# How do Maxwells equations result from the field tensor?

1. Oct 28, 2012

### Azelketh

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.

2. Oct 28, 2012

Staff Emeritus
In what variables are Maxwell's equations usually expressed?

3. Oct 28, 2012

### Azelketh

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.

Last edited: Oct 28, 2012
4. Oct 28, 2012

### dextercioby

I suggest one tiny bit of LaTex: $\partial$, i.e. \partial.

5. Oct 28, 2012

Staff Emeritus
Wouldn't it make more sense to work with E's and B's if you want equations giving you E's and B's?

6. Oct 29, 2012

### jfy4

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.

7. Nov 14, 2012

### Azelketh

I see how I've gone wrong on the last few lines.