# Special relativity: proof of symmetry concept

1. Jun 26, 2010

### Uku

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

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

Now I need to show that:

$$A^{\mu\nu}B_{\mu\nu}=0$$ 0)

2. Relevant equations

We know that an asymmetric tensor can be written as:

$$A^{\mu\nu}=\frac{1}{2}(T^{\mu\nu}-T^{\nu\mu})$$ 1)

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

We can use 1) to write:

$$B_{\mu\nu}=\frac{1}{2}(B_{\mu\nu}-B_{\nu\mu})$$

Now we multiply this by $$A^{\mu\nu}$$:

$$A^{\mu\nu}B_{\mu\nu}=\frac{1}{2}(A^{\mu\nu}B_{\mu\nu}-A^{\mu\nu}B_{\nu\mu})$$ 2)

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

$$A^{\mu\nu}B_{\mu\nu}=\frac{1}{2}(A^{\mu\nu}B_{\mu\nu}-A^{\mu\nu}B_{\mu\nu})=\frac{1}{2}A^{\mu\nu}(B_{\mu\nu}-B_{\mu\nu})=0$$ 4)

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

$$...=\frac{1}{2}(B_{\mu\nu}+B_{\mu\nu})\neq0$$

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

EDIT: Is it just that in 4) $$-B_{\nu\mu}=B_{\mu\nu}$$ ? But that does not work either... because if I plug that in, then I still get B+B

Last edited: Jun 26, 2010
2. Jun 26, 2010

### HallsofIvy

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

Do you understand that $$A^{\mu\nu}B_{\mu\nu}$$ is sum[/b}? That we are summing over $\mu= 1$ to 4 and $\nu= 1$ to 4?

3. Jun 26, 2010

### Uku

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