I know that heat capacity is defined as the amount of energy required to increase the temperature of any given amount of substance by 1 Degree or 1 Kelvin . Since photons also have mass , is it possible that any given photon may have a specific amount of heat capacity ? If yes than how shall I calculate it ?
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5Photons are massless particles – skdys May 23 '17 at 19:13
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Photons have no rest mass but they contain relativistic mass since they carry momentum and energy . – Munj Patel May 23 '17 at 19:14
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8Photons are massless full stop. Anyhow the non-existence of the photon mass is not germane to this discussion. A photon gas can have a well defined heat capacity even though it is massless. See this Wikipedis article. – John Rennie May 23 '17 at 19:27
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It is crucial that you distinguish correctly between heat capacity and specific heat capacity (which is equal to heat capacity per unit mass, as generally with the use of the word specific). If you don't mean to use the word in that sense, it shouldn't be near the words 'heat capacity' at all, particularly in this context. – Emilio Pisanty May 25 '17 at 16:38
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Heat capacity is a thermodynamic variable, i.e. when talking of particles it can be defined within statistical thermodynamics, but not for individual particles.
Since photons also have mass,
A photon is an elementary particle with zero mass, spin and energy $h\nu$.
Is it possible that any given photon may have a specific amount of heat capacity ?
No, it is impossible. In general individual particles cannot have thermodynamic properties which are statistical by construction. More so zero mass particles such as the photon.
Emilio Pisanty
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anna v
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Heat capacity is defined as the amount the energy of a system changes as the temperature is changed while the system remains at constant volume, $$C_V\equiv\left(\frac{\partial E}{\partial T}\right)_V.$$
Photons are massless bosons, which means their density of states is given by the Bose-Einstein distribution, $$n(p) = g[e^{E(p)/T}-1]^{-1}$$ in $\hbar=c=k_B=1$ natural units and where $g=2$ is the degeneracy of the particles, in this case 2 for the two polarizations possible for photons. Therefore the energy of a gas of photons at temperature $T$ is, as $E=p$ for massless particles, $$E=\int d^3x d^3p\,E(p) n(p)=4\pi g V \int_0^\infty p^2 dp\frac{p}{e^{p/T}-1} = 4\pi g V \frac{\pi^4T^4}{15}.$$
Therefore the heat capacity for a gas of photons is $$C_V=\frac{8\pi^5VT^3}{15}.$$
astronautgravity
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