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# When are temperature effects included in collisions?

1. ### geoduck

258
When a proton collides with a proton say at the LHC, is vacuum field theory used? It seems like you shouldn't have to include temperature effects since there are only two particles. But then again, all experiments take place at finite temperature, the ambient temperature of the room?

When a heavy ion collides with a heavy ion, I assume you have to use thermal field theory. But what if you collide them really slowly? Also, at high temperatures, symmetry of a scalar field can be restored, i.e., the Higgs vacuum expectation value can be zero again. Does this mean the particles in heavy ion collisions can be massless?

I guess I'm confused about ambient temperature versus collision temperature, how many particles are required to define a thermal system versus a vacuum system, and also about the Higgs field: if the temperature of one part of the universe is really really high, do particles in the vicinity lose mass?

2. ### mathman

6,490
When two particles collide their momenta determine the physics of the interaction. Ambient temperature is irrelevant.

### Staff: Mentor

Right.

In lead-lead collisions, the collision partners can produce a small, hot volume (with temperatures of the order of 100 MeV), but the environment is always irrelevant.

The LHC is far away from those energies.

4. ### ChrisVer

I am also having this kind of question, for example if someone wants to study the Big Bang Nucleosynthesis, he can find there that the width for the interaction:

$\nu n \rightarrow p e$
is given by:

$\Gamma = \int d \Pi \bar{\delta} |M_{\nu n \rightarrow p e}|^{2} f_{n} f_{\nu} (1-f_{p})(1-f_{e})$
where by $d \Pi$ I mean the product of each momentum phase space, $M$ is the interaction invariant matrix element, $\bar{\delta}$ the appropriate delta function to conserve 4momentum and $f_{i}$ the Fermi-Dirac distribution function.
Why in general don't we take the last into account in other interactions such as the pp collision?
If it's already been answered by someone above, I am sorry but I didn't "get" the answer.

### Staff: Mentor

Your equation here considers the rate of this process in thermal equilibrium, with many neutrinos and neutrons flying around.
Collisions in particle accelerators are far away from this equilibrium, and we are interested in the cross-section instead of the rate (the rate then just follows from geometry).