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I'm deriving the Maxwell equations from this Lagrangian:

$$ \mathscr{L} \, = \, -\frac{1}{4} F^{\mu \nu}F_{\mu \nu} + J^\mu A_\nu \tag{1}$$

My signature is $$(+ - - -)\tag{2}$$ and

$$ F^{\mu \nu} \, = \,\left(\begin{matrix}0 & -E_x & -E_{y} & -E_{z} \\ E_{x} & 0 & -B_{z} & B_{y} \\ E_y & B_{z} & 0 & -B_x \\ E_z & -B_{y} & -B_{x} & 0\end{matrix}\right)\tag{3} $$

My procedure is almost exactly the same as this one: https://physics.stackexchange.com/a/14854/121554

But he has a $+\frac{1}{4}F^{\mu \nu}F_{\mu \nu}$ in the lagrangian.

So, while he obtains the right equation $$\partial_\mu F^{\mu \nu}\, = \, J^\nu,\tag{4}$$ I carry a minus sign till the end and my final equation is

$$ J^\nu \,=\, -\partial_\mu F^{\mu \nu}\, = \, \partial_\mu F^{\nu \mu}\, ; \tag{5} $$

Which is clearly wrong if you write it down explicitely in function of the fields, the charge and the currents.

Is my lagrangian wrong for my metric and my definition of the elctromagnetic field tensor? We spent some time talking about that lagrangian at lesson and the professor gave a lot of importance to that minus sign in order to have positive kinetic term. Am I missing something?

I can write down all my calculations if requested but they are basically the same of the link I provided above.

Community
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RenatoRenatoRenato
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    Note different sign conventions in E&M, cf. e.g. this & this Phys.SE posts. OP's eqs. (1) & (5) are compatible. OP's eqs. (1) & (2) are traditionally incompatible. – Qmechanic Dec 04 '16 at 14:45
  • Does that provide an answer and I'm not getting it? Or, is it just an observation? Thanks, in any case it's a useful observation. As regards my problem, in the first link you provide, there's my convention with the right Maxwell equation, while mine still seems wrong to me. – RenatoRenatoRenato Dec 04 '16 at 15:01
  • Ok I read now about the incompatibility of my metric with my langrangian, I was sure I worked my calculation wrong, instead I started from the wrong starting point. Thanks for the interesting second link too and for your clarification. PS: Lagrange was born Lagrangia, so I'll keep editing the title until someone bans me :P – RenatoRenatoRenato Dec 04 '16 at 15:04

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I suspect the error is in your source term: with reference to Jackson's "Classical electrodynamics", the correct Lagrangian density is $$ {\cal L}=-\frac{1}{16\pi} F_{\alpha\beta}F^{\alpha\beta}-\frac{1}{c}J_\alpha A^\alpha\, , $$ which differ from yours by a sign in the source term. (The other factors $1/16\pi$ and $1/c$ are linked to the use of Gaussian units.) The article you link to also have the same sign for both terms in the Lagrangian density.

ZeroTheHero
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  • Yes, thanks to you and Qmechanic i got this sign wrong. Since the lagrangian was provided by the prof who uses the sign convention I wrote, I assumed that the $\mathscr{L}$ was the right one. – RenatoRenatoRenato Dec 04 '16 at 15:39