Coulomb Gauge and Lorentz Gauge 📂Electrodynamics

Coulomb Gauge and Lorentz Gauge

Overview¹

A relationship exists between the potential and the charge density, current density as follows.

$$ \begin{align*} \nabla ^2 V +\dfrac{\partial }{\partial t}(\nabla \cdot \mathbf{A}) &= -\frac{1}{\epsilon_{0}}\rho \\ \left( \nabla ^2 \mathbf{A}-\mu_{0}\epsilon_{0} \dfrac{\partial ^2 \mathbf{A} }{\partial t^2} \right) -\nabla\left( \nabla \cdot \mathbf{A} +\mu_{0}\epsilon_{0} \dfrac{\partial V}{\partial t}\right) &= -\mu_{0} \mathbf{J} \end{align*} $$

Depending on how assumptions about the potential are made, the expression changes.

Coulomb Gauge

As in magnetostatics, the divergence of the vector potential is made $0$.

$$ \nabla \cdot \mathbf{A}=0 $$

This allows for the representation of the equation regarding charge density solely in terms of scalar potential, thus resulting in the Poisson equation.

$$ \nabla^{2} V = -\frac{1}{\epsilon_{0}}\rho $$

The advantage is that it’s easier to calculate the scalar potential$V$, but the downside is that it’s harder to calculate the vector potential$\mathbf{A}$. The vector potential$\mathbf{A}$ can be calculated using the equation below.

$$ \nabla^{2} \mathbf{A} - \mu_{0} \epsilon_{0} \frac{\partial^{2} \mathbf{A}}{\partial t^{2}} = -\mu_{0} \mathbf{J} + \mu_{0} \epsilon_{0} \nabla \left( \frac{\partial V}{\partial t} \right) $$

Lorenz Gauge

The divergence of the vector potential $\mathbf{A}$ is set as follows.

$$ \nabla \cdot \mathbf{A} = -\mu_{0} \epsilon_{0} \frac{\partial V}{\partial t} $$

Then, the scalar potential$V$ and vector potential$\mathbf{A}$ are separated and expressed as equations of the same form.

$$ \nabla^{2} V - \mu_{0} \epsilon_{0} \frac{\partial^{2} V}{\partial t^{2}} = -\frac{1}{\epsilon_{0}} \rho $$

$$ \nabla^{2} \mathbf{A} - \mu_{0} \epsilon_{0} \frac{\partial^{2} \mathbf{A}}{\partial t^{2}} = -\mu_{0} \mathbf{J} $$

At this point, using the d’Alembertian makes it possible to represent it in a simpler form. The d’Alembertian is defined as follows.

$$ \Box^{2} := \nabla^{2} - \mu_{0} \epsilon_{0} \frac{\partial^{2}}{\partial t^{2}} $$

Using the d’Alembertian,

$$ \Box^{2} V = -\frac{1}{\epsilon_{0}}\rho $$

$$ \Box^{2} \mathbf{A} = -\mu_{0}\mathbf{J} $$