Help with problem in MIT open course general relativity

[DavidL070949]

Hi! Apologies in advance for a somewhat long post.

First, by some background: Long ago in what seems like a galaxy far away, I got a Ph.D. in physics (statistical physics) before switching gears and going into law. The nature of my grad school was such that there wasn't any coursework - you got an office, and advisor and were expected to produce a thesis in a reasonable number of years. Great for learning to be independent but it did result in a number of holes in my education.

Fast forward a number of decades, post-retirement, my interest in physics got rekindled and I began to work my way through the MIT open course on general relativity on my own. Great course! But it can at times be a bit tough to do this solo and I've gotten stuck on Problem set 3, #3(b). There's a set of worked out solutions that I found online. Unfortunately, it contains errors, including (I think) one in the work on this problem, which deals with electromagnetic fields as seen by an observer with 4-velocity U.

I've attached the page with the problem in question. It seems like one way to address this is to plug in the formulas for E and B into the formula given for the generating field tensor and show that the rhs reduces to the lhs, which is what I did. From the problem, it seems like one is supposed to get to the place where the Levi-Civita symbol identity given with the problem should come into play. But the expression I got to only has one index in that part to sum over and the identity needs two. The other way to do this would be to contract the non-field tensor portions of the E and B 4-field formulas with both sides and show that the rhs reduces to the target field. That hasn't gotten me very far either. So, I'm stuck and would appreciate any help/insight anyone can add.

 

[robphy]

But the expression I got to only has one index in that part to sum over and the identity needs two. The other way to do this would be to contract the non-field tensor portions of the E and B 4-field formulas with both sides and show that the rhs reduces to the target field.

Can you show what you did? (Use MathJax.)
Here's the expression from 3b.
$F^{\alpha \beta}=U^\alpha E_{\vec{U}}^\beta-E_{\vec{U}}^\alpha U^\beta+\epsilon^{\alpha \beta}{ }_{\gamma \delta} U^\gamma B_{\vec{U}}^\delta$

 

[PeterDonis]

@DavidL070949 please post your math directly in the thread using LaTeX. There is a LaTeX Guide link at the bottom left of each post window.

 

[DavidL070949]

Thanks. Next task is definitely to figure out how to use LaTeX.

 

[DavidL070949]

And, got it no more attachments.

To the very helpful person who asked me to post my work, I'm working on learning enough about LaTex to do that properly. That may take a bit, but I wanted to thank you in advance for the prompt offer to help.

 

[PeterDonis]

no more attachments

If there's a link to what you're referencing, that's much better, yes.

 

[robphy]

And, got it no more attachments.

To the very helpful person who asked me to post my work, I'm working on learning enough about LaTex to do that properly. That may take a bit, but I wanted to thank you in advance for the prompt offer to help.

If you right-click on the equation in my post, you can see the $\LaTeX$-code used to create it.

 

[Sagittarius A-Star]

As delimiters, you can use double $ or double # on each side. See in below LaTex Guide link under "Delimiting your LaTeX code".

 

[robphy]

Probably useful (especially for conventions --- it's $(-,+,+,+)$; it uses Carroll's text)

https://ocw.mit.edu/courses/8-962-general-relativity-spring-2020/resources/mit8_962s20_pset03/
from
https://ocw.mit.edu/courses/8-962-general-relativity-spring-2020/pages/assignments/

https://ocw.mit.edu/courses/8-962-g...olumes-and-volume-elements-conservation-laws/
from
https://ocw.mit.edu/courses/8-962-general-relativity-spring-2020/video_galleries/video-lectures/

 

[Sagittarius A-Star]

So, I'm stuck and would appreciate any help/insight anyone can add.

What may help is the following hint in the description of problem #3:
$\vec{e}_{\hat 0} = \vec U$
In special relativity, in the restframe of the observer, his 4-velocity has the following components:
$U^\alpha = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix}$

You may right-click ("Show Math as", "TeX Commands") on the following EM field tensor notation if you need it's LaTeX code:

$F^{\alpha\beta}= \begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & B_z & -B_y \\ -E_y & -B_z & 0 & B_x \\ -E_z & B_y & -B_x & 0 \end{pmatrix}$

 

[DavidL070949]

Apologies for the delay, but I'm still getting used to LaTex. Anyway, in answer to how I got to the expression(s) that left me stuck: I subtituted the formulas for the E and B 4-vectors into the expression for the field in terms of those objects and ended up with three terms. Here they are, in order:

Term 1:

$$U^α E^β = U^α F^{βμ} U_μ$$​

Term 2:

$$- E^α U^β = - U^β F^{αμ} U_μ$$
​

Term 3:

$$\epsilon^{\alpha \beta}{ }_{\gamma \delta} U^\gamma B^\delta = - \frac {1} {2} U^γ U_μ F_{νσ} ε^{\alpha \beta}{ }_{\gamma \delta} \epsilon^{δμνσ}$$​

The problem offers the hint that the referenced identity relating to a sum over two of the four indices of the Levi-Civita symbol to Kroncker deltas. But, as you can see, unless I messed up, there's only one index in Term 3 to sum over. Also, the identity in question addresses the product of one LC symbol with all upstairs indices with another LC with the indices downstairs. That's not the case here. Which raises another question: since the LC is technically not a tensor, do you still use the metric to raise and lower indices? I would think not. And, if not, how do you deal with matching the index summations?

I also saw the suggestion to see if using the statement of the problem that, in the instantaneous local inertial reference frame of the moving oberver, U is along one of the basis vectors. Maybe I'm missing something, but wouldn't that be problematic in light of the assertion in part (b) that the equation is valid for any basis?

Finally, another poster pointed me to the MIT lecture resources as an aid. I actually got to this point by listening to those lectures and studying the notes that go with them. Those resources are absolutely terrific. And yet I remain stuck!

Once again, thanks for the help.​

​

 

[robphy]

since the LC is technically not a tensor, do you still use the metric to raise and lower indices?

There is a Levi-Civita tensor ( https://en.wikipedia.org/wiki/Levi-Civita_symbol#Example:_Minkowski_space ),
which is the volume element ( https://en.wikipedia.org/wiki/Volume_element#Volume_element_of_manifolds ).
This is what is used in the definition of the magnetic field associated with a 4-velocity.

The signature-convention and the definitions of the electric and magnetic field given in the problem set agree with those in Wald's General Relativity (p. 64 Eq. 4.2.21 and 4.2.22, and Appendix B.2 for volume elements). (I didn't see these definitions of the fields in Carroll's text.)


I worked out the details, after some tedious calculations (and dummy-variable renaming).
I followed the approach by Geroch that I quoted in my Insight
https://www.physicsforums.com/insig...r-this-Spacetime-Approach-to-Electromagnetism ,
using Geroch's and Wald's conventions (the conventions of your problem)....
[which is different from the conventions used in my Insight (where I used (+---) instead of Wald's (-+++))].

I will give an outline (following Geroch's approach) here... (I can fill in the details later.)

Define $h_{\alpha\beta} = g_{\alpha\beta} + U_\alpha U_\beta$, where $h^\alpha{}_\beta$ is a projection operator orthogonal to $U^\alpha$.

Then, write
\begin{align}
F^{\alpha\beta} &= F^{\mu\nu} g^{\alpha}{}_{\mu} g^{\beta}{}_{\nu}\\
&= F^{\mu\nu} \left( h^{\alpha}{}_{\mu} - U^{\alpha} U_{\mu} \right) \left( h^{\beta}{}_{\nu} - U^{\beta} U_{\nu} \right)\\
& \qquad \mbox{
(use FOIL)}\\
&=
F^{\mu\nu} h^{[\alpha}{}_{[\mu} h^{\beta]}{}_{\nu]}
+ U^{\alpha}E^{\beta}
- E^{\alpha}U^\beta
\end{align}

Over the years, it always bothered me that I got stuck here trying to express
$h^{[\alpha}{}_{[\mu} h^{\beta]}{}_{\nu]}$ in terms of the epsilons so that I can use the definition of the magnetic field.
So, I looked online for some help.
From what I learned, I filled in the details that I needed to see.

I used their two-contraction identity ,
which I write for clarity as
$\epsilon^{\alpha\beta\rho\sigma} \epsilon_{\mu\nu\rho\sigma} = (-1)^s\ 2 \left( \delta^{\alpha}_{\mu} \delta^{\beta}_{\nu} - \delta^{\alpha}_{\nu} \delta^{\beta}_{\mu} \right) \stackrel{s=1}{=} 2 \left( \delta^{\alpha}_{\nu} \delta^{\beta}_{\mu} - \delta^{\alpha}_{\mu} \delta^{\beta}_{\nu} \right)$, where $s$ is the number of minus-signs in the metric).
It was helpful to write this in alternative notations
\begin{align}
\epsilon^{\alpha\beta\rho\sigma} \epsilon_{\mu\nu\rho\sigma}
= (-1)^s\ 2 \left( \delta^{\alpha}_{\mu} \delta^{\beta}_{\nu} - \delta^{\alpha}_{\nu} \delta^{\beta}_{\mu} \right)
= (-1)^s\ 2\ \delta^{\alpha\beta}_{\mu\nu}
= (-1)^s\ 2
\begin{vmatrix}
\delta^\alpha_\mu & \delta^\alpha_\nu \\
\delta^\beta_\mu & \delta^\beta_\nu
\end{vmatrix}
\end{align}

I also found it helpful to use a one-contraction identity, which I write as
\begin{align}
\def\ALPHA{{\color{red}\alpha}}
\def\BETA{{\color{red}\beta}}
\def\MU{{\color{red}\mu}}
\def\NU{{\color{red}\nu}}
\epsilon^{\alpha\beta\gamma\sigma} \epsilon_{\mu\nu\rho\sigma}=(-1)^s\ \delta^{\alpha\beta\gamma}_{\mu\nu\rho}
= (-1)^s
\begin{vmatrix}
\delta^\alpha_\mu & \delta^\alpha_\nu & \delta^\alpha_\rho \\
\delta^\beta_\mu & \delta^\beta_\nu & \delta^\beta_\rho \\
\delta^\gamma_\mu & \delta^\gamma_\nu & \delta^\gamma_\rho
\end{vmatrix}
\end{align}

Here is the first connection to relate the h to the epsilon.
Use the two-contraction identity to show
\begin{align}
\epsilon^{\ALPHA\beta\rho\sigma} \epsilon_{\MU\nu\rho\sigma} U_\beta U^\nu
&=
2 \left( h^{\ALPHA}{}_{\MU} \right)
\end{align}

In forming $h^{[\alpha}{}_{[\mu} h^{\beta]}{}_{\nu]}$, one gets
\begin{align}
h^{[\ALPHA}{}_{[\MU} h^{\BETA]}{}_{\NU]}
&=
\delta^{[\ALPHA|}_{[\MU|} \delta^{|\BETA]}_{|\NU]}
+\delta^{[\ALPHA|}_{[\MU|} U^{|\BETA]} U_{|\NU]}
+ U^{[\ALPHA|} U_{[\MU|} \delta^{|\BETA]}_{|\NU]}
\end{align}.
Then "a miracle occurs".
This (from my online resource) turns out to be equal to $\frac{1}{2}\epsilon^{\ALPHA\BETA\gamma\omega}\epsilon_{\MU\NU\pi\omega} U_\gamma U^\pi$,
shown by using the one-contraction formula.
(It works, but I'm not a fan of this approach.
I prefer some more motivation from the antisymmetrized-hh (using the two-contraction identity)
to a more-recognizable expression using the one-contraction identity.
I wonder if this can be written as a Laplace-expansion of a determinant involving the deltas.)


With this, I finally get
\begin{align}
F^{\alpha\beta} &= F^{\mu\nu} g^{\alpha}{}_{\mu} g^{\beta}{}_{\nu}\\
&=
F^{\mu\nu} h^{[\alpha}{}_{[\mu} h^{\beta]}{}_{\nu]}
+ U^{\alpha}E^{\beta}
- E^{\alpha}U^\beta \\
&=
F^{\mu\nu} \left( \frac{1}{2}\ \epsilon^{\ALPHA\BETA\gamma\omega}\epsilon_{\MU\NU\pi\omega} U_\gamma U^\pi \right)
+ U^{\alpha}E^{\beta}
- E^{\alpha}U^\beta \\
&=
\epsilon^{\ALPHA\BETA}{}_{\gamma\omega} U^\gamma \left( B^\omega \right)
+ U^{\alpha}E^{\beta}
- E^{\alpha}U^\beta
\end{align}

 

[Sagittarius A-Star]

I also saw the suggestion to see if using the statement of the problem that, in the instantaneous local inertial reference frame of the moving oberver, U is along one of the basis vectors. Maybe I'm missing something, but wouldn't that be problematic in light of the assertion in part (b) that the equation is valid for any basis?

The following refers to special relativity.

It is sufficient to prove the validity of the equation of part (b) in only one basis. This would prove automatically it's validity in any basis.

The headline of problem (3) is: “3+1 split of the electromagnetic field". A "3+1" split is observer-dependent.

Only for an observer with $U^μ = (1, 0, 0, 0)$ the vectors, defined in (3a), are fully electric / magnetic:
$E^\mu = (0, E_1, E_2, E_3)$ and $B^\mu = (0, B_1, B_2, B_3)$.

The question in (3a) can be easily answered by showing that the inner products with the 4-velocity are zero. If an inner product it is zero in one basis, it is zero an any basis.

For observer $U^μ = (1, 0, 0, 0)$ the terms of the equation of problem (3b) have the following components:

$F^{\alpha \beta}=U^\alpha E_{\vec{U}}^\beta-E_{\vec{U}}^\alpha U^\beta+\epsilon^{\alpha \beta}{ }_{\gamma \delta} U^\gamma B_{\vec{U}}^\delta=$

$\begin{pmatrix} 0 & E_1 & E_2 & E_3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}- \begin{pmatrix} 0 & 0 &0 & 0 \\ E_1 & 0 & 0 & 0 \\ E_2 & 0 & 0 & 0 \\ E_3 & 0 & 0 & 0 \end{pmatrix}+ \epsilon^{\alpha \beta}{ }_{\gamma \delta} \begin{pmatrix} 0 & B_1 & B_2 & B_3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

$=\begin{pmatrix} 0 & E_1 & E_2 & E_3 \\ -E_1 & 0 & B_3 & -B_2 \\ -E_2 & -B_3 & 0 & B_1 \\ -E_3 & B_2 & -B_1 & 0 \end{pmatrix}$

 

[DavidL070949]

All of this is very helpful indeed! I'm going to try to work through the details on my own, so that I make sure I really understand and learn from the problem and its solution. And you're right about the metric to use. It's the same one that the MIT course uses.

What a great resource PF is!!!

 

[Sagittarius A-Star]

Additional insight: The 4-vector $E_{\vec U}^\alpha$ from problem (3a), multiplied with charge $q$, gives the Lorentz 4-force on a test charge with 4-velocity $U$.

$f^\alpha=qE_{\vec U}^\alpha=qF^{\alpha\beta} U_\beta$.