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How do Maxwells equations result from the field tensor?

  1. Oct 28, 2012 #1
    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.
     
  2. jcsd
  3. Oct 28, 2012 #2

    Vanadium 50

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    In what variables are Maxwell's equations usually expressed?
     
  4. Oct 28, 2012 #3
    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.
     
    Last edited: Oct 28, 2012
  5. Oct 28, 2012 #4

    dextercioby

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    I suggest one tiny bit of LaTex: [itex] \partial [/itex], i.e. \partial.
     
  6. Oct 28, 2012 #5

    Vanadium 50

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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?
     
  7. Oct 29, 2012 #6
    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.
     
  8. Nov 14, 2012 #7
    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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