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Do low energy photons have greater entropy than high energy photons, or is the entropy of all photons the same (and the relationship between high and low entropy radiation is just a function of the number of photons - i.e. a closed system with more photons has higher entropy than the same sized system with fewer photons, regardless of the individual energies of the photons)? Is the entropy related to the number of different energies of a group of photons (i.e. do ten photons with the same wavelength have lower entropy than ten photons with different wavelengths)?

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    $\begingroup$ Interesting question. I can't write a definite answer without research, so I'll add a comment. I don't think you can define the entropy of a single photon (or a single particle for that matter) because entropy is a property of an ensemble. In the case of photons entropy is related to the electronic properties of the material emitting the light (temperature of the electron gas) because photon themselves cannot interact to reach equilibrium. I'm sure there are more rigorous definitions and I hope others will comment. $\endgroup$ – boyfarrell Jun 25 '14 at 1:25
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If you know the precise state of a photon, then the system which contains only the photon has 0 entropy... there is no knowledge that can be gained about the system.

Typically the energy of a system is related to the entropy in a monotonically increasing fashion because in most systems (and a better condensed matter specialist would have more to say here) more energy leads to more possible configurations which the system could occupy. For instance, if a many body quantum system is in a non-degenerate ground state, then again the system has 0 entropy since the precise state of the system is known. However, when all that is known about the system is that it has some average energy $<E>$ above the ground state energy, then there are many states of system which could lead to this measurement.

So your question is not really well defined. But generally, no, higher energy does not imply higher entropy.

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I suggest, the entropy of the photon is a constant $3.72938*10^{-23}$ joule/kelvin.

As entropy is energy divided by temperature we need a temperature before we can estimate the entropy. Black body radiation does have temperature so if we estimate the average energy of a photon in black body radiation at a certain temperature and divide it with the temperature we got a measure of entropy. If that that value is independent of the temperature as it indeed is we got a value for the photon entropy.


The formula for Black body radiation:

$$P(\text{f}):=\frac{8 \pi f^3 h}{c^2 \left(e^{fh/k T}-1\right)}$$

P(f) = energy as a function of frequency in joule / hertz
T = temperature in kelvin
c = the speed of light = $2.998*10^{8}$ meter / second
k = the Boltzmann constant = $1.38064852*10^{-23}$ joule / kelvin
h = the Planck constant = $6.626*10^{-34}$ joule seconds


Now integrate to find the value:

$$S=\frac{\int_0^{\infty } P(f) \, df}{T \int_0^{\infty } \frac{P(f)}{f h} \, df}=\frac{\pi ^4 k}{30 z (3)}≈3.72938*10^{-23}$$

Where z is the Riemann zeta function.

A formidable integral but put into Mathematica, or something similar, it is easily calculated.

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    $\begingroup$ Entropy, in a sense, counts the degeneracy of the macroscopic state in terms of microstates. A better definition of entropy, using your notation, is $S=-\int_\infty^\infty P(f)\ln(P(f))df$.This is the entropy of ensemble of photons. If you have a single photon in thermal equilibrium then this isits entropy. $\endgroup$ – Alexander Jan 17 '19 at 22:27

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