This is a homework problem for a field theory class dealing with an axion model.

Originally, we are given that $$S[a]=\int_Md^4x \frac{1}{2}(\partial_{\mu}a(x))^2$$ has a continuous global invariance $a(x) \to a(x)+\epsilon$ where $\epsilon$ is constant so that we can identify a conserved current according to Noether's theorem. We are then asked to add a term $$S_1[a,A_{\mu}]=\frac{1}{f}\int_M d^4x aF_{\mu\nu}\tilde{F}^{\mu\nu}$$ where $F_{\mu\nu}$ is a $U(1)$ field strength, $\tilde{F}^{\mu\nu}$ its dual tensor, $f$ a constant with units of mass and $a$ is $U(1)$-invariant. We are also told to assume that $F_{\mu\nu}$ drops to zero at infinity.

We are asked to show that $a(x) \to a(x)+\epsilon$ is a symmetry of $S+S_1$ and then compute the new conserved current and show that this current is not gauge invariant.

The hint given in the problem suggest that we should write out $F\tilde{F}$ in terms if $A_{\mu}$ but I am having a hard time with understanding how shifting $a$ changes $A_{\mu}$. I know that the $F\tilde{F}$ term usually comes about from a gauge symmetry but it is not clear to me how to connect that with the given symmetry $a(x) \to a(x)+\epsilon$. I am also quite confused about how to make use of the fact that $a$ is $U(1)$-invariant which makes me think of a multiplication by a phase or a rotation rather than shifting the whole field. Does the field $A_{\mu}$ arise as a gauge field due to the $U(1)$-invariance but has nothing to do with the $\epsilon$ symmetry given? How to consider the variation of $S_1$ under this symmetry then beyond just $$\delta S_1 = \frac{\epsilon}{f}\int d^4x F_{\mu\nu}\tilde{F}^{\mu\nu}$$ or is that all and I should really be worrying about showing that this vanishes by carrying out integration by parts of some sort?

Any help would be very much appreciated - I am really struggling with connecting things conceptually here.

  • 2
    $\begingroup$ $A_\mu$ does not transform under the axionic shift. So your expression for $\delta S_1$ is fine. You need to show that the integrand is a total derivative and see the conditions under which it vanishes. $\endgroup$ – suresh Sep 23 '14 at 9:12
  • $\begingroup$ Thank you, that clears up quite a bit of my confusion. To derive the conserved current after the action has been modified with this term would I then have to consider some arbitrary variation of $A_{\mu}$ since shifting $a$ doesn't affect the gauge field? $\endgroup$ – SallyShears Sep 23 '14 at 12:32
  • $\begingroup$ To derive the Noether current for the axionic shift, you need to use to the standard procedure that must have been taught in your class with $\delta A_\mu=0$ under the axionic shift. $\endgroup$ – suresh Sep 24 '14 at 3:01

Your Answer

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

Browse other questions tagged or ask your own question.