I have read this:

An object without any internal degrees of freedom, like a single photon, can't really have a temperature. But an ensemble of photons can have a temperature. If you put an ensemble of photons in contact with a more familiar object with a well-defined temperature, such as considering the "blackbody" photons inside of an oven with only a very small opening, the photons will come to have a particular distribution of kinetic energies as they interact with the walls of the oven.

How do photons have temperature?

I have not found any experiment that would put the CMB photons into a container and measure how their kinetic energies will be distributed as they interact with the wall, since to do that, you would need a vacuum, that does not have any other particles (there is no perfect vacuum) just the CMB photons.

The only questions I have found somewhat relevant is this:

Before the cosmic neutrino background was formed (when the early universe was >1011 K) neutrinos and anti-neutrinos were produced and destroyed in thermal equilibrium with the rest of the radiation and baryonic matter. That is, the neutrinos had a distribution of energies and momenta that was determined by the temperature of the universe at that time. NB: This is not a blackbody distribution, it is the Fermi-Dirac distribution because neutrinos are spin 1/2 particles with mass.

The problem is, as this one says, this is for massive particles like neutrinos, and photons do not have rest mass.

So the temperature of the CMB reflects the kinetic energies of the particles in the plasma interacting at the time the radiation decoupled.

How can the Cosmic Neutrino Background (CνB) have a temperature? How can any neutrino have a 'temperature'?

And this one says specifically, that this has nothing to do with an actual measurement, but rather a theoretically calculated value of the kinetic energy of plasma particles at the time of decoupling a long time ago.

So the first one says it is possible to measure it with a container (that has a small opening), and the last one says we don't actually measure it.


  1. What do we mean when we say the CMB has a temperature and how do we measure it?

3 Answers 3


The first statement you've found is not suggesting that you need to trap the CMB photons in a box in order to make a measurement from the walls. Instead, it is attempting to explain how the notion of temperature makes sense for an ensemble of photons.

Clearly, a box can have some temperature in the sense which is familiar to us. If this box is held at that temperature and you open a tiny hole in it, the radiation you observe emanating will have a very specific spectrum. That spectrum is uniquely determined by the familiar temperature of the box. Because of that one-to-one correspondence, there is a meaningful sense in which we can speak of the temperature of a photon gas. It is in that sense that we speak of the temperature of the CMB. And that is indeed how we measure it - by measuring its spectrum, and fitting against the spectrum of a blackbody emitter.

The notion of "actual measurement" is, in my view, mostly a semantic one. Are temperature measurements made with a mercury thermometer not "actual measurements" because, strictly, they only measure the volume of some amount of substance? Using the energy spectrum of the CMB as a thermometer is really no less direct.


If you have an object at any temperature above absolute zero, that object will emit photons.

The more closely that object approaches that of a perfect blackbody, the more closely the spectrum of photons reflects a blackbody spectrum.

The neat thing is that for a given surface of a blackbody at temperature $T$, the spectrum is completely determined.

Therefore if we see a blackbody spectrum, we can calculate the temperature that that would produce it. It is common to use the same temperature to refer to both the object that produced the radiation and to the radiation itself.

When we examine the CMB, the spectrum is almost exactly that of blackbody radiation. Other than the tiny fluctuations, the radiation from the CMB is identical to the radiation from a blackbody at 2.7K.


If you take your own personal perfect blackbody, and place it somewhere in space far away from stars and galaxies, then most of the light it absorbs will come from the CMB. And it will radiate away light on its own. At some point, it radiates away just as much energy as it absorbs. It will be in thermal equilibrium with the CMB. Which means they have the same temperature. Whatever temperature your blackbody has at this equilibrium, that's the temperature of the CMB.

  • $\begingroup$ That's a really good explanation, thank you so much! $\endgroup$ Commented Jun 13 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.