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In Schawrtz, Page 116, formula 8.23, he seems to suggest that the square of the Maxwell tensor can be expanded out as follows:

$$-\frac{1}{4}F_{\mu \nu}^{2}=\frac{1}{2}A_{\mu}\square A_{\mu}-\frac{1}{2}A_{\mu}\partial_{\mu}\partial_{\nu}A_{\nu}$$

where:

$$F_{\mu\nu}=\partial_{\mu} A_{\nu} - \partial_{\nu}A_{\mu}$$

For the life of me, I can't seem to derive this. I get close, but always with an extra unwanted term, or two.

Anyone have a hint on the best way to proceed?

Qmechanic
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EthanT
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    Does Schwartz include an integral? If so, he might be integrating out certain terms to the boundary and setting them to zero. – Dhuality Mar 17 '19 at 15:15
  • He does not, and I thought I had derived this in the past sans integral. I'll try that out, though. At the very least, I learn a new way of deriving this – EthanT Mar 17 '19 at 15:37
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    Possible duplicate of Expanding electromagnetic field Lagrangian in terms of gauge field – knzhou Mar 17 '19 at 15:50
  • Yeah, this was really easy keeping it under the integral of S. However, I thought there was a way to achieve the same thing, w/ just tensor manipulation. Maybe I am not remembering correctly, though – EthanT Mar 17 '19 at 16:12
  • Products of zeroth and second derivatives are not generally equal to products of first derivatives in any identity... What you need is a context where a derivative of a product is zero, as $d(x~dx)=dx~dx + x~d^2x.$ Getting the left hand side to vanish in this context might be possible with antisymmetry but looks non-trivial—maybe it amounts to a boundary term in some integral though? – CR Drost Mar 17 '19 at 16:22

3 Answers3

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Your expression is part of a Lagrangian. As the physics remains the same as long as the action remains the same, one can always do partial integration in the action integral over the Lagrangian to derive alternative Lagrangians describing the same physics.

paleonix
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Hint: Try introducing an integral to the expression so it becomes $$-\frac{1}{4}\int F_{\mu\nu}F^{\mu\nu}\text{d}^d x$$ and take the total derivative terms to vanish at infinity. A much more careful argument can be made here in the presence of boundaries, but this should get you started.

Dhuality
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The relation as you state it does not hold. Only the space time integral of both hands of the equation is equal under suitable boundary conditions. So this would be an error.

my2cts
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