Transformation of Electric and Magnetic Fields

Here is a field you could build this afternoon: $\vec E = (0,\ 3.00\times10^{7},\ 0)\ \mathrm{V\,m^{-1}}$ crossed with $\vec B = (0,\ 0,\ 0.500)\ \mathrm{T}$ . Both are real, both measurable, both push on charges. Now walk past it at $6.00\times10^{7}\ \mathrm{m\,s^{-1}}$ — a fifth of the speed of light, brisk but hardly exotic. The electric field is gone. Not small: zero, identically, in every direction. What remains is a pure magnetic field of $0.48988\ \mathrm{T}$ and nothing else. You did not switch the electric field off. You changed your velocity, and a quantity one observer measures as $3.00\times10^{7}\ \mathrm{V\,m^{-1}}$ another measures as exactly nothing. The electric field is not a property of the world; it is a property of the world and your motion through it. The obvious question is when you can get away with this — and that is the real content of this article. Not every field can be boosted away. Which ones can is settled entirely by two numbers you compute in one frame and that every other frame agrees on:

F_{\mu\nu}F^{\mu\nu} = 2\left(B^2 - \frac{E^2}{c^2}\right), \qquad\qquad F_{\mu\nu}\tilde F^{\mu\nu} = -\frac{4}{c}\,\vec E\cdot\vec B .

A frame with no electric field exists if and only if $\vec E\cdot\vec B = 0$ and $B^2 > E^2/c^2$ . If $\vec E\cdot\vec B = 0$ with $E = cB$ exactly — a light wave — neither field can ever be removed, by anybody. The invariants do not merely describe the field; they tell you in advance what you are allowed to do to it. For the field above, $\vec E\cdot\vec B = 0$ and $F_{\mu\nu}F^{\mu\nu} = +0.47997\ \mathrm{T^2} > 0$ : magnetically dominated, so a pure-magnetic frame exists — and the invariant tells you the answer before you boost: $|\vec B'| = \sqrt{0.47997/2} = 0.48988\ \mathrm{T}$ , exactly what the transformation delivers.

Key takeaways

Signature $(+,-,-,-)$ ; $\;\partial^\mu = \big(\tfrac1c\partial_t,\ -\nabla\big)$ ; $\;p\cdot p = +m^2c^2$ ; $\;F^{0i} = -E^i/c$ , $\;F^{ij} = -\epsilon^{ijk}B_k$ .
Everything here is one line applied twice: $\;\boxed{F'^{\mu\nu} = \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta\, F^{\alpha\beta}}$ . There is no separate "law of field transformation" — there is a tensor, and tensors transform.
The four results. Along the boost, nothing moves: $\;\vec E'_\parallel = \vec E_\parallel$ , $\;\vec B'_\parallel = \vec B_\parallel$ . Across it, the fields mix:

\vec E'_\perp = \gamma\big(\vec E + \vec v\times\vec B\big)_\perp, \qquad \vec B'_\perp = \gamma\Big(\vec B - \frac{\vec v\times\vec E}{c^2}\Big)_\perp .

The two invariants — and you must declare $\epsilon^{0123} = +1$ or the second one has no sign: $\;F_{\mu\nu}F^{\mu\nu} = 2(B^2 - E^2/c^2)$ and $\;F_{\mu\nu}\tilde F^{\mu\nu} = -\tfrac{4}{c}\vec E\cdot\vec B$ . Every observer computes the same two numbers.
THE CLASSIFICATION, which is what the invariants buy: a pure-magnetic frame exists iff $\vec E\cdot\vec B = 0$ and $B^2 > E^2/c^2$ — reached at $\;\vec v = (\vec E\times\vec B)/B^2$ , which is subluminal precisely because $B^2 > E^2/c^2$ . A pure-electric frame iff $\vec E\cdot\vec B = 0$ and $E^2/c^2 > B^2$ . If $\vec E\cdot\vec B = 0$ with $E = cB$ (null — a light wave), neither field can be removed in any frame; and if $\vec E\cdot\vec B \ne 0$ , no frame kills either one.
Key idea: a boost does not convert $\vec E$ into $\vec B$ . There is one antisymmetric tensor with six slots, and a boost re-shuffles the same six numbers among the same six slots. Whether you call the result "electric" or "magnetic" is a statement about your velocity, not about the world. What the world actually fixes are the two invariants — and they decide, in advance, which descriptions are available to you at all.

The core insight

The invariants are not decoration on top of the transformation laws; they are the constraint those laws must respect. Six numbers go into a boost and six come out — but not just anywhere: $F_{\mu\nu}F^{\mu\nu}$ and $F_{\mu\nu}\tilde F^{\mu\nu}$ are frozen. That is why the classification exists at all.

Where this lives today

These four formulae are load-bearing infrastructure. A proton bunch at $\gamma = 7000$ carries a field that is, in the lab, a nearly flat pancake of transverse $\vec E$ and $\vec B$ locked in the ratio $E = cB$ — the near-null field of an ultra-relativistic charge. Accelerator beam-beam calculations are done by boosting into the bunch's own frame, where the field is a simple Coulomb field, and boosting back; the Weizsäcker–Williams method takes the same limit seriously and treats a fast charge's field as a beam of real photons. In astrophysics, "is this plasma magnetically or electrically dominated?" is answered by $I_1$ — a number every observer agrees on, and therefore the only well-posed version of the question. And the humblest application is the one that started it all: magnetism is what electrostatics looks like from a moving frame — carried out in full in Magnetism as a Relativistic Effect.

Origins · 2 minElectromagnetic field and how this was discovered

Electromagnetic field

1831–1908 · Faraday's lines · Maxwell's equations · Einstein's asymmetry · Minkowski's tensor

The electromagnetic field is the protagonist of this article, and it took seventy-seven years to become one object. It began as two. Faraday, in 1831, filled the space around a magnet with lines of force — real, physical, strained — and around a charge with a separate set. Two pictures, joined only by his discovery that a changing one produced the other. Maxwell, in 1865, wrote the joining down: four equations in which $\vec E$ and $\vec B$ curl into one another, and out of which fell a wave travelling at a speed he could compute from a capacitor and a coil — and which turned out to be the speed of light. It was the greatest unification in physics to that point, and it still left $\vec E$ and $\vec B$ as two fields with a lot to say to each other. Einstein noticed the remaining seam. He opened the 1905 paper not with clocks or trains but with a complaint about electrodynamics: move a magnet near a conductor and you explain the current one way; move the conductor near the magnet and you explain the same current a completely different way. The asymmetry, he wrote, "does not appear to be inherent in the phenomena." The fix was not a better explanation but a demotion — the electric and magnetic fields are not fundamental, they are components. Minkowski supplied the object they are components of. In 1908, three years after Einstein and less than a year before his own death at forty-four, he recast the theory in four dimensions and produced the antisymmetric tensor $F^{\mu\nu}$ : six slots, six field components, one object. What Faraday drew, Maxwell coupled, Einstein demoted, Minkowski unified — and the price of admission was accepting that the split into "electric" and "magnetic" is a statement about the observer, not the world.

History

1831

— Michael Faraday discovers electromagnetic induction and describes the field in terms of physically real lines of force.

1865

— James Clerk Maxwell publishes "A Dynamical Theory of the Electromagnetic Field," coupling E and B into a single system and predicting electromagnetic waves at the speed of light.

1895

— Hendrik Lorentz derives approximate field transformation laws to first order in v/c, in his theory of electrons.

1905

— Albert Einstein opens "On the Electrodynamics of Moving Bodies" with the magnet-and-conductor asymmetry and derives the exact field transformation laws from the two postulates.

1908

— Hermann Minkowski introduces the antisymmetric field tensor, shows that E and B are components of one four-dimensional object whose transformation law is automatic, and identifies the two invariants that classify the field.

The physical picture

Take a charge drifting along beside a current-carrying wire. It feels a force, and we call it magnetic: $q\vec v\times\vec B$ . Now ride along with the charge. It is at rest, so it cannot be feeling a magnetic force — there is no $\vec v$ . But the force is still there; whether the charge accelerates toward the wire is not a matter of opinion. In this frame the wire must therefore be electrically charged, and the force is electrostatic. The same push, explained two entirely different ways. Both explanations are correct. They are compatible because $\vec E$ and $\vec B$ are not independent things: they are components of one tensor, and changing frames re-partitions the same six numbers between the two names. Magnetism is what electrostatics looks like from a moving frame — the quantitative version, with the length-contracted charge densities that make it work, belongs to Magnetism as a Relativistic Effect, and we will not redo it here. What this article does instead is the general law and its consequence. The law is trivial to state — $F^{\mu\nu}$ is a tensor, so it transforms like one — and everything else follows by turning a crank. The consequence is not trivial: because the transformation is a Lorentz transformation, certain combinations of the fields cannot change, and those frozen combinations dictate what changing frames can achieve. You cannot boost away a light wave's electric field, however fast you run. You can boost away the electric field of the crossed configuration in the overview, at a fifth of $c$ . The difference between those two cases is two numbers.

Figure

THE CLASSIFICATION — what the two invariants let you do. The horizontal axis is the second invariant (E·B); the vertical axis is the first (B² − E²/c²). A field's position on this map is FRAME-INDEPENDENT, because both coordinates are invariants — a boost moves the six field components around but cannot move the dot. TOP AXIS (E·B = 0, magnetically dominated): a PURE-MAGNETIC frame exists. BOTTOM AXIS (E·B = 0, electrically dominated): a PURE-ELECTRIC frame exists. ORIGIN (both zero): the NULL case — a light wave, where NEITHER field can ever be removed by anybody. ANYWHERE OFF THE VERTICAL AXIS (E·B ≠ 0): no frame kills either field

The core equations

There is only one law here, and it is not a law about fields:

\boxed{\;F'^{\mu\nu} \;=\; \Lambda^\mu{}_\alpha\,\Lambda^\nu{}_\beta\, F^{\alpha\beta}\;} \qquad (1)

where: $\Lambda$ is the Lorentz boost matrix. Physical reading: $F^{\mu\nu}$ is a rank-2 tensor, so it transforms with one $\Lambda$ per index. That is the entire physics — everything below is (1) with the indices written out, and there is no additional postulate anywhere. Unpacked for a boost with velocity $\vec v$ , splitting each field into the part along $\vec v$ and the part across it:

\boxed{\;\vec E'_\parallel = \vec E_\parallel\;}, \qquad \boxed{\;\vec B'_\parallel = \vec B_\parallel\;} \qquad (2)

\boxed{\;\vec E'_\perp = \gamma\big(\vec E + \vec v\times\vec B\big)_\perp\;}, \qquad \boxed{\;\vec B'_\perp = \gamma\Big(\vec B - \frac{\vec v\times\vec E}{c^2}\Big)_\perp\;} \qquad (3)

where: $\gamma = (1 - v^2/c^2)^{-1/2}$ . Physical reading: along the motion, nothing changes at all; across it the fields mix, and the mixing is not symmetric — the magnetic correction carries a $1/c^2$ , which is why magnetic effects look "weak" at everyday speeds. Note also that $\vec E + \vec v\times\vec B$ is the Lorentz force per unit charge, hiding in plain sight. The two invariants — the second of which has no sign at all until you declare a convention:

\boxed{\;I_1 \equiv F_{\mu\nu}F^{\mu\nu} = 2\left(B^2 - \frac{E^2}{c^2}\right)\;} \qquad (4) \qquad\qquad \boxed{\;I_2 \equiv F_{\mu\nu}\tilde F^{\mu\nu} = -\frac{4}{c}\,\vec E\cdot\vec B\;} \qquad (5)

where: $\tilde F^{\mu\nu} = \tfrac12\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$ is the dual tensor, and this article uses $\epsilon^{0123} = +1$ . Physical reading: (5)'s coefficient — including its minus sign — is meaningless without that declaration, because the other common choice, $\epsilon_{0123} = +1$ (which gives $\epsilon^{0123} = -1$ in this signature), flips it. State the convention or do not print the number. Both $I_1$ and $I_2$ are fully contracted scalars, hence identical in every inertial frame — and that is what makes the classification possible.

The transformation, and what it cannot change

Derivation AOne component, in full, straight from the tensor law — and then all six for free

The claim that $\vec E$ and $\vec B$ mix under a boost is not an extra postulate bolted onto electromagnetism; it is a corollary of $F^{\mu\nu}$ being a tensor. So we do exactly one component the long way — every index, every sum — and read off the rest by symmetry. Do $E'_y$ and you have done all six.

Step 1 — write down $\Lambda$ and pick a component. For a boost with speed $v$ along $+x$ , the non-zero entries are $\;\Lambda^0{}_0 = \Lambda^1{}_1 = \gamma$ , $\;\Lambda^0{}_1 = \Lambda^1{}_0 = -\gamma\beta$ , and $\;\Lambda^2{}_2 = \Lambda^3{}_3 = 1$ , with $\beta = v/c$ . We want $E'_y$ , which lives in $F'^{02} = -E'_y/c$ . Why: the $y$ -direction is transverse to the boost, so this is where the interesting mixing happens — the parallel components will turn out to be trivial.

Step 2 — expand the double sum, and watch it collapse. From (1), $\;F'^{02} = \Lambda^0{}_\alpha \Lambda^2{}_\beta F^{\alpha\beta}$ . The second factor is the easy one: $\Lambda^2{}_\beta$ is non-zero only for $\beta = 2$ , where it equals $1$ . So the sum over $\beta$ collapses instantly and we are left with $\;F'^{02} = \Lambda^0{}_\alpha F^{\alpha 2}$ . Now the sum over $\alpha$ : $\Lambda^0{}_\alpha$ is non-zero for $\alpha = 0$ (giving $\gamma$ ) and $\alpha = 1$ (giving $-\gamma\beta$ ). Hence

F'^{02} = \gamma\,F^{02} \;+\; (-\gamma\beta)\,F^{12} .

Why: sixteen terms in principle, two in practice. The transverse direction is untouched by $\Lambda$ , so the only mixing is between the time index and the boost-direction index — exactly why $E_y$ ends up entangled with $B_z$ and nothing else.

Step 3 — substitute the components and solve. From The Electromagnetic Field Tensor, $\;F^{02} = -E_y/c$ and $\;F^{12} = -B_z$ . Substituting,

-\frac{E'_y}{c} \;=\; \gamma\left(-\frac{E_y}{c}\right) + (-\gamma\beta)\big(-B_z\big) \;=\; -\frac{\gamma E_y}{c} + \gamma\beta B_z .

Multiply through by $-c$ and use $\beta c = v$ :

\boxed{\;E'_y \;=\; \gamma\big(E_y - v\,B_z\big)\;}

Why: the magnetic field has appeared inside the electric field. Nobody put it there — it fell out of a matrix multiplication, because $E_y$ and $B_z$ occupy slots of the same tensor that $\Lambda$ connects. That is the mechanism, entire.

Step 4 — read off the remaining five, and recognise the pattern. Repeating Step 2 for the other components (or simply relabelling axes) gives $\;E'_z = \gamma(E_z + vB_y)$ , $\;B'_y = \gamma(B_y + vE_z/c^2)$ , $\;B'_z = \gamma(B_z - vE_y/c^2)$ , and — because $\Lambda$ acts trivially on the pair of indices that build them — $\;\boxed{E'_x = E_x}$ and $\;\boxed{B'_x = B_x}$ . Assembled into vectors, with $\vec v = v\hat x$ : since $(\vec v\times\vec B)_y = -vB_z$ and $(\vec v\times\vec E)_z = +vE_y$ , these six statements are exactly

\vec E'_\perp = \gamma\big(\vec E + \vec v\times\vec B\big)_\perp, \qquad \vec B'_\perp = \gamma\Big(\vec B - \frac{\vec v\times\vec E}{c^2}\Big)_\perp .

Why: the parallel components are unchanged and the perpendicular ones mix. The asymmetry between the two lines — a bare $\vec v\times\vec B$ against a $\vec v\times\vec E/c^2$ — is not an aesthetic flaw: it is why a moving charge's magnetic field is $v/c^2$ times its electric field, and therefore why humanity found electrostatics first. *Verified in verify_aq-sr-transformation-electric-magnetic-fields.py: $F'^{\mu\nu} = \Lambda F \Lambda^{\mathsf T}$ computed symbolically for a general boost with symbolic $\vec E$ , $\vec B$ ; all six field components extracted from the result and compared against (2) and (3) entry by entry; $E'_y = \gamma(E_y - vB_z)$ asserted separately as the fully worked component; and the vector forms $\gamma(\vec E + \vec v\times\vec B)_\perp$ and $\gamma(\vec B - \vec v\times\vec E/c^2)_\perp$ checked against the component results for an arbitrary boost direction.*

Go deeper · Derivation B · 2 minThe two invariants, why they cannot move, and why the second one needs a declaration

Derivation BThe two invariants, why they cannot move, and why the second one needs a declaration

A boost shuffles six numbers, but it cannot shuffle them anywhere it likes — and the restriction is exactly what makes this article's classification possible. Two specific combinations of the six are frozen, for a reason that takes one line to see and has nothing to do with electromagnetism.

Step 1 — why anything is invariant at all. $F_{\mu\nu}F^{\mu\nu}$ has every index contracted: one up, one down, summed. Under a boost each $\Lambda$ meets its inverse and they annihilate, $\;\Lambda_\mu{}^\alpha\Lambda^\mu{}_\beta = \delta^\alpha_\beta$ , leaving the original expression untouched. Why: a fully contracted object is a scalar, and a scalar is the same number for everybody. No electromagnetic input was needed — pure index bookkeeping.

Step 2 — evaluate the first one. Lowering both indices flips the electric block and leaves the magnetic block alone (The Electromagnetic Field Tensor, Derivation 2). Contracting the two matrices and collecting terms — each independent component appearing twice, once as $(\mu\nu)$ and once as $(\nu\mu)$ — gives

\boxed{\;I_1 = F_{\mu\nu}F^{\mu\nu} = 2\left(B^2 - \frac{E^2}{c^2}\right)\;}

Why: the relative minus sign is the whole content. $\vec E$ and $\vec B$ enter with opposite signs, so the invariant measures which one wins — and no boost can change the winner.

Step 3 — the second one, and the declaration it demands. The dual tensor is $\;\tilde F^{\mu\nu} = \tfrac12\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$ , and contracting it with $F$ gives $\;I_2 = F_{\mu\nu}\tilde F^{\mu\nu} = -\tfrac{4}{c}\,\vec E\cdot\vec B$ — with $\epsilon^{0123} = +1$ . Why: $\epsilon$ is antisymmetric, so fixing its overall sign is a choice, and the choice propagates straight into $I_2$ . The other common convention, $\epsilon_{0123} = +1$ , means $\epsilon^{0123} = -1$ here, and flips $I_2$ to $+\tfrac{4}{c}\vec E\cdot\vec B$ . Neither is wrong; printing the number without saying which is. What is convention-independent — and all the classification needs — is whether $\vec E\cdot\vec B$ is zero.

Step 4 — what the two mean together. $I_1$ answers "which field dominates?" and $I_2$ answers "are they perpendicular?" — and both answers are the same in every frame. Why: if you measure $\vec E\cdot\vec B \ne 0$ , then every observer who ever lives measures $\vec E\cdot\vec B \ne 0$ , so no observer anywhere can see a pure electric or a pure magnetic field. The question "can I boost this field away?" is answered before you attempt a single boost, by two numbers computed in whatever frame you already occupy. *Verified in verify_aq-sr-transformation-electric-magnetic-fields.py: $I_1$ and $I_2$ computed symbolically from the field tensor and its dual with `LeviCivita`, and confirmed to equal $2(B^2 - E^2/c^2)$ and $-\tfrac4c\vec E\cdot\vec B$ with $\epsilon^{0123} = +1$ ; both asserted unchanged under an explicit symbolic boost with arbitrary $\vec E$ , $\vec B$ , $\beta$ ; and the opposite $\epsilon$ convention shown to flip $I_2$ 's sign while leaving $I_1$ alone.*

Go deeper · Derivation C · 2 minTHE CLASSIFICATION: exactly when a field can be boosted away, and how fast you must run

Derivation CTHE CLASSIFICATION: exactly when a field can be boosted away, and how fast you must run

Now the question the invariants were built to answer. Given a field, does there exist an inertial frame in which the electric field vanishes entirely? One in which the magnetic field does? The answer is a clean three-way classification, and it drops out with almost no work — the sign that the invariants were the right tool.

Step 1 — ask what the invariants would have to be. Suppose a frame exists in which $\vec E' = \vec 0$ . Then in that frame, $\;I_1 = 2\big(B'^2 - 0\big) = 2B'^2 \ge 0$ and $\;I_2 = -\tfrac4c\,\vec 0\cdot\vec B' = 0$ . But $I_1$ and $I_2$ are invariants, so they have those values in every frame, including yours. Hence, necessarily: $\;\boxed{\vec E\cdot\vec B = 0}$ and $\;\boxed{B^2 \ge E^2/c^2}$ . Why: a two-line proof of a necessary condition, requiring no boost at all. The invariants let you interrogate a frame you have never visited, using only numbers measured in the one you occupy.

Step 2 — show the conditions are also sufficient, by producing the velocity. Suppose $\vec E\cdot\vec B = 0$ and $B^2 > E^2/c^2$ . Try

\boxed{\;\vec v \;=\; \frac{\vec E\times\vec B}{B^2}\;}

Its speed is $\;v = EB\sin 90^\circ / B^2 = E/B$ (the fields are perpendicular), so $\;v/c = E/(cB) < 1$ precisely because $B^2 > E^2/c^2$ — the condition is exactly what makes the required velocity subluminal. Substituting into (3), the transverse electric field becomes $\gamma\big(\vec E + \vec v\times\vec B\big)$ , and $\;\vec v\times\vec B = \big[(\vec E\times\vec B)\times\vec B\big]/B^2 = -\vec E$ (using $\vec E\cdot\vec B = 0$ ), so $\;\vec E' = \vec 0$ identically. Why: the magnetic-dominance condition is not an extra hypothesis — it IS the statement that the necessary boost is slower than light. If $E > cB$ you would need $v > c$ , which is the whole reason an electrically dominated field cannot be boosted to pure magnetic.

Step 3 — the mirror case, and the case between. By the identical argument with the roles swapped: a pure-electric frame exists iff $\;\vec E\cdot\vec B = 0$ and $E^2/c^2 > B^2$ , reached by $\;\vec v = (\vec E\times\vec B)/(E^2/c^2)$ . And if $\vec E\cdot\vec B = 0$ with $E = cB$ exactly, then $I_1 = I_2 = 0$ : the null case. Why: in the null case the required speed is exactly $c$ , and no massive observer can get there. So neither field can be removed, by anybody, ever. This is not a technicality — it is the electromagnetic plane wave, i.e. light, and "you cannot boost away a light beam's electric field" is "you cannot catch up with light," wearing a different hat.

Step 4 — and if $\vec E\cdot\vec B \ne 0$ ? Then $I_2 \ne 0$ in every frame, and Step 1's necessary condition fails for both targets. No frame kills either field. Why: the best you can do is find a frame in which $\vec E'$ and $\vec B'$ are parallel — but both are present, everywhere, for everyone. The classification is complete and decided entirely by two numbers, and a field like Worked Example 1's ( $\vec E\cdot\vec B = 5.00\times10^{7}\ \mathrm{V\,T\,m^{-1}}$ ) is permanently, unavoidably, both at once. *Verified in verify_aq-sr-transformation-electric-magnetic-fields.py: the necessary conditions derived from the invariants; $\vec v = (\vec E\times\vec B)/B^2$ shown symbolically to give $\vec E' = \vec 0$ whenever $\vec E\cdot\vec B = 0$ ; $|\vec v| < c \iff B^2 > E^2/c^2$ asserted; the null case shown to require exactly $v = c$ ; and the full Worked Example 2 boost carried out numerically, giving $\vec E' = \vec 0$ to machine precision and $|\vec B'| = \sqrt{I_1/2} = 0.48988\ \mathrm{T}$ .*

Figure

WORKED EXAMPLE 2, drawn — the boost that annihilates an electric field. LEFT: the lab frame, with E = 3.00×10⁷ V/m crossed with B = 0.500 T. Because E·B = 0 and I₁ = +0.47997 T² is POSITIVE, the field is magnetically dominated and the classification guarantees a pure-magnetic frame exists. MIDDLE: the required velocity is v = (E×B)/B² = 6.00×10⁷ m/s, only 0.2001c — and it is subluminal precisely BECAUSE B² > E²/c². RIGHT: in that frame the electric field is EXACTLY zero and a 0.48988 T magnetic field remains — the value the invariant predicted in advance, √(I₁/2)

Limiting cases

Field	$\vec E\cdot\vec B$	$B^2 - E^2/c^2$	What frames exist
Magnetically dominated	$0$	$> 0$	a pure-magnetic frame, at $\vec v = (\vec E\times\vec B)/B^2$ , $\;v = E/B < c$
Electrically dominated	$0$	$< 0$	a pure-electric frame, at $\vec v = (\vec E\times\vec B)c^2/E^2$
A light wave ( $E = cB$ )	$0$	$0$	NULL — neither field removable, ever. You cannot catch light
Generic	$\ne 0$	any	neither field removable; best available is $\vec E' \parallel \vec B'$
$v \ll c$ ( $\gamma \to 1$ )	—	—	$\vec E' \approx \vec E + \vec v\times\vec B$ — the pre-relativistic lab

Common trapTwo traps: the ε convention, and "the fields turn into each other"

1. $I_2$ has no sign until you declare $\epsilon^{0123}$ . This article uses $\;\epsilon^{0123} = +1$ , giving $\;F_{\mu\nu}\tilde F^{\mu\nu} = -\tfrac{4}{c}\vec E\cdot\vec B$ . The other common convention sets $\epsilon_{0123} = +1$ ; in signature $(+,-,-,-)$ raising all four indices costs three factors of $(-1)$ , so that means $\epsilon^{0123} = -1$ and $I_2$ comes out $+\tfrac{4}{c}\vec E\cdot\vec B$ . Both are correct. Printing the number without saying which is not. Fortunately the classification depends only on whether $\vec E\cdot\vec B$ vanishes, which is convention-independent — so nothing physical is at stake, only arithmetic.

2. A boost does not "turn $\vec E$ into $\vec B$ ." The phrasing is everywhere and it is actively misleading. There is one tensor with six slots, and a boost re-expresses those same six numbers in a new basis — it does not convert one field into another, and it certainly does not create field out of motion. The tell: if $F^{\mu\nu} = 0$ in one frame it is zero in all frames, because that is a tensor equation. Boost as hard as you like through empty space and nothing appears.

Worked examples

1 · Boost a generic field and watch both invariants refuse to move

A region contains $\vec E = (1.00\times10^{8},\ 2.00\times10^{8},\ 0)\ \mathrm{V\,m^{-1}}$ and $\vec B = (0.500,\ 0,\ 1.00)\ \mathrm{T}$ . Boost along $+x$ at $0.6c$ and compute everything in both frames.

Step 1 — the invariants, before. $\;\vec E/c = (0.33356,\ 0.66713,\ 0)\ \mathrm{T}$ , so $\;E^2/c^2 = \mathbf{0.55633\ \mathrm{T^2}}$ , and $\;B^2 = 0.500^2 + 1.00^2 = \mathbf{1.25\ \mathrm{T^2}}$ . Also $\;\vec E\cdot\vec B = \mathbf{5.00\times10^{7}\ \mathrm{V\,T\,m^{-1}}}$ . Hence

I_1 = 2(1.25 - 0.55633) = \mathbf{1.3873\ \mathrm{T^2}}, \qquad I_2 = -\frac{4}{c}(5.00\times10^{7}) = \mathbf{-0.66713\ \mathrm{T^2}} \;\;(\epsilon^{0123} = +1).

Both non-zero ⇒ by the classification, no frame kills either field — known before boosting.

Step 2 — the boost. $\;\beta = 0.6$ , $\;\gamma = \mathbf{1.25}$ , $\;v = \mathbf{1.7987\times10^{8}\ \mathrm{m\,s^{-1}}}$ .

Step 3 — apply (2) and (3) component by component. Parallel ( $x$ ): $\;E'_x = E_x = \mathbf{1.00\times10^{8}\ \mathrm{V\,m^{-1}}}$ and $\;B'_x = B_x = \mathbf{0.500\ \mathrm{T}}$ — unchanged. Transverse:

E'_y = \gamma\big(E_y - vB_z\big) = 1.25\big(2.00\times10^{8} - (1.7987\times10^{8})(1.00)\big) = \mathbf{2.5156\times10^{7}\ \mathrm{V\,m^{-1}}},

B'_z = \gamma\Big(B_z - \frac{vE_y}{c^2}\Big) = 1.25\left(1.00 - \frac{(1.7987\times10^{8})(2.00\times10^{8})}{8.9875\times10^{16}}\right) = 1.25\,(1.00 - 0.40028) = \mathbf{0.74965\ \mathrm{T}} .

(And $E'_z = \gamma(E_z + vB_y) = 0$ , $\;B'_y = 0$ .) The electric field's $y$ -component collapsed by a factor of eight.

Step 4 — recompute the invariants, after. $\;E'^2/c^2 = \mathbf{0.11831\ \mathrm{T^2}}$ and $\;B'^2 = 0.500^2 + 0.74965^2 = \mathbf{0.81198\ \mathrm{T^2}}$ , so

I_1' = 2(0.81198 - 0.11831) = \mathbf{1.3873\ \mathrm{T^2}} \quad\checkmark

And $\;\vec E'\cdot\vec B' = (1.00\times10^{8})(0.500) = \mathbf{5.00\times10^{7}}$ , giving $\;I_2' = \mathbf{-0.66713\ \mathrm{T^2}}$ $\;\checkmark$ .

Step 5 — the boundary number, and the lesson. Every individual field component changed, and both invariants did not. $I_1$ came back at $1.3873\ \mathrm{T^2}$ and $I_2$ at $-0.66713\ \mathrm{T^2}$ , to five figures, from a completely different set of six numbers. That is what "invariant" means in practice — and it is a superb arithmetic check: if your invariants do not match after a boost, your boost is wrong.

2 · Boost the electric field out of existence — and let the invariant predict the answer first

Crossed fields: $\vec E = (0,\ 3.00\times10^{7},\ 0)\ \mathrm{V\,m^{-1}}$ and $\vec B = (0,\ 0,\ 0.500)\ \mathrm{T}$ . Can the electric field be removed? If so, how fast must you run, and what is left?

Step 1 — classify, before doing any work. $\;\vec E\cdot\vec B = \mathbf{0}$ (perpendicular), so $I_2 = 0$ and a pure-field frame is not ruled out. And $\;E/c = 3.00\times10^{7}/2.9979\times10^{8} = \mathbf{0.10007\ \mathrm{T}}$ , so $\;E^2/c^2 = \mathbf{0.010014\ \mathrm{T^2}}$ against $\;B^2 = \mathbf{0.250\ \mathrm{T^2}}$ :

I_1 = 2\,(0.250 - 0.010014) = \mathbf{+0.47997\ \mathrm{T^2}} \;>\; 0 .

Magnetically dominated. By the classification, a pure-magnetic frame exists. We have not boosted anything yet.

Step 2 — the invariant tells you the answer in advance. In the target frame $\vec E' = \vec 0$ , so $\;I_1 = 2B'^2$ , and therefore

|\vec B'| = \sqrt{I_1/2} = \sqrt{0.47997/2} = \sqrt{0.23999} = \mathbf{0.48988\ \mathrm{T}} .

This is a prediction, made before the transformation is performed, using only a number computed in the original frame.

Step 3 — find the velocity. $\;\vec E\times\vec B = (0,\ 3.00\times10^{7},\ 0)\times(0,\ 0,\ 0.500) = (\mathbf{1.50\times10^{7}},\ 0,\ 0)$ , so

\vec v = \frac{\vec E\times\vec B}{B^2} = \frac{1.50\times10^{7}}{0.250}\,\hat x = \mathbf{6.00\times10^{7}\ \mathrm{m\,s^{-1}}}\ \hat x ,

which is $\;\beta = \mathbf{0.20014}$ and $\;\gamma = \mathbf{1.02065}$ . Comfortably subluminal — as Derivation 3 Step 2 guaranteed it must be, because $B^2 > E^2/c^2$ .

Step 4 — do the boost and check. $\;\vec v\times\vec B = (6.00\times10^{7},0,0)\times(0,0,0.500) = (0,\ -3.00\times10^{7},\ 0)$ , so

\vec E'_\perp = \gamma\big(\vec E + \vec v\times\vec B\big) = \gamma\big[(0,\ 3.00\times10^{7},\ 0) + (0,\ -3.00\times10^{7},\ 0)\big] = \mathbf{\vec 0} .

Exactly zero — the two terms cancel identically, not approximately. And the magnetic field: $\;\vec v\times\vec E/c^2 = (0,\ 0,\ 0.020028)\ \mathrm{T}$ , so $\;\vec B' = \gamma(0.500 - 0.020028)\,\hat z = 1.02065 \times 0.47997\,\hat z = \mathbf{0.48988\ \mathrm{T}}\ \hat z$ .

Step 5 — the boundary number, and what it means. The prediction from Step 2 was $0.48988\ \mathrm{T}$ . The transformation delivers $\mathbf{0.48988\ \mathrm{T}}$ . The invariant knew the answer before the boost was computed — which is the entire reason invariants are worth having. Meanwhile: an observer at rest measures a $3.00\times10^{7}\ \mathrm{V\,m^{-1}}$ electric field, and an observer moving at a fifth of the speed of light measures none at all, in any direction, to any precision. Both are right. The electric field is not a fact about the region; it is a fact about the region and your velocity. The magnetic field is not privileged either — an electrically dominated field could have had its magnetic field boosted away just as completely. What survives every observer is $I_1$ and $I_2$ , and nothing else.

Going further

Texts. Griffiths, Introduction to Electrodynamics, 4th ed., §12.3.2 · Jackson, Classical Electrodynamics, 3rd ed., §11.10 · Einstein, "Zur Elektrodynamik bewegter Körper," Ann. Phys. 17, 891 (1905), §6.

Provenance. Signature $(+,-,-,-)$ ; $\;F^{0i} = -E^i/c$ , $\;F^{ij} = -\epsilon^{ijk}B_k$ (derived in The Electromagnetic Field Tensor, not transcribed); $\epsilon^{0123} = +1$ , without which $I_2$ 's sign is undefined; SI units. Every transformation here comes from $F'^{\mu\nu} = \Lambda^\mu{}_\alpha\Lambda^\nu{}_\beta F^{\alpha\beta}$ and is asserted componentwise in the verify script, which also checks both invariants are unchanged under an explicit symbolic boost.

Base defect noted. The AllQFT base for this topic prints the invariant $F_{\mu\nu}\tilde F^{\mu\nu}$ without stating its $\epsilon$ convention, which leaves the sign of the printed result undetermined. This article declares $\epsilon^{0123} = +1$ and prints the sign that follows from it.

Scope. The current-carrying wire, the length contraction of the moving charge density, and the reduction of magnetism to electrostatics belong to Magnetism as a Relativistic Effect — stated here as a conclusion only. The dual tensor $\tilde F^{\mu\nu}$ in full belongs to The Dual Field Tensor; the complete invariant theory to The Electromagnetic Field Invariants; the covariant force law to The Covariant Lorentz Force.

If E·B ≠ 0, is there anything useful I can still do by boosting?

Yes. You cannot remove either field — that is settled by $I_2 \ne 0$ — but you can always find a frame in which $\vec E'$ and $\vec B'$ are parallel. That frame is the natural one to work in. What you can never do is make either field vanish: both are present, everywhere, for every observer.

Why do the parallel components not change?

Because the boost matrix $\Lambda$ acts only on the time index and the index along the motion, leaving the two transverse indices alone. $E_x$ lives in $F^{01}$ , whose indices the boost touches in a way that cancels; $B_x$ lives in $F^{23}$ , where neither index is touched. Both come through unchanged. The transverse components live in slots with one touched index and one untouched — exactly the recipe for mixing.

Which invariant convention should I use for E·B?

Whichever you like, provided you say so. This article uses $\epsilon^{0123} = +1$ , giving $F_{\mu\nu}\tilde F^{\mu\nu} = -\tfrac{4}{c}\vec E\cdot\vec B$ . The other common choice, $\epsilon_{0123} = +1$ , means $\epsilon^{0123} = -1$ in this signature and flips the result to $+\tfrac{4}{c}\vec E\cdot\vec B$ . Neither is more correct; printing the number without declaring the convention is what most sources do, and that is the error. Nothing physical depends on it: the classification cares only whether $\vec E\cdot\vec B$ is zero, which is convention-free.

If a field is magnetically dominated, is it magnetic "really"?

It is magnetically dominated in every frame — that much is objective, because $F_{\mu\nu}F^{\mu\nu} > 0$ is frame-independent. But "really magnetic" is too strong unless $\vec E\cdot\vec B = 0$ as well. With both conditions met there exists one frame in which the field is purely magnetic, and in every other frame there is also an electric field. The honest statement: there is a frame in which it is purely magnetic, and that frame is distinguished but no more "correct" than any other. What is real is the tensor and the two numbers every observer computes from it.

Is magnetism just relativity applied to electrostatics?

Essentially, yes — one of the most satisfying results in physics. The force on a charge moving parallel to a current-carrying wire is magnetic in the lab frame, but in the charge's own rest frame there is no magnetic force at all (it isn't moving), and the same force appears instead as an electrostatic attraction to a wire that is now net-charged, because the positive and negative charge densities Lorentz-contract by different amounts. The two accounts agree exactly, and they must, since the transformation laws here guarantee it. The full derivation belongs to Magnetism as a Relativistic Effect; what this article supplies is the reason the two accounts cannot disagree.

How fast do I have to go to notice this?

Far slower than you would guess, because the effect scales as $vB$ rather than $v/c$ . In Worked Example 2 a boost of $6.00\times10^{7}\ \mathrm{m\,s^{-1}}$ — about $0.20c$ — removes a $3.00\times10^{7}\ \mathrm{V\,m^{-1}}$ electric field completely, while $\gamma$ is only $1.02065$ : the relativistic dilation is a 2% effect and the field change is total. The mixing term $\vec v\times\vec B$ is first order in $v$ ; the $\gamma$ factors everyone associates with relativity are second order and often negligible. Field mixing is a low-speed relativistic effect — which is precisely why the nineteenth century found it experimentally long before anyone suspected relativity existed.