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I can extract the temperaure of the CMB by its width, the same as any other black body radiation.
I guess that, physically, that temperature is the temperature of the object that is emitting the radiation?

Say now I have a laser, monochromatic radiation at $\omega_0$ with a linediwth of $\delta \omega$.
Can I extract a temperature from this spread? What would its physical significance be?

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    $\begingroup$ The distribution of photons in a laser is not a blackbody distribution, while the CMB is. $\endgroup$ – probably_someone Feb 7 '18 at 18:23
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    $\begingroup$ @probably_someone That was noted. In any event, it should be assignable a temperature. $S = k_b \ln \Omega$ and $T = \partial S / \partial E$ $\endgroup$ – cms Feb 7 '18 at 19:48
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The thermodynamic concept of the physical quantity temperature requires that the system you are considering is in equilibrium, or at least approximately in equilibrium, as in the assumed local equilibrium of non-equilibrium thermodynamics. This is not the case for a laser beam with a narrow frequency band around a frequency $\omega_0$, which definitely doesn't have a blackbody energy distribution following Bose-Einstein statistics for a given temperature like the cosmic microwave background radiation.

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  • $\begingroup$ Why isn’t the laser in equilibrium? $\endgroup$ – SuperCiocia Feb 21 '18 at 0:34
  • $\begingroup$ @SuperCiocia- A laser doesn't have a blackbody (or Bose Einstein) distribution of photon energies. From the spread of a laser line. e.g. of a gas laser, you could perhaps get some information about the temperature of the emitting atoms/molecules due to thermal (Doppler) spread of the emission lines. However, this doesn't mean that this is the temperature of the laser light. The temperature of a laser light cannot be defined, much like you cannot define the temperature of a quasi-monoenergetic particle beam. $\endgroup$ – freecharly Feb 21 '18 at 0:49

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