# Help understanding electromagnetism integral from exercise in MTW? [closed]

I was skimming through Misner, Thorne and Wheeler's book Gravitation looking for exercises to challenge myself with and came across the following exercise on page 178:

Verify that the variational principle $\delta I = 0$ gives Maxwell's equations by verifying $A_\mu$, and the Lorentz force law by varying $z^\mu(\tau)$, when \begin{equation} I = -\frac{1}{16\pi}\int F_{\mu\nu}F^{\mu\nu} d^4 x + \frac{1}{2} m \int \frac{dz^\mu}{d\tau}\frac{dz_\mu}{d\tau}d\tau + e\int \frac{dz^\mu}{d\tau} A_\mu (z)d\tau \end{equation} Here $F_{\mu\nu}$ is an abbreviation for $A_{\nu,\mu}-A_{\mu,\nu}$.

I don't understand what this integral represents. I recognize the Faraday tensor, although I don't know why there is a factor of $-\frac{1}{16\pi}$ there. It looked like an interesting problem, but I'd have to know what the terms are before I could tackle it.

Could someone explain term by term what this integral stands for?

## closed as off-topic by JamalS, Rob Jeffries, Brandon Enright, Pranav Hosangadi, Kyle KanosFeb 7 '15 at 19:33

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• The first term is the Maxwell Lagrangian (Lagrangian of the EM field), the second term is just the kinetic energy and the last is coupling of the potential to the particle. – Ryan Unger Feb 7 '15 at 20:03
• More explicitly, the variational equation $$\frac{\delta I}{\delta z^\mu}=0$$ generates the Lorentz force law whereas $$\frac{\delta I}{\delta A^\mu}=0$$ generates (two of) Maxwell's equations with the current defined as $j^\mu=e\dot z^\mu$. – Ryan Unger Feb 7 '15 at 20:14
• I'm curious...what equations did you obtain? I ask because the action I wrote down for this problem was $$S=-\frac{1}{4}\int d^4x \,F^2-m\int d\tau+e\int A_\mu dx^\mu$$ The $\pi$ is a strange convention that some people have, but your action has an extra factor of $1/4$... – Ryan Unger Feb 8 '15 at 0:25
• I don't understand it either. Frankly, I am starting to question this book's editing. I found an exercise in the book to "prove the chain rule" for covariant derivatives, but it actually is the product rule, which seemed like a really sloppy error to me since they said "chain rule" several times not just once. Regarding this equation, if it had been 1/4, I wouldn't have been so confused because it would have looked like a normal classical electrodynamics Lagrangian, but with the 1/16, I wasn't sure. I have never seen the $\pi$ convention. I probably should have just asked about that part. – Stan Shunpike Feb 8 '15 at 0:35
• I just remembered you one asked a question about the current term in the Lagrangian density. Well look at the interaction term $$e\int A_\mu\dot z^\mu d\tau$$ Now a cool trick: $$e\int A_\mu\dot z^\mu d\tau=e\int d^4z \,\delta^4(z-z(\tau))\int \dot z^\mu A_\mu d\tau=\int A^\mu J_\mu\,d^4z$$ if we define $$J^\mu=e\delta^4(z-z(\tau))\int \dot z^\mu \,d\tau$$ Upon calculating the $\delta/\delta A^\mu$ variational integral, we see that a charged point particle creates an EM field because it generates a current. – Ryan Unger Feb 8 '15 at 0:47