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Contraction of an asymmetric tensor?

  1. Nov 11, 2014 #1
    Hey guys,

    So in my notes ive got this statement written:

    If tensor with no symmetry properties, [itex]A^{\mu\nu}[/itex], contracts to [itex]a_{\mu\nu}[/itex], we can write this as [itex]A^{\mu\nu}a_{\mu\nu}=\frac{1}{2}a_{\mu\nu}(A^{\mu\nu}-A^{\nu\mu})[/itex] as [itex]a_{\mu\nu} (A^{\mu\nu}+A^{\nu\mu}) = 0[/itex]. So I dont see how the symmetric part contracts to 0.

    *Note* I do also have written that [itex]a^{\mu\nu}=-a^{\nu\mu}[/itex] but Im not sure if this is relevant.

    I understand that you can decompose the tensor [itex]A^{\mu\nu}[/itex] into the sum of symmetric and anti-symmetric parts, but i dont see why the symmetric part vanishes under contraction.

    If someone could explain I'd be very grateful - thank you!
    Last edited: Nov 11, 2014
  2. jcsd
  3. Nov 11, 2014 #2
    Your notes are inaccurate. What you tried to note down is probably:
    If [itex]a^{\mu\nu}=-a^{\nu\mu}[/itex], then [itex]A^{\mu\nu}a_{\mu\nu}=\frac{1}{2}a_{\mu\nu}(A^{\mu\nu}-A^{\nu\mu})[/itex],
    for any [itex]A^{\mu\nu}[/itex].
  4. Nov 11, 2014 #3
    Whoops that B was meant to be an A -- error fixed! but what do you mean by inaccurate exactly? what part is wrong?
  5. Nov 12, 2014 #4
    The part where you wrote B instead of A ?
  6. Nov 12, 2014 #5
    Yes -- sorry about that!
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