Eq 8.21). He sets w=1/3 in Eq 8.27, which he says holds for radiation by comparing Eq 8.25, 8.26 and 8.15, 8.19. For other perfect fluids, he gets other values of w (Eq 8.31,8.48-8.51).
You say "actually," but I don't see any point of disagreement ... ?
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Dec26-12, 03:53 PM
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Quote Quote by bcrowell View Post
You say "actually," but I don't see any point of disagreement ... ?
Probably half talking to my self. So anyway, it's w=1/3 for radiation, even in an FRW solution. And it is meaningful to talk about pressure, even though we're strictly in a non-equilibrium situation, because the expansion is slow. And it's w=-1 for the cosmological constant treated as a perfect fluid.

?
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Dec26-12, 04:21 PM
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Quote Quote by atyy View Post
Probably half talking to my self. So anyway, it's w=1/3 for radiation, even in an FRW solution. And it is meaningful to talk about pressure, even though we're strictly in a non-equilibrium situation, because the expansion is slow. And it's w=-1 for the cosmological constant treated as a perfect fluid.

?
I don't think the issue of equilibrium is affected in any important way by the fact that the universe is expanding. If you want to define an equation of state, all that matters is that locally, you can grab a sample of the matter and measure its properties. Similarly, I can define the temperature of the air in Los Angeles without worrying about the fact that it's not equilibrated with the air in Chicago. GR and cosmology aren't relevant here at all. The issue here is the properties of matter that are then going to be *inputs* into a cosmological model.
atyy
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Dec26-12, 11:11 PM
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So I think it's basically as bcrowell says that the massless scalar field case w=1 is not in thermal equilibrium.

One can get w=1/3 FRW solutions, which would be consistent with radiation in thermal equilibrium.


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