The Covariant Lorentz Force Law

The Lorentz force $\vec F = q(\vec E + \vec v\times\vec B)$ is the most-used equation in electromagnetism and the least self-explanatory. Why a sum of two utterly different-looking terms? Why does one of them do work and the other never? Why does the cross product appear at all? Write it in spacetime and the questions dissolve:

\frac{\mathrm{d}p^\mu}{\mathrm{d}\tau} = q\,F^{\mu\nu}u_\nu .

That is the entire law. One field tensor, one four-velocity, one invariant charge. The $\nu = i$ terms of the contraction produce $q\vec v\times\vec B$ ; the $\nu = 0$ term produces $q\vec E$ . They were never two forces. They are the space-space and time-space blocks of a single object contracted with the same four-velocity, and the cross product is what a contraction with $F^{ij} = -\varepsilon^{ijk}B_k$ looks like in three-vector language. The rest falls out of one line of index algebra. $F^{\mu\nu}$ is antisymmetric and $u_\mu u_\nu$ is symmetric, so K^\mu u_\mu = q\,F^{\mu\nu}u_\nu u_\mu = 0 identically. The four-force is always orthogonal to the four-velocity — which is exactly the statement that rest mass is conserved, and, unpacked into three-vectors, the statement that the magnetic field can never do work. Both famous facts, one antisymmetry.

Key takeaways

Signature $(+,-,-,-)$ , $\eta = \mathrm{diag}(+1,-1,-1,-1)$ , and therefore $\boxed{p\cdot p = +m^2c^2}$ ; $m$ is the invariant rest mass. Dictionary: $u^\mu = \gamma(c,\vec v)$ , $u_\mu = (\gamma c, -\gamma\vec v)$ , $F^{0i} = -E_i/c$ , $F^{ij} = -\varepsilon^{ijk}B_k$ .
The law: $\boxed{\dfrac{\mathrm{d}p^\mu}{\mathrm{d}\tau} = K^\mu = q\,F^{\mu\nu}u_\nu}$ — one four-vector equation, valid at every speed, in every inertial frame, with no $\gamma$ -factors inserted by hand.
Unpacked: $K^i = \gamma q(\vec E + \vec v\times\vec B)_i$ and $K^0 = \dfrac{\gamma q\,\vec E\cdot\vec v}{c}$ . Dividing by $\gamma$ (because $\mathrm{d}\tau = \mathrm{d}t/\gamma$ ) returns $\mathrm{d}\vec p/\mathrm{d}t = q(\vec E + \vec v\times\vec B)$ and $\mathrm{d}\mathcal E/\mathrm{d}t = q\vec E\cdot\vec v$ . The electric force and the power are the same equation, read in different slots.
Orthogonality, from antisymmetry alone: $K^\mu u_\mu = qF^{\mu\nu}u_\nu u_\mu = 0$ , because antisymmetric contracted with symmetric vanishes. Consequences: $m$ is constant, and $\vec B$ does no work — ever, at any speed.
The base's identity is wrong, and the fix does not touch the physics. The base declares $(-,+,+,+)$ and prints $p^\mu p_\mu = -m^2c^2$ ; in the locked signature it is $+m^2c^2$ . But redo the contraction with the corrected dictionary and $K^i$ and $K^0$ come out bit-for-bit identical. The force law is signature-blind; only the printed invariant flipped. (One of its cards also contradicts itself mid-derivation — see the second warning.)
Key idea: the law and the machine agree. A $7.00\ \mathrm{TeV}$ proton ( $\gamma = 7460.5$ ) in the LHC's $8.33\ \mathrm{T}$ dipoles bends with radius $r = p/(qB) = \mathbf{2803\ m}$ . The ring is $26\,659\ \mathrm{m}$ around, and $2\pi r$ is only $17\,612\ \mathrm{m}$ of that — $\mathbf{66.1\%}$ . That is not an error: it is the fraction of the LHC that is arc, and CERN quotes 66%.

The core insight

Contract an antisymmetric tensor with a four-velocity and you get a four-vector automatically orthogonal to that four-velocity. That one sentence contains the whole of relativistic charged-particle dynamics: the unification of $q\vec E$ with $q\vec v\times\vec B$ , the conservation of rest mass, the fact that magnetic fields steer but never accelerate, and the exact relativistic cyclotron frequency. You do not add relativity to the Lorentz force. You notice that the Lorentz force was already relativistic and had been written down badly.

Where this lives today

Every accelerator on Earth is an application of $r = p/(qB)$ , used in exactly the covariant form derived here, because at LHC energies ( $\gamma = 7460$ ) the Newtonian version is wrong by a factor of seven thousand. The design logic runs backwards through the equation: choose a beam energy and the product $Br$ is fixed; choose a tunnel radius and the dipole field is fixed. The LHC's $8.33\ \mathrm{T}$ is not a round number chosen for elegance — it is what $7\ \mathrm{TeV}$ costs in a $27\ \mathrm{km}$ tunnel that is only $66\%$ arc, and it sits at the edge of what niobium-titanium can do at $1.9\ \mathrm{K}$ . The orthogonality identity is equally practical. Beam-dynamics codes track particles for millions of turns, and $K^\mu u_\mu = 0$ — equivalently, $\gamma$ constant in a pure magnetic field — is the invariant they use to detect that the integrator has gone wrong. A tracker that lets $\gamma$ drift in a dipole is reporting numerical error, not physics. The same antisymmetry that protects the electron's rest mass is what tells the software it has a bug.

The physical picture

A three-force is a bad object. It is not a four-vector, it transforms awkwardly between frames, and Newton's second law written with it is not covariant. The cure is to differentiate momentum with respect to proper time rather than coordinate time, giving the four-force $K^\mu = \mathrm{d}p^\mu/\mathrm{d}\tau$ — a genuine four-vector, because $p^\mu$ is one and $\tau$ is a scalar. Now ask what $K^\mu$ could be for a charge. It must be linear in the field (double the field, double the force), linear in the velocity (a charge at rest still feels $q\vec E$ ; a moving one picks up the magnetic term), and it must be a four-vector. There is essentially one candidate: $q F^{\mu\nu}u_\nu$ . The field tensor is the only object carrying the field with the right rank, and contracting one of its indices with $u_\nu$ leaves exactly one free index. The law is not derived so much as cornered. And then the antisymmetry does the rest. $K^\mu u_\mu = qF^{\mu\nu}u_\mu u_\nu$ contracts an antisymmetric tensor with a symmetric one, so it vanishes identically — any field, any charge, any velocity. Since $p^\mu p_\mu = +m^2c^2$ , differentiating gives $2p_\mu\,\mathrm{d}p^\mu/\mathrm{d}\tau = 2m\,K^\mu u_\mu = 0$ , so $m$ never changes. Rest mass is conserved not because of a conservation law but because of an index symmetry, and the three-dimensional shadow of that same fact is the schoolroom rule that magnetic forces do no work.

Figure

ONE CONTRACTION, TWO FAMOUS FORCES. The field tensor F^{μν} has two blocks: the time-space slots hold −E/c, the space-space slots hold −ε^{ijk}B_k. Contract it with the four-velocity u_ν = (γc, −γv) and each block delivers one of the terms everyone learned separately. The ν = 0 slot of F^{iν} hits u_0 = γc and produces γqE_i. The ν = j slots hit u_j = −γv_j and produce γq(v×B)_i — the cross product IS what a contraction with the antisymmetric magnetic block looks like. And the μ = 0 row produces the POWER, γqE·v/c. Electric force, magnetic force, and the work–energy theorem are three readings of one equation. Finally, because F is antisymmetric and u_μ u_ν is symmetric, K·u = 0 identically: the four-force is always orthogonal to the worldline, so the rest mass never changes and the magnetic field never does work.

The core equations

The law itself — everything else on this page is a consequence:

\boxed{\;\frac{\mathrm{d}p^\mu}{\mathrm{d}\tau} \;=\; K^\mu \;=\; q\,F^{\mu\nu}u_\nu\;} \qquad (1)

where: $p^\mu = (\mathcal E/c, \vec p)$ with $\vec p = \gamma m\vec v$ ; $\tau$ — proper time, a Lorentz scalar; $q$ — the invariant electric charge; $u^\mu = \gamma(c, \vec v)$ . Physical reading: every object here is a tensor of definite rank, so the equation has the same form in every inertial frame and no $\gamma$ -factors need inserting by hand. It is linear in the field and linear in the velocity, and those two requirements plus four-vector covariance leave essentially no other candidate. The dictionary that makes the contraction concrete:

u_\mu = \eta_{\mu\nu}u^\nu = (\gamma c,\ -\gamma\vec v)\ ,\qquad F^{0i} = -\frac{E_i}{c}\ ,\qquad F^{i0} = +\frac{E_i}{c}\ ,\qquad F^{ij} = -\varepsilon^{ijk}B_k \qquad (2)

where: indices are lowered with $\eta = \mathrm{diag}(+1,-1,-1,-1)$ , which flips the spatial part of $u^\mu$ and the time-space block of $F$ . Physical reading: two minus signs are in play and they must not be confused. The one on $\vec v$ comes from lowering $u$ ; the one on $E_i/c$ comes from $F$ . In $K^i = q(F^{i0}u_0 + F^{ij}u_j)$ they land in different terms — and a card that muddles them prints $-\gamma E_i$ where $+\gamma E_i$ belongs, which is exactly what the base does. Unpacking the four components:

\boxed{\;K^i = \gamma\,q\,\big(\vec E + \vec v\times\vec B\big)_i\;}\ ,\qquad \boxed{\;K^0 = \frac{\gamma\,q\,\vec E\cdot\vec v}{c}\;} \qquad\Longrightarrow\qquad \frac{\mathrm{d}\vec p}{\mathrm{d}t} = q\big(\vec E + \vec v\times\vec B\big)\ ,\qquad \frac{\mathrm{d}\mathcal E}{\mathrm{d}t} = q\,\vec E\cdot\vec v \qquad (3)

where: the last step uses $\mathrm{d}\tau = \mathrm{d}t/\gamma$ , which cancels the $\gamma$ on both. Physical reading: the three-force and the power are not two laws — they are the space and time slots of one four-vector. And $\vec B$ is absent from the power, not because it was dropped but because $(\vec v\times\vec B)\cdot\vec v = 0$ identically. The magnetic field cannot do work, at any speed, in any configuration. The orthogonality identity, and what it protects:

K^\mu u_\mu = q\,F^{\mu\nu}u_\nu u_\mu = 0 \qquad\Longrightarrow\qquad \frac{\mathrm{d}}{\mathrm{d}\tau}\big(p^\mu p_\mu\big) = 2m\,K^\mu u_\mu = 0 \qquad\Longrightarrow\qquad \boxed{\;p^\mu p_\mu = +m^2c^2 = \text{const}\;} \qquad (4)

where: the first equality is antisymmetric-contracted-with-symmetric, which vanishes for any $F$ and any $u$ . Physical reading: the rest mass of a charge is protected by an index symmetry. No matter how violently you accelerate it, an electron stays an electron. Note the sign: in this signature $p\cdot p = +m^2c^2$ . The base article prints the opposite sign, which is right only in the opposite signature it silently adopts — see the first warning.

Reading the dynamics out of the contraction

Derivation AThe contraction, done with the signs tracked — and the card the base gets wrong

The goal is to compute $K^\mu = qF^{\mu\nu}u_\nu$ component by component and land on the familiar Lorentz force. The whole difficulty is bookkeeping: two independent minus signs, one from lowering $u$ , one from the field tensor. The base article's derivation collides them, prints two different answers for one quantity in a single card, and ships a literal ellipsis where the algebra should be. Worth doing slowly.

Step 1. Lower the four-velocity. From $u^\mu = \gamma(c, \vec v)$ and $\eta = \mathrm{diag}(+1,-1,-1,-1)$ : u_0 = \eta_{00}u^0 = +\gamma c ,\qquad u_j = \eta_{jj}u^j = -\gamma v_j . Why: the time component does not flip; the spatial ones do. In the opposite signature it is the other way round, which is why signature errors in this subject are so productive of wrong signs.

Step 2. The spatial components. Split the contraction into the $\nu = 0$ piece and the $\nu = j$ pieces: K^i = q\big(F^{i0}u_0 + F^{ij}u_j\big) = q\Big[\Big(\!+\frac{E_i}{c}\Big)\big(\gamma c\big) \;+\; \big(-\varepsilon^{ijk}B_k\big)\big(-\gamma v_j\big)\Big] = q\big[\gamma E_i + \gamma\,\varepsilon^{ijk}v_jB_k\big] . Why: the electric term needs $F^{i0} = +E_i/c$ , not $F^{0i} = -E_i/c$ — antisymmetry has already supplied one minus sign, and $u_0$ supplies none. In the magnetic term, by contrast, two minus signs meet (one from $F^{ij}$ , one from $u_j$ ) and cancel. Miss either and the force comes out backwards.

Step 3. Recognise the cross product. $\varepsilon^{ijk}v_jB_k = (\vec v\times\vec B)_i$ , so \boxed{\;K^i = \gamma\,q\big(\vec E + \vec v\times\vec B\big)_i\;} and dividing by $\gamma$ gives $\mathrm{d}\vec p/\mathrm{d}t = q(\vec E + \vec v\times\vec B)$ , exactly — no approximation, valid at $0.9999c$ . Why: the cross product was never a separate structure. It is what contracting with an antisymmetric $3\times3$ block looks like in three-vector notation, and it appears without being invited.

Step 4. The time component. Only the electric row survives, since $F^{00} = 0$ : K^0 = q\,F^{0j}u_j = q\Big(-\frac{E_j}{c}\Big)\big(-\gamma v_j\big) = \frac{\gamma q\,\vec E\cdot\vec v}{c} . With $p^0 = \mathcal E/c$ this is $\mathrm{d}\mathcal E/\mathrm{d}\tau = \gamma q\vec E\cdot\vec v$ , i.e. $\boxed{\mathrm{d}\mathcal E/\mathrm{d}t = q\vec E\cdot\vec v}$ . Why: here both minus signs land in the same term and cancel — the opposite of Step 2, which is precisely why the two steps are so easy to conflate. And $\vec B$ is absent: the magnetic block never touches $u_0$ , and $(\vec v\times\vec B)\cdot\vec v = 0$ anyway.

Step 5. Read the base's card. Its equation line prints F^{i0}u_0 = \frac{E_i}{c}(-\gamma c) = -\gamma E_i , while the note on the same card says " $F^{i0} = -F^{0i} = -E_i/c$ , so $F^{i0}u_0 = (+E_i/c)(\gamma c)\ \ldots$ signs combine to $+\gamma E_i$ ." The card gives two different answers for one quantity, its note contradicts the component it has just declared, and the ellipsis is in the shipped text. Why: both printed values cannot be right, and in the locked dictionary the answer is $+\gamma E_i$ — Step 2. An unverified derivation drifts toward the answer the author already knows, and stops tracking whether it got there honestly. *Verified in verify_aq-sr-covariant-lorentz-force.py: $F^{\mu\nu}$ is rebuilt from $A^\mu = (\phi/c, \vec A)$ , so the dictionary is derived not assumed; $K^\mu = qF^{\mu\nu}u_\nu$ is then contracted symbolically for arbitrary $\vec E$ , $\vec B$ , $\vec v$ and shown to equal $\gamma q(\vec E + \vec v\times\vec B)_i$ and $\gamma q\vec E\cdot\vec v/c$ ; and $F^{i0}u_0 = +\gamma E_i$ is confirmed, refuting the base card's equation line.*

Go deeper · Derivation B · 2 minOrthogonality, rest-mass conservation, and why the magnetic field can never do work

Derivation BOrthogonality, rest-mass conservation, and why the magnetic field can never do work

Three statements that sound like separate physics — the four-force is orthogonal to the four-velocity, the rest mass of a charge never changes, and magnetic forces do no work — are one statement. The proof uses nothing but the antisymmetry of $F$ , so it holds for every field configuration, however violent.

Step 1. Contract the law with $u_\mu$ : K^\mu u_\mu = q\,F^{\mu\nu}u_\nu u_\mu . The object $u_\mu u_\nu$ is manifestly symmetric under $\mu\leftrightarrow\nu$ , while $F^{\mu\nu}$ is antisymmetric. Relabelling the dummies gives $F^{\mu\nu}u_\mu u_\nu = -F^{\mu\nu}u_\mu u_\nu$ , hence \boxed{\;K^\mu u_\mu = 0\;} Why: a quantity equal to minus itself is zero. No physical input whatsoever — this is true for any antisymmetric tensor and any vector, and it is why the result is so robust.

Step 2. Turn it into rest-mass conservation. Since $p^\mu = m\,u^\mu$ and $p^\mu p_\mu = +m^2c^2$ , \frac{\mathrm{d}}{\mathrm{d}\tau}\big(p^\mu p_\mu\big) = 2\,p_\mu\frac{\mathrm{d}p^\mu}{\mathrm{d}\tau} = 2m\,u_\mu K^\mu = 0 \;\Longrightarrow\; \frac{\mathrm{d}}{\mathrm{d}\tau}\big(m^2c^2\big) = 0 . Why: the rest mass cannot change. An electromagnetic field can pump an electron to $7\ \mathrm{TeV}$ and it is still an electron with $m = 511\ \mathrm{keV}/c^2$ . The mass is protected by an index symmetry, not by a conservation law anyone imposed.

Step 3. Unpack $K^\mu u_\mu = 0$ into three-vectors. With $K^\mu = \gamma(\vec F\cdot\vec v/c,\ \vec F)$ and $u_\mu = (\gamma c, -\gamma\vec v)$ , the contraction is $\gamma^2\vec F\cdot\vec v - \gamma^2\vec F\cdot\vec v = 0$ — an identity, not a constraint. The content appears when you put in the actual force: \vec F\cdot\vec v = q(\vec E + \vec v\times\vec B)\cdot\vec v = q\,\vec E\cdot\vec v + \underbrace{q(\vec v\times\vec B)\cdot\vec v}_{=\,0} . Why: the magnetic term drops out because $\vec v\times\vec B$ is perpendicular to $\vec v$ by construction. Not an approximation, not a low-speed statement, not a special case. A magnetic field changes the direction of $\vec p$ and never its magnitude.

Step 4. Note the sharp consequence. In a pure magnetic field $\mathrm{d}\mathcal E/\mathrm{d}t = 0$ , so $\gamma$ is a constant of the motion and $\mathrm{d}(\gamma m\vec v)/\mathrm{d}t = q\vec v\times\vec B$ becomes linear. The solution is a helix with \omega_c = \frac{qB}{\gamma m},\qquad r = \frac{\gamma m v_\perp}{qB} = \frac{p_\perp}{qB} . Why: the relativistic cyclotron frequency drops with energy, which is why a fixed-frequency cyclotron loses its particles and why synchrotrons must ramp their field in step with $p$ . The radius, by contrast, depends only on momentum — which is why $r = p/(qB)$ is the most-used formula in accelerator design. *Verified in verify_aq-sr-covariant-lorentz-force.py: $K^\mu u_\mu = 0$ is checked symbolically for arbitrary $\vec E$ , $\vec B$ , $\vec v$ ; $(\vec v\times\vec B)\cdot\vec v = 0$ is verified identically; $\mathrm{d}(p^\mu p_\mu)/\mathrm{d}\tau = 0$ is confirmed; and $p^\mu p_\mu = +m^2c^2$ is computed from $p^\mu = m\gamma(c,\vec v)$ with $\gamma = (1 - v^2/c^2)^{-1/2}$ , refuting the base's $-m^2c^2$ .*

Go deeper · Derivation C · 2 minOwned result: the law is signature-blind — only the base's printed identity was wrong

Limiting cases

Case	Result	Meaning
$v \ll c$ ( $\gamma \to 1$ )	$\mathrm{d}\vec p/\mathrm{d}t = q(\vec E + \vec v\times\vec B)$	the Newtonian Lorentz force, unchanged in form
$\vec B = 0$	$K^\mu = \gamma q(\vec E\cdot\vec v/c,\ \vec E)$	pure electric: work is done, $\gamma$ grows
$\vec E = 0$	$K^0 = 0$ , $\gamma = \mathrm{const}$	pure magnetic: no work, ever. Helical motion
$\vec E = 0$ , $\vec v\perp\vec B$	$\omega_c = qB/\gamma m$ , $\;r = p/qB$	relativistic cyclotron. $\omega_c$ falls as energy rises
$\vec v\parallel\vec E$ , $\vec B = 0$	$\mathrm{d}p/\mathrm{d}t = qE$	linear accelerator; $p$ grows linearly in $t$ , $v \to c$
any field, any speed	$K^\mu u_\mu = 0$	$m$ is constant. Antisymmetry, nothing else
$\vec E \perp \vec B$ , $E < cB$	undeflected at $v = E/B$	velocity selector — the frame where $\vec E$ vanishes

Common trapThe base declares the opposite signature and prints p·p = −m²c². The dynamics survive; the identity does not.

The AllQFT base for this page states its signature as $(-,+,+,+)$ , labels its field tensor accordingly, and concludes that $p^\mu p_\mu = -m^2c^2$ is the conserved quantity. In the signature this corpus locks — $(+,-,-,-)$ — the correct identity is $p^\mu p_\mu = +m^2c^2$ , and that is what is printed throughout this article. Two things must be said, and they pull in opposite directions. First, the base's dynamics are not wrong. Its metric and its field tensor flip together, so every sign in $K^\mu = qF^{\mu\nu}u_\nu$ flips twice and cancels: the third derivation carries out both contractions and gets $K^i = \gamma q(\vec E + \vec v\times\vec B)_i$ and $K^0 = \gamma q\vec E\cdot\vec v/c$ either way. Every force, trajectory and energy it predicts is right.

Second, that is exactly what makes the defect dangerous. A reader lifting $p\cdot p = -m^2c^2$ into any $(+,-,-,-)$ article — four-momentum, the field invariants, the stress-energy tensor — will get a sign error with nothing to flag it, because the source page's own calculations all check out. Convention errors do not fail locally. They fail somewhere else, later, in someone else's article.

Common trapOne base card contradicts itself mid-derivation — and ships an ellipsis where the algebra belongs.

The base's derivation of the spatial force contains a single card that manages three defects at once. Its equation line reads F^{i0}u_0 = \frac{E_i}{c}\,(-\gamma c) = -\gamma E_i , while the note on the same card says: " $F^{i0} = -F^{0i} = -E_i/c$ , so $F^{i0}u_0 = (+E_i/c)(\gamma c)\ \ldots$ signs combine to $+\gamma E_i$ ." So: (i) the card gives two different answers, $-\gamma E_i$ and $+\gamma E_i$ , for one quantity; (ii) the note declares $F^{i0} = -E_i/c$ and then immediately uses $+E_i/c$ , with no step between; and (iii) the literal ellipsis `...` sits mid-expression in the shipped text — the algebra was never written, only gestured at. The correct value in the locked dictionary is $+\gamma E_i$ (first derivation, Step 2), so the note's conclusion is right and its equation line is wrong — the worst possible arrangement: the reader who trusts the boxed equation is misled, and the reader who trusts the prose is right for reasons the page never gives. A derivation that does not track its own signs is not a derivation.

Worked examples

1 · A 7 TeV proton in the LHC — and a discrepancy that turns out to be the machine's floor plan

A proton of total energy $\mathcal E = 7.00\ \mathrm{TeV}$ circulates in the LHC's bending dipoles, $B = 8.33\ \mathrm{T}$ . Find $\gamma$ , the bending radius, and the cyclotron frequency — then compare with the real ring.

Step 1 — the Lorentz factor. With $m_pc^2 = 938.272\ \mathrm{MeV}$ , \gamma = \frac{\mathcal E}{m_pc^2} = \frac{7.00\times10^{12}}{938.272\times10^{6}} = \mathbf{7460.52} . The proton is moving at $v/c = \sqrt{1 - \gamma^{-2}} = 0.999999991$ — a fraction of a metre per second short of $c$ .

Step 2 — the momentum. $\mathcal E^2 = (pc)^2 + (m_pc^2)^2$ , and at this $\gamma$ the rest energy is negligible, so $pc \simeq 7.00\ \mathrm{TeV}$ : p = \frac{(7.00\times10^{12})(1.602177\times10^{-19})}{2.997925\times10^{8}} = \mathbf{3.741\times10^{-15}\ kg\,m\,s^{-1}} .

Step 3 — the bending radius. From the second derivation, $r = p/(qB)$ : r = \frac{3.741\times10^{-15}}{(1.602177\times10^{-19})(8.33)} = \mathbf{2803\ m} . Note what did not appear: $\gamma$ and $m$ . The radius depends on momentum alone — which is why accelerator physicists quote fields in "tesla-metres" and never have to ask what particle is in the beam.

Step 4 — the cyclotron frequency, and the check that looks like a failure. f_c = \frac{\omega_c}{2\pi} = \frac{qB}{2\pi\gamma m_p} = \frac{(1.602177\times10^{-19})(8.33)}{2\pi(7460.52)(1.672622\times10^{-27})} = \mathbf{17022\ Hz} . But the LHC's actual revolution frequency is $f_{\text{rev}} = c/C = (2.997925\times10^{8})/(26\,659) = \mathbf{11245\ Hz}$ — 34% lower. Is the law wrong?

Step 5 — no: the ring is not a circle. $f_c$ is the frequency a proton would have if the whole orbit were bent at $8.33\ \mathrm{T}$ . The circumference of that pure circle is $2\pi r = \mathbf{17\,612\ m}$ , but the LHC is $26\,659\ \mathrm{m}$ around, because it contains eight long straight sections for acceleration, focusing, injection and the experiments. The ratio \frac{2\pi r}{C} = \frac{17\,612}{26\,659} = \mathbf{0.661} is the fraction of the machine that is actually arc — and it matches $f_{\text{rev}}/f_c = 11\,245/17\,022 = \mathbf{0.661}$ exactly. CERN quotes the LHC's dipole coverage as 66%. The 34% "discrepancy" is the straight sections, and the covariant Lorentz force reproduces the floor plan of the machine to three significant figures.

2 · The four-force on a proton, and the orthogonality check

A proton moves at $\vec v = 0.600c\,\hat x$ through $\vec E = (2.00\times10^{5},\ 1.00\times10^{6},\ 0)\ \mathrm{V\,m^{-1}}$ and $\vec B = (0,\ 0,\ 0.500)\ \mathrm{T}$ . Compute all four components of $K^\mu$ , verify $K\cdot u = 0$ , and find the power.

Step 1 — kinematics. $\gamma = (1 - 0.600^2)^{-1/2} = \mathbf{1.2500}$ and $v = 0.600c = 1.798755\times10^{8}\ \mathrm{m\,s^{-1}}$ .

Step 2 — the magnetic term. \vec v\times\vec B = (1.798755\times10^{8}\,\hat x)\times(0.500\,\hat z) = -8.993774\times10^{7}\ \hat y\ \ \mathrm{V\,m^{-1}} . Ninety times larger than the applied $E_y$ — at $0.6c$ , a half-tesla field is an enormous effective electric field.

Step 3 — the spatial components. $K^i = \gamma q(\vec E + \vec v\times\vec B)_i$ : K^1 = (1.25)(1.602177\times10^{-19})(2.00\times10^{5}) = \mathbf{4.005442\times10^{-14}\ N}, K^2 = (1.25)(1.602177\times10^{-19})\big(1.00\times10^{6} - 8.993774\times10^{7}\big) = \mathbf{-1.781175\times10^{-11}\ N},\qquad K^3 = 0 .

Step 4 — the time component, and the power. $\vec E\cdot\vec v = (2.00\times10^{5})(1.798755\times10^{8}) = 3.597510\times10^{13}$ , so K^0 = \frac{\gamma q\,\vec E\cdot\vec v}{c} = \mathbf{2.403265\times10^{-14}\ N} ,\qquad \frac{\mathrm{d}\mathcal E}{\mathrm{d}t} = q\,\vec E\cdot\vec v = \mathbf{5.763846\times10^{-6}\ W} = \mathbf{35.98\ TeV\,s^{-1}} . Only $E_x$ contributed — $E_y$ is perpendicular to $\vec v$ , and the huge magnetic term does no work either, because $(\vec v\times\vec B)\cdot\vec v = 0$ exactly.

Step 5 — the check that must come out zero. With $u_\mu = (\gamma c,\ -\gamma\vec v) = (3.747406\times10^{8},\ -2.248443\times10^{8},\ 0,\ 0)$ , K^\mu u_\mu = K^0u_0 + K^1u_1 = (2.403265\times10^{-14})(3.747406\times10^{8}) - (4.005442\times10^{-14})(2.248443\times10^{8}) , and the two terms are $9.006\times10^{-6}$ and $9.006\times10^{-6}$ — equal to every digit carried, so $K\cdot u = 0$ . Note that $K^2$ , by far the largest component, contributes nothing, because $u_2 = 0$ . The orthogonality is exact, guaranteed by antisymmetry, and it is the cheapest possible check that a four-force is right: if $K\cdot u \ne 0$ , you have made a sign error.

Figure

THE MAGNETIC FIELD STEERS BUT NEVER ACCELERATES — and that is antisymmetry, not geometry. A proton at 0.6c (γ = 1.25) in E = (2.00×10⁵, 1.00×10⁶, 0) V/m and B = 0.500 T ẑ. The magnetic term v × B = −8.99 × 10⁷ ŷ V/m is NINETY TIMES the applied E_y — yet it contributes exactly ZERO to the power, because (v × B) · v = 0 identically. The whole energy gain, 5.764 × 10⁻⁶ W = 35.98 TeV/s, comes from E_x alone. And the four-force obeys K·u = 0 to every digit: K⁰u₀ = 9.006 × 10⁻⁶ and K¹u₁ = 9.006 × 10⁻⁶ cancel exactly, while K², the largest component by three orders of magnitude, contributes nothing at all because u₂ = 0. Rest mass is safe. It always is.

Physics of the portrait

The portrait is the law itself. Its point is not that it is compact but that, written covariantly, it stops being a rule to memorise and becomes something with almost no freedom left in it.

Origins · 2 minLorentz force law and how this was discovered

Lorentz force law

1892–present · the force on a moving charge · why E and B were never two forces · and why rest mass is safe

The Lorentz force law gives the force on a charge moving through an electromagnetic field, $\vec F = q(\vec E + \vec v\times\vec B)$ , and it is the bridge between Maxwell's equations — which say what the fields do — and mechanics, which says what the charges do. Neither half is complete without it. Hendrik Antoon Lorentz gave the law its modern form in 1892 as part of his electron theory, in which matter is made of discrete charges and the fields are properties of a stationary aether; the expression had appeared earlier in Heaviside's work, but it is Lorentz's formulation that survived. The historical irony is sharp: Lorentz derived the law inside a theory of a preferred frame, and it turned out to be exactly the frame-independent statement $\mathrm{d}p^\mu/\mathrm{d}\tau = qF^{\mu\nu}u_\nu$ — the aether it was built on was the one thing in the picture that did not survive. What the covariant form reveals is that the two terms were never two forces at all: they are the time-space and space-space blocks of one antisymmetric tensor contracted with one four-velocity. The split between "electric" and "magnetic" is not physics but bookkeeping, and a boost reshuffles it. And the antisymmetry of $F$ , which does all the work, hands over for free the two facts that make charged-particle dynamics tractable at all: the four-force is orthogonal to the four-velocity, so an accelerated electron remains an electron, and a magnetic field can bend a beam through a $27\ \mathrm{km}$ ring without adding a single electron-volt to it.

History

1865

— Maxwell's field equations describe what charges and currents do to the fields, but say nothing about the force the fields exert back on a moving charge.

1889

— Heaviside publishes an expression for the force on a moving charge in an electromagnetic field, an early form of the magnetic term.

1892

— Lorentz, in "La théorie électromagnétique de Maxwell et son application aux corps mouvants", gives the force law its modern form within his electron theory.

1895

— Lorentz's Versuch presents the law together with the coordinate transformations that would later carry his name.

1905

— Einstein's "Zur Elektrodynamik bewegter Körper" shows that the electric-magnetic split of the force is frame-dependent, and that the law is exactly compatible with the relativity principle.

1908

— Minkowski's four-dimensional formulation writes the law as a contraction of the field tensor with the four-velocity, in which form it has been used ever since.

Going further

Texts. Jackson, Classical Electrodynamics, 3rd ed., §§11.9 and 12.1–12.4 · Griffiths, Introduction to Electrodynamics, 4th ed., §12.3 · Rindler, Introduction to Special Relativity, 2nd ed., ch. 7 · Landau & Lifshitz, The Classical Theory of Fields, 4th ed., §§17 and 21.

Provenance. Signature $(+,-,-,-)$ , $\eta = \mathrm{diag}(+1,-1,-1,-1)$ , and therefore $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass, and no velocity-dependent mass appears anywhere. The dictionary $u_\mu = (\gamma c, -\gamma\vec v)$ , $F^{0i} = -E_i/c$ , $F^{i0} = +E_i/c$ , $F^{ij} = -\varepsilon^{ijk}B_k$ is derived from $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ , not assumed. Base-defect repairs (logged): (i) the base declares $(-,+,+,+)$ and prints $p^\mu p_\mu = -m^2c^2$ ; the locked identity is $+m^2c^2$ , and the third derivation shows explicitly that the dynamics are unaffected — every sign flips twice — so the defect is exactly one printed identity and nothing else; (ii) one of its cards prints $F^{i0}u_0 = -\gamma E_i$ in its equation line while its own note argues for $+\gamma E_i$ , declares $F^{i0} = -E_i/c$ and then uses $+E_i/c$ , and ships a literal ellipsis mid-expression. Everything here is recomputed in verify_aq-sr-covariant-lorentz-force.py, which parses this article's own printed digits.

Scope. The construction of $F^{\mu\nu}$ belongs to The Electromagnetic Field Tensor; the field equations to Covariant Maxwell Equations; the classification of fields to Electromagnetic Field Invariants; the general Minkowski four-force to The Four-Force; and radiation from the resulting acceleration to the Liénard–Wiechert Potentials. This page owns the contraction $qF^{\mu\nu}u_\nu$ , the orthogonality identity, and the relativistic cyclotron motion that follows.

Builds on. The Electromagnetic Field Tensor · Covariant Maxwell Equations · The Four-Force · The Four-Velocity. Leads to. The Liénard–Wiechert Potentials · The Electromagnetic Stress-Energy Tensor · The Field of a Uniformly Moving Charge.

Where exactly does the cross product come from? It looks like it was smuggled in.

It was not — it is what $F^{ij}u_j$ is. The space-space block is $F^{ij} = -\varepsilon^{ijk}B_k$ , an antisymmetric $3\times3$ array, and contracting any antisymmetric array with a vector is, in three dimensions, precisely the operation we call a cross product. So $\varepsilon^{ijk}v_jB_k = (\vec v\times\vec B)_i$ is a translation, not a derivation. The cross product is an accident of three dimensions; the contraction is the real object, and it is what survives to four.

If the magnetic force does no work, how does a magnet pick up a nail?

By the electric fields it induces, and by internal forces — not by the $q\vec v\times\vec B$ term acting on free charges. The clean statement is the covariant one: the magnetic part of the Lorentz force on a point charge does no work, because it is perpendicular to that charge's velocity. Energy still flows, but the electric field delivers it somewhere in the accounting, and in magnetised matter the bookkeeping runs through the internal structure of the dipoles. The theorem is exact and narrow, and most paradoxes about it come from widening it illegitimately.

Why differentiate with respect to proper time rather than coordinate time?

Because $\tau$ is a scalar and $t$ is not. $\mathrm{d}p^\mu/\mathrm{d}\tau$ is a four-vector divided by a scalar, so an equation setting it equal to another four-vector is covariant. $\mathrm{d}p^\mu/\mathrm{d}t$ divides a four-vector by a component of a four-vector and is not a tensor of any rank, so an equation built from it has no frame-independent meaning. The $\gamma$ in $K^i = \gamma q(\vec E + \vec v\times\vec B)_i$ is exactly the price of that conversion, and it cancels when you go back to $\mathrm{d}t$ .

Is the charge q really invariant, or does it transform like the mass?

Genuinely invariant, and this is an experimental fact rather than a convention — it is why a hydrogen atom is neutral to better than one part in $10^{21}$ even though its electron moves at $\alpha c$ while its proton barely moves. If charge depended on speed, atoms would be charged. In the covariant law $q$ appears as a bare scalar multiplying a tensor, which is only consistent because of this. Charge density is a different matter — it is the time component of $J^\mu$ , and it transforms.

Does the covariant law include radiation reaction?

No, and that is a genuine limitation rather than a simplification. $\mathrm{d}p^\mu/\mathrm{d}\tau = qF^{\mu\nu}u_\nu$ describes a charge responding to an external field and ignores the field the charge itself radiates, which carries away energy and momentum and reacts back. Including it leads to the Abraham–Lorentz–Dirac equation, which has notoriously pathological solutions (pre-acceleration, runaways) and remains unsettled. For an LHC proton the neglect is superb; for an electron in a strong laser it is not.

Why does the relativistic cyclotron frequency fall with energy, when the radius formula has no γ in it?

Both follow from $r = p/(qB)$ and $v \to c$ . As energy rises, $p$ grows without bound but $v$ saturates at $c$ , so the orbit gets larger while the particle gets no faster — and the time to go round, $2\pi r/v \to 2\pi p/(qBc)$ , grows. That is $\omega_c = qB/(\gamma m)$ falling. The radius formula "has no $\gamma$ " only because $\gamma$ hides inside $p = \gamma m v$ ; the frequency formula exposes it. This is why the cyclotron gave way to the synchrocyclotron and then the synchrotron.