Hi. I thought I had tensors and Lorentz transformations under control, but now I'm in doubt again.(adsbygoogle = window.adsbygoogle || []).push({});

For example, consider the electromagnetic field tensor

[tex] F_{\mu\nu} = \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}

\qquad\text{ so }

F^{\mu\nu} = \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}

[/tex]

in the (-1, 1, 1, 1) metric.

Now we apply a Lorentz transformation, and to keep it simple we take a (counter clockwise) rotation around an angle [itex]\theta[/itex] about the [itex]z[/itex]-axis. Now I thought I'd write this as

[tex] R^\mu_\nu = \begin{pmatrix}

1 & 0 & 0 & 0 \\

0 & \cos\theta & -\sin\theta & 0 \\

0 & \sin\theta & \cos\theta & 0 \\

0 & 0 & 0 & 1 \\

\end{pmatrix}

[/tex]

as it works on a vector and produces a vector ([tex](v')^\mu = R^\mu_\nu v^\nu[/tex]).

Did I get this right? In this case wrong placement of the indices doesn't introduce errors yet, as far as I can see, but this will generally not be the case for boosts (which do not have just zeros in the first column and row).

Now the components of the electric field [itex]E_i = F_{i0}[/itex] transform as

[tex] E_i' = F'_{i0} = R_i^\mu R_0^\nu F_{\mu\nu}. [/tex]

Working out the transformation yields

[tex] E_1' = E_1 \cos\theta - E_2 \sin \theta; \quad

E_2' = E_1 \sin\theta + E_2 \cos \theta; \quad

E_3' = E_3,

[/tex]

which can be written in vector notation as

[tex] \vec E' = \mat R \vec E

\qquad\text{ where }

\mat R = \begin{pmatrix}

\cos\theta & -\sin\theta & 0 \\

\sin\theta & \cos\theta & 0 \\

0 & 0 & 1 \\

\end{pmatrix},

[/tex]

which is, I think, the transformation rule for a vector hence what one would expect.

Similarly, the components of the magnetic field are [itex]B_i = \frac12 \epsilon_{ijk} F^{jk}[/itex].

As raising both the indices on [itex]F_{\mu\nu}[/itex] does not affect the components in the

lower right [itex]3 \times 3[/itex] block -- that is, [itex]F_{ij} = F^{ij}[/itex] for [itex]i, j = 1, 2, 3[/itex] --

we can calculate

[tex] B_i' = \frac12 \epsilon_{ijk} F'^{jk} = \frac12 \epsilon_{ijk} R^j_\mu R^k_\nu F^{\mu\nu}. [/tex]

Explicit calculation yields

[tex] B_1' = B_1 \cos\theta - B_2 \sin\theta; \quad

B_2' = B_1 \sin\theta + B_2 \cos\theta; \quad

B_3' = B_3,

[/tex]

which is exactly the same as the electric field. Yet the magnetic field is not a vector, but a pseudo-vector; therefore I doubt my answer.

I'd like to get this right, especially with the indices etc., before I proceed to boosts, e.g.

[tex] R^\mu_\nu \to \Lambda^\mu_\nu = \begin{pmatrix}

\cosh\theta & \sinh\theta & 0 & 0 \\

\sinh\theta & \cosh\theta & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \\

\end{pmatrix}

[/tex]

Thanks a lot.

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# Electric / magnetic field transformations

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