Electric / magnetic field transformations

[CompuChip]

Hi. I thought I had tensors and Lorentz transformations under control, but now I'm in doubt again.

For example, consider the electromagnetic field tensor
$$F_{\mu\nu} = \begin{pmatrix}<br /> 0 & -E_1 & -E_2 & -E_3 \\<br /> E_1 & 0 & B_3 & -B_2 \\<br /> E_2 & -B_3 & 0 & B_1 \\<br /> E_3 & B_2 & -B_1 & 0 \\<br /> \end{pmatrix} <br /> \qquad\text{ so } <br /> F^{\mu\nu} = \begin{pmatrix}<br /> 0 & E_1 & E_2 & E_3 \\<br /> -E_1 & 0 & B_3 & -B_2 \\<br /> -E_2 & -B_3 & 0 & B_1 \\<br /> -E_3 & B_2 & -B_1 & 0 \\<br /> \end{pmatrix}$$
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 $\theta$ about the $z$-axis. Now I thought I'd write this as
$$R^\mu_\nu = \begin{pmatrix}<br /> 1 & 0 & 0 & 0 \\<br /> 0 & \cos\theta & -\sin\theta & 0 \\<br /> 0 & \sin\theta & \cos\theta & 0 \\<br /> 0 & 0 & 0 & 1 \\<br /> \end{pmatrix}$$
as it works on a vector and produces a vector ($$(v')^\mu = R^\mu_\nu v^\nu$$).
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 $E_i = F_{i0}$ transform as
$$E_i' = F'_{i0} = R_i^\mu R_0^\nu F_{\mu\nu}.$$
Working out the transformation yields
$$E_1' = E_1 \cos\theta - E_2 \sin \theta; \quad<br /> E_2' = E_1 \sin\theta + E_2 \cos \theta; \quad<br /> E_3' = E_3,$$
which can be written in vector notation as
$$\vec E' = \mat R \vec E<br /> \qquad\text{ where }<br /> \mat R = \begin{pmatrix}<br /> \cos\theta & -\sin\theta & 0 \\<br /> \sin\theta & \cos\theta & 0 \\<br /> 0 & 0 & 1 \\<br /> \end{pmatrix},$$
which is, I think, the transformation rule for a vector hence what one would expect.

Similarly, the components of the magnetic field are $B_i = \frac12 \epsilon_{ijk} F^{jk}$.
As raising both the indices on $F_{\mu\nu}$ does not affect the components in the
lower right $3 \times 3$ block -- that is, $F_{ij} = F^{ij}$ for $i, j = 1, 2, 3$ --
we can calculate
$$B_i' = \frac12 \epsilon_{ijk} F'^{jk} = \frac12 \epsilon_{ijk} R^j_\mu R^k_\nu F^{\mu\nu}.$$
Explicit calculation yields
$$B_1' = B_1 \cos\theta - B_2 \sin\theta; \quad<br /> B_2' = B_1 \sin\theta + B_2 \cos\theta; \quad<br /> B_3' = B_3,$$
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.
$$R^\mu_\nu \to \Lambda^\mu_\nu = \begin{pmatrix}<br /> \cosh\theta & \sinh\theta & 0 & 0 \\<br /> \sinh\theta & \cosh\theta & 0 & 0 \\<br /> 0 & 0 & 1 & 0 \\<br /> 0 & 0 & 0 & 1 \\<br /> \end{pmatrix}$$

Thanks a lot.

 

[jostpuur]

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.

Even though it is a pseudo-vector, it is supposed to transform like a vector in rotations. You should check space reflections to see if you get the desired difference in the transformation of E and B.

 

[pervect]

Using GRtensor, I take

$$F_{ab} =\left[ \begin {array}{cccc} 0&-{\it Ex}&-{\it Ey}&-{\it Ez}\\\noalign{\medskip}{\it Ex}&0&{\it Bz}&-{\it By}\\\noalign{\medskip}{<br /> \it Ey}&-{\it Bz}&0&{\it Bx}\\\noalign{\medskip}{\it Ez}&{\it By}&-{<br /> \it Bx}&0\end {array} \right]$$

and
$$L^a{}_b = \left[ \begin {array}{cccc} 1&0&0&0\\\noalign{\medskip}0&\cos \left( \theta \right) &-\sin \left( \theta \right) &0\\\noalign{\medskip}0&<br /> \sin \left( \theta \right) &\cos \left( \theta \right) &0<br /> \\\noalign{\medskip}0&0&0&1\end {array} \right]$$

to compute

$$F^{\prime}_{ab} = F_{cd} L^c{}_a L^d{}_b$$

which is

$$\left[ \begin {array}{cccc} 0&-{\it Ex}\,\cos \left( \theta \right) -{\it Ey}\,\sin \left( \theta \right) &{\it Ex}\,\sin \left( \theta<br /> \right) -{\it Ey}\,\cos \left( \theta \right) &-{\it Ez}<br /> \\\noalign{\medskip}{\it Ex}\,\cos \left( \theta \right) +{\it Ey}\,<br /> \sin \left( \theta \right) &0&{\it Bz}\, \left( \cos \left( \theta<br /> \right) \right) ^{2}+{\it Bz}\, \left( \sin \left( \theta \right) <br /> \right) ^{2}&-{\it By}\,\cos \left( \theta \right) +{\it Bx}\,\sin<br /> \left( \theta \right) \\\noalign{\medskip}-{\it Ex}\,\sin \left( <br /> \theta \right) +{\it Ey}\,\cos \left( \theta \right) &-{\it Bz}\,<br /> \left( \sin \left( \theta \right) \right) ^{2}-{\it Bz}\, \left( <br /> \cos \left( \theta \right) \right) ^{2}&0&{\it By}\,\sin \left( <br /> \theta \right) +{\it Bx}\,\cos \left( \theta \right) <br /> \\\noalign{\medskip}{\it Ez}&{\it By}\,\cos \left( \theta \right) -{<br /> \it Bx}\,\sin \left( \theta \right) &-{\it By}\,\sin \left( \theta<br /> \right) -{\it Bx}\,\cos \left( \theta \right) &0\end {array} \right]$$

I didn't think this quite matched some of the signs in your result, but I thought it might be helpful.

 

[CompuChip]

jostpuur, of course you are right about the (pseudo)-vector remark. Thanks.

pervect: I think you calculated $L^T F L$ whereas I did $L F L^T$. One of us should be wrong then (and it's probably me), which would mean I messed up the indices... just what I was afraid of.

[edit]I found it, there was an error in my Mathematica code (it read

e[i_] := Sum[R[[i, \[Mu]]] R[[1, \[Nu]]] F[[\[Mu], \[Nu]]], {\[Mu], 1, 4}, {\[Nu], 1, 4}]

instead of

e[i_] := Sum[R[[\[Mu], i]] R[[\[Nu], 1]] F[[\[Mu], \[Nu]]], {\[Mu], 1, 4}, {\[Nu], 1, 4}]

-- note the indices of the rotation matrix.
So apparently we agree now.[/edit].