Faraday tensor

Maxwell’s equations already follow relativity, but they aren’t formulated using the language of geometric spacetime objecst which we have been using. We will see how we can represent the equations as well as the Lorentz force using the Faraday tensor.

What one observer considers to be an electric field another observer might consider to be a magnetic field. This is because a charge that is moving in one reference frame may be stationary in another, and so feel no magnetic force. However, all observers must agree on whether there is a force or not — some just might call it electric while others call it magnetic.

We’re looking for a way to pack $\tv E$ and $\tv B$ into a spacetime object. Together, these two have 6 components, which happens to be the number of independent components in an antisymmetric tensor.

We start by writing the Lorentz force as

\tv F = q \left( c \frac{\tv E}{c} + \tv v \times \tv B \right).

This form should make clearer the symmetry between the $\tv E$ and $\tv B$ part: both are multiplied by a speed, and $c$ looks like the timelike component of 4-velocity for a slow-moving object.

Using this hint along with the 4-force, we look for an antisymmetric tensor which satisfies

\frac{\d p^\alpha}{\d\tau} = qF^{\alpha\beta} u_\beta.

Evaluating this component by component, we find

\begin{align*} \frac{\d p^x}{\d\tau} &= q (F^{10} u_0 + F^{12}u_2 + F^{13} u_3) \\ &= \gamma q\left (-cF^{10} + \frac{\d y}{\d t}F^{12} + \frac{\d z}{\d t}F^{13} \right) \\ \frac{\d p^x}{\d t} &= q\left (-cF^{10} + \frac{\d y}{\d t}F^{12} + \frac{\d z}{\d t}F^{13} \right) \\ &= q\left( E^x + \frac{\d y}{\d t}B^z - \frac{\d z}{\d t} B^y \right). \end{align*}

From which we conclude

F^{10} = - \frac{E^x}{c}, \quad F^{12} = B^z, \quad F^{13} = - B^y.

Continuing this analysis for the other components, we find the full Faraday tensor to be

F^{\alpha\beta} \compeq \mat{ 0 & E^x/c & E^y/c & E^z/c \\ -E^x/c & 0 & B^z & -B^y \\ -E^y/c & -B^z & 0 & B^x \\ -E^z/c & B^y & -B^x & 0 }.

Transforming fields

Now that we have a geometric representation of our electric and magnetic fields, we should be able to transform them with the Lorentz transform.

F^{\mu'\nu'} = F^{\alpha\beta} \Lambda^{\mu'}{}_\alpha \Lambda^{\nu'}{}_\beta.

We perform the transformation to find

\begin{align*} E^{x'} &= E^x, & B^{x'} &= B^x,\\ E^{y'} &= \gamma (E^y - vB^z), & B^{y'} &= \gamma(B^y + v E^z/c^2),\\ E^{z'} &= \gamma (E^z + vB^y), \quad& B^{z'} &= \gamma(B^z - vE^y/c^2). \end{align*}

More generally, for some velocity $\tv v$ , the transformed fields are

\begin{align*} \tv E'_\parallel &= \tv E_\parallel, &\quad \tv E'_\perp &= \gamma (\tv E_\perp + \tv v \times \tv B_\perp), \\ \tv B'_\parallel &= \tv B_\parallel, & \tv B'_\perp &= \gamma (\tv B_\perp - \tv v \times \tv E_\perp / c^2). \end{align*}