Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 1 / 35C / iii

Use signature $(-,+,+,+)$, for which $\square =-{\partial }_t^2+\Delta$. With the Lorenz constraint in Proca theory, the Proca equation is $(\square -m^2)A^a=-{\mu }_0{J}^a$. A static solution therefore satisfies $(\Delta -m^2)A^a=-{\mu }_0{J}^a$.

For $R>0$ direct radial differentiation gives $(\Delta -m^2)(e^{-mR}/R)=0$. At the origin its $1/R$ singularity contributes $-4\pi {\delta }_0$: integrating the Laplacian over a small sphere gives limiting outward flux $-4\pi$, while the mass term's integral tends to zero. Thus $G_m(R)=e^{-mR}/(4\pi R)$ obeys $(\Delta -m^2)G_m=-{\delta }_0$. Convolution proves the printed Yukawa potential solutionThe integral requires suitable source decay, for example compact support. For conserved static sources, differentiating the convolution and integrating by parts gives ${\partial }_a{A}^a=0$ as well; its spatial divergence is the convolution with ${\partial }_i{J}^i=0$.

For the usual nonnegative mass parameter this is the decaying static solution, and $m=0$ gives the Coulomb kernel with Lorenz gauge imposed. If the arbitrary real parameter is negative, the printed kernel still gives a particular solution for compact sources, but grows at infinity; the decaying boundary condition selects $e^{-|m|R}/(4\pi R)$ instead. Static sources also allow homogeneous time-dependent solutions, so this construction establishes a static solution, not that every solution is static.