Contraction of an asymmetric tensor?

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Dixanadu
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Hey guys,

So in my notes I've 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 don't see how the symmetric part contracts to 0.

*Note* I do also have written that [itex]a^{\mu\nu}=-a^{\nu\mu}[/itex] but I am 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 don't see why the symmetric part vanishes under contraction.

If someone could explain I'd be very grateful - thank you!
 
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Dixanadu said:
Hey guys,

So in my notes I've 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}-B^{\nu\mu})[/itex] as [itex]a_{\mu\nu} (A^{\mu\nu}+A^{\nu\mu}) = 0[/itex]. So I don't see how the symmetric part contracts to 0.

*Note* I do also have written that [itex]a^{\mu\nu}=-a^{\nu\mu}[/itex] but I am 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 don't see why the symmetric part vanishes under contraction.

If someone could explain I'd be very grateful - thank you!
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].
 
Whoops that B was meant to be an A -- error fixed! but what do you mean by inaccurate exactly? what part is wrong?
 
my2cts said:
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].
Dixanadu said:
Whoops that B was meant to be an A -- error fixed! but what do you mean by inaccurate exactly? what part is wrong?
The part where you wrote B instead of A ?
 
Yes -- sorry about that!