3
$\begingroup$

I've been given the following gauge fixing condition:

$$A_\mu A^\mu = 0 $$

And I've been asked to show if it is a valid gauge fixing condition or not. I believe that it isn't because I've already encountered a term similar to $A_\mu A^\mu$ in the Proca action and I've read that the mass term in that action makes that the Proca action is not Lorentz invariant.

I've tried to show that for a given configuration $(\phi,A)$ in this "gauge" I can't always transform it to a configuration in the Lorentz gauge that I already know that it is well defined.

I haven't been able to show it this far and I've been wondering if there is a mistake in my argument or if the mistake is in my calculations. Is this the right way to prove this? Is this a valid gauge fixing condition? Why?

$\endgroup$
  • $\begingroup$ The Proca action describes massive particle with spin one. There is no concept of gauge invariance for corresponding theory. $\endgroup$ – Name YYY Jan 28 '16 at 18:16
  • 1
    $\begingroup$ hint: try to show whether or not there exists a field $\Omega(x)$ such that $(A_\mu+\partial_\mu\Omega)(A^\mu+\partial^\mu\Omega)=0$. $\endgroup$ – AccidentalFourierTransform Jan 28 '16 at 18:26
  • $\begingroup$ @NameYYY Wrong; There is still a $U(1)$ symmetry if one takes into account the Stueckelberg field that pops up due to the mass. $\endgroup$ – Danu Jan 28 '16 at 18:40
  • $\begingroup$ @Danu : but what to do if one doesn't took it into an account? By having an effective field theory you can't restore fundamental one in general, while the Proca action itself isn't gauge invariant. $\endgroup$ – Name YYY Jan 28 '16 at 20:39
  • $\begingroup$ The Proca action is $U(1)$ invariant. The symmetry transformation is just not the naive one (Googling "Stückelberg field" yields an explicit demonstration of this--- I'd link it but I'm on my phone now). $\endgroup$ – Danu Jan 28 '16 at 21:29
1
$\begingroup$

I believe this is a valid gauge condition. You can use the hint given by AccidentalFourierTransform in a comment and for arbitrary $A^\mu$ find such $\Omega$ that $(A_\mu+\Omega_{,\mu})(A^\mu+\Omega^{,\mu})=0$, as the latter equation can be resolved with respect to $\Omega_{,0}$, so a Cauchy problem can be posed. Of course, the usual mathematical issues of existence of a solution of the Cauchy problem need to be considered, but "at the physical level of mathematical rigor" I don't see any problems with this gauge condition. Some famous people considered this gauge condition and a modified condition with a constant instead of zero, among them Dirac (Proc. Roy. Soc. London A 209, 291 (1951)) and Nambu (Supplement of the Progress of Theoretical Physics, Extra Number, 1968, pp.190-195). By the way, this question did not seem trivial to Nambu, who offered an elegant physical argument suggesting that you can always use this gauge condition, but the argument does not work for zero in the right-hand side.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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