Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

It's been claimed that the LHC's 14 TeV energy produces temperatures comparable to that which occurred very soon after the Big Bang. The well-known $E=1.5kT$ formula from classical statistical mechanics predicts LHC produced temperatures in the $2.44\times 10^{17}$ K range. This temperature is much higher than I expected.

Is there an agreed upon method of calculating temperature equivalents of 14 TeV p-p interactions, and if so, what is the result?

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

Note that the energy is per degree of freedom, so you don't use all the 14 TeV (nor are they actually running at that energy yet).

So, how many degrees of freedom? Good question.

Each proton has three valence quarks, but these are generally agreed to make up a small portion of the mass (and to carry a small portion of the momentum) of the proton. The rest of the mass (or momentum) is carried by particles (quarks and gluons mostly) from the so-called "sea"; these pop into and out-of existence owing to the uncertainty principle.

Worse, when the collision happens there is a great deal of energy available to put the virtual particles of the sea on to (or nearly on to) the mass shell, converting them into real particles, each equipped with their own swarm of ghostly hangers on. Then many of these decay in a very short time.

It is the average energy of this multitude of degrees of freedom which you are trying to measure/calculate, and it is non-trivial.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.