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The word temperature usually refers to the average velocity of massive particles, correct?

And the Cosmic Microwave Background (CMB) has a 'temperature' based on the temperature of a 'black body' that would emit photons of energies corresponding to those seen in the CMB, correct?

But, how can a neutrino or neutrinos have a temperature? What does it correspond to?

Kurt Hikes
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    Have you read the answers to https://physics.stackexchange.com/q/267492/50583? Does thinking of the neutrinos as a neutrino gas help (Compare the question for e.g. hydrogen gas: "How can hydrogen have a temperature?")? – ACuriousMind Jul 10 '22 at 21:21
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    Neutrinos have mass. And they also have average velocity. – Kosm Jul 10 '22 at 21:26
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    Any system of a large number of things can have a temperature, as long as each of the things has an adjustable energy. The things need not be massive particles: they may be wave modes, for example. https://en.wikipedia.org/wiki/Equipartition_theorem – John Doty Jul 10 '22 at 21:42
  • But how can a non-massive particle that is not a photon have a temperature measured in a scale like Kelvin? Photons from a blackbody correspond to an actual, physical temperature... Are these blackbody neutrinos? Since they interacted more with other things prior to 1 second after the Big Bang? – Kurt Hikes Jul 11 '22 at 08:26

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The temperature of a gas is a parameter that reflects the distribution of energy/momentum of the particles. It is not a characteristic of any individual particle.

Before the cosmic neutrino background was formed (when the early universe was $>10^{11}$ 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.

As the universe expanded and cooled, the density fell, and at about 1 second after the big bang, the interaction timescale for the neutrinos became longer than the expansion timescale of the universe. The neutrinos "decoupled" from the other matter and radiation, but their distribution of momentum was preserved, with a characteristic temperature of a few $10^{10}$ K.

Since then, the universe has expanded by a factor of $\sim 10^{10}$ and the momentum of the individual neutrinos with respect to the comoving frame has decreased by a similar amount. (Even though the neutrinos have a small mass, you can think of the process as the expansion stretching their de Broglie wavelengths). Thus the neutrinos still have a momentum distribution, but it is now the equivalent of a much colder gas - about 2K.

ProfRob
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A single particle is not assigned a temperature (which is exhibited by a very large ensemble of particles), it is described by its kinetic energy. So a single neutrino can be described as having a certain amount of kinetic energy- but it in and of itself has no "temperature".

Now if we imagine instead a huge burst of neutrinos released during a supernova collapse inside a supermassive star, that burst will contain a range of kinetic energies which start out being all characteristic of the process which created them and then when the neutrinos interact with themselves and with the matter and radiation surrounding them in the core of that star, that distribution will get averaged into a blackbody distribution with a peak to which a temperature can be ascribed.

niels nielsen
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  • So, these are blackbody neutrinos, corresponding to the temperatures of the massive particles and photons and such that they interacted with just after the Big Bang? Before they stopped interacting (mostly) just one second after the Big Bang? – Kurt Hikes Jul 11 '22 at 08:28
  • @KurtHikes, see weinstein's book the first three minutes for a better explanation than I can provide. it's quite well-written. – niels nielsen Jul 11 '22 at 14:48
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What is the cosmic microwave background radiation "temperature"?

It is the value of temperature found when fitting a black body radiation curve to its spectrum of photons.

cmb

The perfect fit leads to the present cosmological model's hypothesis that it is the result of the loss of energy of the original spectrum from the decoupling of radiation from matter in the Big Bang expansion.

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

In an analogous analysis, the measured spectrum of cosmic neutrinos will have information about the temperature they were produced in the history of the universe. If interested further see this. .

So the temperature of a radiation refers to the black body fit temperature of the particles that produced it.

anna v
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    That agreement between theory and data is better than 1 part in 10,000!!!! I am truly amazed every time I see that graph. – niels nielsen Jul 11 '22 at 14:50