2

The Coulomb Gauge:

$\nabla \cdot A=0\\$

The Lorenz Gauge:

$ \nabla \cdot A= { \mu }_{ 0 }{ \epsilon }_{ 0 }\frac { \partial V }{ \partial t }$

Can both of these gauges be satisfied for some potential?

For example, the potentials:

$V(\vec { r } ,t)\quad =\quad 0\\ \vec { A } (\vec { r } ,t)\quad =\quad \begin{cases} \hat { j } { A }_{ 0 }cos(kx-\omega t)\quad ,\quad x>0 \\ \hat { j } { A }_{ 0 }cos(kx+\omega t)\quad ,\quad x<0 \end{cases}$

These potentials seem to satisfy both gauges. I'm unsure whether this is correct.

Qmechanic
  • 220,844
Rahul Chowdhury
  • 121

2 Answers2

1

A general potential cannot be transformed to satisfy both the Coulomb and Lorenz gauges. There are, however, an infinite number of potentials that can.

Any potential of the form (using the notation of the question)

$$\begin{align} V &= V(\vec{x}); \\ \vec{A} &= \vec{\nabla} \times \vec{F}, \end{align}$$

Where $\vec{F}$ is any vector field, satisfies both gauge conditions.

DavidH
  • 3,275
0

Only in certain simple situation can both gauges be satisfied. $V=0$ is obviously one of them. For a general electric and magnetic field configuration, however, only one gauge can be satisfied.

Walter
  • 1,185