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Field strength tensor / matrix

  1. Jan 27, 2014 #1
    In my note,
    we have written the field strength tensor as:

    [itex]F^{\mu\nu} =\partial ^\mu A^\nu -\partial ^\nu A^\mu =



    \begin{pmatrix}
    0&E_x &E_y&E_z \\
    -E_x&0 &B_z &-B_y \\
    -E_y&-B_z &0 &B_x \\
    -E_z&B_y &-B_x&0
    \end{pmatrix}



    [/itex]

    But if I look into another book or wiki it is written as:


    [itex]
    F^{\mu\nu} =\partial ^\mu A^\nu -\partial ^\nu A^\mu =
    \begin{pmatrix}
    0&-E_x &-E_y&-E_z \\
    E_x&0 &-B_z &B_y \\
    E_y&B_z &0 &-B_x \\
    E_z&-B_y &B_x&0
    \end{pmatrix}


    [/itex]


    Why is it possible to write the field strength tensor in both notations?
    And are both notations really equal?

    THX
    Abby
     
  2. jcsd
  3. Jan 27, 2014 #2

    ChrisVer

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    Gold Member

    It depends on how you define your metric and thus the [itex]∂^{μ}[/itex]
     
  4. Jan 27, 2014 #3
    I think it comes from that $$F^{\alpha \beta} $$ is an antysymmetric tensor. So I can use $$F^{\beta \alpha} =-F^{ \alpha \beta}$$ but I am not sure.
    Need help^^

    Both definitions of $$\partial ^\mu$$ are in both cases equal.
     
  5. Jan 27, 2014 #4

    ChrisVer

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    Gold Member

    Well then try for example to write the first line:
    [itex]F^{0i}=∂^{0}A^{i}-∂^{i}A^{0}[/itex]
    Now I am not sure about minus/plus conventions I would write it:
    [itex]F^{0i}=\frac{∂A^{i}}{∂t}-∇_{i}Φ= Ε^{i}[/itex]
    So that's what I'd use.....
    it has to do I guess with how you define covariant and contravariant vectors.
     
  6. Jan 28, 2014 #5

    samalkhaiat

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    Science Advisor

    Because you can use either one to write the correct form of Maxwell equations:
    [tex]\partial_{ \mu } F_{ 1 }^{ \mu \nu } = - \partial_{ \mu } F_{ 2 }^{ \mu \nu } = - e J^{ \nu }[/tex]

    How can they be equal? [itex]F_{ 1 }^{ \mu \nu } = - F_{ 2 }^{ \mu \nu }[/itex].
     
  7. Jan 28, 2014 #6

    ChrisVer

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    Gold Member

    well the problem is that both notations are practically equivalent- you will just have to define differently the current 4vector....
    In the Lagrangian what appears is the [itex]F_{μν}F^{μν}[/itex], and whether you have them with a minus each, they will lead in the same equations of motion....
     
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