Lorentz Transformation of the Fields

Let us consider the Lorentz transformation of the fields. Clearly $A_\mu$ just transforms like a vector. We could derive the transformed $E$ and $B$ fields using the derivatives of $A_\mu$ but it is interesting to see how the electric and magnetic fields transform.

We know that Maxwell's equations indicate that if we transform a static electric field to a moving frame, a magnetic field will be generated, because there is a current in that frame. Think of the case of a point charge in its rest frame and in a frame in which it is moving. It was clear from E&M that $\vec{E}$ and $\vec{B}$ were not simply parts of 4-vectors.

The Electric and Magnetic fields are part of a rank 2 tensor and so they transform accordingly.

$$F'_{\mu\nu} = B_{\mu\rho}F_{\rho\sigma}B^T_{\sigma\nu}$$

$$F'_{\mu\nu} = \begin{pmatrix}\gamma & i\beta\gamma & 0 & 0 \cr -i\beta\gamma ... ...\beta\gamma & \gamma & 0 & 0\cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0& 1 \cr\end{pmatrix}$$

$$F'_{\mu\nu} = \begin{pmatrix}\gamma & i\beta\gamma & 0 & 0 \cr -i\beta\gamma ... ...amma E_z+i\beta\gamma B_y & -\beta\gamma E_z+\gamma B_y & -B_x & 0\end{pmatrix}$$

$$F'_{\mu\nu} = \begin{pmatrix}-\beta\gamma^2E_x+\beta\gamma^2E_x & i\gamma^2E_... ...amma E_z+i\beta\gamma B_y & -\beta\gamma E_z+\gamma B_y & -B_x & 0\end{pmatrix}$$

$$F'_{\mu\nu} = \begin{pmatrix}0 & iE_x & i\gamma (E_y+\beta B_z) & i\gamma (E_... ...B_x\cr -i\gamma (E_z-\beta B_y) & \gamma(B_y-\beta E_z) & -B_x & 0\end{pmatrix}$$

$$E'_\parallel = E_\parallel$$

$$B'_\parallel = B_\parallel$$

$$E'_\perp = \gamma(E_\perp -\vec{\beta}\times\vec{B})$$

$$B'_\perp = \gamma(B_\perp +\vec{\beta}\times\vec{E})$$

Since we could choose any direction for the $x$ axis that we boosted along, these results for the field transformation are correct for all boosts.