- #71
hellfire
Science Advisor
- 1,051
- 1
If you apply the assumption of adiabatic expansion to a photon gas with [itex]P \propto u/3 = E/3V[/itex]...
[tex]dE + PdV = 0[/tex]
[tex]dE = - \frac{EdV}{3V}[/tex]
[tex]E \propto V^{-1/3}[/tex]
With [itex]V \propto R^3[/itex], this is:
[tex]E \propto R^{-1}[/tex]
For the internal energy per unit volume:
[tex]u \propto R^{-4}[/tex]
But I am confused. This follows merely from the assumption of adiabatic expansion, without taking into consideration the expansion of space. If a photon gas expands adiabatically in a piston of some characteristic lengt L (in static space), it will also increase its wavelengh, because [itex]E \propto L^{-1}[/itex] and [itex]E \propto \lambda^{-1}[/itex]. However, in an expanding space the same relation applies. Why?
[tex]dE + PdV = 0[/tex]
[tex]dE = - \frac{EdV}{3V}[/tex]
[tex]E \propto V^{-1/3}[/tex]
With [itex]V \propto R^3[/itex], this is:
[tex]E \propto R^{-1}[/tex]
For the internal energy per unit volume:
[tex]u \propto R^{-4}[/tex]
But I am confused. This follows merely from the assumption of adiabatic expansion, without taking into consideration the expansion of space. If a photon gas expands adiabatically in a piston of some characteristic lengt L (in static space), it will also increase its wavelengh, because [itex]E \propto L^{-1}[/itex] and [itex]E \propto \lambda^{-1}[/itex]. However, in an expanding space the same relation applies. Why?
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