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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?

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  • $\begingroup$ 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. $\endgroup$ Dec 30, 2021 at 16:26
  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/686047/2451 $\endgroup$
    – Qmechanic
    Dec 31, 2021 at 10:07

1 Answer 1

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The Lorenz Gauge is simply an assumption you make about a specific reference frame. You could choose, as you like, whichever reference frame to have Lorenz Gauge.

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".

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