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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?

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    $\begingroup$ 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 '19 at 17:01
  • $\begingroup$ $F_{01}=-F_{10}$ which shows that your expectation cannot be correct. $\endgroup$
    – my2cts
    Jun 22 '19 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 '19 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 '19 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 '19 at 20:40
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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

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  • $\begingroup$ Oh yes it is correct, I just mistyped it. Why is it 13 and not 01? $\endgroup$ Jun 22 '19 at 15:20
  • $\begingroup$ I edited my question $\endgroup$ Jun 22 '19 at 15:33
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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 $\Lambda^3_{\bar 0}=\beta \gamma$.

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