# Lorenz Gauge different definitions

For the lorenz gauge we can either write:

$$\nabla \vec A(\vec r,t)+\frac{1}{c^2}\frac{\partial \phi(\vec r,t)}{\partial t}=0$$

If we also consider the following invariant transformations:

$$\vec A(\vec r,t)'= \vec A(\vec r,t) + \nabla f(\vec r,t)$$

$$\phi(\vec r,t)'=\phi(\vec r,t) - \frac{\partial f(\vec r,t)}{\partial t}.$$

Then we can have:

$$\nabla \vec A(\vec r,t)'+\frac{1}{c^2}\frac{\partial \phi(\vec r,t)'}{\partial t}=g(\vec r,t)$$

And from here by substituting the transformations in this final equation we find:

$$\square f(\vec r,t)= g(\vec r,t)$$

But we can also assume the following:

$$\nabla \vec A(\vec r,t)'+\frac{1}{c^2}\frac{\partial \phi(\vec r,t)'}{\partial t}=0$$

And from here we find out that:

$$\square f(\vec r,t)= -(\nabla \vec A(\vec r,t) +\frac{1}{c^2}\frac{\partial \phi(\vec r,t)}{\partial t}$$

So from this you can see that the condition $$\nabla \vec A(\vec r,t)+\frac{1}{c^2}\frac{\partial \phi(\vec r,t)}{\partial t}=0$$ sometimes is applied on the non-transformed potential $$\vec A(\vec r,t)$$ and sometimes is applied in the transformed potential $$\vec A(\vec r,t)'$$

So, when I need to solve a random exercise, which case should I consider?

• I think it would depend on the exercise you are facing. I would recommend to use the one that makes calculus easier. This should be the utility of gauge invariance. Dec 30, 2021 at 16:26
• Related post by OP: physics.stackexchange.com/q/686047/2451 Dec 31, 2021 at 10:07

Once you've chosen the gauge and labeled it $$A_{lorenz},\phi_{lorenz}$$, other set of $$A,\phi$$ or $$A',\phi'$$ or whichever name they have, as long as they are related to $$A_{lorenz},\phi_{lorenz}$$ by an nontrivial $$f$$, are not in Lorenz Gauge.
Just choose some label for $$A,\phi$$, and call that label "Lorenz Label", and stick to it. Every other set of $$A,\phi$$ with different labels are "non-Lorenz".