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By using Lagrangian for Yukawa interaction,

$$ L = -\frac{1}{c}A_{\alpha}j^{\alpha} + \frac{1}{8 \pi c}(\partial_{\alpha}A_{\beta})(\partial^{\alpha}A^{\beta}) + \frac{m^{2}c}{\hbar^{2}}A_{\alpha}A^{\alpha} $$

we can get equation of field dynamics,

$$ 4 \pi j_{\beta} = \frac{\mu^{2}c^{2}}{\hbar^{2}}A_{\beta} + \partial^{2}A_{\beta}, \quad j_{\beta} = \frac{v_{\beta}}{c} Q \delta (\mathbf r - \mathbf r_{0}(t)). $$

How to solve it (define $A_{\beta}$)?

I tried solve equation by using Fourier transform $$ f(r^{\alpha}) -> f(w^{\alpha}) = \int e^{-ir_{\alpha}w^{\alpha}}d^{4}x, $$ and got $$ w_{\alpha}w^{\alpha}A_{\mu}(w_{\alpha}) + \frac{\mu^{2}c^{2}}{\hbar^{2}}A_{\mu}(w_{\alpha}) = 4 \pi \int \limits_{-\infty}^{\infty} e^{-ictw^{0} + i(\mathbf r_{0}(t) \cdot \mathbf w)}dt $$.

I cannot take the integral from the right part, because I don't know how the particle moves (i.e. the dependence $\mathbf r = \mathbf r_{0}(t)$). Can you help me?

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Probably Green function for the homogeneous equation is the way to go, then plug in your $j^{\alpha}$ – Jorge Mar 28 '13 at 15:24
That looks like the lagrangian for a massive vector boson (except the kinetic term isn't antisymmetrized for some reason) to me. The yukawa interaction is between a scalar particle and a fermion. – Jerry Schirmer Mar 28 '13 at 15:52
For the static case the equation has the solution $$ \varphi = Q\frac{e^{-\frac{\mu c r}{\hbar}}}{r}. $$ – user8817 Mar 28 '13 at 16:09

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