12
$\begingroup$

According to the standard model, the electroweak symmetry is unbroken at high temperatures, and therefore all gauge bosons are massless then. But since fermions are said to acquire mass by a different mechanism, I'm wondering if they too become massless.

So are fermions massless at high temperatures?

$\endgroup$
6
$\begingroup$

Yes, your expectations seem reasonable when thinking of the Higgs mechanism induced mass, as explained by @nmoy.

However, note that one needs to be careful in defining what one means by "mass" at high temperatures. A theory at finite (non-zero) temperature breaks Lorentz invariance. There are multiple ways to think of this:

  1. There is a preferred frame, where the heat bath is at rest.
  2. In the Matsubara formalism, one Wick-rotates and compactifies the time direction into a circle of circumference $\beta = 1/T$. This explicitly breaks the equivalence between space and time directions.

Without Lorentz invariance, one needs to be careful about the conserved quantities being considered. Roughly, one can imagine that at high temperatures, the "time/thermal circle" becomes small and we reduce to an effective 3-dimensional theory, a la Kaluza-Klein. This theory will have rotational invariance in 3d (rather than 3+1) and we could talk about the 3d effective-masses of the reduced fields (an infinite number of them). In such a case, all the fermion fields in that theory will have a large mass parametrically proportional to the temperature ($m_{\text{3d, fermions}} \sim n \pi T$).

So, even though they get no mass from the Higgs mechanism, those fields seem to have an effective mass simply due to their interaction (thermalization) with the heat bath.

$\endgroup$
  • $\begingroup$ Nice argument! Can you give some references for these stuff? $\endgroup$ – Rexcirus Dec 24 '14 at 14:00
  • $\begingroup$ Take a look at physics.stackexchange.com/questions/131197/… $\endgroup$ – Siva Dec 24 '14 at 20:04
  • $\begingroup$ For comments on finite temperature field theory in general, there are tons of reviews or lecture notes or even a few textbooks. I think some of them should mention this point paranthetically. $\endgroup$ – Siva Dec 24 '14 at 20:13
5
$\begingroup$

In the Standard Model, fermions are given their mass through yukawa terms, described by the Lagrangian: $$\mathcal{L}_{\mathrm{F}} = \overline{\psi} \gamma^{\mu} D_{\mu} \psi + y_{\psi} \overline{\psi} \phi \psi$$

Where $y_\psi$ is the yukawa coupling and $\phi$ is the Higgs field. At this stage, much like the gauge bosons, the fermions do not yet have mass. This is due to the fact that a mass term for fermions breaks gauge invariance and can only be generated spontaneously by symmetry breaking. Essentially, the higgs field $\phi$ takes on a non-zero vacuum expectation value of $\langle\phi\rangle = v$ (meaning its ground state energy is not zero) which generates a mass term of the form: $$y_\psi v~\bar{\psi}\psi = m\bar{\psi}\psi$$ As the temperature increases beyond the electroweak symmetry breaking threshold (say, in the very early universe for example) this vacuum expectation value of the Higgs field approaches zero, meaning the mass of the fermion approaches zero. This means that before electroweak symmetry breaking, there may have existed massless fermions. However, this does assume that the Standard model is still valid at these high energies, something that is not certain.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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