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The equation of state of radiation 
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#1
Oct1609, 09:09 AM

P: 54

Why does the EOS of radiation set to 1/3? Where does this come from?



#2
Oct1609, 10:07 AM

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P: 1,276

Heuristic argument:
P = N * F/A P = N * dp_x/dt / A P = N * dp_x / (L / v_x) / A And v^2 = v_x^2 + v_y^2 + v_z^2 = 3v_x^2 > v = sqrt(3) v_x We can make a similar argument for the components of momentum to get an overall factor of 3. P = N/V * <pv> / 3 Then for a photon gas <pv> is the energy, so we have P = Energy per particle * Number / Volume / 3 = energy density / 3 Another way I have seen it derived, is to take the EM Stress tensor, show that it must be traceless, and compare that with the general identity (for a perfect fluid) [tex]T^{\mu}_{\mu} = \rho + 3p[/tex] 


#3
Oct1609, 08:44 PM

P: 54

If there is a layer of photon gas surrounding a black hole‘s surface, will the pressure of this gas still be isotropic? In other words, the energy tensor is still
[tex]\begin{display} T^{\mu}_{\nu}=\left( \begin{array}{cccc} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \\ \end{array} \right) \end{display}[/tex] , isn't it? Or the g11 is not equal to g22,g33 any more? 


#4
Oct1709, 04:02 PM

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P: 1,276

The equation of state of radiation
I would think so, but I'm not terribly certain.



#5
Oct2309, 04:33 AM

P: 54

Does somebody know that which book gives the detail derivation of this formula—the energymomentum tensor of radiation, namely
[tex] \begin{display} T^{\mu}_{\nu}=\left( \begin{array}{cccc} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \\ \end{array} \right) \end{display} [/tex] ? 


#6
Oct2309, 05:11 AM

Sci Advisor
P: 1,253

That is actually the form for any perfect fluid (i.e. one in which we can neglect viscosity and voritcity). Any GR textbook will have some amount of explanation about the derivation of this. I find 'Gravitation' by Hartle an excellent introductory textbook, but others will have this info also.



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