# I Lorenz Gauge

1. Apr 23, 2017

### Silviu

Can someone explain to me (or point me towards a source) how is the Lorenz Gauge derived? I am reading the Griffiths book and from what I understand we can do the transformation $A' = A + \nabla \lambda$ and at the same time $V' = V - \frac{\partial \lambda}{\partial t}$ and B and E remain unchanged. This is pretty obvious for the Coulomb gauge (in finding $\lambda$ to satisfy the condition). But for the Lorenz gauge if you try to find $\lambda$ from the first equation, it will be a function of V and when you plug it in $V' = V - \frac{\partial \lambda}{\partial t}$, V' will be different from V so the V you used in $\nabla A = -\mu_0 \epsilon_0 \frac{\partial V}{\partial t}$ is not anymore the same? So how do you prove it?

2. Apr 23, 2017

### Twigg

Suppose $\vec{A}$ and $V$ are the potentials in the Coulomb gauge, and that you have already solved them from the relevant form of Maxwell's equations. Now, you want to prove that you can choose a value of $\lambda$ in the gauge transformation formulas you gave such that the resulting $\vec{A}'$ and $V'$ satisfy the Lorenz gauge condition: $div(\vec{A}') + \frac{1}{c^{2}} \frac{\partial V'}{\partial t} = 0$. If you substitute in the gauge transformation formulas and the Coulomb gauge condition, $div(\vec{A}) = 0$, then you will find $\Box \lambda = -\frac{1}{c^{2}} \frac{\partial V'}{\partial t}$ if my algebra isn't messed up, where the $\Box$ is the wave equation operator.