# Coulomb Gauge invariance, properties of Lambda

1. Apr 15, 2013

### tylerscott

1. The problem statement, all variables and given/known data
A gauge transformation is defined so as to leave the fields invariant. The gauge transformations are such that $\vec{A}=\vec{A'}+\nabla\Lambda$ and $\Phi=\Phi'-\frac{\partial\Lambda}{\partial t}$. Consider the Coulomb Gauge $\nabla\cdot\vec{A}=0$. Find out what properties the function $\Lambda$ must satisfy in order for the Coulomb Gauge to be satisfied.

2. Relevant equations
Lorentz Gauge Condition
$\nabla\vec{A}+\frac{1}{c^{2}}\frac{\partial\Phi}{\partial t}=0$

3. The attempt at a solution
$\nabla\cdot\vec{A}=\nabla\left(\vec{A'}+\nabla\Lambda\right)=0$
$\nabla\cdot\vec{A'}+\nabla^{2}\Lambda=0$
so
$\nabla\cdot\vec{A'}+\nabla^{2}\Lambda+\frac{1}{c^{2}}\frac{\partial }{\partial t}(\Phi'-\frac{\partial\Lambda}{\partial t})=0$
and
$\nabla\cdot\vec{A'}+\frac{1}{c^{2}}\frac{\partial\Phi'}{\partial t}=\frac{1}{c^{2}}\frac{\partial^{2}\Lambda}{\partial t^{2}}-\nabla^{2}\Lambda$

So in order for these to be invariant, $\frac{1}{c^{2}}\frac{\partial^{2}\Lambda}{\partial t^{2}}-\nabla^{2}\Lambda=0$

So $\Lambda$ must satisfy wave equation.

Now, I feel like I'm missing some of the properties of the gauge, so any help into more insight into $\Lambda$ would be appreciated.

2. Apr 15, 2013

### BruceW

You shouldn't be solving for the Lorentz gauge. The question only mentions the Coulomb gauge. That is the one you need to satisfy. The question says:
$$\vec{A}=\vec{A'}+\nabla\Lambda$$
So from here, if you want both A and A' to satisfy the Coulomb gauge, then what does that tell you about lambda? (I'll admit, the question wasn't very specific about what it was asking for).

3. Apr 15, 2013

### tylerscott

I don't think I understand what that tells me about lambda... More insight please? Haha

4. Apr 15, 2013

### BruceW

If a general vector V is going to satisfy the Coulomb gauge, then you need:
$$\nabla \cdot \vec{V} = 0$$
right, and the question is saying that it wants both A and A' to satisfy the Coulomb gauge. Also they give you the equation:
$$\vec{A}=\vec{A'}+\nabla\Lambda$$
So using the given information, can you get an equation that contains only lambda (not A or A').