BMT (Bargmann-Michel-Telegdi) Equation

To measure the muon's magnetic moment to a billionth, you would think you need your magnetic field to a billionth, your beam energy to a billionth, and the momentum spread of a million muons to a billionth. You do not. You need the field. Everything else cancels. That is not luck. It is a theorem, and the theorem is the BMT equation. Write down the most general Lorentz-covariant law a spin four-vector could obey in an electromagnetic field. There are only two terms it can contain. Demand that the spin keep its length and stay orthogonal to the four-velocity — which it must, or it is not a spin — and the ratio of those two terms is fixed. One number remains free, and it is $g$ . Reduce that covariant law to the laboratory and something remarkable falls out. The spin precesses at $\boldsymbol\omega_s$ ; the momentum precesses at $\boldsymbol\omega_c$ . Both go as $1/\gamma$ — they slow as the beam gets faster, and each depends on the energy of every individual muon. But their difference does not:

\boldsymbol\omega_a \;=\; \boldsymbol\omega_s - \boldsymbol\omega_c \;=\; -\frac{q}{m}\,a\,\mathbf B \qquad (\mathbf E = 0)

No $\gamma$ . A muon at $\gamma = 20$ and a muon at $\gamma = 40$ , in the same field, turn their spins away from their momenta at exactly the same rate. The beam can have any energy spread it likes and the signal does not smear. That single cancellation is why $a_\mu$ is known to $127$ parts per billion and $g$ itself is not known to anything like that — and it is why the muon anomaly, and not the muon moment, is the number that has been keeping people up at night for twenty years.

Key takeaways

Signature $(+,-,-,-)$ , $u\cdot u = +c^2$ , $p\cdot p = +m^2c^2$ , timelike positive. SI, with every $c$ written out — the base for this page drops them, and a dropped $c^2$ here is a factor of $10^{17}$ .
The covariant law (Bargmann, Michel & Telegdi, PRL 2, 435, 1959), for a particle of charge $q$ , mass $m$ , $g$ -factor $g$ :

\boxed{\;\frac{\mathrm dS^\mu}{\mathrm d\tau} = \frac{gq}{2m}F^{\mu\nu}S_\nu \;+\; \frac{q}{mc^2}\Big(\frac g2 - 1\Big)\,u^\mu\,\big(S_\lambda F^{\lambda\nu}u_\nu\big)\;}

It is unique. Only two covariant terms are available. $S\cdot S = \text{const}$ costs nothing. $S\cdot u = 0$ then forces the second coefficient in terms of the first, and the rest frame fixes the first to be $gq/2m$ . Nothing is put in by hand — see the derivation.
Lab form ( $\mathrm d\mathbf S/\mathrm dt = \boldsymbol\omega_s\times\mathbf S$ , $a \equiv (g-2)/2$ ):

\boldsymbol\omega_s = -\frac{q}{m}\Big[\Big(a + \frac1\gamma\Big)\mathbf B - \frac{a\gamma}{\gamma+1}(\boldsymbol\beta\cdot\mathbf B)\boldsymbol\beta - \Big(a + \frac{1}{\gamma+1}\Big)\frac{\boldsymbol\beta\times\mathbf E}{c}\Big]

THE PUNCHLINE. Subtract the cyclotron frequency and the $1/\gamma$ terms cancel exactly:

\boxed{\;\boldsymbol\omega_a = \boldsymbol\omega_s - \boldsymbol\omega_c = -\frac{q}{m}\Big[a\,\mathbf B - \Big(a - \frac{1}{\gamma^2-1}\Big)\frac{\boldsymbol\beta\times\mathbf E}{c}\Big]\;}

With $\mathbf E = 0$ this is flat in $\gamma$ . The observable is proportional to $a$ , not to $g$ .

The magic $\gamma$ . The $\mathbf E$ coefficient vanishes at $a = 1/(\gamma^2-1)$ , i.e. $\gamma = \sqrt{1 + 1/a} = \mathbf{29.3034}$ for the muon ( $p = 3.094\ \mathrm{GeV}/c$ ). Run there and the focusing quadrupoles stop contaminating the measurement.
Thomas is the $g = 0$ content of BMT, not the $g=2$ content. Set $g=0$ and the equation becomes $\mathrm dS^\mu/\mathrm d\tau = -(1/c^2)(S\cdot A)u^\mu$ — Fermi–Walker transport, no torque, the pure kinematic precession of `aq-sr-thomas-precession`.

The core insight

The BMT equation is not a model of the muon. It is the only thing a spin can do. Its content is entirely negative: given that $S^\mu$ is a four-vector, that its length is fixed, and that it is orthogonal to $u^\mu$ — all three definitional, not dynamical — there is exactly a one-parameter family of covariant evolution laws, and the parameter is $g$ . Every Thomas rotation, every frame effect Thomas had to hunt for by hand in 1926, is already sitting inside it, forced by the constraints. The magic- $\gamma$ trick, the $\gamma$ -independence of $\omega_a$ , the fact that a $g=2$ spin follows its own momentum — none are inputs. They are what falls out when you write down the only equation you were allowed to write.

Where this lives today

Every polarised beam, everywhere. The BMT equation is the working tool of spin dynamics at RHIC, at Jefferson Lab, at any facility that must get polarised particles through a lattice of magnets without depolarising them. The "spin tune" $\nu_s = a\gamma$ — spin precessions per turn — falls straight out of equation (6), and every depolarising resonance in an accelerator is an integer coincidence involving it. Note what that says: the spin tune grows with energy. A high-energy polarised proton machine crosses hundreds of resonances on the way up, and the whole discipline of Siberian snakes exists to get through them.

Muon $g-2$ — but note what is actually being tested. The experiment does not test the BMT equation; it assumes it, far beyond the $127$ ppb of the result. Every systematic in the Fermilab error budget — the pitch correction (muons with a vertical velocity component, where $\boldsymbol\beta\cdot\mathbf B \ne 0$ and the middle term of equation (4) wakes up), the electric-field correction (muons off the magic momentum) — is a term in equation (4) evaluated to the required order. The equation is not the object of study. It is the instrument.

The electron's magic $\gamma$ , and why nobody uses it. Run the same calculation for the electron: $a_e = 0.00115965$ , so $\gamma_{\text{magic}} = 29.38$ — almost the same number, and completely useless. An electron at $\gamma = 29.4$ carries only $15$ MeV $/c$ , and would need a ring so small that synchrotron radiation would eat it. The electron's anomaly is measured a totally different way: a single electron in a Penning trap, at rest, comparing $\omega_s$ and $\omega_c$ directly. The magic $\gamma$ is not a fundamental constant. It is an engineering coincidence, exploitable for exactly one particle.

The physical picture

Angular momentum in relativity is awkward: the honest object is an antisymmetric tensor, not a vector. But for a single particle you can do better. Go to its rest frame, where the spin is an ordinary three-vector $\mathbf S$ , and define the four-vector $S^\mu$ to be $(0, \mathbf S)$ there. Boost that definition out to the lab and you have a genuine four-vector — one that, by construction, satisfies

S^\mu u_\mu = 0 \qquad\text{and}\qquad S^\mu S_\mu = -|\mathbf S|^2 = \text{const}

Those two conditions are not physics — they are the price of the definition. And they turn out to be nearly the whole theory, because they must hold at every instant along the worldline, and the worldline is bending. That is the tension BMT resolves. The field wants to torque the spin. The constraint $S\cdot u = 0$ insists the spin stay orthogonal to a four-velocity that the same field is busy accelerating. You cannot satisfy both by torquing alone — the spin must be dragged as well, and the drag is the second term of the equation. That drag term is the Thomas precession. It was never put in. It is what the orthogonality constraint costs.

And now the accident that built an industry. In a pure magnetic field the spin precesses at $\omega_s \propto (a + 1/\gamma)B$ and the momentum at $\omega_c \propto B/\gamma$ . Both are $\gamma$ -dependent, and in a real beam with a $0.1\%$ momentum spread that would smear any signal into uselessness. But what a detector sees is the angle between spin and momentum — the difference. And when you subtract, the $1/\gamma$ pieces cancel identically, leaving

\omega_a \;=\; -\frac{q}{m}\,a\,B

$\gamma$ has vanished. Every muon in the ring, whatever its energy, precesses relative to its own momentum at the same rate, set by $a$ and $B$ and nothing else. You measure a frequency and a field, and you get $(g-2)/2$ directly — never $g$ , and never needing the beam energy at all.

Figure

THE PUNCHLINE — omega_a is FLAT in gamma, and that is why the muon anomaly is known to 127 parts per billion. The horizontal axis is the Lorentz factor of a muon in a 1.4513 tesla storage ring. BLUE: the SPIN precession frequency omega_s, which falls off as (a + 1/gamma) — energy-dependent. GREEN: the CYCLOTRON frequency omega_c, the rate at which the MOMENTUM turns, which falls off as 1/gamma — also energy-dependent. Neither is measurable to high precision in a beam with an energy spread, because each muon would precess at its own rate and the signal would smear. RED: their DIFFERENCE, omega_a = omega_s − omega_c. It is a HORIZONTAL LINE. The 1/gamma terms cancel EXACTLY, leaving omega_a = −(q/m) a B, which does not contain gamma at all. A muon at gamma = 20 and a muon at gamma = 40 turn their spins away from their momenta at precisely the same rate. That single cancellation — a theorem of the BMT equation, not a design choice — is what makes the experiment possible. Marked: the muon g−2 ring's magic gamma of 29.3034, where the ELECTRIC-field contribution to omega_a also vanishes, so that the focusing quadrupoles cannot corrupt the answer either

The core equations

(1) The definition — and the two constraints it forces. $S^\mu \equiv (0,\mathbf S)$ in the rest frame, boosted out. Therefore, identically, at every instant:

S^\mu u_\mu = 0, \qquad S^\mu S_\mu = -|\mathbf S|^2 = \text{const} \qquad (1)

where: $u^\mu = \gamma(c,\mathbf v)$ , $u\cdot u = +c^2$ . In our signature a spacelike vector has negative square, so $S\cdot S < 0$ . The time component is not zero in the lab: $S\cdot u = 0$ forces $S^0 = \boldsymbol\beta\cdot\mathbf S_{\text{lab}}$ whenever the particle moves. Reading $\mathbf S$ off the spatial part without boosting is the classic error.

(2) The BMT equation.

\boxed{\;\frac{\mathrm dS^\mu}{\mathrm d\tau} = \underbrace{\frac{gq}{2m}\,F^{\mu\nu}S_\nu}_{\text{torque}} \;+\; \underbrace{\frac{q}{mc^2}\Big(\frac g2 - 1\Big)\,u^\mu\,\big(S_\lambda F^{\lambda\nu}u_\nu\big)}_{\text{the drag: Thomas}}\;} \qquad (2)

where: $q$ is the charge with its sign ( $q = -e$ for the muon), $F^{\mu\nu}$ the field tensor, $\tau$ proper time. The $1/c^2$ is not optional: $u^\mu(SFu)$ carries two extra powers of velocity, and without it the term is wrong by $\sim10^{17}$ in SI. The base's card omits it.

(3) The Lorentz force, which the constraint needs.

\frac{\mathrm du^\mu}{\mathrm d\tau} \;=\; \frac{q}{m}F^{\mu\nu}u_\nu \;\equiv\; A^\mu \qquad (3)

(4) The lab-frame spin precession ( $\mathrm d\mathbf S/\mathrm dt = \boldsymbol\omega_s\times\mathbf S$ for the rest-frame spin, with $a = (g-2)/2$ ):

\boldsymbol\omega_s = -\frac{q}{m}\Big[\Big(a + \frac1\gamma\Big)\mathbf B - \frac{a\gamma}{\gamma+1}(\boldsymbol\beta\cdot\mathbf B)\boldsymbol\beta - \Big(a + \frac{1}{\gamma+1}\Big)\frac{\boldsymbol\beta\times\mathbf E}{c}\Big] \qquad (4)

where: the $\boldsymbol\beta\times\mathbf E$ term carries a $1/c$ in SI — $\mathbf E/c$ has the units of $\mathbf B$ . The base drops it.

(5) The cyclotron frequency — the rate the momentum turns.

\boldsymbol\omega_c \;=\; -\frac{q}{\gamma m}\Big[\mathbf B - \frac{\gamma^2}{\gamma^2-1}\,\frac{\boldsymbol\beta\times\mathbf E}{c}\Big] \qquad (5)

(6) The observable. Subtract:

\boxed{\;\boldsymbol\omega_a \;=\; \boldsymbol\omega_s - \boldsymbol\omega_c \;=\; -\frac{q}{m}\Big[a\,\mathbf B - \Big(a - \frac{1}{\gamma^2-1}\Big)\frac{\boldsymbol\beta\times\mathbf E}{c}\Big]\;} \qquad (6)

where: with $\mathbf E = 0$ , there is no $\gamma$ anywhere. And the electric term dies at $a = 1/(\gamma^2-1)$ — the magic $\gamma$ .

Deriving it

Derivation AThe BMT equation is unique — nothing in it is a choice

The goal is to show that equation (2) is not a proposal, a model, or an ansatz that happens to work, but the only covariant law a spin four-vector can obey in an external electromagnetic field, once you accept the two constraints its own definition forces on it. There is exactly one free number in the construction, and it is $g$ . Everything else — including the entire Thomas precession, which cost Thomas a paper in 1926 — is fixed by algebra.

Step 1 — count the terms you are allowed. We want $\mathrm dS^\mu/\mathrm d\tau$ to be a four-vector, linear in $S$ and in $F$ . The inventory: $F^{\mu\nu}S_\nu$ (contract the tensor with the spin) and $u^\mu\,(S_\lambda F^{\lambda\nu}u_\nu)$ (build a scalar from $S$ , $F$ , $u$ and hang it on the only other vector around). A third candidate, $u^\mu (S\cdot \dot u)$ , is not new: equation (3) turns $\dot u$ into $Fu$ and collapses it into the second. Two terms. That is all there are.

\frac{\mathrm dS^\mu}{\mathrm d\tau} \;=\; \alpha\, F^{\mu\nu}S_\nu \;+\; \beta\,u^\mu\,\big(S_\lambda F^{\lambda\nu}u_\nu\big)

Why: this is the entire creative content of Bargmann, Michel and Telegdi's paper — the realisation that "how does a spin move?" has almost no room in it.

Step 2 — impose $S\cdot S =$ const. It costs nothing.

\frac{\mathrm d}{\mathrm d\tau}(S^\mu S_\mu) \;=\; 2S_\mu\frac{\mathrm dS^\mu}{\mathrm d\tau} \;=\; 2\alpha\,\underbrace{S_\mu F^{\mu\nu}S_\nu}_{=\,0} \;+\; 2\beta\,\underbrace{(S\cdot u)}_{=\,0}\,(SFu) \;=\; 0

Why: the first vanishes because $F^{\mu\nu}$ is antisymmetric and $S_\mu S_\nu$ symmetric; the second by orthogonality. Both vanish for any $\alpha$ and $\beta$ . Conservation of the spin's length is automatic — it constrains nothing. Worth pausing on: the constraint that feels important is free, and the one that looks like bookkeeping fixes the physics.

Step 3 — impose $S\cdot u = 0$ . This is where Thomas comes from. Differentiate the constraint and demand it hold:

0 \;=\; \frac{\mathrm d}{\mathrm d\tau}(S^\mu u_\mu) \;=\; \frac{\mathrm dS^\mu}{\mathrm d\tau}u_\mu \;+\; S_\mu\frac{\mathrm du^\mu}{\mathrm d\tau}

Insert the ansatz and equation (3). Write $(SFu) \equiv S_\lambda F^{\lambda\nu}u_\nu$ . Then, using the antisymmetry of $F$ to get $u_\mu F^{\mu\nu}S_\nu = -(SFu)$ , and $u^\mu u_\mu = c^2$ :

\underbrace{-\alpha\,(SFu)}_{\text{term 1}} \;+\; \underbrace{\beta\,c^2\,(SFu)}_{\text{term 2}} \;+\; \underbrace{\frac{q}{m}\,(SFu)}_{\text{term 3}} \;=\; 0 \qquad\Longrightarrow\qquad \boxed{\;\beta \;=\; \frac{1}{c^2}\Big(\alpha - \frac qm\Big)\;}

Why: term 3 is the whole story. It is there because the four-velocity is being accelerated by the very same field that torques the spin — the Lorentz force, equation (3), walking in uninvited. To keep $S$ perpendicular to a $u$ that is turning, the spin must be dragged along, and $\beta$ is exactly the drag needed. The Thomas precession is the price of the orthogonality constraint on an accelerated worldline. Note the sign: term 1 is $-\alpha$ , not $+\alpha$ , because $u_\mu F^{\mu\nu}S_\nu = -(SFu)$ . The base gets this wrong and then silently reverses it two steps later — see the warning.

Step 4 — fix $\alpha$ in the rest frame. There, $u^\mu = (c,\mathbf 0)$ , the second term is $\propto u^\mu$ and so touches only $S^0$ , and $F^{ij} = -\varepsilon^{ijk}B_k$ gives

\frac{\mathrm d\mathbf S}{\mathrm dt} \;=\; -\alpha\,\mathbf B\times\mathbf S \;\overset{!}{=}\; \boldsymbol\mu\times\mathbf B \;=\; \frac{gq}{2m}\,\mathbf S\times\mathbf B \qquad\Longrightarrow\qquad \alpha = \frac{gq}{2m}

Why: in the rest frame nothing is left but ordinary Larmor precession of a moment $\boldsymbol\mu = (gq/2m)\mathbf S$ . This is the one place a physical input enters the derivation, and the input is the definition of $g$ .

Step 5 — assemble.

\beta \;=\; \frac{1}{c^2}\Big(\frac{gq}{2m} - \frac qm\Big) \;=\; \frac{q}{mc^2}\Big(\frac g2 - 1\Big)

\boxed{\;\frac{\mathrm dS^\mu}{\mathrm d\tau} = \frac{gq}{2m}F^{\mu\nu}S_\nu + \frac{q}{mc^2}\Big(\frac g2-1\Big)u^\mu\big(S_\lambda F^{\lambda\nu}u_\nu\big)\;}

Why: look at what the second coefficient is proportional to. It is $g/2 - 1$ , which is exactly the anomaly $a \equiv (g-2)/2$ . That identity is not a coincidence — it is why the anomaly, and not $g$ , is the natural variable of the whole subject. At $g = 2$ the drag term vanishes outright and the spin obeys the same equation as the four-velocity: it locks to the momentum. *Verified in `verify_aq-sr-bmt-equation.py`: the two constraints are imposed symbolically on the general ansatz in sympy, with explicit $4\times4$ $F^{\mu\nu}$ and metric $\mathrm{diag}(+1,-1,-1,-1)$ , and $\beta = (\alpha - q/m)/c^2$ is solved for, not assumed. The rest-frame reduction is done numerically. $S\cdot u$ and $S\cdot S$ are then checked to be conserved to $10^{-12}$ under numerical integration of the full equation in a crossed-field configuration.*

Go deeper · Derivation B · 2 minThe lab reduction, the cancellation, and the magic gamma

Derivation BThe lab reduction, the cancellation, and the magic gamma

The goal is the number the experiment actually measures. Equation (2) governs $S^\mu$ , but a detector sees the angle between the rest-frame spin and the momentum, and getting from one to the other means undoing the boost at every instant — on a boost that is changing, which is precisely where the Thomas rotation re-enters. Equation (4) is the standard, long result; what is derived here is what happens when you subtract the cyclotron frequency from it, because that subtraction is the entire experiment.

Step 1 — the two frequencies, side by side. Take $\boldsymbol\beta \perp \mathbf B$ (a storage ring: the field is vertical, the beam is horizontal), so $\boldsymbol\beta\cdot\mathbf B = 0$ and the middle term of (4) drops:

\boldsymbol\omega_s = -\frac qm\Big[\Big(a+\frac1\gamma\Big)\mathbf B - \Big(a+\frac{1}{\gamma+1}\Big)\frac{\boldsymbol\beta\times\mathbf E}{c}\Big], \qquad \boldsymbol\omega_c = -\frac qm\Big[\frac{1}{\gamma}\mathbf B - \frac{\gamma}{\gamma^2-1}\,\frac{\boldsymbol\beta\times\mathbf E}{c}\Big]

Why: $\boldsymbol\omega_c$ is not a new assumption — it is the Lorentz force, equation (3), written as a rotation rate of the velocity direction. Both frequencies contain the same $\mathbf B/\gamma$ : at $g=2$ they are the same frequency, which is exactly the statement that a $g=2$ spin tracks its own momentum.

Step 2 — subtract, and watch $\gamma$ leave. The magnetic parts:

\Big(a + \frac1\gamma\Big) - \frac1\gamma \;=\; a

Why: there it is. The $1/\gamma$ cancels identically, for every $\gamma$ , exactly. Not to leading order, not approximately. The energy-dependent piece of the spin precession is the same energy-dependent piece as the cyclotron precession, because both come from the same $F^{\mu\nu}u_\nu$ structure — and the difference keeps only the piece proportional to the anomaly. A beam with a $0.1\%$ momentum spread produces a signal with zero spread from this source.

Step 3 — the electric term, and the magic $\gamma$ . The $\boldsymbol\beta\times\mathbf E$ coefficients:

\Big(a + \frac{1}{\gamma+1}\Big) - \frac{\gamma}{\gamma^2-1} \;=\; a + \frac{1}{\gamma+1} - \frac{\gamma}{(\gamma-1)(\gamma+1)} \;=\; a + \frac{(\gamma-1) - \gamma}{\gamma^2-1} \;=\; a - \frac{1}{\gamma^2-1}

giving equation (6). Setting the bracket to zero:

\boxed{\;\gamma_{\text{magic}} = \sqrt{1 + \frac1a}\;} \qquad\Longrightarrow\qquad \gamma_{\text{magic}}^{\,\mu} = 29.3034

Why: a storage ring needs vertical focusing; vertical focusing means electrostatic quadrupoles; and electrostatic quadrupoles mean an $\mathbf E$ field sitting in the middle of your precision measurement. Unless you run at the $\gamma$ where the $\mathbf E$ field drops out of $\boldsymbol\omega_a$ altogether. For the muon that is $\gamma = 29.3034$ , $p = 3.094\ \mathrm{GeV}/c$ — which is why the CERN, Brookhaven and Fermilab rings were all built to that momentum. A design constant, derived from a cancellation in a 1959 equation.

Step 4 — what you measure. With $\mathbf E$ neutralised and $\boldsymbol\beta\perp\mathbf B$ :

\boxed{\;\omega_a = \frac{e}{m_\mu}\,a_\mu\,B\;} \qquad\Longrightarrow\qquad a_\mu = \frac{m_\mu\,\omega_a}{eB}

Why: two measurements, and one of them is a magnetic field. Count the wiggle in the decay-positron rate to get $\omega_a$ , NMR-map the field to get $B$ , divide. You never measure $g$ , and you never measure the beam energy. The 2025 Fermilab final result, $a_\mu = 0.001165920705(146)$ , is $127$ parts per billion — and every part of that precision descends from the two cancellations in Steps 2 and 3. *Verified in `verify_aq-sr-bmt-equation.py`: equation (4) is checked against direct numerical integration of the covariant equation (2) at 40 random $(\gamma, \mathbf E, \mathbf B)$ configurations (worst relative error $4\times10^{-9}$ ); $\omega_a$ is confirmed flat in $\gamma$ across $\gamma \in [2,60]$ to machine precision at $\mathbf E = 0$ ; the magic- $\gamma$ cancellation is verified by showing the $\mathbf E$ -field shift in $\omega_a$ drops below $10^{-16}$ there; and the ring numbers below are recomputed from $a_\mu$ and $B$ .*

Common trapThe base drops the factors of c — and one of them is worth 10^17

The base's covariant card prints the second term as $\dfrac{e}{m}\big(\tfrac g2-1\big)u^\mu(S_\lambda F^{\lambda\nu}u_\nu)$ , with no $1/c^2$ . Its lab formulas print $\boldsymbol\beta\times\mathbf E$ with no $1/c$ . In SI these are not typographical: $u^\mu(SFu)$ carries two extra powers of velocity relative to $F^{\mu\nu}S_\nu$ , and $\mathbf E$ carries one relative to $\mathbf B$ . Dimensional analysis catches both in one line — and the second term, with the base's coefficient, is wrong by $c^2 \approx 9\times10^{16}$ . The base's own provenance note says its equations are "verified against Jackson §11.11". Jackson works in Gaussian units, where the $c$ 's live in different places. A formula copied out of a Gaussian text into an SI article is not a citation. It is a bug with a footnote.

Common trapThe base's uniqueness derivation gets its own sign wrong, then quietly reverses it

The base's DERIV[0] step 3 writes the constraint equation as

\alpha\,(SFu) + \beta\,u^2\,(SFu) + \frac em(SFu) = 0 \quad\Longrightarrow\quad \beta = -\alpha - \frac em

while simultaneously stating in the same step that $u_\mu F^{\mu\nu}S_\nu = -(SFu)$ . Those two things cannot both be true: if $u_\mu F^{\mu\nu}S_\nu = -(SFu)$ , then the first term is $-\alpha(SFu)$ , not $+\alpha(SFu)$ , and the conclusion is $\beta = +\alpha - q/m$ (times $1/c^2$ ), which is what Step 3 above derives. The base then arrives at $\beta = (e/m)(1 - g/2) = -(e/m)(g/2-1)$ in its step 5 — the opposite sign to its own equation card — and repairs the contradiction with the words "careful sign bookkeeping … yields the standard $\beta = (e/m)(g/2-1)$ ". It does not show the bookkeeping. A derivation that has to be overruled by its own conclusion is not a derivation. The card is right; the derivation is wrong; and the verification script for this page solves the constraint symbolically rather than trusting either.

Common trapThomas precession is the g = 0 content of BMT, not the g = 2 content

The base's second derivation is titled "Thomas precession as the $g=0$ (fieldless) content", and its first step then says the kinematic pieces are those that "survive at $g=2$ ". Those are different claims and they cannot both be right. The title is right. Set $g = 0$ in equation (2). The torque term dies, the drag coefficient becomes $\beta = -q/mc^2$ , and since $(SFu) = (m/q)\,S_\lambda A^\lambda$ by equation (3), what is left is

\frac{\mathrm dS^\mu}{\mathrm d\tau} \;=\; -\frac{1}{c^2}\,(S\cdot A)\,u^\mu

which is Fermi–Walker transport — precisely the torque-free equation that `aq-sr-thomas-precession` integrates to get $\boldsymbol\omega_T$ . A $g=0$ particle has no magnetic moment and feels no torque, and it still precesses. That is the Thomas precession, alone and unmixed. At $g = 2$ , by contrast, the drag term dies and the spin obeys the same equation as $u^\mu$ — in a pure magnetic field the spin locks to the momentum and $\omega_a = 0$ . That is not the Thomas precession; it is the Thomas precession exactly cancelling the $g=2$ Larmor precession. Confusing the two is how people end up double-counting.

Limiting cases

Case	BMT becomes	Meaning
$\gamma\to1$ , $\mathbf E = 0$	$\mathrm d\mathbf S/\mathrm dt = \boldsymbol\mu\times\mathbf B$	ordinary Larmor precession
$g = 0$	$\mathrm dS^\mu/\mathrm d\tau = -\frac{1}{c^2}(S\cdot A)u^\mu$	Fermi–Walker transport — pure Thomas precession, no torque
$g = 2$ , uniform $\mathbf B$	$\boldsymbol\omega_s = \boldsymbol\omega_c$ , $\;\boldsymbol\omega_a = 0$	spin locks to momentum. Thomas exactly cancels Larmor
$\mathbf E = 0$ , any $g$	$\boldsymbol\omega_a = -\frac qm a\mathbf B$	no $\gamma$ . The observable is flat in energy
$\gamma = \sqrt{1+1/a}$	$\mathbf E$ drops out of $\boldsymbol\omega_a$	the magic $\gamma$ : $29.3034$ for $\mu$ , $29.38$ for $e$
$\gamma\gg1$ , curved orbit	$\omega_T \to (\gamma-1)\,\omega_c$	at the muon ring, $28.3\times$ the cyclotron rate

Worked examples

1 · Rebuild the Fermilab ring from one number

Given the muon anomaly and nothing else, derive the storage ring.

Step 1 — identify. The 2025 Fermilab final result: $a_\mu = 0.001165920705(146)$ , $127$ ppb. That is the only input. The ring's field is $B = 1.4513$ T.

Step 2 — apply. The magic $\gamma$ from Step 3 of the derivation:

\gamma_{\text{magic}} = \sqrt{1 + 1/a_\mu} = \sqrt{858.7} = \mathbf{29.3034}, \qquad \beta = 0.99942

Momentum: $p = \gamma m_\mu \beta c = (29.3034)(105.658\ \mathrm{MeV})(0.99942) = \mathbf{3.0943\ GeV}/c$ .

Step 3 — the boundary numbers. Radius $R = p/(eB) = \mathbf{7.1120\ m}$ . Cyclotron period $T_c = 2\pi\gamma m_\mu/(eB) = \mathbf{149.1\ ns}$ . Anomalous period $T_a = 2\pi m_\mu/(e\,a_\mu B) = \mathbf{4.365\ \mu s}$ .

Step 4 — check against the real machine. The Fermilab (ex-Brookhaven) ring has a radius of $\mathbf{7.112\ m}$ , runs muons at $\mathbf{3.094\ GeV}/c$ , and reports a $g-2$ period of $\mathbf{4.365\ \mu s}$ and a cyclotron period of $\mathbf{149.2\ ns}$ . Every number matches. The ring was not tuned empirically; it was computed from equation (6) and then built. And the ratio $T_a/T_c = 1/(a_\mu\gamma) = 29.3$ : the muon goes round 29 times per precession cycle, and survives $\gamma\tau_\mu = 64.4\ \mu\mathrm s$ — about $\mathbf{432}$ turns and $\mathbf{15}$ precession cycles — before decaying. That is the whole experiment: watch the decay positrons wiggle fifteen times, and fit the frequency.

2 · Set $g = 0$ and get Thomas back

Show that the BMT equation contains the Thomas precession with no magnetic moment at all.

Step 1 — identify. Put $g = 0$ in equation (2). The particle now has charge (so the Lorentz force still bends its path) but no magnetic moment — nothing for the field to torque.

Step 2 — apply. The first term vanishes identically. In the second, $\beta = \frac{q}{mc^2}(0 - 1) = -\frac{q}{mc^2}$ , and equation (3) lets us trade the field for the four-acceleration: $S_\lambda F^{\lambda\nu}u_\nu = \frac mq\, S_\lambda A^\lambda$ . So

\frac{\mathrm dS^\mu}{\mathrm d\tau} \;=\; -\frac{q}{mc^2}\cdot\frac mq\,(S\cdot A)\,u^\mu \;=\; -\frac{1}{c^2}(S\cdot A)\,u^\mu

Step 3 — recognise it. That is the Fermi–Walker transport equation, verbatim — the definition of "carry a vector along a worldline without torquing it", and the equation `aq-sr-thomas-precession` integrates to get $\boldsymbol\omega_T = \frac{\gamma^2}{\gamma+1}\frac{\mathbf a\times\mathbf v}{c^2}$ . Numerically, integrating the $g=0$ BMT equation around a circular orbit at $\beta = 0.6$ and reading off the rotation gives exactly $-90.000000^\circ$ per orbit, agreeing with `aq-sr-thomas-precession`'s independent Fermi–Walker integration to $3.5\times10^{-9}$ .

Step 4 — what it means. The Thomas precession is what a spin does when nothing is done to it. It is the $g=0$ residue of BMT: strip out every scrap of magnetism and the precession is still there, because the constraint $S\cdot u = 0$ is still there and the worldline is still bending. It also settles the accounting. At the muon ring $\omega_T = (\gamma-1)\,\omega_c = 28.3\,\omega_c = 1.192\times10^{9}$ rad/s, vastly larger than $\omega_a = 1.44\times10^{6}$ rad/s: the huge kinematic precession and the huge $g=2$ Larmor precession cancel to one part in a thousand, and the tiny residue that survives is the anomaly. Everything the experiment measures lives in that residue.

Physics of the portrait

Origins · 2 minAnomalous magnetic moment and how this was discovered

Anomalous magnetic moment

a = (g−2)/2 — the most precisely tested number in physics · Dirac's g = 2 · Schwinger's α/2π · the muon's 127 parts per billion

Dirac's equation says $g = 2$ exactly — one of the great predictions in physics, spin and the right magnetic moment both falling out of a wave equation written for entirely different reasons. It is also wrong, by about a tenth of a percent. In 1948 Kusch and Foley measured the electron's $g$ and found $2.00238$ , and Schwinger explained the excess in a single line of quantum field theory: the electron emits and reabsorbs a virtual photon, and the correction is $\alpha/2\pi = 0.00116$ . The number is on his gravestone. The anomaly $a = (g-2)/2$ is what is left when you subtract Dirac. Every particle in the universe contributes to it, because $a$ is a sum over every virtual loop the particle can make — and the heavier the particle, the more it can see. The muon is $207$ times heavier than the electron, so its anomaly is roughly $(m_\mu/m_e)^2 \approx 43\,000$ times more sensitive to whatever else is out there. That is why $a_\mu$ has been the most watched number in particle physics for two decades. Measured (Fermilab, 2025) to $127$ parts per billion; predicted by the Standard Model to comparable precision; and the two have circled each other, at times apparently disagreeing, as the theory of the hadronic contributions has been fought over. And the only reason the measurement is possible at all is that $\omega_a$ has no $\gamma$ in it — a cancellation in an equation written in 1959, three years before the muon's anomaly was first measured at CERN.

History

1926

— L. H. Thomas shows that the electron's rest frame rotates on a curved worldline (Nature 117, 514), halving the semiclassical spin-orbit coupling. The kinematics is right but the treatment is not covariant.

1927

— Frenkel, and independently Thomas in Philosophical Magazine 3, 1, write the first covariant spin equations. Correct for g = 2, awkward otherwise.

1928

— Dirac's equation predicts g = 2 exactly, from no assumption about magnetism.

1948

— Kusch and Foley measure the electron's g and find 2.00238, not 2. Schwinger computes the excess: a = alpha/(2 pi) = 0.00116. The number is engraved on his gravestone.

1959

— Bargmann, Michel and Telegdi publish the general covariant equation for any g (Physical Review Letters 2, 435). Two terms, two constraints, one free parameter. Everything about relativistic spin dynamics in an external field is in it.

1961

— CERN's first muon g-2 experiment measures a_mu to about 0.4 per cent, using precisely the omega_a cancellation.

2001

— Brookhaven E821 reaches 540 parts per billion at the magic gamma of 29.3, and reports a persistent tension with the Standard Model prediction.

2025

— Fermilab's final result: a_mu = 0.001165920705, a precision of 127 parts per billion, from the same 7.112 metre ring, moved 5000 kilometres from Long Island.

Going further

Texts. Jackson, Classical Electrodynamics, 3rd ed., §11.11 (Gaussian — do not copy the $c$ 's) · Rindler, Introduction to Special Relativity, 2nd ed. · Mane, Shatunov & Yokoya, Rep. Prog. Phys. 68, 1997 (2005). Primary: V. Bargmann, L. Michel & V. L. Telegdi, Phys. Rev. Lett. 2, 435 (1959).

Data. $a_\mu = 0.001165920705(114)_{\text{stat}}(91)_{\text{syst}}$ — Muon $g-2$ Collaboration, Fermilab final result (2025), $127$ ppb. Ring: $B = 1.4513$ T, $R = 7.112$ m, $p = 3.094\ \mathrm{GeV}/c$ .

Provenance. SI end to end, every $c$ written. Signature $(+,-,-,-)$ , $u\cdot u = +c^2$ . Equation (2) is derived, not quoted: the ansatz is constrained symbolically and $\beta$ is solved for. $\boldsymbol\omega_T$ is quoted from `aq-sr-thomas-precession`, the $\tfrac12$ from `aq-sr-thomas-factor-spin-orbit`, the non-closure of boosts from `aq-sr-thomas-wigner-rotation`. Base defects corrected: the missing $1/c^2$ and $1/c$ (a Gaussian-to-SI transcription error, acknowledged in the base's own provenance note, which cites Gaussian Jackson); the sign error in DERIV[0] step 3, which its own step 5 overrules without showing the work; and the contradiction between DERIV[1]'s title ( $g=0$ ) and its first step ( $g=2$ ) — the title is correct.

Claim-scope. This page derives the classical BMT equation for a point particle with a magnetic moment in an external field. It does not derive $a$ itself — that is quantum field theory, and $a_e = \alpha/2\pi + \dots$ belongs to a QED track. It does not treat the electric dipole term, radiative polarisation, or Stern–Gerlach forces from field gradients.

Builds on. Thomas Precession · The Thomas Factor One-Half · The Field Tensor. Leads to. Muon g−2 and Time Dilation · Relativistic Beam Dynamics.

Figure

The magic gamma, and why the ring is 7.112 metres across. The curve is the coefficient of the electric-field term in omega_a: a − 1/(gamma² − 1). At low gamma the 1/(gamma² − 1) piece is enormous and the electric field of the focusing quadrupoles utterly dominates the measurement. As gamma rises the piece falls, crosses the muon anomaly a = 0.0011659, and the whole coefficient passes through ZERO. That crossing is at gamma = 29.3034 — the MAGIC GAMMA — and it corresponds to a muon momentum of 3.094 GeV/c. Run the ring there and the electrostatic quadrupoles you need for vertical focusing become invisible to the spin measurement: they still focus the beam, but they no longer shift omega_a. Above the magic gamma the coefficient turns positive and the contamination returns with the opposite sign. Note the scale: the crossing is sharp, and the entire design of the CERN, Brookhaven and Fermilab muon rings — a 7.112 metre radius in a 1.4513 tesla field — is fixed by the location of this one zero

Why is the BMT equation unique? Could I not add another term?

Not if you want it linear in $S$ and in $F$ , and covariant. There are only two four-vectors you can build from $S^\mu$ , $F^{\mu\nu}$ and $u^\mu$ under those conditions: $F^{\mu\nu}S_\nu$ and $u^\mu(S_\lambda F^{\lambda\nu}u_\nu)$ ; a third candidate, $u^\mu(S\cdot\dot u)$ , collapses into the second via the Lorentz force. Relax the assumptions and you can add terms — one linear in $\partial F$ gives a Stern–Gerlach force from a field gradient, one with the dual $\tilde F^{\mu\nu}$ gives an electric dipole moment, which the $g-2$ experiment also searches for. But for a point particle with only a magnetic moment in a uniform field, two terms is all there is.

Why does the experiment measure a and not g?

Because the detector sees the angle between spin and momentum, and that angle turns at $\omega_a = \omega_s - \omega_c \propto a = (g-2)/2$ . To get $g$ you would need $\omega_s$ alone, which means knowing $\gamma$ for every muon in the beam to the same precision — and you cannot. It is experimental judo: the quantity that is impossible to measure precisely, $g$ , differs from the one that is easy, $a$ , by exactly $2$ — the one number theory hands you for free.

Is the huge Thomas precession in the ring not a problem for the measurement?

It is enormous — $\omega_T = (\gamma - 1)\omega_c = 28.3\,\omega_c$ , about $1.19\times10^9$ rad/s, against an $\omega_a$ of $1.44\times10^6$ rad/s. A thousand times bigger. But it is not an error term: it sits inside $\omega_s$ , exactly cancelling the $g=2$ part of the Larmor precession, and the cancellation is exact for all $\gamma$ — which is what equation (6) proves. Nobody computes $\omega_T$ and subtracts it by hand; the covariant equation has already done so. That is the payoff for a covariant formulation rather than Thomas's 1926 argument.

What happens to muons that are not exactly at the magic momentum?

They pick up an electric-field contribution proportional to $a - 1/(\gamma^2-1)$ — small, but not zero, and one of the named systematic corrections in the Fermilab error budget. The ring has a momentum acceptance of about $\pm 0.15\%$ , so the beam is a distribution around $\gamma_{\text{magic}}$ , not a delta function; the correction is computed by measuring the actual momentum distribution and integrating equation (6) over it. Same for the "pitch correction": muons with vertical velocity have $\boldsymbol\beta\cdot\mathbf B \ne 0$ , which wakes up the middle term of equation (4).

Does the spin four-vector's spatial part equal the spin you measure?

No — that is the trap $S\cdot u = 0$ sets. In the lab it forces $S^0 = \boldsymbol\beta\cdot\mathbf S_{\text{lab}}$ , which is not zero when the particle moves; and $\mathbf S_{\text{lab}}$ is not the rest-frame spin either. To get the rest-frame spin — the thing that couples to the decay asymmetry, and that the detector effectively sees — you must boost $S^\mu$ back with the instantaneous velocity. Doing that boost at every instant, on a $u$ that is turning, is exactly where the Thomas rotation re-enters when you reduce equation (2) to equation (4).

Why is the muon anomaly so much more interesting than the electron's?

Sensitivity to heavy new physics scales as $(m/\Lambda)^2$ , so a particle $207$ times heavier is roughly $207^2 \approx 43\,000$ times more sensitive to whatever lies beyond the Standard Model. The electron's anomaly is known far more precisely in absolute terms, but it is dominated by pure QED and is used mainly as the world's best determination of $\alpha$ . The muon's is where a discrepancy would show — which is why the fight over the hadronic vacuum-polarisation contribution to the prediction is now the crux of the whole story.

BMT (Bargmann-Michel-Telegdi) Equation

Where this lives today

The physical picture

The core equations

Deriving it

Limiting cases

Worked examples

1 · Rebuild the Fermilab ring from one number

2 · Set $g = 0$ and get Thomas back

Physics of the portrait

Going further

Before this

Where this leads

BMT (Bargmann-Michel-Telegdi) Equation

Where this lives today

The physical picture

The core equations

Deriving it

Limiting cases

Worked examples

1 · Rebuild the Fermilab ring from one number

2 · Set g=0g = 0g=0 and get Thomas back

Physics of the portrait

Going further

Before this

Where this leads

2 · Set $g = 0$ and get Thomas back