Coulomb gauge

Consider a gauge in which we have $\vec \nabla \cdot \vec A(\vec r, t) = 0$. In this gauge the equation for the scalar potential becomes

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

which is same as the equation for the electric potential in electrostatics even if the charge density as well as the potential may be time dependent. The solution is

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

Thus this gauge is called Coulomb gauge.

Above we have seen that the scalar potential is same as the electric potential and it is instantaneous. That is, the potential at time t is given by the charge distribution at the same time t. This is an apparant violation of relativity. We shall show later that this apparant inconsistancy with the relativity theory can be resolved.

In Coulomb gauge, the equation for the vector potential becomes

$$\nabla^2 \vec A(\vec r,t) - \frac{\kappa \mu}{c^2} \frac{\p... ...mu}{c} \frac{\partial}{\partial t} \vec \nabla \phi(\vec r,t).$$

So, the equation is not decoupled and the scalar potential appears as a source term for the vector potential. We can however show that, using the continuity equation for charge density, this term can be written as

$$\frac{\kappa \mu}{c} \frac{\partial}{\partial t} \vec \nabla \phi(\vec r,t) = \frac{\kappa \mu}{c} \vec \nabla \int d^3r' \frac{\partial}{\partial t} \rho(\vec r',t) \frac{1}{\vert\vec r - \vec r'\vert}$$

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

$$= \kappa \vec j_l(\vec r. t)$$ (166)

where $\vec j_l$ is the longitudinal component of the current¹⁵. This can be seen from the fact that

$$\vec \nabla \times \vec j_l(\vec r,t) = \vec \nabla \times \v... ...ot \vec j(\vec r', t) \frac{1}{\vert\vec r - \vec r'\vert} = 0.$$

Thus, the source term is just the transverse component of the current and the equation satisfied by the vector potential is

$$\Big ( \nabla^2 - \frac{\kappa \mu}{c^2} \frac{\partial ^2}{\partial t ^2} \Big ) \vec A(\vec r,t) = - \frac{4 \pi \mu}{c} \vec j_{tr}(\vec r, t)$$ (167)

where $\vec j_{tr}$ is the transverse part of the current which satisfies $\vec \nabla \cdot \vec J_{tr} = 0$.

We therefore find that both the vector potential and its source, the electric current are transverse in Coulomb gauge. Another thing to note is that the scalar potential for localised charge distributions vanish faster than $r^{-1}$ as the distance between the charges and the point of observation goes to infinity. The vector potential, on the other hand, is a solution of wave equation and therefore, as we shall see later, it vanishes as $r^{-1}$ at large distances from the sources.