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?

  • $\begingroup$ So the question is about total derivative terms? $\endgroup$
    – Qmechanic
    May 1, 2020 at 15:15
  • $\begingroup$ Indeed, as well as why the Lagrangian seems to vanish under the equations of motion. $\endgroup$
    – The One
    May 1, 2020 at 15:45
  • $\begingroup$ Related : Squaring the E&M (Maxwell) field strength tensor. $\endgroup$
    – Frobenius
    May 2, 2020 at 18:31
  • $\begingroup$ @The One : Is your field $A$ complex or real ? If your field $A$ would be complex, your formula would be wrong. Do you agree ? en.wikipedia.org/wiki/Proca_action $\endgroup$ Jan 10, 2021 at 12:29
  • 1
    $\begingroup$ @Mathieu Krisztian The fields A_\mu under question are real. $\endgroup$ Nov 28, 2021 at 21:30

1 Answer 1


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."

  • $\begingroup$ I see, is that because under Gauss's theorem these total derivative terms just add extra terms outside the integral to the action? $\endgroup$
    – The One
    May 1, 2020 at 15:48
  • $\begingroup$ Yes, since we usually neglect boundary terms, total derivatives in the Lagrangian can be omitted. $\endgroup$ May 2, 2020 at 11:55
  • $\begingroup$ Very well. Thank you both for your answers. $\endgroup$
    – The One
    May 3, 2020 at 9:38
  • $\begingroup$ 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. $\endgroup$ Mar 19 at 22:47
  • $\begingroup$ 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. $\endgroup$ Mar 19 at 22:49

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