0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Commented Dec 30, 2021 at 16:26
  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/686047/2451 $\endgroup$
    – Qmechanic
    Commented Dec 31, 2021 at 10:07

1 Answer 1

1
$\begingroup$

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".

$\endgroup$
3

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.