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In the Coulomb gauge for the Maxwell potential we have $$ A^0 = 0 \\ \partial_i A^i = 0 $$

Under an infinitesimal Lorentz Transformation with parameter $\epsilon$, we have

$$ A^\mu(x) \rightarrow A'^\mu(x) = A^\mu(x) + \epsilon^{\mu}_{\ \ \nu}A^\nu(x)$$

But then the Coulomb gauge no longer holds as $$ A'^0 = 0 + \epsilon^{0}_{\ \ \nu}A^\nu(x) \neq 0$$

So we add a supplementary gauge term $\lambda(x,\epsilon)$

$$ A^\mu(x) \rightarrow A'^\mu(x) = A^\mu(x) + \epsilon^{\mu}_{\ \ \nu}A^\nu(x) + \frac{\partial}{\partial x_\mu}\lambda(x,\epsilon)$$

If I perform this Lorentz Transformation now, can I say Coulomb gauge now holds due to the $\lambda$? So we have $$ A'^0 = 0 \\ \partial_i A'^i = 0 $$ Is this the entire point of the supplementary gauge term?

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  • $\begingroup$ God yes - exams are getting to me. Edit made. $\endgroup$ – Tweej Apr 21 '16 at 19:26
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    $\begingroup$ That is correct. $\endgroup$ – Prahar Apr 21 '16 at 19:32

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