
#1
Dec2612, 05:47 AM

P: 11

Hi,
I have a very simple question. Consider a free scalar field in the realm of GR. Then its stress energy tensor, in a RobertonWalker 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! 



#2
Dec2612, 09:58 AM

Emeritus
Sci Advisor
PF Gold
P: 5,500

Possibly useful:
http://www.physicsforums.com/showthread.php?t=134682 http://faculty.washington.edu/mrdepi...rk_Energy2.pdf [itex]\rho = (1/2)\dot{\phi}^2+V(\phi)[/itex] [itex]p = (1/2)\dot{\phi}^2V(\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 RobertonWalker 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? 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. 



#3
Dec2612, 10:35 AM

P: 11

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 



#4
Dec2612, 10:48 AM

P: 11

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 



#5
Dec2612, 11:40 AM

Emeritus
Sci Advisor
PF Gold
P: 5,500

A uniform electric field doesn't represent a thermalequilibrium state of the electromagnetic field, so the above logic fails, and we don't get a stressenergy 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. 



#6
Dec2612, 12:30 PM

Sci Advisor
P: 8,006

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? 



#7
Dec2612, 12:59 PM

Emeritus
Sci Advisor
PF Gold
P: 5,500

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 stressenergy 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. 



#8
Dec2612, 01:19 PM

Sci Advisor
P: 8,006





#9
Dec2612, 02:31 PM

Emeritus
Sci Advisor
PF Gold
P: 5,500





#10
Dec2612, 03:53 PM

Sci Advisor
P: 8,006

? 



#11
Dec2612, 04:21 PM

Emeritus
Sci Advisor
PF Gold
P: 5,500





#12
Dec2612, 11:11 PM

Sci Advisor
P: 8,006

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. 


Register to reply 
Related Discussions  
derive the fundamental equation from and equation of state  Classical Physics  0  
what condtions u need for zero gravity state?  Classical Physics  1  
Fractal SpaceTime and Microphysics: Towards a Theory of Scale Relativity  Beyond the Standard Model  27  
gravity in bound state  Beyond the Standard Model  13  
The state of magnetism and its possible correlation to gravity...  Quantum Physics  1 