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Special relativity: proof of symmetry concept

  1. Jun 26, 2010 #1

    Uku

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    1. The problem statement, all variables and given/known data

    I am given a symmetric tensor A, meaning [tex]A^{\mu\nu}=A^{\nu\mu}[/tex] and I am given an asymmetric tensor B, meaning [tex]B_{\mu\nu}=-B_{\nu\mu}[/tex]

    Now I need to show that:

    [tex]A^{\mu\nu}B_{\mu\nu}=0[/tex] 0)

    2. Relevant equations

    We know that an asymmetric tensor can be written as:

    [tex]A^{\mu\nu}=\frac{1}{2}(T^{\mu\nu}-T^{\nu\mu})[/tex] 1)

    3. The attempt at a solution
    This is what I have written down from the class:

    We can use 1) to write:

    [tex]B_{\mu\nu}=\frac{1}{2}(B_{\mu\nu}-B_{\nu\mu})[/tex]

    Now we multiply this by [tex]A^{\mu\nu}[/tex]:

    [tex]A^{\mu\nu}B_{\mu\nu}=\frac{1}{2}(A^{\mu\nu}B_{\mu\nu}-A^{\mu\nu}B_{\nu\mu})[/tex] 2)

    Here comes the point which I am confused: the lecturer has written that we can use
    [tex]\mu\leftrightarrow\nu[/tex] 3) to change the indexes on 2) such as:

    [tex]A^{\mu\nu}B_{\mu\nu}=\frac{1}{2}(A^{\mu\nu}B_{\mu\nu}-A^{\mu\nu}B_{\mu\nu})=\frac{1}{2}A^{\mu\nu}(B_{\mu\nu}-B_{\mu\nu})=0[/tex] 4)

    Right, but here is the question. I know that tensor B is antisymmetric, meaning that
    [tex]B_{\mu\nu}=-B_{\nu\mu}[/tex]!! How can I then use 3) to change the indexes? That should give me a change of signs in 4)

    [tex]...=\frac{1}{2}(B_{\mu\nu}+B_{\mu\nu})\neq0[/tex]

    So... how is 0) properly shown? Have I missed some principle at 3)?

    EDIT: Is it just that in 4) [tex]-B_{\nu\mu}=B_{\mu\nu}[/tex] ? But that does not work either... because if I plug that in, then I still get B+B
     
    Last edited: Jun 26, 2010
  2. jcsd
  3. Jun 26, 2010 #2

    HallsofIvy

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    Staff Emeritus
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    That's an "anti-symmetric" tensor. Any tensor that is not symmetric is an "asymmetric" tensor.

    Do you understand that [tex]A^{\mu\nu}B_{\mu\nu}[/tex] is sum[/b}? That we are summing over [itex]\mu= 1[/itex] to 4 and [itex]\nu= 1[/itex] to 4?
     
  4. Jun 26, 2010 #3

    Uku

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    Yes I do. I figure you gave me a hint there, I'll look at the problem again.
     
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