# How do Maxwells equations result from the field tensor?

by Azelketh
Tags: equations, field, maxwells, result, tensor
 P: 36 Hi, I've been trying to solve problem 2.1 a in Peskin and schroeder, an introduction to QFT. The problem is to derive Maxwells equations for free space, which I have almost managed to do, using the Euler- lagrange euqation And the definition of the field tensor as $$F_{μv} = d_μ A_v - d_v A_μ$$ So I have managed to get to; $$0=d_μ F^{μv}$$ But I am unable to see how this shows Maxwells equations. Any points would be appreciated. Thanks.
 Mentor P: 15,405 In what variables are Maxwell's equations usually expressed?
 P: 36 My apologies for a less than comprehensive post at first; Well, usually in the vector form of E and B or the 4 vector A, where; $$A= ( \phi, \vec{A} )$$ where$$\vec{A}$$is the magnetic vector $$\phi$$ is the scalar electric potential. $$E = -∇\phi - \frac{\partial\vec{A}}{\partial t}$$ And, $$B= ∇ X \vec{A}$$ It's clear to me that 2 of maxwells equations result directly from this definition; $$∇.B = ∇. ( ∇ X \vec{A} ) = 0$$ and $$∇ X E = ∇ X ∇\phi - ∇ X \frac{\partial\vec{A}}{\partial t}$$$$∇ X E = ∇(∇ X \phi) - \frac{\partial (∇ X\vec{A} )}{\partial t}$$$$∇ X E = 0 - \frac{\partial ( B )}{\partial t}$$$$∇ X E = - \frac{\partial B}{\partial t}$$ Which leaves $$∇.E = 0$$ and $$∇ X B = \frac{\partial E}{\partial t}$$ to be found from $$0=\partial_μ F^{μv}$$ So I have tried putting in $$F_{μv} = \partial_μ A_v - \partial_v A_μ$$ To give; $$0=\partial_μ ( \partial_μ A_v - \partial_v A_μ )$$ $$0=\partial_μ \partial_μ A_v - \partial_μ \partial_v A_μ$$ so I then tried expanding this out, hoping that some terms would cancel and that I would recognize others and perhaps then they would be close to the E and B field formulation that I am more familier with. This yielded; $$\partial_μ \partial_μ A_v= \frac{\partial^2 \phi}{\partial t^2} + \frac{\partial^2 \vec(A)}{\partial t^2} + ∇^2 \phi + ∇^2 \vec{A}$$ And $$\partial_μ \partial_v A_μ = \partial_v \partial_μ A_μ = \frac{\partial^2 \phi}{\partial t^2} - \frac{\partial \vec(∇A)}{\partial t} - \frac{\partial ∇\phi}{\partial t} + ∇^2 \vec{A}$$ which when combined gives; $$0= \frac{\partial^2 \phi}{\partial t^2} + \frac{\partial^2 \vec(A)}{\partial t^2} + ∇^2 \phi + ∇^2 \vec{A} - ( \frac{\partial^2 \phi}{\partial t^2} - \frac{\partial \vec(∇A)}{\partial t} - \frac{\partial ∇\phi}{\partial t} + ∇^2 \vec{A} )$$ $$0= \frac{\partial^2 \vec(A)}{\partial t^2} + ∇^2 \phi + \frac{\partial \vec(∇A)}{\partial t} + \frac{\partial ∇\phi}{\partial t}$$ which is where I am scratching my head..... EDIT:replaced d's with $$\partial$$ as per dextercioby's suggestion.
I suggest one tiny bit of LaTex: $\partial$, i.e. \partial.
 PF Gold P: 647 at the end of post #3, like in the last 4 lines, you have vector things added to scalar things, which is no good. And on the left hand side of those lines there is a free index $\nu$ I think... Those are vectors. You can write it in a vector format, but the index is not summed over.