Strong interaction and the Lagrangian for electromagnetic interaction

Question

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 $A_{\alpha}A^{\alpha}$.

Answer 1

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 $\mu$.