# Misner, Thorne, and Wheeler: Exercise 5.1 (stress-energy tensor symmetry)

• A
Mike Karr
TL;DR Summary
It seems that the stress-energy tensor is both symmetric and non-symmetric.
I am a beginner in GR, working my through Gravitation by the above authors. If there is a better place to ask this question, please let me know.

I understand (from section 5.7) that the stress-energy tensor is symmetric, and from equation 5.23 (p. 141), it is explicitly symmetric. But evaluating equation 5.22 in a Lorentz coordinate frame, the last term is clearly symmetric, but the first term involves ##F^{\nu}_{\alpha}##, which is non-symmetric, at least according to equation 5.3 (p. 73). In particular, it is symmetric in the 0th column and row and anti-symmetric in the rest of the matrix. Multiplied by the anti-symmetric ##F^{\mu\alpha}## does not help the situation. So I can't get from 5.22 to 5.23.

I suppose my problem is understanding "in a Lorentz frame", but I thought all definitions of F and ##\nu## *are* in a Lorentz frame. Where am I going wrong?

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ergospherical
The first term is symmetric for the simple reason that ##S^{\mu \nu} \equiv F^{\mu \alpha} {F^{\nu}}_{\alpha} = {F^{\mu}}_{\alpha} F^{\nu \alpha} = F^{\nu \alpha}{F^{\mu}}_{\alpha} = S^{\nu \mu}##.

vanhees71
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Summary:: It seems that the stress-energy tensor is both symmetric and non-symmetric.

Multiplied by the anti-symmetric $F^{\mu\alpha}$ does not help the situation.
As @ergospherical showed, yes it does. The index ##\alpha## is not a free index and you therefore cannot consider if the expression is symmetric under exchange ##\mu \leftrightarrow \alpha## because doing so does not make any sense.

vanhees71
Mike Karr
As @ergospherical showed, yes it does. The index ##\alpha## is not a free index and you therefore cannot consider if the expression is symmetric under exchange ##\mu \leftrightarrow \alpha## because doing so does not make any sense.
Of course. Sorry to bother you with something so trivial.

cianfa72
The first term is symmetric for the simple reason that ##S^{\mu \nu} \equiv F^{\mu \alpha} {F^{\nu}}_{\alpha} = {F^{\mu}}_{\alpha} F^{\nu \alpha}##
For the last term above I found ##F^{\mu \alpha} {F^{\nu}}_{\alpha} = {F^{\mu}}_{\gamma} \eta^{\alpha \gamma} {F^{\nu}}_{\alpha} = {F^{\mu}}_{\gamma} F^{\nu \gamma}## , then renaming the dummy index ##\gamma \leftrightarrow \alpha## we get
$$S^{\mu \nu} = {F^{\mu}}_{\alpha} F^{\nu \alpha} = F^{\nu \alpha}{F^{\mu}}_{\alpha} = S^{\nu \mu}$$
Btw, just from a formal point of view, is it required to switch the indexes of the metric tensor ##\eta^{\alpha \gamma}## in order to raise the ##\alpha## index of ##{F^{\nu}}_{\alpha}## turning it into ##F^{\nu \gamma}## ? Thank you.

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Btw, just from a formal point of view, is it required to switch the indexes of the metric tensor ηαγ in order to raise the α index of Fνα turning it in Fνγ ?
That is kind of a moot point since the metric is symmetric by definition. Most people will do this without blinking.

cianfa72
That is kind of a moot point since the metric is symmetric by definition. Most people will do this without blinking.
yes of course, however from what you said the process of raising or lowering a tensor index by the metric tensor formally seems to require that.

Mentor
is it required to switch the indexes of the metric tensor ##\eta^{\alpha \gamma}##
What does "switch the indexes" even mean? The labels you put on the indexes are arbitrary.

cianfa72
What does "switch the indexes" even mean? The labels you put on the indexes are arbitrary.
If you see for example here https://en.wikipedia.org/wiki/Raising_and_lowering_indices the process of raising or lowering a tensor index involves as dummy/summing index the second index of the metric tensor.

Said that, from a practical point of view since the metric tensor is symmetric, there is actually no business to switch the indexes as in the above #5.

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ergospherical
It's irrelevant for tensors over a real vector space. You do have to worry about the order with spinorial tensors, whose indices are by convention lowered by contracting over the left index of ##\epsilon_{AB}## and vice versa, e.g. ##v_A = \epsilon_{BA} v^B = - \epsilon_{AB} v^B##. (One consequence of this is that swapping the heights of a pair of contracted indices introduces a minus sign, ##u_A v^A = - u^A v_A##).

dextercioby
Mentor
If you see for example here https://en.wikipedia.org/wiki/Raising_and_lowering_indices the process of raising or lowering a tensor index involves as dummy/summing index the second index of the metric tensor.
Assuming that is correct (Wikipedia is not always a reliable source), why is that a problem?

cianfa72
As I said in #9 there is no problem at all ! As @ergospherical clarified for tensors over real vector spaces it is actually irrelevant.

Mike Karr
Of course. Sorry to bother you with something so trivial.
Sorry, I got unconfused too quickly. I don't get the correct answer for ## T^{00} ##. Let's look at the two terms of equation 5.22. The last one is simple, I think:

## - \frac{1}{4} \eta^{\mu\nu} F_{\alpha\beta} F^{\alpha\beta}
= - \frac{1}{4} \eta^{\mu\nu} 2(E_x^2 + E_y^2 + E_z^2 + B_x^2 + B_y^2 + B_z^2)
= - \frac{1}{2} \eta^{\mu\nu} (E^2 + B^2) ##

Let's look at the ## \mu = 0 = \nu ## of the first term. The ## \alpha ##'s (second index) run across the rows of the matrices on pp. 73-74, so:

## F^{0\alpha} F^0_\alpha = (0, -E_x, -E_y, -E_z) \cdot (0, E_x, E_y, E_z)
= - E_x^2 - E_y^2 - E_z^2 = - E^2 ##

Since ## \eta^{00} = -1 ## (p. 53), the sum of these two terms is ## \frac{1}{2} ( - E^2 + B^2) ##, so dividing by ## 4 \pi ## gives:

## \frac{1}{8 \pi} ( - E^2 + B^2) ##

But the answer given by Ex. 5.1 is:

## \frac{1}{8 \pi} (E^2 + B^2) ##

What am I doing wrong?

ergospherical
In a local Lorentz frame ##\eta = \mathrm{diag}(-1,\mathbf{I})## so\begin{align*}
4\pi T^{00} &= F^{0\alpha} {F^0}_{\alpha} + \frac{1}{4} F_{\alpha \beta} F^{\alpha \beta}
\end{align*}Recall that ##E_i = F_{0i}## and ##B_i = -(1/2) \epsilon_{ijk} F^{jk}##. That means ##F^{0\alpha} {F^0}_{\alpha} = E^i E_i = E^2## and also \begin{align*}
F_{\alpha \beta} F^{\alpha \beta} &= 2F_{0i} F^{0i} + F_{ij} F^{ij} \\
&= -2E_i E^i + (\epsilon_{ijk} B^k)(\epsilon_{ijl} B_l)
\end{align*}Use the epsilon-delta identity: ##\epsilon_{ijk} \epsilon_{ijl} = \delta_{jj} \delta_{kl} - \delta_{jl} \delta_{kj} = 2\delta_{kl}##. Then\begin{align*}
F_{\alpha \beta} F^{\alpha \beta} &= -2E^2 + 2B^k B_l \delta_{kl} = 2(-E^2 + B^2)
\end{align*}Overall: ##4\pi T^{00} = E^2 + \frac{1}{2}(-E^2 + B^2) = \frac{1}{2}(E^2 + B^2)##.

Mike Karr
ergospherical, thanks for your answer. What I was doing wrong was using the wrong components in ##F^{\alpha\beta} ##---I swapped the 0th row and 0th column. But your answer is much more clever, using formulas for ##E_i## and especially for ##B_i## that I would never have thought of. Also, I did not know of the epsilon-delta identity. Is it in MTW? Again, thanks for the help.

ergospherical
Also, I did not know of the epsilon-delta identity. Is it in MTW?
not sure, but the general formula is a fiddly determinant: https://en.wikipedia.org/wiki/Levi-Civita_symbol#Product

the most useful version to memorise is a single contracted index ##\epsilon_{ijk} \epsilon_{ilm} = \delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl}## (mnemonic: something like "product of same positions minus product of opposite positions")