Understanding and Proving the Antisymmetry of the Electromagnetic Field Tensor

[ehrenfest]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

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

 

[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 ?

 

[ehrenfest]

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

 

[dextercioby]

What equality of the mixed partial are you taling about ?

 

[ehrenfest]

What equality of the mixed partial are you taling about ?

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?

 

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

 

[ehrenfest]

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

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

 

[dextercioby]

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

 

[ehrenfest]

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

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

But why does that help us?

 

[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.

 

[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?

 

[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 ?

 

[ehrenfest]

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 ?

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?