# Remaining Gauge Freedom in Lorenz gauge

If one fixes the gauge in Electrodynamics to fulfill the Lorenz gauge $\partial_\mu A^\mu=0$, then the gauge scalar field $\chi$ has to fulfill (eq 3 page 5): $$\partial^\mu\partial_\mu \chi=-\partial_\mu A^\mu~~~~~~~~~~~~~~(1)$$

If the 4 potential transforms as $A^\mu \longrightarrow A^\mu+\partial^\mu \chi$.

I think I got that point. There exists a solution to (1) for $\chi$, since it is the solution for an inhomogeneous wave equation with well behaved source fields. Is that correct?

I have read that the gauge is not fully fixed by the Lorenz gauge. Does that mean that (1) does not determine $\chi$ completely? If yes what is the fully left freedom? And where does this condition $$\partial^\mu\partial_\mu \chi=0~~~~~~~~~~~~~~(2)$$

comes into play, which can be found here (on page 6). Right now it seems to me that $\chi$ then has to fulfill (1) and (2) which would be a bit weird..?

So my clear question is:

• How is it possible that a gauge freedom of the type (2) is still left by demanding (1)?
• And if this should be wrong and one only demands (1) what is the left gauge freedom then?
• – Qmechanic Jan 29 '17 at 16:46
• thanks for commenting :) I read that post before but I did not get the answer.. Basically my question is how to see that (2) is arising from the left gauge freedom.. It does not make sense to me that $\chi$ has to fulfill (1) and (2).. – Mr Puh Jan 29 '17 at 16:49

Lorenz gauge reads: $$\partial_\mu A^\mu = 0.$$

In slide 5 what they want to say is that in general $A_\mu$ does not satisfy Lorenz gauge fixing, but you can always choose a $\chi$ such that $A'_\mu = A_\mu + \partial_\mu \chi$ does: $$0 = \partial_\mu A'^\mu = \partial_\mu A^\mu + \partial_\mu\partial^\mu \chi \qquad (1)$$ is indeed an equation for $\chi$ given some $A^\mu$.

This condition does not uniquely determine the function $\chi$. Consider some function $\xi$ such that $\partial_\mu\partial^\mu \xi =0$. A gauge transformation generated by $\xi$ does not change the 4-divergence of $A_\mu$: $$\partial^\mu A'_\mu = \partial^\mu (A_\mu + \partial_\mu \xi) = \partial^\mu A_\mu$$ This is called a 'residual gauge freedom'.

So, if you start with a generic potential $A_\mu$, you can perform a gauge transformation so that it satisfies Lorenz gauge by solving (1) and finding $\chi$; but this is just up to a function $\xi$: $$\partial_\mu A'^\mu = \partial_\mu ( A^\mu + \partial^\mu \chi + \partial^\mu \xi) = \partial_\mu ( A^\mu + \partial^\mu \chi ) + \partial^\mu \partial_\mu\xi = \partial^\mu \partial_\mu \xi = 0$$ In other words if you pick a solution $\chi$ of (1) and you add a function $\xi$, the sum will still solve (1).

Indeed in those slides they are not saying that the gauge transformation function satisfies both equations. They are considering two gauge transformations; the first fixing the divergence as in my (1), the second leaving that equation invariant.

• I got it thanks! So $A^\mu$ still has 'another' gauge freedom left, which is now restricted to be a gradient and fulfill $\partial_\mu \partial^\mu \xi$. And such a $\xi$ exists since it is just the solution to a normal wave equation. Do you know why the Lorentz condition does not fix the gauge completely (I mean one could impose further conditions)? Has this leftover freedom any physical interpretation? – Mr Puh Jan 29 '17 at 17:28
• @Mr Puh: Note that $A^\mu$ itself isn't restricted to be a gradient and satisfy the wave equation. We shift the original $A^\mu$ by a gradient to make it divergenceless: $\partial_\mu A^\mu = 0$. One can then ask whether it's possible to shift $A^\mu$ again whilst preserving this condition. In fact it is possible to do so, since we can shift by the gradient of any function that satisfies the wave equation, without spoiling the Lorenz gauge condition. – gj255 Jan 29 '17 at 18:47
• @MrPuh yes you got the point (@gj255: Mr Puh did not write that $A_\mu$ is a gradient). Why we have a residual gauge freedom: well, we've just proved it :-) Remember however that $A_\mu$ is not a physical field, but a mathematical construction; gauge freedom indicates a redundancy in our mathematical description; in this case we get that the EM field has fewer degrees of freedom than it looks. We start with a four component vector ($A_\mu$), but we do not have four dof's: two can be killed via gauge fixing, in this case by imposing Lorenz gauge and some additional constraint (eg $A_3=0$) – user139175 Jan 29 '17 at 19:01