CompuChip
Sep23-07, 05:42 AM
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.
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.