# Relativity and electromagnetic field

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?

• 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 $3$.) Jun 22, 2019 at 17:01
• $F_{01}=-F_{10}$ which shows that your expectation cannot be correct. Jun 22, 2019 at 17:26
• @G.Smith I thought I did understand it. Why is it nonzero for 0 and 3? That's my question. Jun 22, 2019 at 18:08
• Look at eqn 2.45. The only nonzero elements are in positions 00, 11, 22, 33, 03, and 30. Jun 22, 2019 at 20:02
• 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. Jun 22, 2019 at 20:40

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

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

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 $$\Lambda^3_{\bar 0}=\beta \gamma$$.