Special relativity the EM stress energy tensor

[peterjaybee]

Homework Statement

Using the expression below for the stress energy tensor of the em field, show that it has zero trace.

Homework Equations

T^{\mu\nu}=F^{\mu}_{ \alpha}F^{\alpha\nu}+\frac{1}{4}\eta^{\mu\nu}F_{\beta\gamma}F^{\beta\gamma}

F is the faraday tensor and eta is the minkowski metric.

The Attempt at a Solution

I started by trying to calculate T^{00} with the aim of then calculating the other diagonal components i.e. 11, 22 and 33. But I did not get anywhere. I couldn't get my head around the summations.

 

[nickjer]

Well, you could definitely solve it that way, but you would be creating a lot of unnecessary work. Think of it this way, the trace of your tensor is given by:

\eta_{\mu\nu}T^{\mu \nu} = T

So just multiply the right side by the minkowski metric. You will also notice that:

\eta_{\mu \nu} \eta^{\mu \nu} = 4

for 4 dimensions (3 space and 1 time).

 

[peterjaybee]

Thanks, for your reply. I think I can get the answer but there is one bit I just don't get. Ill come back to that at the end.

I get
\eta_{\mu\nu}T^{\mu\nu}=\eta_{\mu\nu}F^{\mu}_{ \alpha}F^{\alpha\nu}+\frac{1}{4}\eta_{\mu\nu}\eta^{\mu\nu}F_{\beta\gamma}F^{\beta\gamma}

=F_{ \nu\alpha}F^{\alpha\nu}+F_{\beta\gamma}F^{\beta\gamma}

Then using the antisymetry properties of the faraday tensor this can be written as

=-F_{\alpha\nu}F^{\alpha\nu}+F_{\beta\gamma}F^{\beta\gamma}

This is then zero because the the indicies are being summed over and as such are dummy indicies implying we can change \beta and \gamma to \alpha and \nu, thus the LHS goes to zero.

My issue with this is I do not understand why multiplying the stress energy tensor by the metric gives the trace of the metric.

If you had not told me that this gives the trace i would have said that
\eta_{\mu\nu}T^{\mu\nu}=T_{\nu}^{\nu}
which I assume is another matrix as opposed to the trace (i.e. a number).

It is the same with the metric multiplication
\eta_{\mu\nu}\eta^{\mu\nu}=\eta_{\nu}^{\nu}
Again I cannot see how this gives 4.

I think I am missing some key thing in my understanding of the topic that will make this obvious.

 

[Cyosis]

And T_{\nu}^{\nu}=T_{0}^{0}+T_{1}^{1}+T_{2}^{2}+T_{3}^{3}, which is the summation of all diagonal entries, which is the definition of the...?

 

[peterjaybee]

ok, but if that is the case how do you get
\eta_{\mu\nu}\eta^{\mu\nu}=\eta_{\nu}^{\nu}=4

surely because \eta^{\mu\nu}=diag(1,-1,-1,-1)

then

\eta_{\nu}^{\nu}=-2

 

[vela]

There's a difference between \eta^{\mu\nu} and \eta^\mu{}_\nu = \eta^{\mu a}\eta_{a\nu}. The latter is represented by the identity matrix.

 

[Fredrik]

See also the first comment in this post, and don't forget the definition of matrix multiplication.