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

The Lagrangian for electromagnetic field has the following expression: $$ L = -\frac{1}{c^{2}}A_{\alpha}j^{\alpha} - \frac{1}{8 \pi c}(\partial_{\alpha} A_{\beta})(\partial^{\alpha}A^{\beta}) $$

(I used Lorentz calibration $\partial_{\alpha} A^{\alpha} = 0 $).

If I add the summand $\frac{\mu^{2}}{8 \pi c}A_{\alpha}A^{\alpha}$, I'll get an equations for field (which characterized by some 4-vector $A^{\alpha}$ (not electromagnetic (!!!))) of strong interaction and (for static case) the expression for Yukawa potential. So what is the physical meaning of summand written above?

This term is somehow characteristic of the mass of the interaction carriers, but I don't understand the physical meaning of $A_{\alpha}A^{\alpha}$.

share|cite|improve this question
up vote 1 down vote accepted

If you take your Lagrangian, including the $A^\alpha A_\alpha$ and vary it with respect to $A^\alpha$, you will get the classical equation of motion:

$\partial_\beta \partial^\beta A^\alpha + \mu^2 A^\alpha = 0 $.

If you use a plane wave as a trial solution for this: $A^\alpha = e^{i p\cdot x} \epsilon^\alpha $ where the $\epsilon^\alpha$'s are polarization vectors that obey your gauge (calibration) condition, you will get:

$(p^2 -\mu^2)e^{i p\cdot x} \epsilon^\alpha= 0 $.

Which enforces:

$(p^2 -\mu^2)= 0 $.

Expanding out the four momentum we get:

$E^2-|\vec{p}|^2-\mu^2 = 0 $.

After rearranging we get:

$E^2 =|\vec{p}|^2+\mu^2 $

which is the dispersion relation for a relativistic particle of mass $\mu$.

share|cite|improve this answer

Your Answer


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