The Electromagnetic Field Tensor

Count the room. A rank-2 tensor in four dimensions has $4 \times 4 = 16$ components. Demand that it be antisymmetric — $F^{\mu\nu} = -F^{\nu\mu}$ — and the four diagonal entries are forced to zero, while the remaining twelve pair off into six independent numbers. Six. Exactly six. Now count electromagnetism. $E_x, E_y, E_z, B_x, B_y, B_z$ . Six. That is not a coincidence, and it is not numerology. It is the whole point. An antisymmetric rank-2 tensor built from the four-potential has exactly enough room for the electric and magnetic fields and not one slot left over — no scalar hiding in the trace, no seventh component we have failed to notice. $\vec E$ and $\vec B$ are not two fields that happen to be related; they are the timelike and spacelike halves of one object, split apart by the accident of which frame you happen to be standing in. Asking whether a field "is really electric or really magnetic" is exactly as meaningful as asking whether a stick is "really along $x$ or really along $y$ ." The object is

F^{\mu\nu} \;=\; \partial^\mu A^\nu - \partial^\nu A^\mu ,

and the single most-botched line in the entire subject is what its $0i$ components are. In signature $(+,-,-,-)$ with $A^\mu = (\varphi/c,\ \vec A)$ the answer is

\boxed{\;F^{0i} \;=\; -\,\frac{E^i}{c}\;} \qquad\text{and, with both indices down,}\qquad \boxed{\;F_{0i} \;=\; +\,\frac{E^i}{c}\;}

— and the minus sign on the upper-index form is not a convention you may take or leave. Get it backwards and $\nabla\!\cdot\!\vec E = \rho/\varepsilon_0$ comes out as $\nabla\!\cdot\!\vec E = -\rho/\varepsilon_0$ : two protons attract. Worked Example 2 prices that error at $2.3071\times10^{-10}\ \mathrm{N}$ , in the wrong direction. The AllQFT base article for this very topic prints $F^{0i} = +E^i/c$ in two separate places — and then prints a matrix, in the same article, whose top row reads $-E_x/c$ . It contradicts itself. We will derive the sign from scratch, in four lines, and never have to remember it again.

Key takeaways

Signature $(+,-,-,-)$ ; $\;\partial_\mu = \big(\tfrac1c\partial_t,\ +\nabla\big)$ , $\;\partial^\mu = \big(\tfrac1c\partial_t,\ -\nabla\big)$ ; $\;p\cdot p = +m^2c^2$ . Every sign below comes from that minus on the raised spatial index.
The object: $\boxed{F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu}$ — manifestly antisymmetric, manifestly gauge-invariant, a genuine rank-2 tensor.
The counting, which is the payoff: antisymmetry kills the 4 diagonal entries and pairs the other 12, leaving $\;\boxed{(16 - 4)/2 = 6}$ independent components. Electromagnetism has exactly 6 field components. The tensor has room for $\vec E$ and $\vec B$ and nothing else.
The signs — derive, never transcribe: $\;F^{0i} = -E^i/c$ , $\;F_{0i} = +E^i/c$ , $\;F^{ij} = F_{ij} = -\epsilon^{ijk}B_k$ . The base prints $F^{0i} = +E^i/c$ twice, while its own matrix prints $-E_x/c$ .
The check that catches you: $\partial_\mu F^{\mu 0} = \mu_0 J^0$ unpacks to $\nabla\!\cdot\!\vec E = +\rho/\varepsilon_0$ only with $F^{0i} = -E^i/c$ . Flip it and Gauss's law flips: like charges attract.
The invariant: $\;F_{\mu\nu}F^{\mu\nu} = 2\big(B^2 - E^2/c^2\big)$ — the same number in every frame. Worked Example 1 gets $\mathbf{12.277\ \mathrm{T^2}}$ for a real lab field, two independent ways.
Key idea: $\vec E$ and $\vec B$ are not two fields that transform into each other. They are the six components of ONE antisymmetric tensor — and antisymmetry is what guarantees there are exactly six, no more and no fewer. What a boost does is not convert one field into another; it re-slices a single object along a different axis, exactly as rotating your head re-slices a stick into different amounts of $x$ and $y$ .

The core insight

Antisymmetry is doing all the work. It is what forces the diagonal to vanish, what halves the off-diagonal count, and what leaves precisely six slots. Six is also the number of independent field components electromagnetism has. If nature had needed seven, or five, no antisymmetric tensor in four dimensions could have held them — and the fact that it is six is the structural reason $\vec E$ and $\vec B$ can be, and must be, one object.

Where this lives today

$F^{\mu\nu}$ is not a pedagogical repackaging of Maxwell; it is the object every modern formulation is written in. The Lagrangian of classical electrodynamics is $\;\mathcal L = -\tfrac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu} - J_\mu A^\mu$ , and varying it returns $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ — the whole inhomogeneous half of Maxwell, from one scalar. The stress–energy tensor that tells you how much momentum a light beam carries is built from $F$ quadratically, and the entire Standard Model is written by generalising this one construction: replace $\partial^\mu A^\nu - \partial^\nu A^\mu$ with its non-abelian cousin and you have the gluon and the weak bosons. The counting argument generalises too. In $n$ dimensions an antisymmetric rank-2 tensor has $n(n-1)/2$ components — which is $6$ in four dimensions, and is also exactly the dimension of the Lorentz group (three rotations, three boosts). Not an accident either: $F^{\mu\nu}$ transforms in the adjoint representation of that group, and the six field components and the six generators are the same six slots seen from two directions.

Origins · 2 minMichael Faraday and how this was discovered

Michael Faraday

1791–1867 · the field concept · lines of force · the unification of electricity and magnetism

Michael Faraday could not do the mathematics. Apprenticed to a bookbinder at fourteen, largely self-taught, he never learned calculus and wrote almost no equations in a career that produced electromagnetic induction, the laws of electrolysis, the magneto-optical effect that bears his name, and the diamagnetism of matter. What he had instead was a picture — and the picture turned out to be the truth. Faraday's picture was lines of force. He refused to believe in action at a distance, and insisted that the space between a magnet and a piece of iron was not empty but strained, filled with something real that carried the influence from one to the other. He drew these lines obsessively, watched iron filings arrange themselves along them, and in 1852 wrote of "the physical character of the lines of magnetic force" as though they were as concrete as ropes. The mathematicians of his day thought this a crutch for a man who could not integrate. It was, instead, the field. Maxwell said so explicitly: he set out to put Faraday's ideas into mathematical form, and found that "Faraday's methods... were also mathematical, though not exhibited in the conventional form of mathematical symbols." What Faraday saw in the filings — that the field is a thing, present at every point, with a structure of its own, and not merely a bookkeeping device for forces between distant charges — is precisely what $F^{\mu\nu}$ is. The tensor is Faraday's lines of force, written in the only language that survives a change of observer. Faraday believed electricity and magnetism were one phenomenon, and spent decades hunting for connections between light, gravity, and magnetism because he could not accept that nature was compartmented. He was right about the first pair, in a way he could not have made precise: $\vec E$ and $\vec B$ are not two fields but one tensor, and the frame-dependence he could not have imagined is exactly what makes them look like two. It is entirely fitting that $F^{\mu\nu}$ is still, in much of the literature, called the Faraday tensor.

History

1831

— Michael Faraday discovers electromagnetic induction and develops the concept of lines of force, insisting the field between bodies is physically real.

1852

— Faraday publishes "On the Physical Character of the Lines of Magnetic Force," arguing the field is a thing in its own right rather than a bookkeeping device.

1865

— James Clerk Maxwell publishes "A Dynamical Theory of the Electromagnetic Field," giving Faraday's picture its mathematical form and predicting electromagnetic waves.

1905

— Albert Einstein opens "On the Electrodynamics of Moving Bodies" with the magnet-and-conductor asymmetry: the same current gets two different explanations depending on which body is called stationary.

1908

— Hermann Minkowski recasts electrodynamics in four-dimensional form, introducing the antisymmetric field tensor and showing that E and B are components of a single object.

1912

— Max Abraham and Arnold Sommerfeld develop the tensor notation into the form still used today.

The physical picture

Before relativity, $\vec E$ and $\vec B$ were two vector fields, coupled by Maxwell's equations but conceptually separate: one pushed charges, the other pushed currents. Einstein's 1905 opening paragraph is about the discomfort this causes. A magnet and a conductor in relative motion produce a current, and the explanation depends entirely on which one you call stationary — an electric field from a changing flux in one telling, a magnetic force on moving charges in the other. Same current, same ammeter, two completely different stories. That asymmetry, Einstein wrote, "does not appear to be inherent in the phenomena." It is not. The two stories are two slicings of a single object. $F^{\mu\nu}$ is a rank-2 tensor field on spacetime, and $\vec E$ and $\vec B$ are the names we give to its components in whatever frame we happen to occupy. A boost does not turn $\vec E$ into $\vec B$ any more than turning your head turns $x$ into $y$ ; it re-expresses one unchanged geometric object in a new basis. The magnet-and-conductor asymmetry dissolves because there was never anything asymmetric there — only two coordinate descriptions of one tensor. The tensor is manifestly antisymmetric by construction: $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ swaps sign when you swap $\mu$ and $\nu$ . That gives it, for free, both properties that make it the right object: six independent components, and exact gauge invariance (a curl cannot see a gradient — see Gauge Transformations and the Lorenz Gauge). What remains is bookkeeping: which component goes where, and with what sign.

Figure

THE COUNTING ARGUMENT — why the electromagnetic field tensor has room for exactly E and B, with nothing left over. A rank-2 tensor in 4D has 16 components. ANTISYMMETRY (Fᵘᵛ = −Fᵛᵘ) forces the 4 diagonal entries to zero and pairs the remaining 12 into 6 independent numbers. Electromagnetism has exactly 6 field components: Ex Ey Ez Bx By Bz. Six slots, six fields — no scalar in the trace, no seventh component, nothing missing

The core equations

\boxed{\;F^{\mu\nu} \;=\; \partial^\mu A^\nu - \partial^\nu A^\mu\;} \qquad\Longrightarrow\qquad F^{\mu\nu} = -F^{\nu\mu} \qquad (1)

where: $A^\mu = (\varphi/c,\ \vec A)$ and $\partial^\mu = \big(\tfrac1c\partial_t,\ -\nabla\big)$ . Physical reading: the antisymmetry is not imposed, it is built in — swap the labels and the expression negates itself. So is gauge invariance: replace $A^\mu \to A^\mu + \partial^\mu\chi$ and the extra term is $\partial^\mu\partial^\nu\chi - \partial^\nu\partial^\mu\chi = 0$ . The components, derived in Derivation 2 and never to be transcribed:

\boxed{\;F^{0i} = -\frac{E^i}{c}\;}, \qquad \boxed{\;F_{0i} = +\frac{E^i}{c}\;}, \qquad \boxed{\;F^{ij} = F_{ij} = -\epsilon^{ijk}B_k\;} \qquad (2)

where: $\epsilon^{ijk}$ is the three-dimensional Levi-Civita symbol with $\epsilon^{123} = +1$ . Physical reading: lowering an index with $\eta = \mathrm{diag}(+1,-1,-1,-1)$ costs a factor $(-1)$ for each spatial index. The $0i$ entries carry one spatial index, so they flip sign. The $ij$ entries carry two, so the signs cancel and the magnetic block is unchanged. That asymmetry between the electric and magnetic blocks is the single fact that generates every sign confusion in the literature. Written out, with $\mu$ the row and $\nu$ the column:

F^{\mu\nu} = \begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ +E_x/c & 0 & -B_z & +B_y \\ +E_y/c & +B_z & 0 & -B_x \\ +E_z/c & -B_y & +B_x & 0 \end{pmatrix}, \qquad F_{\mu\nu} = \begin{pmatrix} 0 & +E_x/c & +E_y/c & +E_z/c \\ -E_x/c & 0 & -B_z & +B_y \\ -E_y/c & +B_z & 0 & -B_x \\ -E_z/c & -B_y & +B_x & 0 \end{pmatrix} \qquad (3)

where: the two matrices differ only in the electric row and column. The magnetic $3\times3$ block is identical in both. Physical reading: if you can remember one thing about lowering indices on $F$ , remember that — the E block flips, the B block does not. The invariant, and the equation that fixes the sign:

\boxed{\;F_{\mu\nu}F^{\mu\nu} \;=\; 2\left(B^2 - \frac{E^2}{c^2}\right)\;} \qquad (4) \qquad\qquad \boxed{\;\partial_\mu F^{\mu\nu} = \mu_0 J^\nu\;} \qquad (5)

where: (4) is a fully contracted scalar — the same number in every inertial frame. Physical reading: it is not merely a curiosity but the first hard, frame-independent fact you can extract from a field, and Worked Example 1 evaluates it to $12.277\ \mathrm{T^2}$ for a real lab field. (5) is the inhomogeneous half of Maxwell's equations in one line; unpacking it at $\nu = 0$ gives Gauss's law with the correct plus sign — but only if $F^{0i} = -E^i/c$ . This is the check that catches a sign error before it eats your physics.

The object, its six components, and its signs

Derivation AAntisymmetry forces exactly six components — and electromagnetism has exactly six fields

This is the structural heart of the subject, and it is a counting argument that takes four steps. The claim is not "we noticed $\vec E$ and $\vec B$ fit into a tensor." The claim is that an antisymmetric rank-2 tensor in four dimensions has exactly six independent components — no fewer, no more — and that this is precisely the number of independent field components electromagnetism possesses. The fit is exact, and an exact fit is a reason.

Step 1 — start with 16 and kill the diagonal. A general rank-2 tensor $T^{\mu\nu}$ in four dimensions has $4 \times 4 = 16$ components. Antisymmetry says $F^{\mu\nu} = -F^{\nu\mu}$ . Set $\nu = \mu$ : $F^{\mu\mu} = -F^{\mu\mu}$ , hence $\;\boxed{F^{\mu\mu} = 0}$ for each of the four values of $\mu$ — no summation. Why: the diagonal is not merely small or conventionally chosen; it is forced to vanish. Four components gone, twelve remain. And note what this kills: there is no trace, so there is no scalar hiding inside $F$ — no extra field we have overlooked.

Step 2 — pair off the rest. The twelve off-diagonal entries come in pairs $(\mu\nu)$ and $(\nu\mu)$ , and antisymmetry means the second is minus the first: knowing $F^{01}$ is knowing $F^{10}$ . So the twelve carry only $12/2 = 6$ pieces of independent information. Total: $\;\boxed{(16 - 4)/2 = 6}$ . Why: six. The lower triangle of the matrix contains no new information at all — it is the upper triangle with every sign flipped. That is why writing the matrix out looks redundant: it is redundant, by construction.

Step 3 — split the six by index type. The six independent slots divide naturally into two groups of three. Three of them have one time index and one space index: $F^{01}, F^{02}, F^{03}$ . Three have two space indices: $F^{12}, F^{13}, F^{23}$ . Why: this split is not arbitrary — it is the only division a Lorentz-invariant object admits, because "time" and "space" are the only two kinds of index there are. And it is exactly the split between the electric field (three components) and the magnetic field (three components). The tensor tells you, before you have put a single field into it, that electromagnetism must consist of one three-vector tied to the time direction and one three-vector living purely in space.

Step 4 — check the fit, and check that nothing is left over. Electromagnetism has $E_x, E_y, E_z, B_x, B_y, B_z$ : six numbers. The tensor has six slots. Every slot is used and every field has a home. Why: this is the payoff, and it is worth being precise about what it rules out. There is no seventh component we have failed to notice, and no scalar in the trace (Step 1 removed it). When you boost a field, the six numbers are rearranged among themselves — they cannot spill into a slot that does not exist. That is the deepest reason $\vec E$ and $\vec B$ are one object rather than two: there is exactly enough room for them, and no room for anything else. *Verified in verify_aq-sr-electromagnetic-field-tensor.py: the diagonal proved zero from antisymmetry alone; the count $(16-4)/2 = 6$ carried out on the explicit index set; the six independent slots enumerated and shown to partition into three $0i$ and three $ij$ ; and $F^{\mu\nu}$ built from six arbitrary functions shown to reproduce exactly $\vec E$ and $\vec B$ with nothing unassigned.*

Go deeper · Derivation B · 2 minThe F-zero-i sign, derived from scratch in four lines — and why you must never transcribe it

Derivation BThe F-zero-i sign, derived from scratch in four lines — and why you must never transcribe it

Here is the line that the literature, and the AllQFT base article for this very topic, gets wrong more often than any other in relativistic electrodynamics. We derive it. It takes four steps and it depends on exactly one thing: the minus sign that raising a spatial index puts on the gradient.

Step 1 — write down the two ingredients, honestly. In signature $(+,-,-,-)$ with $\eta = \mathrm{diag}(+1,-1,-1,-1)$ : the four-potential is $A^\mu = (\varphi/c,\ A_x, A_y, A_z)$ , and the raised derivative is $\;\partial^\mu = \eta^{\mu\nu}\partial_\nu = \big(\tfrac1c\partial_t,\ -\nabla\big)$ . Why: $\partial_\mu = \partial/\partial x^\mu$ has a lower index by construction, so raising it costs a factor of $\eta^{ii} = -1$ on each spatial slot. This single minus sign generates every sign in the field tensor. It is the only thing you need to remember, and it is not a convention — given the signature, it is forced.

Step 2 — compute $F^{01}$ directly. By definition, $\;F^{01} = \partial^0 A^1 - \partial^1 A^0$ . Substitute: $\;\partial^0 = \tfrac1c\partial_t$ and $A^1 = A_x$ , giving the first term $\tfrac1c\,\partial_t A_x$ . And $\;\partial^1 = -\partial_x$ with $A^0 = \varphi/c$ , giving the second term $-\big(-\partial_x\big)\big(\varphi/c\big) = +\tfrac1c\,\partial_x\varphi$ . So $\;F^{01} = \tfrac1c\big(\partial_t A_x + \partial_x\varphi\big)$ . Why: every symbol here is fixed by Step 1. There is no freedom left.

Step 3 — recognise what is in the bracket. The electric field in terms of potentials is $\;\vec E = -\nabla\varphi - \partial_t\vec A$ , so its $x$ -component is $\;E_x = -\partial_x\varphi - \partial_t A_x$ . The bracket in Step 2 is exactly minus that: $\;\partial_t A_x + \partial_x\varphi = -E_x$ . Therefore

\boxed{\;F^{01} \;=\; -\,\frac{E_x}{c}\;}

and by the same computation on the other two axes, $\;\boxed{F^{0i} = -E^i/c}$ . Why: the minus is not optional. It arrived from $\partial^1 = -\partial_x$ in Step 2 and it is now sitting in the answer. Anyone who writes $F^{0i} = +E^i/c$ in this signature, with this $A^\mu$ , has dropped it — and Worked Example 2 shows what that costs.

Step 4 — lower the indices and watch the electric block flip. $F_{\mu\nu} = \eta_{\mu\alpha}\eta_{\nu\beta}F^{\alpha\beta}$ . For a $0i$ entry: one factor of $\eta_{00} = +1$ and one of $\eta_{ii} = -1$ , so $\;F_{0i} = (+1)(-1)F^{0i} = \boxed{+E^i/c}$ . For an $ij$ entry: two factors of $(-1)$ , which cancel, so $\;\boxed{F_{ij} = F^{ij} = -\epsilon^{ijk}B_k}$ — unchanged. Why: the electric block flips sign under index lowering and the magnetic block does not, because they carry different numbers of spatial indices. This is the entire content of the difference between the two matrices in (3), and it is why "the sign of $F$ 's electric entries" is a meaningless question until you say whether the indices are up or down. Half the contradictions in the literature are two authors who are both right, about different objects. *Verified in verify_aq-sr-electromagnetic-field-tensor.py: $F^{\mu\nu}$ built symbolically from an arbitrary $A^\mu(t,x,y,z)$ with $\partial^\mu = (\tfrac1c\partial_t, -\nabla)$ ; all sixteen entries compared against the matrix in (3); $F^{01} = -E_x/c$ and $F_{01} = +E_x/c$ asserted separately; $F^{12} = F_{12} = -B_z$ asserted; and the whole $F_{\mu\nu} = \eta F^{\mu\nu} \eta$ lowering carried out and compared entry-by-entry.*

Go deeper · Derivation C · 2 minUnpack ν = 0 and out falls Gauss's law — with the plus sign, but only if the sign is right

Derivation CUnpack ν = 0 and out falls Gauss's law — with the plus sign, but only if the sign is right

The tensor would be an elegant filing cabinet and nothing more if Maxwell's equations did not come out of it. They do, and the time component is the one that audits your signs: it returns Gauss's law, and it returns it with the correct sign if and only if $F^{0i} = -E^i/c$ .

Step 1 — write the covariant field equation and set $\nu = 0$ . The inhomogeneous Maxwell equations are $\;\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ , with $J^\nu = (c\rho,\ \vec J\,)$ . Taking $\nu = 0$ : $\;\partial_\mu F^{\mu 0} = \mu_0 J^0 = \mu_0 c\rho$ . Why: $\nu = 0$ selects the equation whose source is the charge density rather than the current density — so we should expect Gauss's law, and the only question is what sign it comes out with.

Step 2 — expand the sum over $\mu$ . $\;\partial_\mu F^{\mu 0} = \partial_0 F^{00} + \partial_i F^{i0}$ . The first term dies: $F^{00} = 0$ by antisymmetry. For the second, antisymmetry gives $\;F^{i0} = -F^{0i} = -\big(-E^i/c\big) = +E^i/c$ . So $\;\partial_\mu F^{\mu 0} = \partial_i\big(E^i/c\big) = \tfrac1c\,\nabla\!\cdot\!\vec E$ . Why: note the double negative. Antisymmetry flips $F^{0i}$ to $F^{i0}$ , and the sign we derived in Derivation 2 flips it back. The result is a plus. And $\partial_i$ is a lower-index derivative — just $\partial/\partial x^i$ , with no metric factor — so no further minus creeps in.

Step 3 — equate and clean up. $\;\tfrac1c\nabla\!\cdot\!\vec E = \mu_0 c\rho$ , so $\;\nabla\!\cdot\!\vec E = \mu_0 c^2 \rho$ . And since $\;\mu_0\varepsilon_0 = 1/c^2$ , we have $\mu_0 c^2 = 1/\varepsilon_0$ — numerically $\mathbf{1.1294\times10^{11}\ \mathrm{m\,F^{-1}}}$ — giving

\boxed{\;\nabla\!\cdot\!\vec E \;=\; +\,\frac{\rho}{\varepsilon_0}\;}

Why: Gauss's law, with a plus. Positive charge is a source of electric field; field lines point outward; like charges repel. Every one of those statements lives in that plus sign.

Step 4 — now break it deliberately. Suppose you had transcribed $F^{0i} = +E^i/c$ from a text that uses a different signature, as the base for this article did. Then $F^{i0} = -E^i/c$ , Step 2 returns $\;-\tfrac1c\nabla\!\cdot\!\vec E$ , and Step 3 delivers $\;\nabla\!\cdot\!\vec E = -\rho/\varepsilon_0$ . Why: a positive charge would now be a sink. Field lines would point inward, and two protons a nanometre apart would attract with a force of $2.3071\times10^{-10}\ \mathrm{N}$ instead of repelling with it (Worked Example 2). The error does not stay quietly in the notation — it inverts the observable universe. This is why you derive the sign and never copy it, and it is why the $\nu = 0$ unpacking is worth doing once by hand: it is a free audit of everything upstream. *Verified in verify_aq-sr-electromagnetic-field-tensor.py: $\partial_\mu F^{\mu 0} = \tfrac1c\nabla\!\cdot\!\vec E$ derived symbolically from the potentials, and $\nabla\!\cdot\!\vec E = +\rho/\varepsilon_0$ obtained with the plus; the same computation re-run with the base's $F^{0i} = +E^i/c$ and shown to yield $\nabla\!\cdot\!\vec E = -\rho/\varepsilon_0$ ; $\mu_0 c^2 = 1/\varepsilon_0 = 1.1294\times10^{11}$ checked numerically; and $\nu = i$ shown to give the Ampère–Maxwell law.*

Figure

THE TWO MATRICES, side by side — and the one rule you need. Raising or lowering an index costs a factor of (−1) for EACH SPATIAL index. The 0i entries carry ONE spatial index, so the ELECTRIC block FLIPS SIGN. The ij entries carry TWO, so the minus signs cancel and the MAGNETIC block is UNCHANGED. That is the whole difference between the two matrices, and it is the source of nearly every sign contradiction in the literature

Limiting cases

Situation	The tensor	What it means
Pure electrostatics ( $\vec B = \vec 0$ )	only the $0i$ entries survive: $F^{0i} = -E^i/c$	$F_{\mu\nu}F^{\mu\nu} = -2E^2/c^2 < 0$ . Electric-dominated in every frame
Pure magnetostatics ( $\vec E = \vec 0$ )	only the $ij$ block survives	$F_{\mu\nu}F^{\mu\nu} = +2B^2 > 0$ . Magnetic-dominated in every frame
A light wave ( $E = cB$ , $\vec E \perp \vec B$ )	both blocks, exactly balanced	$F_{\mu\nu}F^{\mu\nu} = 0$ . Null — and no boost can ever remove either field
No field at all	$F^{\mu\nu} = 0$	zero in one frame ⇒ zero in all frames (a tensor equation)
Boost	components rearrange among themselves	six in, six out — the invariant $F_{\mu\nu}F^{\mu\nu}$ does not move

Common trapThe base article for this topic prints the wrong sign — twice — and then contradicts itself in its own matrix

This is not a hypothetical trap. The AllQFT base article for The Electromagnetic Field Tensor prints

F^{0i} \;=\; +\frac{E^i}{c} \qquad\text{— WRONG in this signature —}

in two separate places: in its equation card, and again in its "Pure electrostatics" limiting-case card. And in the very same article it prints the matrix for $F^{\mu\nu}$ whose top row begins $\;0,\ -E_x/c,\ \dots$ — the correct sign. The two cannot both be right, and the matrix is the one that is.

How to tell, in ten seconds, without trusting anybody: unpack $\partial_\mu F^{\mu 0} = \mu_0 J^0$ . With $F^{0i} = -E^i/c$ you get $\nabla\!\cdot\!\vec E = +\rho/\varepsilon_0$ . With $F^{0i} = +E^i/c$ you get $\nabla\!\cdot\!\vec E = -\rho/\varepsilon_0$ , which says a proton is a sink of electric field and that two protons attract. Gauss's law is the audit, and it takes four lines (Derivation 3).

Why the error is so common: in signature $(-,+,+,+)$ — used by Misner–Thorne–Wheeler and much of the general-relativity literature — the raised gradient is $\partial^\mu = \big(-\tfrac1c\partial_t,\ +\nabla\big)$ , and $F^{0i} = +E^i/c$ is then perfectly correct. Authors copy a line across a signature boundary without re-deriving it. Never transcribe a sign. Derive it, in four lines, every time — and then audit it against Gauss's law.

Worked examples

1 · Build the tensor for a real field, and get its invariant two independent ways

A laboratory region contains $\vec E = (1.00\times10^{8},\ 0,\ 0)\ \mathrm{V\,m^{-1}}$ and $\vec B = (2.00,\ 0,\ 1.50)\ \mathrm{T}$ . Construct $F^{\mu\nu}$ , lower its indices, and compute the invariant.

Step 1 — convert $\vec E$ to tensor units. Every electric entry appears as $E^i/c$ : $\;E_x/c = (1.00\times10^{8})/(2.9979\times10^{8}) = \mathbf{0.33356\ \mathrm{T}}$ , and $E_y/c = E_z/c = 0$ . The tensor's natural unit for the electric field is the tesla — which is already a hint that $\vec E$ and $\vec B$ are the same kind of thing.

Step 2 — fill the six independent slots. From (2): $\;F^{01} = -E_x/c = \mathbf{-0.33356}$ , $\;F^{02} = F^{03} = 0$ . And $\;F^{12} = -B_z = \mathbf{-1.50}$ , $\;F^{13} = +B_y = \mathbf{0}$ , $\;F^{23} = -B_x = \mathbf{-2.00}$ (all in tesla). The other ten entries are the diagonal (zero) and the lower triangle (minus these).

Step 3 — lower both indices. Only the electric row and column flip: $\;F_{01} = +0.33356$ , while the magnetic block is untouched — $\;F_{12} = -1.50$ , $\;F_{23} = -2.00$ , exactly as before.

Step 4 — the invariant, computed two ways. From the formula: $\;E^2/c^2 = (0.33356)^2 = \mathbf{0.11127\ \mathrm{T^2}}$ and $\;B^2 = 2.00^2 + 1.50^2 = \mathbf{6.25\ \mathrm{T^2}}$ , so

F_{\mu\nu}F^{\mu\nu} = 2\left(B^2 - \frac{E^2}{c^2}\right) = 2\,(6.25 - 0.11127) = \mathbf{12.277\ \mathrm{T^2}} .

Independently, contracting the two matrices entry by entry — $\sum_{\mu\nu} F_{\mu\nu}F^{\mu\nu}$ , all sixteen terms — returns $\mathbf{12.277\ \mathrm{T^2}}$ as well. The two routes agree, which is the check that the index-lowering was done correctly.

Step 5 — the boundary number, and what it buys. The invariant is positive, and positive means magnetically dominated: $B^2 > E^2/c^2$ . Every inertial observer in the universe, whatever they measure for $\vec E$ and $\vec B$ separately, will compute $12.277\ \mathrm{T^2}$ from their own numbers. Here $\vec E\cdot\vec B = 2.00\times10^{8}\ \mathrm{V\,T\,m^{-1}} \ne 0$ , so the second invariant does not vanish either — and the full classification (which fields can be boosted to purely magnetic, which to purely electric) needs both. That is derived in The Transformation of Electric and Magnetic Fields.

2 · What the wrong sign costs: two protons that attract

Take a proton at the origin. Evaluate Gauss's law at $r = 1.00\ \mathrm{nm}$ using the tensor — first with the correct $F^{0i} = -E^i/c$ , then with the base's $F^{0i} = +E^i/c$ — and compare the force on a second proton.

Step 1 — the correct unpacking. From Derivation 3, $\;\partial_\mu F^{\mu 0} = \tfrac1c\nabla\!\cdot\!\vec E = \mu_0 c\rho$ , hence $\;\nabla\!\cdot\!\vec E = \mu_0 c^2\rho = +\rho/\varepsilon_0$ , using $\mu_0 c^2 = 1/\varepsilon_0 = \mathbf{1.1294\times10^{11}\ \mathrm{m\,F^{-1}}}$ .

Step 2 — integrate over a sphere. Gauss's theorem gives $\;\oint\vec E\cdot\mathrm d\vec A = q/\varepsilon_0 = (1.6022\times10^{-19})/(8.8542\times10^{-12}) = \mathbf{1.8095\times10^{-8}\ \mathrm{V\,m}}$ . On a sphere of radius $1.00\ \mathrm{nm}$ this means $\;4\pi r^2 E = 1.8095\times10^{-8}$ , so

E = \frac{1.8095\times10^{-8}}{4\pi(1.00\times10^{-9})^2} = \mathbf{1.4400\times10^{9}\ \mathrm{V\,m^{-1}}}, \quad\text{pointing } \textbf{outward}.

Step 3 — the force on a second proton. $\;F = eE = (1.6022\times10^{-19})(1.4400\times10^{9}) = \mathbf{2.3071\times10^{-10}\ \mathrm{N}}$ , directed away from the first proton. Two protons repel. This is not in dispute.

Step 4 — now use the base's sign. With $F^{0i} = +E^i/c$ , antisymmetry gives $F^{i0} = -E^i/c$ , so Step 1 returns $\;\partial_\mu F^{\mu 0} = -\tfrac1c\nabla\!\cdot\!\vec E = \mu_0 c\rho$ , i.e. $\;\nabla\!\cdot\!\vec E = -\rho/\varepsilon_0$ . The flux integral is unchanged in magnitude but reversed in sign: $\vec E$ points inward, and the second proton is pulled toward the first with the same $2.3071\times10^{-10}\ \mathrm{N}$ .

Step 5 — the boundary number, and the lesson. The magnitude is identical; only the direction is wrong. A sign error in $F^{0i}$ does not produce a small error — it produces the exactly-wrong universe, in which like charges attract, atoms do not exist, and every electrostatics problem you have ever solved is inverted. The error is invisible in the magnitude and fatal in the physics, which is precisely why it survives so long in print. Derive the sign. Then audit it against Gauss's law, which costs four lines and catches this immediately.

Going further

Texts. Griffiths, Introduction to Electrodynamics, 4th ed., §12.3.2 · Jackson, Classical Electrodynamics, 3rd ed., §11.9 · Rindler, Introduction to Special Relativity, 2nd ed., ch. 7 · Minkowski, "Die Grundgleichungen für die elektromagnetischen Vorgänge in bewegten Körpern," Nachr. Ges. Wiss. Göttingen (1908).

Provenance. Signature $(+,-,-,-)$ ; $\eta = \mathrm{diag}(+1,-1,-1,-1)$ ; $\partial^\mu = (\tfrac1c\partial_t, -\nabla)$ ; $A^\mu = (\varphi/c, \vec A)$ ; SI units. Every sign in this article is derived, in Derivation 2, from $\partial^i = -\partial_i$ — none is transcribed.

Base defect found and repaired. The AllQFT base article prints $F^{0i} = +E^i/c$ in two places (its equation card and its "Pure electrostatics" limiting-case card), while printing a matrix in the same article whose first row correctly reads $-E_x/c$ . The article is internally inconsistent. The correct value in this signature is $F^{0i} = -E^i/c$ , audited against Gauss's law in Derivation 3 and asserted in the verify script. The base's sign is correct in signature $(-,+,+,+)$ , which is almost certainly where it was copied from.

Scope. The full set of covariant Maxwell equations belongs to The Covariant Maxwell Equations; the dual tensor $\tilde F^{\mu\nu}$ and the homogeneous half to The Dual Field Tensor; the boost transformation laws to The Transformation of Electric and Magnetic Fields; both invariants and the field classification to The Electromagnetic Field Invariants.

Is F⁰ⁱ = −Eⁱ/c or +Eⁱ/c? I have seen both.

Both are printed, and both are correct — in different signatures. In $(+,-,-,-)$ with $A^\mu = (\varphi/c, \vec A)$ and $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ , the raised gradient is $\partial^\mu = (\tfrac1c\partial_t, -\nabla)$ and the answer is $F^{0i} = -E^i/c$ . In $(-,+,+,+)$ the raised gradient is $(-\tfrac1c\partial_t, +\nabla)$ and the answer is $F^{0i} = +E^i/c$ . The error is not in either book; it is in copying a line from one book into the other's conventions. Derive it in four lines (Derivation 2), then audit against Gauss's law (Derivation 3).

Why is the electric field divided by c in the tensor?

Because the tensor's slots must all carry the same units, and in SI $\vec E$ and $\vec B$ do not: $\vec E$ is in volts per metre, $\vec B$ in tesla, and $1\ \mathrm{T} = 1\ \mathrm{V\,s\,m^{-2}}$ . Dividing $\vec E$ by $c$ converts it to tesla — for the field in Worked Example 1, $E_x/c = 0.33356\ \mathrm{T}$ against $B_z = 1.50\ \mathrm{T}$ , directly comparable. This is the SI system announcing, through its units, that $\vec E$ and $\vec B$ are the same kind of quantity. In Gaussian units the $c$ disappears entirely and the two have identical dimensions.

Is Fᵘᵛ gauge-invariant?

Yes, exactly and identically. Replace $A^\mu \to A^\mu + \partial^\mu\chi$ for any smooth $\chi$ and the extra term is $\partial^\mu\partial^\nu\chi - \partial^\nu\partial^\mu\chi$ , which vanishes because partial derivatives commute. Not an approximation, and it does not depend on $\chi$ being small: a curl cannot see a gradient. It is also why $F^{\mu\nu}$ , rather than $A^\mu$ , is the object appearing in every measurable quantity — forces, energies, the stress–energy tensor, both invariants.

Does a boost turn an electric field into a magnetic one?

Not in the way the phrase suggests. A boost does not convert one field into another; it re-expresses one unchanged tensor in a new basis, and the six components get redistributed among the same six slots. The analogy is a stick on a table: rotate your head and its $x$ - and $y$ -components change, but you have not converted length-along- $x$ into length-along- $y$ — you have re-described one object. That is also why the field cannot spill into a seventh slot: antisymmetry says there isn't one. What cannot change is $F_{\mu\nu}F^{\mu\nu}$ , which every observer computes to the same number.

Why does the ν = 0 component of ∂ᵤFᵘᵛ = μ₀Jᵛ give Gauss's law and not something else?

Because $\nu = 0$ selects the equation whose source is $J^0 = c\rho$ — the charge density rather than the current density. Expanding, $\partial_\mu F^{\mu 0} = \partial_0 F^{00} + \partial_i F^{i0}$ ; the first term dies by antisymmetry ( $F^{00} = 0$ ), and the second is $\partial_i(E^i/c) = \tfrac1c\nabla\!\cdot\!\vec E$ . Setting that equal to $\mu_0 c\rho$ and using $\mu_0 c^2 = 1/\varepsilon_0$ gives $\nabla\!\cdot\!\vec E = \rho/\varepsilon_0$ . The three spatial components give Ampère–Maxwell. One tensor equation, both inhomogeneous Maxwell equations.

What does the invariant F_{μν}Fᵘᵛ actually tell you?

It is the first frame-independent fact you can state about a field, and it sorts fields into three classes. $F_{\mu\nu}F^{\mu\nu} = 2(B^2 - E^2/c^2)$ : positive means magnetically dominated ( $B > E/c$ ), negative electrically dominated, and zero means the two are exactly balanced — which is what a light wave in vacuum does, where $E = cB$ always. No boost can change the sign, because no boost can change the number. Worked Example 1 gets $+12.277\ \mathrm{T^2}$ for a real lab field. What that classification buys is derived in The Transformation of Electric and Magnetic Fields, and needs the second invariant too.