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.

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