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.