# Field tensor

1. Aug 28, 2007

### ehrenfest

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

Zwiebach 44

My book defines

$$T_{\lambda \mu \nu} = \partial_{\lambda} F_{\mu \nu} + \partial_{\mu} F_{ \nu \lambda} + \partial_{\nu} F_{\lambda \mu }$$

where F is the electromagnetic field tensor
and says that it is identically zero due to Maxwell's. It then asks me to prove that it is antisymmetric?

2. Relevant equations

3. The attempt at a solution

Why would I need to prove that when it is identically zero due to Maxwell's equations.

2. Aug 28, 2007

### dextercioby

Just don't take into account the Maxwell equations. Think of F as a simple 2-form. How would you go about proving that T is a 3-form ?

3. Aug 28, 2007

### ehrenfest

Actually, even if you forget about Maxwell's equations, it is still identically zero due to the equality of mixed partials?

4. Aug 28, 2007

### dextercioby

What equality of the mixed partial are you taling about ?

5. Aug 28, 2007

### ehrenfest

So,

$$T_{\lambda \mu \nu} = \partial_{\lambda} F_{\mu \nu} + \partial_{\mu} F_{ \nu \lambda} + \partial_{\nu} F_{\lambda \mu } = \partial_{\lambda} (\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} ) + \partial_{\mu} (\partial_{\nu} A_{\lambda} - \partial_{\lambda} A_{\nu} ) + \partial_{\nu} (\partial_{\lambda} A_{\mu} - \partial_{\mu} A_{\lambda} ) }$$ which is identically 0 due to the commutativity of partial derivates without even assuming Maxwell's equation are true. I am confused about what is left to prove?

6. Aug 29, 2007

### dextercioby

What you wrote there is nothing but d^{2}=0. But just forget for a moment that F=dA. (Just the way you forgot that $\delta F= 0$). Prove that T is completely antisymmetric.

HINT: $$g^{\lambda\mu}T_{\lambda\mu\nu}=...?$$.

7. Aug 29, 2007

### ehrenfest

Sorry. This is probably a silly question--but what is g ?

8. Aug 30, 2007

### dextercioby

The metric tensor on the flat Minkowski space. Sorry, i guess Zwiebach uses $\eta^{\lambda\mu}$.

9. Aug 30, 2007

### ehrenfest

$$g^{\lambda\mu}T_{\lambda\mu\nu}= \sum_{v=0}^3 -T_{00v} + T_{11v} + T_{22v} + T_{33v}$$

But why does that help us?

10. Aug 30, 2007

### dextercioby

Oh, deer (!)... Do you agree that

$$g^{\lambda\mu}T_{\lambda\mu\nu}=\partial^{\mu}F_{\mu\nu}+\partial^{\mu}F_{\nu\mu}+\partial_{\nu}F^{\mu}{}_{\mu}$$

??

If so, use the fact that F is antisymmetric and you're 1/3 done. The other 2/3 can be done in the same way, just pick the other 2 possible pair of indices.

11. Aug 30, 2007

### ehrenfest

OK. So, now I have that

$$(g^{\lambda \nu} + g^{\nu \mu} + g^{\lambda\mu}) T_{\lambda\mu\nu}=\partial_{\nu}F^{\ mu}{}_{\mu} + \partial_{\lambda}F^{\ nu}{}_{\nu} + \partial_{\mu}F^{\lambda}{}_{\lambda}$$

and I can find a similar expression involving $$T_{\mu\lambda\nu}$$ but now how do I get the negative relationship between them?

12. Aug 30, 2007

### dextercioby

Hmm. There's no need to shuffle the indices of T. Only of g. Now do you see that each contraction of g and T is identically zero ?

13. Aug 30, 2007

### ehrenfest

I was changing the indices of T in order to show that it is antisymmetric. I thought we wanted to show that if you switched two indices, the sign flips.

No. I do not see that. What do you mean "each contraction of g and T"? Is the equation I wrote in my previous post incorrect?