Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Peskin and Schroeder's chapter 12 about the renormalization group, it is stated that the parameter $\rho_m=m^2/M^2$, where $m$ is the mass and $M$ is the renormalization scale, is proportional to $T-T_c$. But this assertion is not derived in any detail, except for a handwaving one-line explanation "$\rho_m$ is just the parameter that one adjust finely to bring the system to the critical temperature". Could someone elaborate this? In particular, why doesn't the parameter depend on the square or some other power of $T-T_c$?

share|cite|improve this question

with some extra background knowledge, it is actually not a real handwaving but the full answer to your question.

You have omitted the previous sentence which says "In the statistical interpretation of scalar field theory, $\rho_m$ is just the parameter one must adjust finely to bring the system to the critical temperature."

At this moment, the reader is probably supposed to know Landau's phenomenological theory of the second-order phase transitions:

At any rate, there is also the fourth-order term (in the order parameter $\Psi$ which is interpreted as the scalar field $\phi$ in QFT) but the coefficient $r$ of the quadratic term - which gets interpreted as the mass term in QFT - is proportional to $(T-T_c)$.

Why is it so? Because the action, or free energy, may be viewed as a result of a calculation at temperature $T$. The coefficients in the action - or free energy will be smooth around the critical temperature because nothing special happens at the level of the "action" at this point. So $dr/dT$ is finite. The phase transition - and all the complicated and sometimes fractional powers around the critical point - only arise once you try to find solutions to the system defined by the action - or free energy. The "phase" - whether the spins like to create domains etc. - is not "directly" encoded in the coefficients; the theory is always the same. Its solutions may be qualitatively different for different values of parameters.

That's what happens generically. Of course that you could try to imagine a system in which $dr/dT$ is either zero or infinite (different power laws) near the critical temperature $T_C$. But you would need some extra fine-tuning of the material or theory to obtain $dr/dT=0$, and even if you succeeded, such a behavior would be a higher-order phase transition. I don't think it's possible to have $dr/dT$ being infinite.

Cheers LM

share|cite|improve this answer

Your Answer


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