advanced25 min studyLesson 264 in Physics

Electromagnetic Lorentz invariants

Use the two electromagnetic invariants to classify fields, rule out impossible rest frames, and identify when a purely electric or magnetic description exists.

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What you’ll understand

  • Evaluate both electromagnetic invariants with correct SI units.
  • Determine when one field can vanish in an inertial frame.
  • Distinguish a nonzero null field from a vanishing field tensor.

Transforming all six field components can be unnecessary. Two scalar combinations already tell us which transformations are possible and which are forbidden. They serve the same diagnostic role as an interval's sign in special relativity, but describe an electromagnetic tensor rather than an event separation. We will use these invariants to classify crossed fields and explain why chasing a light wave never produces a frame in which the wave has only an electric field.

Contract the tensor with the metric

Retain signature (+,−,−,−)(+,-,-,-), x0=ctx^0=ct, F0i=−Ei/cF^{0i}=-E_i/c, and Fij=−ϵijkBkF^{ij}=-\epsilon^{ijk}B_k. Contracting both tensor indices gives an invariant. Every independent antisymmetric entry appears twice because both index orders are summed. The time-space entries contribute negatively after index lowering, while the purely spatial entries contribute positively. The result compares magnetic field squared with electric field squared divided by c squared. Both terms have units of tesla squared, making the subtraction physically meaningful in SI.

There is a second independent invariant, the ordinary spatial dot product E⋅B\mathbf E\cdot\mathbf B. It is unchanged by proper Lorentz transformations, including the boosts used here. Under a spatial reflection it changes sign because the electric field is a polar vector and the magnetic field an axial vector; it is therefore called a pseudoscalar. For our classification only whether it vanishes is needed. Do not replace either invariant by the sum of field magnitudes: the sum contributes to energy density and is observer dependent.

I1=12FμνFμν=B2−E2c2,I2=E⋅BI_1=\frac12F^{\mu\nu}F_{\mu\nu}=B^2-\frac{E^2}{c^2},\qquad I_2=\mathbf E\cdot\mathbf B
I1 has units T² and I2 has units (V/m) T. Both are invariant under proper Lorentz transformations; I2 is a pseudoscalar under parity. E and B are measured at the same event.↔ Scroll wide equations horizontally.

Ask whether one field can disappear

If a frame has no electric field, its second invariant is zero and its first invariant is positive unless the entire tensor vanishes. Therefore nonzero I2I_2 immediately rules out both a purely electric and a purely magnetic frame. When the fields are perpendicular and I1>0I_1>0, the magnetic field is strong enough that a subluminal boost can remove the electric field. The necessary boost points along the original Poynting direction. Substituting into the field transformation proves sufficiency as well as necessity for this crossed-field case.

For perpendicular fields with I1<0I_1<0, a different subluminal boost removes the magnetic field. The remaining electric magnitude follows from the invariant without repeating the transformation. These statements are local: a boost chosen from the fields at one event may not remove a component everywhere in a nonuniform field. A single global inertial frame requires the appropriate velocity to be the same throughout the region. Fields that are parallel with a nonzero dot product cannot be reduced to a single kind by any inertial boost.

VE′=0=E×BB2(E<cB),VB′=0=c2E×BE2(E>cB),E⋅B=0\mathbf V_{E'=0}=\frac{\mathbf E\times\mathbf B}{B^2}\quad(E<cB),\qquad \mathbf V_{B'=0}=c^2\frac{\mathbf E\times\mathbf B}{E^2}\quad(E>cB),\qquad \mathbf E\cdot\mathbf B=0
These frame velocities apply to nonzero perpendicular fields. The inequalities ensure speeds below c. They transform to a magnetic-only or electric-only field at the event, not automatically throughout spacetime.↔ Scroll wide equations horizontally.

The null case retains a real field

When both invariants vanish, either the tensor is zero or the field is nonzero and null. A plane electromagnetic wave in vacuum is the familiar nonzero example: its electric and magnetic vectors are perpendicular and their magnitudes satisfy E=cBE=cB. Trying either elimination velocity now requires speed c. No inertial observer reaches that speed. The zero invariants do not indicate zero energy density, zero force on a charge, or zero radiation. They indicate a special algebraic class whose electromagnetic energy cannot be brought to rest by an inertial boost.

A useful workflow is to compute both invariants before and after any proposed component transformation. A mismatch diagnoses an error, often a missing gamma, an incorrect boost sign, or a lost factor of c. Matching invariants is necessary but not sufficient to validate every component: many different tensors share the same two numbers. The invariants classify fields coarsely and constrain possible frames; they do not determine spatial direction, the complete tensor, or its derivatives. Maxwell's equations still supply information that these local scalars cannot contain.

Predict the magnetic-only magnitude

A crossed field has B=5 mTB=5\,\mathrm{mT} and E=0.6cBE=0.6cB. The electric field points along y and the magnetic field along z. Find both invariants, the boost eliminating E, and the remaining field.

  1. Perpendicularity gives I2=0I_2=0. The first invariant is I1=(1−0.36)B2=0.64(25×10−6)=16×10−6 T2I_1=(1-0.36)B^2=0.64(25\times10^{-6})=16\times10^{-6}\,\mathrm{T^2}, which is positive.

  2. The elimination velocity is E×B/B2=0.6cx^\mathbf E\times\mathbf B/B^2=0.6c\hat{\mathbf x}. Its speed is allowed, so a magnetic-only frame exists. Here γ=1.25\gamma=1.25.

  3. In that frame E′=0E'=0, hence B′2=I1B'^2=I_1. Taking the positive magnitude gives B′=4 mTB'=4\,\mathrm{mT}. Direct transformation also gives γ(1−0.36)B=0.8B\gamma(1-0.36)B=0.8B, confirming the answer.

The result

The field is magnetically dominated: an observer moving along positive x at 0.6c measures a 4 mT magnetic field and no electric field.

Explore the model

Interactive lab

Interactive lab · spacetime

An invariant is stronger than component appearances

Reset and set BOTH β = 0 and ct = 0.4, then for a fresh comparison Reset and set β = 0.8 and ct = 0.4. Predict a negative interval in both runs for the fixed x = 0.6 event, even though its transformed time changes sign. Inspect paused readouts or Step.

An invariant is stronger than component appearancesA fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Current readings: Lorentz factor γ: 1.12; Event x′: 0.168; Event ct′: 0.817. All parameters are adjustable with the labeled controls below.−4.69−2.3502.354.69−2.2−1.101.12.2x or x′ · c = 1ct or ct′EE′ct² − x² = 0.64
Original eventTransformed eventLight cone
Lorentz factor γ
1.12
Event x′
0.168
Event ct′
0.817
t = 0 s

A fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThis illustrates the logic of invariant classification using an event interval. It does not compute electromagnetic invariants, field energies, or a wave rest frame.

Check your understanding

A nonzero field has E perpendicular to B and E = cB. Which conclusion follows?

Try it yourself

Let E=5cby^\mathbf E=5cb\hat{\mathbf y} and B=3bz^\mathbf B=3b\hat{\mathbf z}, where b=1 mTb=1\,\mathrm{mT}. Find the boost eliminating B and the surviving electric magnitude.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Classify the sign of I1 first. For the electric-only frame use the velocity with E squared in the denominator.

Compare with the solution

Here I2=0I_2=0 and I1=(9−25)b2=−16b2I_1=(9-25)b^2=-16b^2. The electric-only frame has V=c2EB/E2=(3/5)cV=c^2EB/E^2=(3/5)c along positive x. Then γ=1.25\gamma=1.25, Bz′=1.25(3−0.6×5)b=0B'_z=1.25(3-0.6\times5)b=0, and Ey′=1.25(5−0.6×3)cb=4cbE'_y=1.25(5-0.6\times3)cb=4cb. With c=3.00×108 m s−1c=3.00\times10^8\,\mathrm{m\,s^{-1}}, the remaining electric magnitude is 1.20×106 V m−11.20\times10^6\,\mathrm{V\,m^{-1}}, consistent with the invariant.

Take it with you

  • Field invariants constrain frame changes without fixing every component.
  • A zero dot product and magnetic or electric dominance permit one field to disappear locally.
  • A nonzero null field has zero invariants but can carry energy and momentum.

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