# Electromagnetic field tensor: change in kinetic energy

Required math: algebra

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 4; Problems 4.7.

In the last post, we introduced the electromagnetic field tensor ${F^{ij}}$:

$\displaystyle F^{ij}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (1)$

In terms of ${F^{ij}}$, the electric and magnetic (Lorentz) force laws for a charge ${q}$ can be combined into a single equation:

$\displaystyle \frac{dp^{i}}{d\tau}=qF^{ij}\eta_{ja}u^{a} \ \ \ \ \ (2)$

where ${u^{a}=\gamma\left[1,v_{x},v_{y},v_{z}\right]}$ is the four-velocity. The three spatial components give the force law, but what about the time component? The time component of the four-momentum is the relativistic energy which, for a particle of rest mass ${m}$ is ${\gamma m}$. If we expand this in a Taylor series for small ${v}$, we get

$\displaystyle \gamma m=m+\frac{1}{2}mv^{2}+\ldots \ \ \ \ \ (3)$

Thus the relativistic energy is the rest mass plus the Newtonian kinetic energy (plus higher order terms). In the small-${v}$ limit, ${\gamma\rightarrow1}$ and the ${t}$ component of 2 therefore is

 $\displaystyle \frac{d}{d\tau}\left(m+\frac{1}{2}mv^{2}\right)$ $\displaystyle =$ $\displaystyle q\left[\gamma E_{x}v_{x}+\gamma E_{y}v_{y}+\gamma E_{z}v_{z}\right]\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle q\gamma\mathbf{v}\cdot\mathbf{E}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle q\mathbf{v}\cdot\mathbf{E} \ \ \ \ \ (6)$

Since the rest mass doesn’t change, this equation is saying that the rate of change of kinetic energy is ${q\mathbf{v}\cdot\mathbf{E}}$. Does this make sense?

The force on a charge in an electric field is ${q\mathbf{E}}$, so the work done by this field in moving the charge through a distance ${d\mathbf{r}}$ is ${q\left(d\mathbf{r}\cdot\mathbf{E}\right)}$. This work accelerates the charge, thus increasing its kinetic energy. The rate at which the kinetic energy increases is therefore ${q\frac{d}{dt}\left(d\mathbf{r}\cdot\mathbf{E}\right)=q\mathbf{v}\cdot\mathbf{E}}$.

Since the magnetic force on a charge is always perpendicular to the direction of motion, magnetic forces do no work, so there is no contribution to the kinetic energy from the magnetic field.