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This page says cosmic photons have an energy of 2.78 kelvin, so how much energy would the particles that show up in a bubble chamber have?

Cees Timmerman
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The cosmic-ray tag wiki already explains this:

Usually referring to stable high energy elementary particles - cosmic rays were found to arrive the Earth from galactic and extragalactic sources and interact with the atmosphere. The energies of this particles are one of their prominent features and extend over multiple orders of magnitude, at least up to $10^{20}eV$.

So going by this table, that's widely about 1.0E20 / 1E9 = 1.0E11 GeV, so 1.0E11 * 1.16E13 = $1.16^{24}K$. Enough to boil 5E-7 liters of 25 degree Celsius water.

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Cees Timmerman
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  • Of course, a single particle does not have a temperature. The cosmic microwave background is a thermal distribution of many photons with a temperature of 2.78K. – Jon Custer Aug 23 '16 at 22:14
  • What about a basketball-sized amount of them, all heading this way with the same energy? – Cees Timmerman Aug 23 '16 at 22:20
  • Temperature is a statistical measurement. All particles having the same energy is not a thermal distribution, so you cannot define a temperature. They have energy, sure, but not a temperature. – Jon Custer Aug 23 '16 at 22:24
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    But this is not a typical "particle that shows up in a bubble chamber". By many orders of magnitude. – dmckee --- ex-moderator kitten Aug 24 '16 at 14:02
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    There is no reason you can't use Kelvin as a energy scale (the size of the unit is just set by Boltzmann's constant), but it is rarely done unless there is a system that can reasonable be described as having a temperature involved somehow. Particle energies are not generally described that way, unless you are talking about thermal neutrons (which average about 1/40 eV of kinetic energy at room temperature). – dmckee --- ex-moderator kitten Aug 24 '16 at 14:06