Electromagnetic Invariants and Field Classification

Electric and magnetic fields depend on the observer, but two contractions of the electromagnetic tensor do not. They decide whether a subluminal boost can remove one field, make the two fields parallel, or do neither. The invariant test should come before any guessed boost.

Required background. Special Relativity supplies Lorentz transformations, invariant contractions, and index raising and lowering; Calculus and Linear Algebra supplies dot products, cross products, and vector identities. Helpful background. Ordinary Differential Equations supplies limiting and initial-value reasoning; Differential Forms and Integration supplies the Hodge dual and orientation.

The two algebraic invariants

Use the chapter conventions

$$F_{0i}=-E_i, \qquad F_{ij}=\epsilon_{ijk}B^k, \qquad (\star F)_{\mu\nu} =\frac12F^{\rho\sigma}\epsilon_{\rho\sigma\mu\nu}.$$

The two independent quadratic contractions are

$$I=F_{\mu\nu}F^{\mu\nu}, \qquad J=F_{\mu\nu}(\star F)^{\mu\nu}.$$

Raising a temporal index changes its sign, while raising spatial indices does not. Summing the antisymmetric pairs gives

$$\boxed{ I=2(B^2-E^2), \qquad J=4\,\mathbf E\mathbin{\cdot}\mathbf B.}$$

The distinction between the two contractions matters. $I$ is a Lorentz scalar. $J$ is a pseudoscalar: it is invariant under proper orientation-preserving Lorentz transformations and changes sign under an improper transformation. Its sign also reverses if the chosen spacetime orientation is reversed. The conditions $J=0$ and $J^2>0$ are independent of that orientation choice. The alternating tensor and the restriction to the proper Lorentz group are discussed in Tong 2015, §5.3.5, printed pp. 111–113, PDF.

Classification and simplest frames

The local algebraic classification is compact:

Invariant condition	Field type	Simplest inertial frame
$J=0$ and $I\lt0$	Electric-dominated	$\mathbf B'=0$
$J=0$ and $I\gt0$	Magnetic-dominated	$\mathbf E'=0$
$J\ne0$	Mixed non-null	$\mathbf E'$ and $\mathbf B'$ parallel and both nonzero
$I=J=0$ and $F\ne0$	Nonzero null	Neither field can be removed by a subluminal boost
$F=0$	Trivial	Both fields vanish

For a nonzero null field, every inertial observer finds

$$E'^2=B'^2, \qquad \mathbf E'\mathbin{\cdot}\mathbf B'=0.$$

This is the local algebraic pattern of an ideal plane wave, but the converse is not automatic. Vanishing invariants at one event do not prove that the field solves the vacuum Maxwell equations or forms a globally propagating plane wave.

For any non-null field, choose a frame in which the fields are parallel. Their canonical amplitudes obey

$$\begin{aligned} E_{\mathrm{can}}^2 &=\frac{\sqrt{I^2+J^2}-I}{4},\\ B_{\mathrm{can}}^2 &=\frac{\sqrt{I^2+J^2}+I}{4},\\ E_{\mathrm{can}}B_{\mathrm{can}} &=\frac{J}{4}. \end{aligned}$$

Take $E_{\mathrm{can}}\ge0$ and treat $B_{\mathrm{can}}$ as a signed parallel amplitude so the last equation holds. In the purely magnetic case $E_{\mathrm{can}}=0$, the sign of $B_{\mathrm{can}}$ is not fixed by the two quadratic invariants. These relations reproduce the purely electric and magnetic limits and show why $J\ne0$ forces both fields to remain nonzero.

Constructing the simplifying boost

The general boost transformations are given in Tong 2015, §5.3.2, printed pp. 106–108, PDF. Choose $\mathbf v$ parallel to $\mathbf E\times\mathbf B$, so $\mathbf v$ is perpendicular to both fields. The longitudinal correction terms then vanish and

$$\mathbf E'=\gamma(\mathbf E+\mathbf v\times\mathbf B), \qquad \mathbf B'=\gamma(\mathbf B-\mathbf v\times\mathbf E).$$

For a non-null field, the velocity

$$\boxed{ \mathbf v= \frac{2\,\mathbf E\times\mathbf B} {E^2+B^2+\sqrt{(E^2-B^2)^2 +4(\mathbf E\mathbin{\cdot}\mathbf B)^2}}}$$

makes $\mathbf E'$ and $\mathbf B'$ parallel. To see why it is physical, define

$$D=(E^2-B^2)^2+4(\mathbf E\mathbin{\cdot}\mathbf B)^2.$$

The identity

$$(E^2+B^2)^2-D=4\lvert\mathbf E\times\mathbf B\rvert^2$$

gives

$$v^2=\frac{E^2+B^2-\sqrt D}{E^2+B^2+\sqrt D}\lt1$$

whenever $D>0$. For a nonzero null field, $D=0$ and the same expression demands $v=1$, which no inertial observer can attain.

The formula contains the eliminating boosts as special cases. If $\mathbf E\mathbin{\cdot}\mathbf B=0$ and $E\gt B$, then

$$\mathbf v=\frac{\mathbf E\times\mathbf B}{E^2}, \qquad v=\frac{B}{E}\lt1,$$

and $\mathbf B'=0$. If $B\gt E$, then

$$\mathbf v=\frac{\mathbf E\times\mathbf B}{B^2}, \qquad v=\frac{E}{B}\lt1,$$

and $\mathbf E'=0$. Tong summarizes these pure-field frame conditions in Tong 2015, §5.4.2, printed p. 118, PDF.

Operational meaning and limitations

The fields $\mathbf E$ and $\mathbf B$ are observer-dependent projections of $F$. A momentarily comoving test charge measures the electric part through force per unit charge; magnetic effects appear for moving charges or magnetic dipoles. The invariants classify the tensor at an event without selecting an observer.

This classification assumes fields measured in an orthonormal inertial frame. Coordinate components in a non-orthonormal basis are not automatically the measured three-vectors. The analysis is also algebraic and local: it says nothing by itself about sources, boundary conditions, or how the field evolves.

Common pitfalls

Calling $J$ an unrestricted scalar. It is a pseudoscalar. Its numerical sign assumes an orientation, although its vanishing does not.

Equating zero invariants with zero field. A nonzero null field has $I=J=0$. Check $F$ itself.

Using orthogonality alone. The condition $\mathbf E\mathbin{\cdot}\mathbf B=0$ is necessary for eliminating one field, but the sign of $I$ decides which field can be removed by a subluminal boost.

Exercises

Crossed laboratory fields

Let $\mathbf E=E_{\mathrm L}\widehat{\mathbf x}$ and $\mathbf B=(E_{\mathrm L}/2)\widehat{\mathbf y}$. Classify the field and find a boost that removes $\mathbf B$.

Solution

Here

$$J=0, \qquad I=2\left(\frac{E_{\mathrm L}^2}{4}-E_{\mathrm L}^2\right) =-\frac32E_{\mathrm L}^2\lt0.$$

The field is electric-dominated. Since

$$\mathbf E\times\mathbf B =\frac{E_{\mathrm L}^2}{2}\widehat{\mathbf z},$$

choose $\mathbf v=\tfrac12\widehat{\mathbf z}$. Then $\mathbf v\times\mathbf E=\mathbf B$, so $\mathbf B'=0$. The invariant fixes the remaining magnitude:

$$E'^2=E^2-B^2=\frac34E_{\mathrm L}^2.$$

A genuinely mixed field

Let $\mathbf E=(2,0,0)$ and $\mathbf B=(1,1,0)$ in common units. Find $I$, $J$, the parallelizing boost, and the canonical amplitudes.

Solution

The dot and cross products are

$$\mathbf E\mathbin{\cdot}\mathbf B=2, \qquad \mathbf E\times\mathbf B=(0,0,2).$$

Therefore

$$I=2(2-4)=-4, \qquad J=8, \qquad \sqrt{I^2+J^2}=4\sqrt5.$$

The field is mixed. The boost formula gives

$$\mathbf v =\frac{4}{6+2\sqrt5}\widehat{\mathbf z} =\frac{3-\sqrt5}{2}\widehat{\mathbf z},$$

which is subluminal. The canonical amplitudes satisfy

$$E_{\mathrm{can}}^2=1+\sqrt5, \qquad B_{\mathrm{can}}^2=\sqrt5-1, \qquad E_{\mathrm{can}}B_{\mathrm{can}}=2.$$

Both remain nonzero, as required by $J\ne0$.

Lorentz Force and Electromagnetic Stress-Energy next turns the same tensor into force, energy density, energy flux, and momentum flux.

References

• David Tong, Electromagnetism (PDF), §§5.3.2, 5.3.5, and 5.4.2, printed pp. 106–108, 111–113, and 118, University of Cambridge Mathematical Tripos lecture notes, Lent Term 2015.