# Lorenz and coulumb gauge

1. May 30, 2013

### rida

i am searching the physical significance of lorenz and coloumb gauge but cant find anything besides that coloumb gauge is used for finding time independent potentials while lorenz is for finding time dependent potentials. and i'll be really grateful if u give me applications of then in any field.

2. May 30, 2013

### WannabeNewton

Hi rida! Gauge freedom in the classical Maxwell field equations is not a physical degree of freedom in the sense that you are not doing anything physical by fixing a gauge; they are just choices of functions that allow you to put the field equations in some convenient form for the problem at hand while still yielding the same physical electromagnetic field.

3. May 31, 2013

### rida

but my teacher has asked for significance and applications, and from your point there wont be any of them.
moreover my teacher has told us that lorenz gauge can give us pure scalar and vector potentials. but coloumb gauge can not. is it true? and what is the logic behind it

4. Jun 1, 2013

### vanhees71

I don't understand exactly the problem given by your teacher.

The electromagnetic field is given by $\vec{E}$ and $\vec{B}$, and these are the physically fields which give the force acting on charges according to Lorentz's Law, and that's how they are operationally defined (in the limit of negligible test charges).

For the fully time-dependent electromagnetic field you need a scalar and a vector potential. These are introduced due to the homogeneous Maxwell equations, i.e.,
$$\frac{1}{c} \partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{B}=0.$$
From these equations you draw the conclusion that there are a scalar field $\Phi$ and a vector field $\vec{A}$ such that
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
For a given electromagnetic field $(\vec{E},\vec{B})$, the potentials are not uniquely defined since obviously if $(\Phi,\vec{A})$ are a set of potentials that give the em. field, also
$$\Phi'=\Phi+\frac{1}{c} \partial_{\chi}, \quad \vec{A}'=\vec{A} - \nabla \chi$$
give the same em. field for an arbitrary scalar field $\chi$. Thus you can give one scalar constraint to simplify your task to solve for $\Phi$ and $\vec{A}$ from the remaining inhomogeneous Maxwell equations,
$$\vec{\nabla} \cdot \vec{E}=\rho, \quad \vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\vec{j}.$$
The first inhomogeneous equation (Gauß's Law), written in terms of the potential reads
$$-\left (\frac{1}{c} \partial_t \vec{\nabla} \cdot \vec{A} + \Delta \Phi \right )=\rho.$$
If you choose the Lorenz-gauge condition
$$\frac{1}{c} \partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0,$$
then you have
$$\left (\frac{1}{c^2} \partial_t^2 - \Delta \right ) \Phi:=\Box \Phi=\rho.$$
For the Ampere-Maxwell Law you get in terms of the potentials
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A}) + \frac{1}{c} \left (\frac{1}{c} \partial_t^2 \vec{A}+\nabla \partial_t \Phi \right)=\vec{j}.$$
If you use the Lorenz-gauge condition again in
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}=-\frac{1}{c} \nabla \partial_t \Phi - \Delta \vec{A},$$
you get
$$\Box \vec{A}=\vec{j}.$$
As you see, the Lorenz gauge has the advantage that the equations for the scalar potential and the three components of the vector field decouple to four wave equations, and that's why this is a convenient gauge to describe the radiation from given charge and current distributions.

You can of course use any other gauge to solve for the inhomogeneous equations. For some tasks other gauges can be more convenient. The physical outcome, i.e., the em. field is of course always the same, independent of the gauge condition chosen.