# Electric / magnetic field transformations

1. Sep 23, 2007

### 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
$$F_{\mu\nu} = \begin{pmatrix} 0 & -E_1 & -E_2 & -E_3 \\ E_1 & 0 & B_3 & -B_2 \\ E_2 & -B_3 & 0 & B_1 \\ E_3 & B_2 & -B_1 & 0 \\ \end{pmatrix} \qquad\text{ so } F^{\mu\nu} = \begin{pmatrix} 0 & E_1 & E_2 & E_3 \\ -E_1 & 0 & B_3 & -B_2 \\ -E_2 & -B_3 & 0 & B_1 \\ -E_3 & B_2 & -B_1 & 0 \\ \end{pmatrix}$$
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
$$R^\mu_\nu = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta & 0 \\ 0 & \sin\theta & \cos\theta & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$
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
$$E_1' = E_1 \cos\theta - E_2 \sin \theta; \quad E_2' = E_1 \sin\theta + E_2 \cos \theta; \quad E_3' = E_3,$$
which can be written in vector notation as
$$\vec E' = \mat R \vec E \qquad\text{ where } \mat R = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \\ \end{pmatrix},$$
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
$$B_1' = B_1 \cos\theta - B_2 \sin\theta; \quad B_2' = B_1 \sin\theta + B_2 \cos\theta; \quad B_3' = B_3,$$
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.
$$R^\mu_\nu \to \Lambda^\mu_\nu = \begin{pmatrix} \cosh\theta & \sinh\theta & 0 & 0 \\ \sinh\theta & \cosh\theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$

Thanks a lot.

2. Sep 23, 2007

### jostpuur

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.

3. Sep 23, 2007

### pervect

Staff Emeritus
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}{ \it Ey}&-{\it Bz}&0&{\it Bx}\\\noalign{\medskip}{\it Ez}&{\it By}&-{ \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& \sin \left( \theta \right) &\cos \left( \theta \right) &0 \\\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 \right) -{\it Ey}\,\cos \left( \theta \right) &-{\it Ez} \\\noalign{\medskip}{\it Ex}\,\cos \left( \theta \right) +{\it Ey}\, \sin \left( \theta \right) &0&{\it Bz}\, \left( \cos \left( \theta \right) \right) ^{2}+{\it Bz}\, \left( \sin \left( \theta \right) \right) ^{2}&-{\it By}\,\cos \left( \theta \right) +{\it Bx}\,\sin \left( \theta \right) \\\noalign{\medskip}-{\it Ex}\,\sin \left( \theta \right) +{\it Ey}\,\cos \left( \theta \right) &-{\it Bz}\, \left( \sin \left( \theta \right) \right) ^{2}-{\it Bz}\, \left( \cos \left( \theta \right) \right) ^{2}&0&{\it By}\,\sin \left( \theta \right) +{\it Bx}\,\cos \left( \theta \right) \\\noalign{\medskip}{\it Ez}&{\it By}\,\cos \left( \theta \right) -{ \it Bx}\,\sin \left( \theta \right) &-{\it By}\,\sin \left( \theta \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.

Last edited: Sep 23, 2007
4. Sep 23, 2007

### 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.

I found it, there was an error in my Mathematica code (it read
Code (Text):
e[i_] := Sum[R[[i, \[Mu]]] R[[1, \[Nu]]] F[[\[Mu], \[Nu]]], {\[Mu], 1, 4}, {\[Nu], 1, 4}]
Code (Text):
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 apparantly we agree now.[/edit].

Last edited: Sep 23, 2007