1
$\begingroup$

Suppose the the Lagrangian $\mathscr{L}$ of the free electromagnetic field is augmented with the term $$F_{\mu\nu}\tilde{F}^{\mu\nu}=\partial_{\mu}(\epsilon^{\nu\nu\lambda\rho}A_\nu F_{\lambda\rho}).$$ Since this term is a total divergence it can be converted to a surface integral over $A_\nu F_{\lambda\rho}$.

Now, if $F_{\mu\nu}$ is assumed to vanish at the boundary (at infinity) it is sufficient to argue that such a term drops out in the equation of motion (EOM) this abelian theory even if $A_\mu$ is nonzero.

Questions

  1. Is $F_{\mu\nu}\to 0$ at infinity sufficient to derive the EOM in absence of the $F\tilde{F}$ term? Did anywhere in the derivation of the EOM we need to use that $A_\mu$ itself have to vanish at infinity/boundary?

  2. As a related question, physically, why do we need $F_{\mu\nu}$ to vanish at infinity?

Next, if we augment the Lagrangian of a nonabelian Yang-Mills (YM) theory with a $F\tilde{F}$ term, unlike the previous case, it does affect the EOM unless $A^a_\mu$ itself are assumed to vanish at infinity.

  1. Do we assume $A_\mu^a=0$ at infinity only for a nonabelian YM theory or do we always have to assume $A_\mu=0$ irrespective of whether the theory is abelian or nonabelian?
$\endgroup$
2
$\begingroup$

You cannot really argue about the value of $A$ "at infinity" because a gauge field is not really defined as a single global field - rather, one should think of it as living in local patches, see also this answer of mine.

The integral of $F\wedge F$ is an invariant of the gauge principal bundle - its second Chern number - and is zero only if the bundle is trivial. That is, this term contributes nothing only if $A$ is globally defined. So your goal of discarding this term is misguided because this term actually has a prominent physical meaning - for example it is the value of the chiral anomaly. Note also that you will often hear about $A_\mu$ being "pure gauge" at infinity. This is an idiom physicists use to avoid the language of bundles and of the gauge field as being locally defined. The same language occurs in the context of discussing "large gauge transformations", see this answer of mine.

The above should suffice as an answer to your second and third sub-questions - the value of $A$ at infinity doesn't really exist, and if we pretend it does, then it doesn't necessarily vanish - it is just pure gauge, so that $F$ vanishes.

Which brings us around to your first subquestion: $F$ is a physical field. We almost always assume that physical fields fall off sufficiently fast at infinity so that the theory does not contain infinite energy and so that interactions do not have finite strength at infinite distance, see also this question on the electric field vanishing at infinity.

As for whether or not $A$ vanishes "at infinity" for the purpose of the derivation of the equations of motion: The e.o.m. are local, and at each point only a single local gauge field $A$ enters into them. All these local fields can trivially be said to fall off towards infinity because they are only defined as non-zero on a subset of the space(time), anyway (note that this conflicts with the usage above where the $A$ are "pure gauge at infinity" in the non-trivial case. That's why that usage is confusing.). In the one case where we can't say this - when a single $A$ is globally defined - then that $A$ truly falls off at infinity.

$\endgroup$
  • $\begingroup$ 1. So your goal of discarding this term is misguided because this term actually has a prominent physical meaning even for a U(1) theory? 2. In Srednicki's QFT, he says that the $F\tilde{F}$ term does not affect the EOM of even nonabelian YM theories. This seems to suggest that we need $A_\mu$ to vanish at infinity to derive the Euler-Lagrange EOM. 3. I know that the $F\tilde{F}$ term has rich physics but I'm interested in the classical EOM. Srednicki's comment suggests that we need to assume $A_\mu=0$ to derive the EOM. Is that true? @ACuriousMind $\endgroup$ – SRS Nov 1 '18 at 17:33
  • $\begingroup$ @SRS I added an additional paragraph on why one can claim that $A$ falls off during the derivation of the e.o.m. $\endgroup$ – ACuriousMind Nov 1 '18 at 17:55
  • $\begingroup$ So the conclusion is as follows. $F\tilde{F}$ term does not affect the EOM, irrespective of whether the theory is abelian or nonabelian YM. This is because in the derivation of EOM one needs to assume $A_\mu$ to fall off at infinity. EOM does not capture the rich physics contained in $F\tilde{F}$ term. Is this fine? $\endgroup$ – SRS Nov 1 '18 at 18:14

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.