How do Maxwells equations result from the field tensor?

AI Thread Summary
The discussion focuses on deriving Maxwell's equations from the field tensor in the context of quantum field theory, specifically referencing problem 2.1a from Peskin and Schroeder. The user successfully applies the Euler-Lagrange equation and the definition of the field tensor but struggles to connect their results to Maxwell's equations. They express the equations in terms of the electric field (E) and magnetic field (B) and identify that two of Maxwell's equations can be derived directly from the field tensor. However, they encounter difficulties in deriving the remaining equations, particularly in combining terms correctly. The conversation highlights the importance of maintaining consistency in variable types when working through the equations.
Azelketh
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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
<br /> F_{μv} = d_μ A_v - d_v A_μ<br />
So I have managed to get to;
<br /> 0=d_μ F^{μv}<br />
But I am unable to see how this shows Maxwells equations.
Any points would be appreciated.
Thanks.
 
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In what variables are Maxwell's equations usually expressed?
 
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;
<br /> A= ( \phi, \vec{A} )<br />

where\vec{A}is the magnetic vector
\phi is the scalar electric potential.

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

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

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

<br /> 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;
<br /> \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}<br />
And
<br /> \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}<br />
which when combined gives;
<br /> 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} )<br />
<br /> 0= \frac{\partial^2 \vec(A)}{\partial t^2} + ∇^2 \phi + \frac{\partial \vec(∇A)}{\partial t} + \frac{\partial ∇\phi}{\partial t}<br />
which is where I am scratching my head...

EDIT:replaced d's with \partial as per dextercioby's suggestion.
 
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I suggest one tiny bit of LaTex: \partial, i.e. \partial.
 
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?
 
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.
 
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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