I have a question regarding Misner, Charles W.; Thorne, Kip S.; Wheeler, John Archibald (1973), Gravitation ISBN 978-0-7167-0344-0. It is a book about Einstein's theory of gravitation.

At the page 79, in the Chapter 3.4. about the Maxwell's equation, we assume that we are in a Rocket with velocity $\beta^j$ with $j$ being x, y and z. But we change to a system of coordinates so that the rocket is moving in the z direction.

The first result introduced is the calculation of the electric field along the z after the Lorentz tranformation:

$\bar E_z = F_\bar {30} =\Lambda^\alpha_{\bar 3}\Lambda^\beta_{\bar 0} F_{\alpha \beta} = (1-\beta^2) F_{30} = E_z$

I have no problem with that. However, in the calculation along the x axis:

$\bar E_x = F_\bar {10} =\Lambda^\alpha_{\bar 1}\Lambda^\beta_{\bar 0} F_{\alpha \beta} = \gamma F_{10} + \beta \gamma F_{13} = \gamma(E_x-\beta B_y)$

However, I don't understand how come $F_{13}$ appears, I would have said that $F_{01}$ appears. So what am I missing?

  • 1
    $\begingroup$ Do you understand what $\Lambda^\alpha_{\bar 3}\Lambda^\beta_{\bar 0} F_{\alpha \beta}$ means? For which values of $\beta$ is $\Lambda^\beta_{\bar 0}$ nonzero? ($0$ and $3$.) $\endgroup$
    – G. Smith
    Jun 22, 2019 at 17:01
  • $\begingroup$ $F_{01}=-F_{10}$ which shows that your expectation cannot be correct. $\endgroup$
    – my2cts
    Jun 22, 2019 at 17:26
  • $\begingroup$ @G.Smith I thought I did understand it. Why is it nonzero for 0 and 3? That's my question. $\endgroup$ Jun 22, 2019 at 18:08
  • $\begingroup$ Look at eqn 2.45. The only nonzero elements are in positions 00, 11, 22, 33, 03, and 30. $\endgroup$
    – G. Smith
    Jun 22, 2019 at 20:02
  • $\begingroup$ In case you are confused about contracted tensor expressions, $\Lambda^\alpha_{\bar 3}\Lambda^\beta_{\bar 0} F_{\alpha \beta}$ would be, for a general Lorentz transformation, a sum of 16 terms. You sum over $\alpha$ from $0$ to $3$ and over $\beta$ from $0$ to $3$. But for a Lorentz boost along one of the coordinate axes, many of these terms are zero. $\endgroup$
    – G. Smith
    Jun 22, 2019 at 20:40

2 Answers 2


Actually the last of eqs 3.4 on page 79 on my "Gravitation" reads

$\bar E_x = F_\bar {10} =\Lambda^\alpha_{\bar 1}\Lambda^\beta_{\bar 0} F_{\alpha \beta} = \gamma F_{10} + \beta \gamma F_{13} = \gamma(E_x-\beta B_y)$

which is perfectly correct.

Please note that the $ \Lambda^\alpha_{\bar \beta}$ that they are using is given by eq 2.45 on page 69 (boost in the z direction).

Sure you got the original MTW ? :-)

PS After the OP edited the question...

In the equation $\bar E_x = F_\bar {10} =\Lambda^\alpha_{\bar 1}\Lambda^\beta_{\bar 0} F_{\alpha \beta} $ the $\alpha$ and $\beta$ indices are summed over, giving in general 4*4 terms. However most of these terms are 0 because $\Lambda$ has mostly empty components (see eq 2.45 right, on page 69). For example if $\alpha=0$ and $\beta=1$ the term $\Lambda^0_{\bar 1}\Lambda^1_{\bar 0} F_{0 1} $ is null since $\Lambda^0_{\bar 1}$ and $\Lambda^1_{\bar 0} $ are null

  • $\begingroup$ Oh yes it is correct, I just mistyped it. Why is it 13 and not 01? $\endgroup$ Jun 22, 2019 at 15:20
  • $\begingroup$ I edited my question $\endgroup$ Jun 22, 2019 at 15:33

I simply misread the tensor.

The first term is the product of $\Lambda^1_{\bar 1}=1$ and $\Lambda^0_{\bar 0}=\gamma$, while the second is the product of $\Lambda^1_{\bar 1} = 1$ and $\Lambda^3_{\bar 0}=\beta \gamma$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.