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How do Maxwells equations result from the field tensor?

by Azelketh
Tags: equations, field, maxwells, result, tensor
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Azelketh
#1
Oct28-12, 10:13 AM
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
[tex]
F_{μv} = d_μ A_v - d_v A_μ
[/tex]
So I have managed to get to;
[tex]
0=d_μ F^{μv}
[/tex]
But I am unable to see how this shows Maxwells equations.
Any points would be appreciated.
Thanks.
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Vanadium 50
#2
Oct28-12, 10:27 AM
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In what variables are Maxwell's equations usually expressed?
Azelketh
#3
Oct28-12, 11:18 AM
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;
[tex]
A= ( \phi, \vec{A} )
[/tex]

where[tex] \vec{A} [/tex]is the magnetic vector
[tex]\phi[/tex] is the scalar electric potential.

[tex]
E = -∇\phi - \frac{\partial\vec{A}}{\partial t}
[/tex]
And,
[tex]
B= ∇ X \vec{A}
[/tex]
It's clear to me that 2 of maxwells equations result directly from this definition;
[tex]
∇.B = ∇. ( ∇ X \vec{A} ) = 0
[/tex]
and
[tex]
∇ X E = ∇ X ∇\phi - ∇ X \frac{\partial\vec{A}}{\partial t}
[/tex][tex]
∇ X E = ∇(∇ X \phi) - \frac{\partial (∇ X\vec{A} )}{\partial t}
[/tex][tex]
∇ X E = 0 - \frac{\partial ( B )}{\partial t}
[/tex][tex]
∇ X E = - \frac{\partial B}{\partial t}

[/tex]

Which leaves
[tex]
∇.E = 0
[/tex]
and
[tex]
∇ X B = \frac{\partial E}{\partial t}
[/tex]
to be found from
[tex]
0=\partial_μ F^{μv}[/tex]

So I have tried putting in
[tex]
F_{μv} = \partial_μ A_v - \partial_v A_μ
[/tex]
To give;
[tex]
0=\partial_μ ( \partial_μ A_v - \partial_v A_μ )[/tex]

[tex]
0=\partial_μ \partial_μ A_v - \partial_μ \partial_v A_μ [/tex]
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;
[tex]
\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}
[/tex]
And
[tex]
\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}
[/tex]
which when combined gives;
[tex]
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} )
[/tex]
[tex]
0= \frac{\partial^2 \vec(A)}{\partial t^2} + ∇^2 \phi + \frac{\partial \vec(∇A)}{\partial t} + \frac{\partial ∇\phi}{\partial t}
[/tex]
which is where I am scratching my head.....

EDIT:replaced d's with [tex]\partial[/tex] as per dextercioby's suggestion.

dextercioby
#4
Oct28-12, 02:47 PM
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How do Maxwells equations result from the field tensor?

I suggest one tiny bit of LaTex: [itex] \partial [/itex], i.e. \partial.
Vanadium 50
#5
Oct28-12, 07:46 PM
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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?
jfy4
#6
Oct29-12, 09:24 AM
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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 [itex]\nu[/itex] I think... Those are vectors. You can write it in a vector format, but the index is not summed over.
Azelketh
#7
Nov14-12, 10:49 AM
P: 36
I see how I've gone wrong on the last few lines.

Thanks for your help vanadium 50 , jfy4 and dextercioby!


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