5

In his book Quantum Field Theory and the Standard Model, Matthew D. Schwartz derives the Lagrangian for the massive spin 1 field (section 8.2.2). In eq. (8.23) he finds this to be \begin{align} \mathcal L&=\frac{1}{2}A_\mu\square A^\mu-\frac{1}{2}A_\mu\partial^\mu\partial_\nu A^\nu+\frac{1}{2}m^2A_\mu A^\mu,\tag{8.23} \end{align} where $\square = \partial_\mu\partial^\mu$. In the very same equation, he equates this to the Proca Lagrangian \begin{align} \mathcal L=\mathcal L_\mathrm{Proca}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m^2A_\mu A^\mu,\tag{8.23} \end{align} where $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$.

I fail to understand however, how the first Lagrangian can be rewritten to this Proca Lagrangian. My attempt was to rewrite the first term of the Proca Lagrangian into something that resembles the first two terms of the first Lagrangian above. It involves the product rule \begin{align} -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}&=-\frac{1}{4}(2\partial_\mu A_\nu\partial^\mu A^\nu-2\partial_\mu A_\nu\partial^\nu A^\mu)\\ &=-\frac{1}{4}(2\partial_\mu[A_\nu\partial^\mu A^\nu]-2A_\nu\partial_\mu\partial^\mu A^\nu-2\partial_\mu[A_\nu\partial^\nu A^\mu]+2A_\nu\partial_\mu\partial^\nu A^\mu)\\ &=\frac{1}{2}A_\mu\square A^\mu-\frac{1}{2}A_\mu\partial^\mu\partial_\nu A^\nu+\frac{1}{2}\partial_\mu(A_\nu\partial^\nu A^\mu)-\frac{1}{2}\partial_\mu(A_\nu\partial^\mu A^\nu), \end{align} having applied some relabelling in the second term of the final expression. The first two terms in this final expression are the first two terms in the Lagrangian, but then I'm stuck with the final two terms. Could someone explain to me what I'm missing here?

Also, the equation of motion for the Proca Lagrangian are \begin{align} (\square+m^2)A_\mu=0\tag{8.18}\\ \partial_\mu A^\mu=0. \end{align} Substituting this in the first Lagrangian would make it vanish. How does that make sense?

Qmechanic
  • 201,751
The One
  • 75
  • 7

1 Answers1

8

This somewhat sloppy statement means that the two Lagrangian expressions are the same up to a total derivative. Such a total derivative does not contribute to the action $A=\int d^4x \cal L$, for suitable conditions at the edge of the integration domain, and therefore is considered to be without physical consequence.

A word of caution is needed. Lagrangians differing by a total derivative produce the same equations of motion but different Noether energy-momentum distributions. As energy-momentum is the source of gravitation, such Lagrangians lead to physically distinct theories. The gravitational effects are so small that will probably never be experimentally accessible and so this physical distinction is ignored. Feynman in his Lectures II, ch. 27.4 said this about it:

"It is interesting that there seems to be no unique way to resolve the indefiniteness in the location of the field energy. It is sometimes claimed that this problem can be resolved by using the theory of gravitation in the following argument. In the theory of gravity, all energy is the source of gravitational attraction. Therefore the energy density of electricity must be located properly if we are to know in which direction the gravity force acts. As yet, however, no one has done such a delicate experiment that the precise location of the gravitational influence of electromagnetic fields could be determined. That electromagnetic fields alone can be the source of gravitational force is an idea it is hard to do without. It has, in fact, been observed that light is deflected as it passes near the sun—we could say that the sun pulls the light down toward it. Do you not want to allow that the light pulls equally on the sun? Anyway, everyone always accepts the simple expressions we have found for the location of electromagnetic energy and its flow. And although sometimes the results obtained from using them seem strange, nobody has ever found anything wrong with them—that is, no disagreement with experiment. So we will follow the rest of the world—besides, we believe that it is probably perfectly right."

my2cts
  • 24,097
  • I see, is that because under Gauss's theorem these total derivative terms just add extra terms outside the integral to the action? – The One May 01 '20 at 15:48
  • Yes, since we usually neglect boundary terms, total derivatives in the Lagrangian can be omitted. – QuirkyTurtle98 May 02 '20 at 11:55
  • Very well. Thank you both for your answers. – The One May 03 '20 at 09:38
  • Using different non-gravitational Lagrangians differing by a total time derivative does not imply working with distinct theories or with different distributions of energy and momentum in GR. The energy-momentum tensor $T_{\mu\nu}$ referred to in Einstein's equations is not necessarily the Noether charge, but instead it is the accepted "real" distribution of energy and momentum, which is unique and the same for all the various possible equivalent choices of Lagrangians and canonical energy-momentum tensors. – Ján Lalinský Mar 19 '24 at 22:47
  • The real energy-momentum tensor has to be defined by us, and then validity of Einstein's equations with it checked via observations/experiments. This has nothing to do with the freedom of choice of Lagrangian implied by the irrelevance of the total time derivative. Definition of energy and momentum is an independent physical definition, it is not completely determined by which Lagrangian or Hamiltonian we use. – Ján Lalinský Mar 19 '24 at 22:49
  • @JánLalinský 'The real energy-momentum tensor has to be defined by us'. I disagree. It has to be consistent mathematically and it has to agree with experiment. As different choices predict different experimental outcomes and experiments are at this point lacking, any mathematically consistent choice is acceptable for now. – my2cts Mar 20 '24 at 11:19
  • @JánLalinský How can an 'accepted "real" distribution of energy and momentum' be 'unique and the same for all the various possible equivalent choices of Lagrangians and canonical energy-momentum tensors'? This is not science but dogma. Nature has the last word, not a gremium of cigar smoking grey old men. – my2cts Mar 20 '24 at 11:20
  • @my2cts Energy and momentum expressions regularly used are definitions that have to obey some constraints (transformation properties, local conservation). These constraints do not determine the expressions uniquely. The correct unique expressions may be found in future due to additional constraints we will want the definitions to obey, e.g. the Einstein equations. If we could measure weak gravity accurately, we could use Einstein's equations to define $T$ and check if it is consistent with the usual Poynting and other formulae for matter, or their alternatives, but this is not possible today. – Ján Lalinský Mar 20 '24 at 14:23
  • @JánLalinský True. And until we have this information only the divergency of energy-momentum is constrained by experiment. And then there is still the matter of how to divide energy and momentum between matter and field. This also impacts the field energy-momentum distribution. I’referring to EM symmetrisation. – my2cts Mar 20 '24 at 14:53
  • Yes we agree on this. My original point was that many different Lagrangians and corresponding canonical energy-momentum tensors are physically equivalent and consistent with the same theory where there is one and only correct energy-momentum tensor. Choice of one specific Lagrangian and its implied canonical energy-momentum tensor (the Noether charge implied by that choice) bring us no more closer to what the correct energy-momentum tensor is. – Ján Lalinský Mar 20 '24 at 16:35
  • There we disagree. Although only experiment can decide, ad hoc modification of the EM tensor by the Belinfante-Rosenfeld symmetrisation can only lead us away from the correct result. – my2cts Mar 23 '24 at 08:34