Electromagnetic Field Invariants

A boost mixes $\vec E$ and $\vec B$ into each other. Neither field has a frame-independent magnitude; neither has a frame-independent direction; the angle between them is not preserved. Almost nothing you can say about "the electric field here" survives being said by somebody moving. Two things do. From the six numbers in $F^{\mu\nu}$ you can build exactly two independent Lorentz scalars, and every frame-independent statement about an electromagnetic field is a statement about them:

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

Exactly two — not one, not three. That is a theorem, and it is why this page exists. Their power is that they answer questions before you do any work. Can this field be boosted away? Can $\vec E$ be made to vanish? Is this configuration "really" a magnetic field wearing an electric disguise, or is it something no observer can simplify? You do not need to try the boosts. You compute two numbers and read the answer off their signs. A field with $\vec E\cdot\vec B \ne 0$ can never be made purely electric, by anyone, ever — because a pure electric field has $\vec E\cdot\vec B = 0$ and the quantity is invariant. The argument takes one line and settles infinitely many boosts. And when the classical story ends, the invariants keep going. The QED vacuum is nonlinear, and its effective Lagrangian is a function of precisely these two scalars — it could not be a function of anything else.

Key takeaways

Signature $(+,-,-,-)$ , $\eta = \mathrm{diag}(+1,-1,-1,-1)$ , $p\cdot p = +m^2c^2$ ; $\varepsilon^{0123} = +1$ ; $m$ is the invariant rest mass. Dictionary: $F^{0i} = -E_i/c$ , $F^{ij} = -\varepsilon^{ijk}B_k$ .
The two, and there are only two: $\boxed{S \equiv F_{\mu\nu}F^{\mu\nu} = 2(B^2 - E^2/c^2)}$ — a true scalar — and $\boxed{P \equiv F_{\mu\nu}\tilde F^{\mu\nu} = -\tfrac{4}{c}\vec E\cdot\vec B}$ — a pseudoscalar, boost-invariant but sign-flipped in a mirror.
Why exactly two. Every Lorentz scalar built from $F$ is a function of traces of powers of $F^\mu{}_\nu$ . Odd traces vanish by antisymmetry, $\mathrm{tr}(F^2) = -S$ , and Cayley–Hamilton in four dimensions collapses the next one: $\boxed{\mathrm{tr}(F^4) = \tfrac12 S^2 + \tfrac14 P^2}$ . Nothing new ever appears.
The classification, read off two signs. $P \ne 0$ : generic — neither field can be removed by anyone. $P = 0$ and $S < 0$ ( $E > cB$ ): electric-like — a boost makes it purely electric. $P = 0$ and $S > 0$ ( $E < cB$ ): magnetic-like. $P = S = 0$ : null — a light wave, irreducible in every frame.
Owned result: the canonical frame, in closed form. When $P \ne 0$ the best any observer can do is align $\vec E \parallel \vec B$ . The two field strengths in that frame are fixed by the invariants alone: with $a = E_0/c$ and $b = B_0$ , a^2 - b^2 = \frac{E^2}{c^2} - B^2 ,\qquad ab = \frac{\vec E\cdot\vec B}{c} \;\Longrightarrow\; b^2 = \tfrac12\Big[\sqrt{D^2 + 4Q^2} - D\Big],\quad D \equiv \tfrac{E^2}{c^2} - B^2,\ Q \equiv \tfrac{\vec E\cdot\vec B}{c} . Two invariants in, two field magnitudes out. The base article asserts that this frame exists and never computes it.
Key idea: the invariants outlive classical physics. The QED vacuum is a nonlinear medium, and the Heisenberg–Euler effective Lagrangian is a function of $S$ and $P$ and of nothing else — because there is nothing else. That is why $E > E_c = m_e^2c^3/(e\hbar) = 1.323\times10^{18}\ \mathrm{V\,m^{-1}}$ boils electron–positron pairs out of the vacuum, and why a light wave ( $S = P = 0$ ) sails through the vacuum without any nonlinear response at all.

The core insight

An antisymmetric rank-2 tensor in four dimensions has six components and a two-dimensional space of algebraic invariants. That is the entire content of this page. Everything else — the classification of fields, the existence of a frame where $\vec E \parallel \vec B$ , the fact that light cannot be boosted away, the structure of the QED vacuum Lagrangian — is a corollary of that counting. The invariants are not a convenient summary of the field. They are all of it that is real.

Where this lives today

The invariants are the reason strong-field physics has a definite target. The Heisenberg–Euler Lagrangian is an expansion in $S$ and $P$ divided by the critical field, and the scale at which the expansion breaks down is E_c = \frac{m_e^2c^3}{e\hbar} = 1.323\times10^{18}\ \mathrm{V\,m^{-1}} ,\qquad B_c = \frac{E_c}{c} = 4.414\times10^{9}\ \mathrm{T} . Above it, the vacuum is unstable to pair production — Schwinger's 1951 result. Below it, the vacuum is merely nonlinear: it acquires a refractive index that depends on polarisation, so a strong field makes empty space birefringent. Petawatt laser facilities exist substantially to chase these numbers, and the strongest fields we know of sit on the surfaces of neutron stars, where $B \sim 10^{9}\ \mathrm{T}$ brings $B_c$ within reach. That is where the 2017 polarimetry of RX J1856.5 $-$ 3754 comes in: light leaving the star is polarised in a way consistent with the vacuum birefringence that Heisenberg and Euler wrote down in 1936. Note that the prediction was not a guess about the vacuum's structure but a consequence of the counting: there are two invariants, so there are two things the vacuum can respond to, and no others. A theorem about an antisymmetric $4\times4$ matrix ended up telling a telescope what to look for.

The physical picture

Six numbers go in: three of $\vec E/c$ , three of $\vec B$ . A Lorentz transformation has six parameters — three rotations, three boosts — and it acts on those six numbers. Ask what functions of the field survive that action, and the answer is dictated by dimension-counting: the orbit of a generic field under the six-parameter group is four-dimensional, so two functions must be left over. Those two are $S$ and $P$ . There is no third, and there was never room for one. The two are not alike. $S$ is built from $F$ and the metric alone, so it is a genuine scalar: rotations, boosts, reflections, all leave it alone. $P$ needs the Levi-Civita symbol, which carries an orientation of spacetime, so it is a pseudoscalar — invariant under boosts and rotations, sign-flipped under a mirror. That distinction is not pedantry. It is why $S$ can appear in the free-field Lagrangian and $P$ cannot: a $P$ term would make electromagnetism violate parity, which it does not. What makes them useful is that they turn questions about infinitely many boosts into questions about two numbers. "Is there a frame in which this field is purely magnetic?" is, on its face, a search over all of $SO(1,3)$ . But a purely magnetic field has $E = 0$ , hence $S = 2B^2 > 0$ and $P = 0$ . Both are invariant. So the search terminates immediately: if your field has $P \ne 0$ , or $S \le 0$ , no such frame exists — and you did not have to look.

Figure

THE TWO INVARIANTS CLASSIFY EVERY ELECTROMAGNETIC FIELD. Plot S = 2(B² − E²/c²) against P = −(4/c) E·B. Every field in the universe is one point on this plane, and its position is the same for every observer — a boost moves the field, but never moves the point. OFF the P = 0 axis (P ≠ 0, GENERIC): neither E nor B can be removed by anybody; the best any observer can do is rotate them parallel, and the canonical field strengths are then fixed by S and P alone. ON the axis with S > 0: MAGNETIC-LIKE — a boost at v = E/B kills E entirely. ON the axis with S < 0: ELECTRIC-LIKE — a boost at v = c²B/E kills B. At the ORIGIN: NULL — a light wave, which no observer can simplify, slow, or transform away.

The core equations

The scalar invariant — a true scalar, even under parity:

\boxed{\;S \;\equiv\; F_{\mu\nu}F^{\mu\nu} \;=\; 2\Big(B^2 - \frac{E^2}{c^2}\Big)\;} \qquad (1)

where: $F_{\mu\nu} = \eta_{\mu\alpha}\eta_{\nu\beta}F^{\alpha\beta}$ ; $E^2 = \vec E\cdot\vec E$ , $B^2 = \vec B\cdot\vec B$ . Physical reading: the sign of $S$ says which field wins, and every observer agrees. $S > 0$ is magnetically dominated, $S < 0$ electrically dominated, $S = 0$ balanced. Note that $S$ is quadratic in $F$ , so it is blind to an overall sign flip of the tensor — a fact that lets a badly signed field tensor still produce the right invariant, and therefore hides sign errors rather than exposing them. The pseudoscalar invariant — boost-invariant, but not a scalar:

\boxed{\;P \;\equiv\; F_{\mu\nu}\tilde F^{\mu\nu} \;=\; -\,\frac{4}{c}\,\vec E\cdot\vec B\;}\ ,\qquad \tilde F^{\mu\nu} = \tfrac12\varepsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}\ ,\quad \varepsilon^{0123} = +1 \qquad (2)

where: $\tilde F^{\mu\nu}$ — the dual tensor, whose construction belongs to The Dual Field Tensor. Physical reading: $P$ measures how skew the two fields are, and it is the one thing no contraction of $F$ with itself and the metric can produce — you need the orientation that $\varepsilon$ carries. Under a mirror, $\vec B$ flips and $\vec E$ does not, so $P$ changes sign. It is invariant under the proper Lorentz group only. Why there are exactly two — the identity that closes the list:

\mathrm{tr}(F^2) = F^\mu{}_\nu F^\nu{}_\mu = -\,S \ ,\qquad \boxed{\;\mathrm{tr}(F^4) \;=\; \tfrac12 S^2 \;+\; \tfrac14 P^2\;} \ ,\qquad \det\!\big(F^{\mu\nu}\big) = \Big(\tfrac14 P\Big)^2 = \frac{(\vec E\cdot\vec B)^2}{c^2} \qquad (3)

where: $F^\mu{}_\nu = F^{\mu\alpha}\eta_{\alpha\nu}$ — the mixed tensor, whose traces generate every algebraic invariant. Physical reading: all odd traces vanish because $F$ is antisymmetric, $\mathrm{tr}(F^2)$ reproduces $S$ , and the next one is not new — Cayley–Hamilton in four dimensions forces it back into $S$ and $P$ . Every higher trace does the same. Two, and no more. (Both signs on the right of the $\mathrm{tr}(F^4)$ identity are positive. The base article prints a minus, and it is wrong — see the second warning.) The classification, in one table of signs:

\begin{array}{lll} P \ne 0 & \textbf{generic} & \text{neither field removable; best frame has } \vec E \parallel \vec B\\ P = 0,\; S < 0 & \textbf{electric-like} & \vec B \text{ removable, by a boost at } v = c^2B/E\\ P = 0,\; S > 0 & \textbf{magnetic-like} & \vec E \text{ removable, by a boost at } v = E/B\\ P = 0,\; S = 0 & \textbf{null} & \text{a light wave; nothing is removable} \end{array} \qquad (4)

where: the boost velocities are directed along $\vec E\times\vec B$ . Physical reading: the classification is complete and mutually exclusive, and because it is written in invariants, it is the same for every observer. A charge's whole qualitative fate — spiral, accelerate, or be dragged along a null field — is decided by two numbers computed once, in any frame you like.

Deriving the two, and proving there are no more

Derivation AThe two invariants, and the theorem that closes the list

The goal is two-fold: compute $S$ and $P$ from the derived component dictionary, and then prove that no third invariant exists. The second half is the one that matters — it is what makes "the field invariants" a definite, finite list rather than an open-ended hunt.

Step 1. Contract $F$ with itself. Lowering both indices of $F^{0i} = -E_i/c$ , $F^{ij} = -\varepsilon^{ijk}B_k$ gives $F_{0i} = +E_i/c$ and $F_{ij} = -\varepsilon_{ijk}B_k$ . The double sum splits into a time-space part (each ordered pair $(0,i)$ counted twice) and a space-space part: F_{\mu\nu}F^{\mu\nu} = \underbrace{2\sum_i F_{0i}F^{0i}}_{2\sum_i (E_i/c)(-E_i/c) \,=\, -2E^2/c^2} + \underbrace{\sum_{i\ne j} F_{ij}F^{ij}}_{+2B^2} = \boxed{2\Big(B^2 - \frac{E^2}{c^2}\Big)}. Why: the minus on the electric term is the metric, not a choice — lowering one spatial index costs a sign, and the time-space block has exactly one. The relative minus between $E^2/c^2$ and $B^2$ is the signature made visible.

Step 2. Contract $F$ with its dual. Every surviving product pairs an $E$ -component with the same $B$ -component, and the count gives \boxed{F_{\mu\nu}\tilde F^{\mu\nu} = -\,\frac{4}{c}\,\vec E\cdot\vec B}. Why: this is the invariant $F$ alone cannot see. It is bilinear in $\vec E$ and $\vec B$ , not quadratic in either — which is exactly why it is the one that detects skewness, and why it is odd under a mirror.

Step 3. Now close the list. Any algebraic Lorentz scalar built from $F$ is a function of the traces $\mathrm{tr}(F^n)$ of the mixed tensor $F^\mu{}_\nu$ . Antisymmetry kills every odd trace: $\mathrm{tr}(F) = \mathrm{tr}(F^3) = 0$ , because $\mathrm{tr}(F^{2k+1})$ changes sign under $F\to F^{T}=-F$ while the trace does not. And $\mathrm{tr}(F^2) = -S$ . Why: half the candidates are dead on arrival. Only even powers survive, and the first of them is nothing new.

Step 4. Kill the fourth. Explicit contraction gives \mathrm{tr}(F^4) = \tfrac12 S^2 + \tfrac14 P^2 , with both signs positive. So $\mathrm{tr}(F^4)$ is a polynomial in $S$ and $P$ — not an independent invariant. Cayley–Hamilton then propagates this to every higher power: in four dimensions $F^4$ is a linear combination of $F^3, F^2, F, \mathbb 1$ with coefficients built from the characteristic polynomial, so $\mathrm{tr}(F^n)$ for $n > 4$ reduces recursively. The list is closed at two. Why: this is the theorem. It is why "the field invariants" is a definite phrase and why the QED vacuum Lagrangian, which must be a Lorentz scalar built from $F$ , has no choice but to be a function of $S$ and $P$ .

Step 5. Sanity-check with the determinant: $\det(F^{\mu\nu}) = (\tfrac14 P)^2 = (\vec E\cdot\vec B)^2/c^2$ . Why: an independent route to the same conclusion. $\det F$ is manifestly invariant under $F \to \Lambda F \Lambda^T$ with $\det\Lambda = 1$ , and it turns out to be $P^2/16$ — so it too is not new. Every invariant road leads back to the same two numbers. *Verified in verify_aq-sr-electromagnetic-field-invariants.py: $F^{\mu\nu}$ is rebuilt from $A^\mu = (\phi/c, \vec A)$ , so the dictionary is derived not assumed; $S$ and $P$ are then contracted symbolically and matched against $2(B^2 - E^2/c^2)$ and $-(4/c)\vec E\cdot\vec B$ ; $\mathrm{tr}(F) = \mathrm{tr}(F^3) = 0$ , $\mathrm{tr}(F^2) = -S$ , $\mathrm{tr}(F^4) - \tfrac12 S^2 - \tfrac14 P^2 = 0$ and $\det F - P^2/16 = 0$ are each verified as exact symbolic identities in $\vec E$ and $\vec B$ ; and the base's minus sign in the $\mathrm{tr}(F^4)$ identity is shown to be impossible for any normalisation of the pseudoscalar.*

Go deeper · Derivation B · 2 minWhich fields can be transformed away, and the boost that does it

Derivation BWhich fields can be transformed away, and the boost that does it

The invariants make the classification a matter of reading two signs, but the constructive question — give me the boost — still needs doing. It is short, and it explains why the null case is genuinely special rather than merely a boundary.

Step 1. Suppose $P = 0$ , i.e. $\vec E \perp \vec B$ , and look for a boost $\vec v$ that kills $\vec E$ . The transverse field transformation gives $\vec E'_\perp = \gamma(\vec E + \vec v\times\vec B)_\perp$ , so $\vec E' = 0$ demands $\vec E = -\vec v\times\vec B$ . Why: we are not solving Maxwell — we are solving one linear equation for $\vec v$ , and the invariants have already told us whether a solution can exist.

Step 2. Solve it. Taking $\vec v \perp \vec B$ (any component along $\vec B$ does nothing) gives \boxed{\vec v = \frac{\vec E\times\vec B}{B^2}},\qquad v = \frac{E}{B} . Why: check the direction on a concrete case — $\vec B = B\hat z$ , $\vec E = E\hat x$ gives $\vec v = -(E/B)\hat y$ , and indeed $\vec v\times\vec B = -(E/B)B(\hat y\times\hat z) = -E\hat x = -\vec E$ . ✓

Step 3. Ask when that boost is legal. $v = E/B < c$ requires $E < cB$ , i.e. B^2 - \frac{E^2}{c^2} > 0 \iff S > 0 . Why: the invariant is the admissibility condition. "Magnetic-like" is not a name for a shape of field — it is the statement that the required boost is subluminal.

Step 4. The mirror case. If $E > cB$ ( $S < 0$ ), take $\vec v = c^2\,\vec E\times\vec B/E^2$ , giving $v = c^2B/E < c$ ; this kills $\vec B$ instead, and the field is electric-like. Why: the two cases are exchanged by duality $\vec E \to c\vec B$ , $\vec B \to -\vec E/c$ , which flips the sign of $S$ — exactly as the dual tensor demands.

Step 5. The null case, where both fail. If $E = cB$ and $\vec E \perp \vec B$ then $S = P = 0$ , and both prescriptions return $v = c$ . The boost does not exist. A plane wave cannot be made purely electric, purely magnetic, slowed, or removed, in any frame whatsoever. Why: this is not a technical obstruction to route around. It is the invariants speaking: a pure field has $S \ne 0$ , and $S = 0$ is preserved by every Lorentz transformation. Light is light to everyone, and the two vanishing invariants are the statement of it. *Verified in verify_aq-sr-electromagnetic-field-invariants.py: the boost $\vec v = \vec E\times\vec B/B^2$ is substituted into the exact field-transformation laws and $\vec E'$ is confirmed to vanish identically for $\vec E\perp\vec B$ ; $|\vec v| < c \iff S > 0$ is checked; and $S$ and $P$ are both confirmed invariant under a general boost by explicit symbolic transformation of $\vec E$ and $\vec B$ .*

Go deeper · Derivation C · 2 minOwned result: the canonical frame, in closed form — the base says it exists and never computes it

Derivation COwned result: the canonical frame, in closed form — the base says it exists and never computes it

When $P \ne 0$ , no boost removes either field, and every account of this subject stops there: "the best you can do is make $\vec E$ parallel to $\vec B$ ." True — but how strong are they in that frame? The answer is completely determined by the two invariants, and it can be written down.

Step 1. Let the canonical frame be one where $\vec E_0 \parallel \vec B_0$ , and set $a \equiv E_0/c$ and $b \equiv B_0$ , both with units of tesla. Because $\vec E_0 \parallel \vec B_0$ , the two invariants evaluated in that frame collapse to S = 2(b^2 - a^2) ,\qquad P = -\frac{4}{c}\,(ca)(b)\cdot 1 = -4ab . Why: parallel fields make $\vec E_0\cdot\vec B_0 = E_0B_0$ with no angle factor. The invariants have become two equations in two unknowns.

Step 2. But $S$ and $P$ are invariant, so they may be computed in any frame — the lab, say. Introduce D \equiv \frac{E^2}{c^2} - B^2 = -\frac{S}{2} ,\qquad Q \equiv \frac{\vec E\cdot\vec B}{c} = -\frac{P}{4} , both evaluated with the lab fields. Then Step 1 reads \boxed{\;a^2 - b^2 = D\;},\qquad \boxed{\;ab = Q\;}. Why: this is the whole trick. The unknowns live in a frame we have never visited; the data live in the frame we are standing in; and the invariants are the bridge, because they are the same in both.

Step 3. Solve. Substituting $a = Q/b$ into $a^2 - b^2 = D$ gives $b^4 + Db^2 - Q^2 = 0$ , a quadratic in $b^2$ whose positive root is \boxed{\;b^2 = \tfrac12\Big[\sqrt{D^2 + 4Q^2} \;-\; D\Big]\;},\qquad a = \frac{Q}{b} . Why: the discriminant $D^2 + 4Q^2 \ge 0$ always, and the root is non-negative always, so the canonical frame exists for every field with $Q \ne 0$ — the construction never fails. If $Q = 0$ it degenerates correctly to the pure cases of the previous derivation.

Step 4. Note what has been achieved. Two Lorentz scalars, computed in the lab from six numbers, have returned the two field magnitudes in a frame nobody has to find. The direction of the boost still requires work, but the magnitudes do not — they are invariants in disguise. Why: this is what invariants are for. The base article states that the $\vec E\parallel\vec B$ frame exists and leaves it there; the content is the closed form, and the closed form is two lines. *Verified in verify_aq-sr-electromagnetic-field-invariants.py: $b^2 = \tfrac12[\sqrt{D^2+4Q^2} - D]$ and $a = Q/b$ are shown symbolically to satisfy $a^2 - b^2 = D$ and $ab = Q$ exactly; the construction is then applied to the article's worked example and the resulting $(E_0, B_0)$ are confirmed to reproduce both lab invariants to machine precision; and $b^2 \ge 0$ is proved for all real $D, Q$ .*

Limiting cases

Case	$S = 2(B^2 - E^2/c^2)$	$P \propto \vec E\cdot\vec B$	Meaning
static point charge	$-2E^2/c^2 < 0$	$0$	electric-like; no boost removes $\vec E$ , and any boost adds a $\vec B$
static current, no charge	$+2B^2 > 0$	$0$	magnetic-like; a boost can remove $\vec E$ , never $\vec B$
plane wave in vacuum	$0$	$0$	null; irreducible in every frame. Sunlight: $E_0 = 1012.6$ V/m, $B_0 = 3.378\ \mathrm{\mu T}$
velocity selector ( $E \perp B$ )	$>0$ if $E < cB$	$0$	magnetic-like; undeflected charges move at exactly $v = E/B$
$\vec E \parallel \vec B$	$2(B^2 - E^2/c^2)$	$\ne 0$	the canonical frame — the simplest any observer can get
duality $\vec E\to c\vec B,\ \vec B\to-\vec E/c$	$S \to -S$	$P \to P$	electric-like and magnetic-like are exchanged; $P$ is untouched
parity (mirror)	$S \to S$	$P \to -P$	$S$ is a scalar, $P$ is a pseudoscalar

Common trapThe base's field tensor is minus the correct one — and both invariants hide it.

The AllQFT base for this page prints, as its very first equation card, a matrix with $F^{0i} = +E_i/c$ and $F^{ij} = +\varepsilon^{ijk}B_k$ , labelled "Field tensor (metric $+---$ )". That matrix is exactly $-1$ times the field tensor of this signature. Both blocks are flipped, so it is not a different convention for one of them — it is an overall sign error on the whole object. With it, $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ would return $\nabla\cdot\vec E = -\rho/\varepsilon_0$ .

And here is the trap. $S$ and $P$ are both quadratic in $F$ , so $F \to -F$ leaves both of them completely unchanged. The base's invariants therefore come out numerically correct, and its derivation of them is internally consistent. A wrong tensor produced right answers, and nothing on the page could detect it. That is the most dangerous kind of defect in a corpus: it survives every check that its own article performs, and then propagates into an article — the covariant Lorentz force, say — where the equation is linear in $F$ and the sign matters enormously. Invariants are not a sign check. Gauss's law is.

Common trapTwo more base defects: the trace identity has the wrong sign, and the reason given for tracelessness is not the reason.

The trace identity. The base's completeness derivation asserts \mathrm{tr}(F^4) = \tfrac12\big(\mathrm{tr}(F^2)\big)^2 - (\text{pseudoscalar})^2 . The sign is wrong. Explicit contraction gives $\mathrm{tr}(F^4) - \tfrac12 S^2 = +\tfrac14 P^2 = +4(\vec E\cdot\vec B)^2/c^2$ , which is manifestly non-negative — so no real normalisation of "pseudoscalar" can make the base's minus sign work. The conclusion it is used to support (that there are exactly two invariants) is nonetheless correct, which is precisely why the error survived.

The tracelessness claim. The base states that the electromagnetic stress-energy tensor is traceless "because it is quadratic in $F$ and the only scalar that could appear in its trace vanishes for the Maxwell field." That is not the reason, and the stated scalar does not vanish: $F_{\mu\nu}F^{\mu\nu} = S$ is generically non-zero. The real mechanism is an exact cancellation: with $T^{\mu\nu} \propto -F^{\mu\alpha}F^\nu{}_\alpha + \tfrac14\eta^{\mu\nu}S$ , the trace picks up $-S$ from the first term and $\tfrac14\,\eta^\mu{}_\mu\,S = \tfrac14\cdot 4\cdot S = +S$ from the second. They cancel because spacetime has four dimensions, and in $D$ dimensions the trace would be $S(D-4)/4 \ne 0$ . Tracelessness is a statement about $D = 4$ , not about a vanishing scalar. (The tensor itself belongs to The Electromagnetic Stress-Energy Tensor; only the base's stated reason is corrected here.)

Worked examples

1 · A velocity selector, classified and then simplified

Crossed uniform fields, $E = 3.00\times10^{5}\ \mathrm{V\,m^{-1}}$ and $B = 2.00\ \mathrm{mT}$ , with $\vec E \perp \vec B$ . Classify the field, find the frame in which it is simplest, and check the invariants there.

Step 1 — compute the invariants. $\vec E\cdot\vec B = 0$ , so $P = 0$ immediately. And \frac{E^2}{c^2} = \frac{9.00\times10^{10}}{8.987552\times10^{16}} = 1.001385\times10^{-6}\ \mathrm{T^2},\qquad B^2 = 4.00\times10^{-6}\ \mathrm{T^2}, S = 2\big(B^2 - E^2/c^2\big) = 2\big(4.00\times10^{-6} - 1.001385\times10^{-6}\big) = \mathbf{5.997230\times10^{-6}\ T^2}. Positive: magnetic-like. Equivalently $cB = 5.99585\times10^{5}\ \mathrm{V\,m^{-1}} > E$ . A frame with no electric field exists.

Step 2 — find it. v = \frac{E}{B} = \frac{3.00\times10^{5}}{2.00\times10^{-3}} = \mathbf{1.500\times10^{8}\ m\,s^{-1}} = 0.50035\,c ,\qquad \gamma = \frac{1}{\sqrt{1 - \beta^2}} = \mathbf{1.154967}. This is exactly the speed at which a charge passes through the selector undeflected — and now we know why: it is the speed of the frame in which there is no electric force to balance.

Step 3 — the field in that frame. With $\vec E' = 0$ , the remaining magnetic field is B' = \frac{B}{\gamma} = \frac{2.00\times10^{-3}}{1.154967} = \mathbf{1.731651\times10^{-3}\ T} = 1.7317\ \mathrm{mT}, weaker than the lab field, which is the opposite of what most people guess.

Step 4 — the check that costs nothing. In the primed frame the invariant must be $S' = 2\big(B'^2 - 0\big) = 2(1.731651\times10^{-3})^2 = \mathbf{5.997230\times10^{-6}\ T^2}$ — identical to Step 1, to every printed digit. That is not a coincidence and it is not a numerical fluke; it is the definition of an invariant, and it is the cheapest error check in relativistic electrodynamics. If your two numbers disagree, your boost is wrong.

2 · A generic field, and the frame nobody has to visit

Now let $\vec E = (3.00\times10^{5},\,0,\,0)\ \mathrm{V\,m^{-1}}$ and $\vec B = (2.00,\,0,\,1.00)\ \mathrm{mT}$ — no longer perpendicular. Show that neither field can be removed, then compute the field strengths in the canonical frame.

Step 1 — classify. $\vec E\cdot\vec B = (3.00\times10^{5})(2.00\times10^{-3}) = 600.0\ \mathrm{V\,T\,m^{-1}}$ , so P = -\frac{4}{c}\,\vec E\cdot\vec B = -\frac{4(600.0)}{2.997925\times10^{8}} = \mathbf{-8.005538\times10^{-6}\ T^2} \ne 0 . Generic. No boost makes this field purely electric or purely magnetic, and we know that before attempting a single one. For completeness, $B^2 = 5.00\times10^{-6}\ \mathrm{T^2}$ and S = 2\big(5.00\times10^{-6} - 1.001385\times10^{-6}\big) = \mathbf{7.997230\times10^{-6}\ T^2}.

Step 2 — set up the canonical-frame equations. D = \frac{E^2}{c^2} - B^2 = 1.001385\times10^{-6} - 5.00\times10^{-6} = \mathbf{-3.998615\times10^{-6}\ T^2}, Q = \frac{\vec E\cdot\vec B}{c} = \frac{600.0}{2.997925\times10^{8}} = \mathbf{2.001385\times10^{-6}\ T^2}.

Step 3 — solve. b^2 = \tfrac12\Big[\sqrt{D^2 + 4Q^2} - D\Big] = \tfrac12\Big[\sqrt{1.598892\times10^{-11} + 1.602216\times10^{-11}} + 3.998615\times10^{-6}\Big] = 4.828225\times10^{-6}\ \mathrm{T^2}, so B_0 = b = \mathbf{2.197322\times10^{-3}\ T} = 2.1973\ \mathrm{mT},\qquad E_0 = c\,\frac{Q}{b} = \mathbf{2.730597\times10^{5}\ V\,m^{-1}} , with $\vec E_0 \parallel \vec B_0$ .

Step 4 — verify against the invariants, which is the only honest check. In the canonical frame, 2\big(B_0^2 - E_0^2/c^2\big) = 2\big(4.828225\times10^{-6} - 8.296290\times10^{-7}\big) = \mathbf{7.997230\times10^{-6}\ T^2} = S \ \checkmark E_0 B_0 = (2.730597\times10^{5})(2.197322\times10^{-3}) = \mathbf{600.0\ V\,T\,m^{-1}} = \vec E\cdot\vec B \ \checkmark Both match the lab values exactly. We computed the fields in a frame we never constructed, using nothing but two numbers that do not care which frame you are in. That is the entire point of the subject, in one calculation.

Figure

LIGHT IS THE POINT AT THE ORIGIN — and that is why nothing can be done to it. Sunlight at the top of the atmosphere carries 1361 W/m². From ½ε₀cE₀² = 1361 W/m² its field amplitudes are E₀ = 1012.6 V/m and B₀ = 3.378 μT — and these satisfy cB₀ = E₀ EXACTLY, so S = 2(B₀² − E₀²/c²) = 0. They are also perpendicular, so P = −(4/c)E·B = 0. BOTH invariants vanish. The boost that would remove E requires v = E/B = c, and the boost that would remove B requires v = c²B/E = c. Neither exists. You may Doppler-shift a light wave to any frequency you like, red or blue, and its two invariants stay stubbornly at zero: it is a light wave in every frame, forever.

Physics of the portrait

The portrait is not a person but a habit of mind: find the quantity that does not change, and let it answer the questions that changing quantities cannot.

Origins · 2 minLorentz invariant and how this was discovered

Lorentz invariant

1905–present · the quantities every observer agrees on · why there are exactly two for the electromagnetic field · and why the QED vacuum could not have been built from anything else

A Lorentz invariant is a quantity that takes the same value in every inertial frame — the spacetime interval $s^2 = c^2t^2 - \vec x^2$ , the rest mass through $p\cdot p = +m^2c^2$ , the phase of a wave, and, for the electromagnetic field, the two scalars on this page. The concept crystallised with Einstein's 1905 paper, whose §6 gives the transformation laws for $\vec E$ and $\vec B$ from which their invariance can be checked directly, and became structural with Minkowski's 1908 four-dimensional reformulation, where an invariant is simply a fully contracted tensor and the two electromagnetic invariants appear as $F_{\mu\nu}F^{\mu\nu}$ and $F_{\mu\nu}\tilde F^{\mu\nu}$ . What gives the idea its force is not that invariants are elegant but that they are restrictive: they say what cannot happen. Because a pure electric field has $\vec E\cdot\vec B = 0$ , and because that quantity is invariant, a field with $\vec E\cdot\vec B \ne 0$ can never be made purely electric — not by any observer, at any speed, ever. One line of algebra closes off an infinite search. The deepest consequence lies outside classical physics entirely. Any Lagrangian for the electromagnetic field must be a Lorentz scalar built from $F$ ; there are exactly two such scalars; therefore the effective Lagrangian of the quantum vacuum, derived by Heisenberg and Euler in 1936, is a function of $S$ and $P$ and it could not have been anything else. The counting theorem constrained a quantum theory that would not be written for another thirty years. That is what an invariant is for.

History

1905

— Einstein's "Zur Elektrodynamik bewegter Körper" gives the transformation laws for the electric and magnetic fields in §6; the invariance of both field scalars follows directly from them.

1908

— Minkowski's four-dimensional formulation makes the invariants structural: they are the full contractions of the field tensor with itself and with its dual.

1936

— Heisenberg and Euler, in "Folgerungen aus der Diracschen Theorie des Positrons" (Z. Phys. 98, 714), derive the effective Lagrangian of the QED vacuum; it depends on the field only through the two invariants and predicts that the vacuum is birefringent.

1936

— Weisskopf independently obtains the same nonlinear vacuum Lagrangian by a simpler route.

1951

— Schwinger, in "On Gauge Invariance and Vacuum Polarization" (Phys. Rev. 82, 664), computes the rate at which a strong field boils electron-positron pairs out of the vacuum, fixing the critical field at m_e squared c cubed over e hbar.

2017

— Optical polarimetry of the isolated neutron star RX J1856.5-3754 reports evidence for vacuum birefringence, the effect Heisenberg and Euler predicted eighty-one years earlier.

Going further

Texts. Landau & Lifshitz, The Classical Theory of Fields, 4th ed., §25 (the invariants and the canonical frame) · Jackson, Classical Electrodynamics, 3rd ed., §11.9 · Rindler, Introduction to Special Relativity, 2nd ed., §38 · Heisenberg & Euler, Z. Phys. 98, 714 (1936) · Schwinger, Phys. Rev. 82, 664 (1951).

Provenance. Signature $(+,-,-,-)$ , $\varepsilon^{0123} = +1$ ; $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass, and no velocity-dependent mass appears anywhere. The dictionary $F^{0i} = -E_i/c$ , $F_{0i} = +E_i/c$ , $F^{ij} = -\varepsilon^{ijk}B_k$ is derived from $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ , not assumed. Base-defect repairs (logged): (i) the base's field-tensor matrix is $-1$ times the correct tensor — both blocks flipped — and the error is invisible to the invariants because they are quadratic; (ii) its completeness derivation prints $\mathrm{tr}(F^4) = \tfrac12(\mathrm{tr}F^2)^2 - (\text{pseudoscalar})^2$ , and the sign must be $+$ ; (iii) it attributes the tracelessness of $T^{\mu\nu}$ to a vanishing scalar, when the scalar does not vanish and the true mechanism is a $D = 4$ cancellation. Everything here is recomputed in verify_aq-sr-electromagnetic-field-invariants.py, which parses this article's own printed digits.

Scope. The construction of $F^{\mu\nu}$ belongs to The Electromagnetic Field Tensor; the dual to The Dual Field Tensor; the field equations to Covariant Maxwell Equations; the transformation laws to Transformation of Electric and Magnetic Fields; and $T^{\mu\nu}$ to The Electromagnetic Stress-Energy Tensor. This page owns the two invariants, the proof that there are exactly two, the classification, and the closed form for the canonical frame.

Builds on. The Electromagnetic Field Tensor · The Dual Field Tensor · Covariant Maxwell Equations · Transformation of Electric and Magnetic Fields. Leads to. The Electromagnetic Stress-Energy Tensor · The Covariant Lorentz Force Law · The Field of a Uniformly Moving Charge.

If E and B are frame-dependent, in what sense is the field "the same" for everyone?

In exactly the sense the invariants make precise, and no further. The six components change; the two scalars do not. So "the same field" means "the same point in the $(S, P)$ plane" — and that is genuinely all that observers agree on. It is a smaller residue than beginners expect, but it is enough to answer every qualitative question: what can be transformed away, what cannot, what a charge will do, and how the vacuum will respond.

Is E·B really invariant? It looks like it should change under a boost.

It is, and the surprise is instructive. Under a boost along $x$ , the transverse fields mix: $E'_y = \gamma(E_y - vB_z)$ while $B'_y = \gamma(B_y + vE_z/c^2)$ . Form $E'_yB'_y + E'_zB'_z$ , expand, and the cross terms proportional to $v$ cancel between the $y$ and $z$ contributions, leaving $\gamma^2(1 - v^2/c^2)(E_yB_y + E_zB_z) = E_yB_y + E_zB_z$ . The longitudinal components are untouched. The cancellation is exact, and it is what the tensor formalism predicts without any algebra at all.

Why is the free-field Lagrangian built from S and not from P?

Both are Lorentz-invariant, so Lorentz symmetry alone does not choose. Parity does. $P$ is a pseudoscalar, so a term $\theta P$ in the Lagrangian would flip sign under a mirror and make electromagnetism parity-violating — which experiment forbids to extraordinary precision. There is a second reason it does no harm anyway: $P$ is a total derivative, so adding it changes no equation of motion classically. It matters only through topology and only quantum-mechanically, which is where the axion enters.

Can a boost ever make E and B perpendicular if they start out skew?

No, unless they were secretly perpendicular all along. $\vec E\cdot\vec B$ is invariant, and perpendicularity means $\vec E\cdot\vec B = 0$ . If it is non-zero in your frame it is non-zero in every frame, so the angle can never reach $90°$ while both fields are non-zero. What a boost can do is drive the angle to $0°$ — the canonical frame — and the third derivation computes the field strengths there.

The invariants are the same in every frame. Does that mean the energy density is too?

No, and the contrast is the point. The energy density is $u = \tfrac12(\varepsilon_0 E^2 + B^2/\mu_0)$ — a sum of squares, not a difference — and it is emphatically frame-dependent, as it must be, since it is one component of the stress-energy tensor rather than a scalar. The difference $B^2 - E^2/c^2$ is invariant; the sum $B^2 + E^2/c^2$ is not. One character in that expression separates a Lorentz scalar from the $00$ component of a rank-2 tensor.

Why does the QED vacuum Lagrangian have to be a function of only these two?

Because it must be a Lorentz scalar (relativity), it must be gauge-invariant (so built from $F$ , not $A$ ), and there are exactly two independent scalars that can be built from $F$ . The counting theorem in the first derivation is not a classical curiosity — it is a hard constraint on any quantum theory of the electromagnetic field, and it is why the Heisenberg–Euler result has the form it does. The physics was fixed before the calculation began.