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,
$$ \Box^{2} V = -\frac{1}{\epsilon_{0}}\rho $$
$$ \Box^{2} \mathbf{A} = -\mu_{0}\mathbf{J} $$
See Also
David J. Griffiths, 기초전자기학(Introduction to Electrodynamics, 김진승 역) (4th Edition1 2014), p476-478 ↩︎