# Lagrangian for electromagnetism

1. Nov 19, 2007

### jdstokes

In A. Zee's quantum field theory in a nutshell he assumes familiarity with Maxwell's lagrangian $\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ where $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ with A the vector potential.

Although I've seen the magnetic vector potential, I've never seen the lagrangian formalism in either electrodynamics or lagrangian/hamiltonian dynamics courses.

Could anyone point me in the direction of a suitable reference to allow me to familiarise myself with this?

Thanks

2. Nov 19, 2007

### Avodyne

3. Nov 19, 2007

### siddharth

Try chapter 12 in Griffiths, for a brief introduction into the field tensor and the four-vector potential. For an introduction to the lagrangian formalism in electrodynamics, try Goldstein's book on classical mechanics.

4. Nov 20, 2007

### jdstokes

I find Zee's notation a little bit confusing here. It seems like he is writing $\partial_\mu$ to mean $(\partial_t,\nabla)$ and at the same time writing e.g. $A_\mu = (V,-\mathbf{A})$ and thus $A^\mu = (V,\mathbf{A})$. Is this standard or am I misunderstanding his notation?

This is the only way I could get Maxwell's equations out of

$F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$.

$F^{0i} = \partial^0 A^i - \partial^i A^0 = -E^i$. etc

Last edited: Nov 20, 2007
5. Nov 20, 2007

### jdstokes

After checking in another QFT text by Ryder it seems like this is indeed standard notation.

6. Nov 20, 2007

### nrqed

yes, for $\partial_\mu$ the sign is opposite to the other vectors. That's because
$$\partial_\mu \equiv \frac{\partial}{\partial x^\mu}$$

7. Jul 16, 2009

### jacobrhcp

never mind I'll make my own topic

8. Jul 16, 2009

### turin

The ultimate E&M reference: J. D. Jackson, "Classical Electrodynamics", 3rd ed., Chap. 12, Sec. 7.
R. Shankar, "Principles of Quantum Mechanics", 2nd ed., Chap. 18, Sec. 5, Subsec. "Field Quantization".
(more advanced) Peskin & Shroeder, "An Introduction to Quantum Field Theory", Chap. 15.

9. Jul 16, 2009

### turin

It is perhaps standard, but it is certainly just a convention. For example, in the "East Coast Metric" (η=diag(-1,+1,+1,+1)), that could be changed to $A_\mu = (-V,\mathbf{A})$, and for implicit metric: $\partial_\mu=\partial^\mu=(\nabla,ic\partial_t)$, $A_\mu=A^\mu=(\mathbf{A},icV)$.