# Electromagnetic field tensor: conservation of mass

Required math: algebra

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 4; Problems 4.6.

As we’ll study in more detail a bit later, the electric and magnetic fields can be combined into a single tensor known as the electromagnetic field tensor ${F^{ij}}$:

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

We can see from its definition that this tensor is anti-symmetric, that is, that ${F^{ij}=-F^{ji}}$. For any anti-symmetric tensor we can show that

$\displaystyle F^{ij}\eta_{ia}\eta_{jb}u^{a}u^{b}=0 \ \ \ \ \ (2)$

In this equation, ${\eta_{ij}}$ is the metric tensor in flat space and ${u^{a}}$ is the four-velocity, but in fact the formula is valid for any tensors ${\eta}$ and ${u}$, provided that ${F}$ is anti-symmetric. The proof involves a bit of index-switching.

 $\displaystyle F^{ij}\eta_{ia}\eta_{jb}u^{a}u^{b}$ $\displaystyle =$ $\displaystyle -F^{ji}\eta_{ia}\eta_{jb}u^{a}u^{b}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -F^{ij}\eta_{ja}\eta_{ib}u^{a}u^{b}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -F^{ij}\eta_{jb}\eta_{ia}u^{b}u^{a} \ \ \ \ \ (5)$

In the second line, we swapped the dummy indexes ${i}$ and ${j}$, and in the third line we swapped ${a}$ and ${b}$. The result shows that the original quantity is equal to its negative, which means it must be zero.

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

$\displaystyle \frac{dp^{i}}{d\tau}=qF^{ij}\eta_{ja}u^{a} \ \ \ \ \ (6)$

where ${u^{a}=\gamma\left[1,v_{x},v_{y},v_{z}\right]}$ is the four-velocity.

For example, if ${i=1}$ we get

 $\displaystyle \frac{dp^{1}}{d\tau}$ $\displaystyle =$ $\displaystyle q\gamma\left(E_{x}+v_{y}B_{z}-v_{z}B_{y}\right) \ \ \ \ \ (7)$

In the non-relativistic limit, ${\gamma\rightarrow1}$ and this is the ${x}$ component of the force law ${\frac{d\mathbf{p}}{dt}=q\mathbf{E}+q\mathbf{v}\times\mathbf{B}}$. We’ll explore some of the other properties of this tensor later.

Since the square of the four-momentum of a particle is the negative of its mass squared (${\mathbf{p}\cdot\mathbf{p}=\gamma^{2}m^{2}\left(-1+v^{2}\right)=-m^{2}}$), this should be conserved for a charged particle moving in an electromagnetic field. (Its total momentum is, of course, not conserved since the fields exert a force on the particle.)

We have

 $\displaystyle \frac{d\left(\mathbf{p}\cdot\mathbf{p}\right)}{d\tau}$ $\displaystyle =$ $\displaystyle \frac{d}{d\tau}\left(\eta_{ij}p^{i}p^{j}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta_{ij}\left[\frac{dp^{i}}{d\tau}p^{j}+p^{i}\frac{dp^{j}}{d\tau}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\eta_{ij}\frac{dp^{i}}{d\tau}p^{j}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2q\eta_{ij}F^{ik}\eta_{ka}u^{a}p^{j}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2qmF^{ik}\eta_{ij}\eta_{ka}u^{a}u^{j}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (13)$

In the third line, we used the fact that ${\eta_{ij}=\eta_{ji}}$ and swapped ${i}$ and ${j}$ in the second term. The fourth line uses 6 and the last line uses 2.

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