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

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  • #1
Mike Karr
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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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  • #2
ergospherical
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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}##.
 
  • #3
Orodruin
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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.
 
  • #4
Mike Karr
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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.
 
  • #5
cianfa72
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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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Orodruin
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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.
 
  • #7
cianfa72
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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.
 
  • #8
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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.
 
  • #9
cianfa72
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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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  • #10
ergospherical
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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##).
 
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  • #12
cianfa72
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As I said in #9 there is no problem at all ! As @ergospherical clarified for tensors over real vector spaces it is actually irrelevant.
 
  • #13
Mike Karr
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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?
 
  • #14
ergospherical
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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)##.
 
  • #15
Mike Karr
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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.
 
  • #16
ergospherical
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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")
 

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