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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