2
$\begingroup$

How are we still left with one Residual Gauge Freedom in the choice of Electromagnetic Potential after having already exploited the Gauge Freedom once. As is mentioned in Halzen and Martin Section 6.9.

Doesn't choosing another scalar to induce Gauge Transformation spoils the Lorenz Gauge condition we have earlier?

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ Please don't rely on people knowing the section of the book you're talking about, but include all relevant context in the question. How exactly did you "exploit the gauge freedom once" and what exactly is the residual gauge freedom supposed to be left? $\endgroup$
    – ACuriousMind ♦
    Commented Sep 24, 2016 at 11:46

1 Answer 1

Reset to default
5
$\begingroup$

When there are residual gauge freedoms, then it means that you are not fixing the gauge completely. For example, for the gauge $$\nabla\cdot \vec{A}=0,\ A_0=0,$$ there is no residual gauge. But for some conditions such as $\partial^\mu\cdot A_\mu=0,$ the gauge isn't fixed completely. Because, if you take $A_\mu\rightarrow A_\mu+\partial_\mu\chi$ with $\partial^2\chi=0$, then $A_\mu+\partial_\mu\chi$ is also a gauge. To fix the gauge completely, you should assign more conditions, in our example, we can take, say $A_0=0$.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ What do you mean by the condition $\chi = 0$? The gauge conditions have to be phrased in terms of $A$. If I hand you an $A$ with $\partial^\mu A_\mu = 0$, how are you going to decide whether "$\chi = 0$" is fulfilled or not? $\endgroup$
    – ACuriousMind ♦
    Commented Jun 6, 2017 at 11:46
  • $\begingroup$ @ACuriousMind I think my statement is some sloppy here. $\chi$ is a relative parameter (field) which only has meanings when we have one complete gauge in mind. I meant we can take $\partial^\mu A_\mu=0$ plus an extra condition, say A_0=0 (essentially reduces to the Coulomb gauge). $\endgroup$
    – Wein Eld
    Commented Jun 6, 2017 at 19:55

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