Green function for electromagnetic field

If we work in Lorentz gauge, the equations satisfied by the electromagnetic potentials are

$$\Big ( \nabla^2 - \frac{\kappa \mu}{c^2} \frac{\partial^2}{\partial t^2} \Big ) \phi(\vec r, t) = - \frac{4 \pi}{\kappa} \rho(\vec r,t)$$ (180)

$$\Big ( \nabla^2 - \frac{\kappa \mu}{c^2} \frac{\partial^2}{\partial t^2} \Big ) \vec A(\vec r, t) = - \frac{ 4 \pi \mu}{c} \vec j(\vec r, t).$$ (181)

These are essentially four equations, three for the three components of the vector potential and one for the scalar potential. Using the retarded Green function obtained for massless scalar field which was discussed in the preceeding subsection, we can write down the solutions for vector and scalar potentials as

$$\vec A(\vec r, t) = \int d^3r' dt' G_r(\vert\vec r - \vec r'\vert, t-t') \vec j(\vec r', t')$$

$$= \int d^3 r' \frac{1}{\vert\vec r - \vec r'\vert} \vec j(\vec r' , t')$$ (182)

and

$$\phi(\vec r, t) = \int d^3r' dt' G_r(\vert\vec r - \vec r'\vert, t-t') \rho(\vec r', t')$$

$$= \int d^3 r' \frac{1}{\vert\vec r - \vec r'\vert} \rho(\vec r' , t')$$ (183)

However, strictly speaking, the four solutions are not independent. One needs to ensure that the potentials should satisfy the gauge condition $\vec \nabla \cdot \vec A(\vec r, t) + \frac{\partial}{\partial t} \phi(\vec r, t) = 0$. One can show that the Gauge condition is indeed satisfied by the solution since the charge and current density satisfies the continuity equation, $\vec \nabla \cdot \vec j(\vec r, t) + \frac{\partial}{\partial t} \rho(\vec r, t) = 0$.

If we work in Coulomb gauge, the equations satisfied by the vector and scalar fields are

$$\nabla^2 \phi(\vec r, t) = - \frac{4 \pi}{\kappa} \rho(\vec r,t)$$ (184)

$$\Big ( \nabla^2 - \frac{\kappa \mu}{c^2} \frac{\partial^2}{\partial t^2} \Big ) \vec A(\vec r, t) = - \frac{ 4 \pi \mu}{c} \vec j_{tr}(\vec r, t).$$ (185)

where $\vec j_{tr}(\vec r, t)$ is the transverse component of the current ( $\vec \nabla \cdot \vec j_{tr} (\vec r, t) = 0$ ). The scalar potential satisfies ( static ) Laplace's equation and solutions of this equation has been discussed in details in electrostatics section. For the vector potential, we use the transverse Green function $G_{tr}(\vec r - \vec r', t - t')$ which satisfies the equation

$$\Big ( \nabla^2 - \frac{\kappa \mu}{c^2} \frac{\partial^2}{\parti... ...t') = - \frac{4 \pi \mu}{c} \delta_{tr}^{i,j} (\vec r - \vec r') \delta(t - t')$$ (186)

where $\delta_{tr}^{i,j}(\vec r - \vec r')$ is the transverse diadic $delta$-function ¹⁷ which satisfies the equation $\sum_i \frac{\partial}{\partial r_i} \delta_{tr}^{i,j} \delta(\vec r - \vec r') = 0$.

When there are no free charges $\rho(\vec r, t)=0$, the electromagnetic potentials in Lorentz gauge also satisfy the Coulomb gauge condition $\vec \nabla \cdot \vec A(\vec r, t) = 0$ since the scalar potential vanishes. In that case, we can solve the problem in Lorentz gauge, which is simpler to do, and the solution would still be satisfying the Coulomb gauge condition.