Relativistic Force and Newton's Second Law

Push an object due north. It accelerates due north. That sentence has been the silent assumption behind every free-body diagram ever drawn, and in relativity it is false. Take a proton moving at $0.9c$ along $x$ , and apply a force at $45^\circ$ to its motion. The acceleration does not come out at $45^\circ$ . It comes out at $\mathbf{79.24^\circ}$ — nearly perpendicular to the velocity, a full $34.24^\circ$ away from the force that caused it. Nothing is broken and nothing is subtle: the force and the acceleration simply are not parallel, and no choice of "mass" can make them so. Choose the angle to maximise the disagreement and at this speed you can open a gap of $\mathbf{42.90^\circ}$ between where you push and where the particle goes. The reason is one line of calculus. Newton's second law survives relativity untouched in the form Newton actually wrote it — force is the rate of change of momentum, $\vec F = \mathrm{d}\vec p/\mathrm{d}t$ — but the momentum is now $\vec p = \gamma m\vec v$ , and $\gamma$ depends on the speed. So differentiating hits two things: the velocity, and the $\gamma$ riding on it. The first term gives an acceleration along $\vec a$ ; the second an extra kick along $\vec v$ . Two directions, one derivative, and the sum points somewhere else entirely. The historical response was to invent two masses — a "longitudinal mass" $\gamma^3m$ and a "transverse mass" $\gamma m$ — and physics spent forty years with that vocabulary before abandoning it, because it is a bad description of a good result. The factors are real and exactly right. They are not masses: they are direction-dependent response coefficients, and $m$ — the one invariant rest mass — sits underneath both.

Key takeaways

Newton's second law is not repealed; $\vec F = m\vec a$ is. The surviving form is the one Newton wrote: $\boxed{\vec F = \mathrm{d}\vec p/\mathrm{d}t}$ with $\vec p = \gamma m\vec v$ , and $m$ the invariant rest mass.
Differentiating gives $\boxed{\vec F = \gamma m\,\vec a + \gamma^3 m\,\dfrac{(\vec v\cdot\vec a)}{c^2}\,\vec v}$ — two terms, two directions. $\vec F$ and $\vec a$ are not parallel.
Inverting it is more useful: $\boxed{\vec a = \dfrac{1}{\gamma m}\left[\vec F - \dfrac{(\vec F\cdot\vec v)}{c^2}\vec v\right]}$ — the longitudinal part of $\vec F$ is shaved by $\gamma^2$ , the transverse part passes through untouched. The only aligned cases are $\;F_\parallel = \gamma^3 m\,a_\parallel$ and $\;F_\perp = \gamma m\,a_\perp$ ; $\gamma^3$ and $\gamma$ are response coefficients, not masses.
Owned result: $\tan\theta_a = \gamma^2\tan\theta_F$ . Force at angle $\theta_F$ to the velocity, acceleration at $\theta_a$ — and the misalignment is maximal at $\tan\theta_F = 1/\gamma$ , where it equals $\boxed{\Delta_{\max} = 2\arctan\gamma - 90^\circ}$ . At $\beta = 0.9$ that is $42.90^\circ$ ; as $\gamma\to\infty$ it tends to a full right angle.
Key idea: it costs $\gamma^2$ times more force to speed a particle up than to steer it sideways by the same amount. That single ratio is why particle accelerators are built the way they are — you bend a beam with modest magnets and accelerate it with enormous cavities.

The core insight

The velocity-dependence lives in the momentum, not the mass. Accept $\vec p = \gamma m\vec v$ , refuse to touch $m$ , and everything else is the product rule: $\mathrm{d}(\gamma m\vec v)/\mathrm{d}t$ has a term where the derivative hits $\vec v$ and a term where it hits $\gamma$ — and the second points along $\vec v$ regardless of where you pushed. The whole strangeness of relativistic force — non-parallel acceleration, two "stiffnesses", a speed limit that is never imposed — is that one extra term.

Where this lives today

Every circular accelerator on Earth is a monument to the $\gamma^2$ gap between the two response coefficients. Bending is transverse, costs $\gamma m$ , and is governed by $B\rho = p/e$ — expensive, but linear in momentum and achievable with magnets. Accelerating is longitudinal, costs $\gamma^3 m$ , and the same field that would double a slow particle's speed barely nudges a fast one. The two jobs need entirely different hardware — superconducting dipoles for the ring, RF cavities for the push — and no engineering will let one do the other's work. That division is not an accident of technology; it is equation (4).

And the reason the speed of light is never exceeded — in a laboratory, a supernova shock, or a plasma wakefield — is on this page. Nobody has to forbid it. The longitudinal coefficient $\gamma^3 m$ diverges, the acceleration quenches, and the energy that would have gone into speed goes into momentum instead. The speed limit is a consequence of the force law, not an axiom sitting beside it.

The physical picture

Why should pushing a fast particle forwards be harder than pushing it sideways? Because "forwards" changes its speed, and speed is what $\gamma$ depends on. Push transversely and, to first order, you only turn the velocity vector: the speed is unchanged, so $\gamma$ is unchanged, and the $\gamma$ in $\vec p = \gamma m\vec v$ sits there as a constant multiplier. The particle responds with a plain inertia of $\gamma m$ . Push longitudinally and you change the speed itself — so $\gamma$ is also changing, and changing in the direction that opposes you, since $\gamma$ grows without bound as $v \to c$ . You fight the inertia and the growth of $\gamma$ , and the two compound into $\gamma^3$ .

Figure

PUNCHLINE — force and acceleration are NOT parallel. A proton at β = 0.9 (γ = 2.294) is pushed at 45° to its motion; it accelerates at 79.24°, a full 34.24° away from the force. The rule is tan θₐ = γ² tan θ_F, so the transverse component passes through while the longitudinal component is shaved by γ². The misalignment is largest at θ_F = 23.55°, where it reaches 42.90° — and it grows toward a full right angle as γ increases

The core equations

Newton's second law is kept; only the momentum is upgraded.

\vec{F} \;=\; \frac{\mathrm{d}\vec{p}}{\mathrm{d}t}, \qquad \vec{p} = \gamma m\vec{v}, \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \qquad (1)

where: $m$ — the invariant rest mass, a Lorentz scalar, identical for every observer at every speed. Physical reading: this is Newton's own formulation — the Principia states the law as a proportionality between impressed force and the change in the "quantity of motion", not as $\vec F = m\vec a$ . Nothing about (1) is relativistic except the $\gamma$ inside $\vec p$ . Differentiate, using $\mathrm{d}\gamma/\mathrm{d}t = \gamma^3(\vec v\cdot\vec a)/c^2$ :

\boxed{\;\vec{F} \;=\; \gamma m\,\vec{a} \;+\; \gamma^{3} m\,\frac{(\vec{v}\cdot\vec{a})}{c^{2}}\,\vec{v}\;} \qquad (2)

where: $\vec a = \mathrm{d}\vec v/\mathrm{d}t$ is the ordinary three-acceleration. Physical reading: two terms, two directions. The first is along $\vec a$ , the second along $\vec v$ . Unless $\vec a$ is parallel or perpendicular to $\vec v$ , their sum lies along neither — so $\vec F \nparallel \vec a$ , and no scalar $M$ gives $\vec F = M\vec a$ . The second term is the price of $\gamma$ depending on speed. Solving (2) for $\vec a$ gives the form you actually use, and it is prettier:

\boxed{\;\vec{a} \;=\; \frac{1}{\gamma m}\left[\vec{F} \;-\; \frac{(\vec{F}\cdot\vec{v})}{c^{2}}\,\vec{v}\right]\;} \qquad (3)

where: the bracket subtracts a piece of $\vec F$ along the motion. Physical reading: an instruction. Take the force; shrink its longitudinal component by $1/\gamma^2$ (which is what subtracting $(\vec F\cdot\vec v)\vec v/c^2$ does, since $1 - \beta^2 = 1/\gamma^2$ ); leave the transverse component alone; divide by $\gamma m$ . The particle is stiff along its motion and soft across it, and the stiffness ratio is exactly $\gamma^2$ . The two aligned special cases follow immediately, and are the only ones in which $\vec F \parallel \vec a$ :

F_{\parallel} = \gamma^{3} m\,a_{\parallel} \qquad\qquad F_{\perp} = \gamma m\,a_{\perp} \qquad (4)

where: $\parallel$ and $\perp$ are relative to $\vec v$ . Physical reading: $\gamma^3$ and $\gamma$ are direction-dependent response coefficients — how much force a unit of acceleration costs, in that direction, at that speed. Historically they were called "longitudinal mass" and "transverse mass"; that language is deprecated (see the warning). The relations themselves are exactly correct, with $m$ the invariant mass. Finally, energy. The work–energy theorem needs no repair at all:

\boxed{\;\vec{F}\cdot\vec{v} \;=\; \frac{\mathrm{d}E}{\mathrm{d}t}\;}, \qquad E = \gamma mc^{2} \qquad (5)

where: $E$ is the total relativistic energy. Physical reading: power is still force dotted into velocity — identically, no correction factor. A force perpendicular to $\vec v$ therefore does zero work: it changes direction without changing speed or energy, which is what a bending magnet does, and why it steers a 7 TeV beam without adding a single eV.

Force, acceleration, and the angle between them

Derivation ADifferentiating gamma m v gives two terms, and they point in different directions

Everything in this article is the product rule applied to $\gamma m\vec v$ , taken seriously. The fact that makes relativistic dynamics feel alien — you push one way and accelerate another — is not deep; it is just $\gamma$ being a function of speed. Deriving it once retires the mystery permanently and hands over both the forward law (2) and the inverse law (3).

Step 1. Get the derivative of $\gamma$ . With $\gamma = (1 - v^2/c^2)^{-1/2}$ and $v^2 = \vec v\cdot\vec v$ , the chain rule gives $\dfrac{\mathrm{d}\gamma}{\mathrm{d}t} = \gamma^3\,\dfrac{\vec v\cdot\vec a}{c^2}$ . Why: note the $\gamma^3$ appearing here, before any talk of "longitudinal mass". Every $\gamma^3$ in this subject traces back to this one line — it is the derivative of a $-1/2$ power, nothing more.

Step 2. Apply the product rule to (1). $\;\vec F = \dfrac{\mathrm{d}}{\mathrm{d}t}(\gamma m\vec v) = m\left(\dfrac{\mathrm{d}\gamma}{\mathrm{d}t}\vec v + \gamma\vec a\right) = \gamma m\vec a + \gamma^3m\dfrac{(\vec v\cdot\vec a)}{c^2}\vec v$ , which is equation (2). Why: $m$ never gets differentiated, because $m$ is invariant. That is the whole discipline of this subject in one clause. The $\gamma$ is what carries the speed-dependence, and putting it on the $m$ instead is precisely the relativistic-mass error.

Step 3. See the non-parallelism directly. The result is a vector along $\vec a$ plus a vector along $\vec v$ ; if $\vec v\cdot\vec a \ne 0$ and $\vec a \nparallel \vec v$ , those are different directions, so $\vec F$ points along neither. Why: no scalar can therefore relate $\vec F$ and $\vec a$ . Any $\vec F = M\vec a$ must fail, whatever velocity-dependence you allow $M$ — the obstruction is directional, not numerical.

Step 4. Invert it. Dot (2) with $\vec v$ : $\;\vec F\cdot\vec v = \gamma m(\vec a\cdot\vec v) + \gamma^3m\dfrac{(\vec v\cdot\vec a)v^2}{c^2} = m(\vec v\cdot\vec a)\,\gamma\big(1 + \gamma^2\beta^2\big) = \gamma^3 m\,(\vec v\cdot\vec a)$ , using the identity $1 + \gamma^2\beta^2 = \gamma^2$ . So $\;\vec v\cdot\vec a = \dfrac{\vec F\cdot\vec v}{\gamma^3 m}$ . Substitute back into (2) and rearrange: $\;\gamma m\vec a = \vec F - \dfrac{(\vec F\cdot\vec v)}{c^2}\vec v$ , which is equation (3). Why: the identity $1 + \gamma^2\beta^2 = \gamma^2$ is doing all the work here and is worth memorising — it is just $\gamma^2(1-\beta^2) = 1$ rearranged, and it is the reason $\gamma^3$ collapses so neatly.

Step 5. Read off the aligned cases. If $\vec F \parallel \vec v$ , the bracket in (3) becomes $\vec F(1-\beta^2) = \vec F/\gamma^2$ , so $\vec a = \vec F/(\gamma^3 m)$ . If $\vec F \perp \vec v$ , the subtracted term vanishes and $\vec a = \vec F/(\gamma m)$ . Why: these are the only two cases in which force and acceleration are parallel — which is exactly why they are the only two that were ever given names. *Verified in verify_aq-sr-relativistic-force.py: $\mathrm{d}\gamma/\mathrm{d}t = \gamma^3(\vec v\cdot\vec a)/c^2$ symbolically; equation (2) from the product rule; the identity $1+\gamma^2\beta^2 = \gamma^2$ ; equation (3) as the exact inverse of (2) (composed both ways, symbolically, in 3D); and both aligned cases $\gamma^3m$ , $\gamma m$ .*

Go deeper · Derivation B · 2 minThe angle between force and acceleration, and the worst case

Derivation BThe angle between force and acceleration, and the worst case

Knowing that $\vec F$ and $\vec a$ disagree is worth something; knowing by how much is worth more. This is the article's owned structural result: it reduces the whole business to one statement about tangents, and produces a maximum misalignment in closed form, evaluable at any speed.

Step 1. Set up. Let $\vec v = v\hat{x}$ and let $\vec F$ make an angle $\theta_F$ with it: $\vec F = F(\cos\theta_F\,\hat x + \sin\theta_F\,\hat y)$ . Why: two dimensions suffice — the third has nothing to do, since equation (3) only ever singles out the direction $\vec v$ .

Step 2. Apply the inverse law (3). The subtracted term is $\dfrac{(\vec F\cdot\vec v)}{c^2}\vec v = F\cos\theta_F\,\beta^2\,\hat x$ , so $\;\gamma m\,\vec a = F\cos\theta_F(1-\beta^2)\,\hat x + F\sin\theta_F\,\hat y = \dfrac{F\cos\theta_F}{\gamma^2}\hat x + F\sin\theta_F\,\hat y$ . Why: this displays the mechanism nakedly. The $\hat y$ (transverse) component of $\vec F$ survives untouched; the $\hat x$ (longitudinal) component is divided by $\gamma^2$ . The acceleration is the force with its forward part suppressed.

Step 3. Take the ratio. $\;\tan\theta_a = \dfrac{a_y}{a_x} = \dfrac{F\sin\theta_F}{F\cos\theta_F/\gamma^2} = \gamma^2\tan\theta_F$ , so $\boxed{\tan\theta_a = \gamma^2\tan\theta_F}$ . Why: the mass has cancelled, the force magnitude has cancelled, and $\gamma^2$ is left standing alone. The misalignment depends on the speed and the push angle — and on nothing else at all.

Step 4. Put a number on it. At $\beta = 0.9$ : $\gamma = 2.294$ , $\gamma^2 = 5.263$ . Push at $\theta_F = 45^\circ$ , where $\tan\theta_F = 1$ : then $\tan\theta_a = 5.263$ and $\theta_a = \mathbf{79.24^\circ}$ . The acceleration lies $\mathbf{34.24^\circ}$ away from the force. Why: a third of a right angle, from an entirely ordinary force on an entirely ordinary particle. This is not a small correction.

Step 5. Maximise the disagreement. Maximise $\Delta(\theta_F) = \arctan(\gamma^2\tan\theta_F) - \theta_F$ . Setting $\mathrm{d}\Delta/\mathrm{d}\theta_F = 0$ gives $\gamma^2\sec^2\theta_F = 1 + \gamma^4\tan^2\theta_F$ , which reduces to $\tan^2\theta_F = 1/\gamma^2$ , i.e. $\boxed{\tan\theta_F = 1/\gamma}$ . There $\tan\theta_a = \gamma$ , so $\theta_a = \arctan\gamma$ , and since $\arctan\gamma + \arctan(1/\gamma) = 90^\circ$ : $\boxed{\Delta_{\max} = 2\arctan\gamma - 90^\circ}$ . Why: closed form, correct at both ends. At $\gamma = 1$ (Newton) it gives $0$ — force and acceleration always parallel. As $\gamma\to\infty$ it tends to $90^\circ$ : an ultra-relativistic particle can be made to accelerate almost perpendicular to the force pushing it.

Step 6. Evaluate at $\beta = 0.9$ . $\;\theta_F = \arctan(1/2.294) = \mathbf{23.55^\circ}$ , $\;\theta_a = \arctan(2.294) = \mathbf{66.45^\circ}$ , $\;\Delta_{\max} = \mathbf{42.90^\circ}$ . Why: the worst push is not the $45^\circ$ one — it is a shallower push at $23.55^\circ$ , which the $\gamma^2$ then levers up past $66^\circ$ . *Verified in verify_aq-sr-relativistic-force.py: $\tan\theta_a/\tan\theta_F = \gamma^2$ symbolically; the stationarity condition reduced to $\tan\theta_F = 1/\gamma$ ; the closed form $\Delta_{\max} = 2\arctan\gamma - 90^\circ$ checked against a brute-force numerical maximisation over $\theta_F$ at six values of $\beta$ ; and the printed numbers $\gamma = 2.294$ , $\gamma^2 = 5.263$ , $\gamma^3 = 12.07$ , $\theta_a = 79.24^\circ$ , $\Delta = 34.24^\circ$ , $\theta_F^{\max} = 23.55^\circ$ , $\Delta_{\max} = 42.90^\circ$ .*

Go deeper · Derivation C · 1 minThe work-energy theorem survives untouched, and a magnetic force does no work

Derivation CThe work-energy theorem survives untouched, and a magnetic force does no work

Given what happened to $\vec F = m\vec a$ , you would expect the work–energy theorem to acquire a $\gamma$ somewhere. It does not: $\vec F\cdot\vec v = \mathrm{d}E/\mathrm{d}t$ holds exactly, with no correction factor, provided $E$ is the total energy $\gamma mc^2$ . That robustness is why energy methods stay the most reliable way to solve relativistic problems — and it has an immediate corollary about magnets.

Step 1. Differentiate the energy. $\;\dfrac{\mathrm{d}E}{\mathrm{d}t} = mc^2\,\dfrac{\mathrm{d}\gamma}{\mathrm{d}t} = mc^2\cdot\gamma^3\dfrac{(\vec v\cdot\vec a)}{c^2} = \gamma^3 m\,(\vec v\cdot\vec a)$ , reusing Step 1 of the first derivation. Why: again, $m$ is not differentiated.

Step 2. Compare with the power delivered. Step 4 of the first derivation gave $\;\vec F\cdot\vec v = \gamma^3 m\,(\vec v\cdot\vec a)$ — the identical expression. Hence $\boxed{\vec F\cdot\vec v = \mathrm{d}E/\mathrm{d}t}$ . Why: the $\gamma^3$ that made the force law ugly is the same $\gamma^3$ that makes the energy law clean, and it cancels out of the comparison.

Step 3. Take the transverse case. If $\vec F \perp \vec v$ then $\vec F\cdot\vec v = 0$ , so $\mathrm{d}E/\mathrm{d}t = 0$ : the energy is constant, hence $\gamma$ is constant, hence the speed is constant. The force turns the velocity without changing its magnitude. Why: this is why a magnetic field can steer a beam. The magnetic force $q\vec v\times\vec B$ is always perpendicular to $\vec v$ , so it can never do work — true in Newtonian physics too, but with real teeth here: an accelerator's bending magnets and its RF cavities do genuinely different jobs and cannot substitute for one another.

Step 4. Take the longitudinal case, and watch the speed limit appear. With $\vec F \parallel \vec v$ and $F$ constant, $a = F/(\gamma^3m) \to 0$ as $\gamma\to\infty$ : the acceleration is quenched and the particle asymptotes to $c$ without reaching it — while the power $\vec F\cdot\vec v \to Fc$ stays finite and the energy grows without bound. Why: the energy is going somewhere, and it is not going into speed. $p = \gamma mv$ grows without limit while $v$ saturates. The speed limit is not imposed on the dynamics; it falls out of it. *Verified in verify_aq-sr-relativistic-force.py: $\mathrm{d}(\gamma mc^2)/\mathrm{d}t - \vec F\cdot\vec v = 0$ symbolically for arbitrary $\vec v(t)$ , $\vec a(t)$ in 3D; $\vec F\perp\vec v \Rightarrow \mathrm{d}|\vec v|/\mathrm{d}t = 0$ ; and $a_\parallel = F/(\gamma^3 m)\to 0$ as $\beta\to1$ at fixed $F$ .*

Figure

The two response coefficients. γ³ (longitudinal — speeding up) and γ (transverse — steering) plotted against β, with m the INVARIANT rest mass throughout. They are equal only at rest. By β = 0.9 the longitudinal coefficient is 12.07 and the transverse is 2.294 — a ratio of exactly γ² = 5.26. This gap is why an accelerator bends its beam with modest magnets and accelerates it with enormous RF cavities: steering stays cheap while speeding up becomes ruinous

Limiting cases

Case	Result	Meaning
$\beta \to 0$ , $\gamma \to 1$	$\vec F \to m\vec a$	Newton recovered. Both coefficients collapse to $m$ ; force and acceleration realign
$\vec F \parallel \vec v$ / $\vec F \perp \vec v$	$F = \gamma^3 m\,a$ / $F = \gamma m\,a$	stiff (speeding up) vs soft (steering). Transverse does no work
general $\vec F$	$\tan\theta_a = \gamma^2\tan\theta_F$	force and acceleration not parallel
worst case	$\Delta_{\max} = 2\arctan\gamma - 90^\circ$ at $\tan\theta_F = 1/\gamma$	$42.90^\circ$ at $\beta = 0.9$ ; $\to 90^\circ$ as $\gamma\to\infty$
$\beta \to 1$ , $F$ fixed and longitudinal	$a = F/(\gamma^3m) \to 0$ , but $\mathrm{d}E/\mathrm{d}t \to Fc$	the speed limit emerges — finite power, unbounded energy, all of it into momentum
stiffness ratio	$\gamma^3m / \gamma m = \gamma^2$	why accelerators bend cheaply and accelerate expensively

Common trapThe old names — 'longitudinal mass' and 'transverse mass' — are deprecated. The formulas are not.

You will meet $m_\parallel = \gamma^3 m$ and $m_\perp = \gamma m$ in older texts, called the longitudinal and transverse mass. The relations $F_\parallel = \gamma^3 m\,a_\parallel$ and $F_\perp = \gamma m\,a_\perp$ are exactly right and are used throughout this article. It is the naming that is wrong, and it is wrong in a way that misleads. A mass is a single scalar property of a body. These are two different numbers for the same body at the same instant, depending on which way you push — so they cannot both be "the mass", and neither is. Worse, the names suggest a general force can be handled by picking the appropriate mass, which is false: at a general angle no scalar $M$ satisfies $\vec F = M\vec a$ , because $\vec F$ and $\vec a$ are not even parallel. The two-mass vocabulary papers over exactly the phenomenon it was invented to describe. The modern statement keeps every formula and drops the vocabulary: $m$ is the invariant rest mass, one number, the same at all speeds; $\gamma^3$ and $\gamma$ are direction-dependent response coefficients belonging to the force law, not to the particle. Okun's 1989 Physics Today essay is the standard argument, and the community made the change. Read the old names historically, never as current usage.

Worked examples

1 · Why the LHC's dipoles are 8.33 tesla — the transverse coefficient, in a real machine

A proton of momentum $p$ and charge $e$ is bent in a circle of radius $\rho$ by a uniform magnetic field $B$ perpendicular to its motion. Find the relation between $B$ , $\rho$ and $p$ — and check it against the LHC's actual magnets.

Step 1 — identify. The magnetic force $\vec F = e\,\vec v\times\vec B$ is always perpendicular to $\vec v$ — the transverse case of equation (4): $F_\perp = \gamma m\,a_\perp$ , with centripetal acceleration $a_\perp = v^2/\rho$ .

Step 2 — apply. $\;evB = \gamma m\,v^2/\rho$ . Cancel a power of $v$ : $\;eB = \gamma m v/\rho$ . But $\gamma m v$ is exactly the relativistic momentum $p$ , so $\;\boxed{B\rho = p/e}$ — the magnetic rigidity.

Step 3 — notice what vanished. Neither $\gamma$ nor $m$ appears separately; only the combination $p = \gamma mv$ . A bending magnet measures momentum and nothing else. That is no coincidence: the transverse response coefficient is exactly $\gamma m$ , which is exactly what turns $mv$ into $p$ .

Step 4 — put the LHC in. In practical units $p\,[\mathrm{GeV}/c] = 0.299792458 \times B\,[\mathrm{T}] \times \rho\,[\mathrm{m}]$ . The LHC's main dipoles run at a nominal $B = 8.33\ \mathrm{T}$ with bending radius $\rho = 2803.95\ \mathrm{m}$ , giving $\;p = 0.299792458 \times 8.33 \times 2803.95 = \mathbf{7002\ \mathrm{GeV}/c} = \mathbf{7.00\ \mathrm{TeV}/c}$ — the LHC's design beam momentum, from one line of relativistic dynamics.

Step 5 — the boundary number. To double the beam energy, double $B\rho$ : double the field or double the ring. $8.33\ \mathrm{T}$ is already at the practical ceiling of niobium–titanium superconductor, so a higher-energy machine needs a new conductor or a much larger tunnel. That engineering fork is a direct consequence of $B\rho = p/e$ .

2 · Push a proton at 45 degrees and watch it go somewhere else

A proton moves at $\beta = 0.9$ along $x$ . A force of magnitude $F$ is applied at $45^\circ$ to its velocity. Find the direction of the acceleration. Then find the push angle that maximises the disagreement, and the size of that disagreement.

Step 1 — identify. Equation (3), and only equation (3). Both $F$ and $m$ cancel — the answer is a direction, depending on nothing but $\beta$ and $\theta_F$ .

Step 2 — get the Lorentz factors. $\gamma = 1/\sqrt{1 - 0.81} = 1/\sqrt{0.19} = \mathbf{2.294}$ , so $\gamma^2 = \mathbf{5.263}$ and $\gamma^3 = \mathbf{12.07}$ . The story is already visible: speeding this proton up costs $12.07\,m$ per unit acceleration, steering it only $2.294\,m$ — a factor $\gamma^2 = 5.26$ .

Step 3 — apply the tangent rule. $\tan\theta_a = \gamma^2\tan45^\circ = 5.263$ , so $\theta_a = \arctan(5.263) = \mathbf{79.24^\circ}$ — a misalignment of $\mathbf{34.24^\circ}$ , swung toward the transverse direction, because that is where the particle is willing to move.

Step 4 — find the worst case. Maximum misalignment at $\tan\theta_F = 1/\gamma = 0.4360$ , i.e. $\theta_F = \mathbf{23.55^\circ}$ . There $\tan\theta_a = \gamma = 2.294$ , so $\theta_a = \arctan(2.294) = \mathbf{66.45^\circ}$ .

Step 5 — the boundary number. $\Delta_{\max} = 66.45^\circ - 23.55^\circ = \mathbf{42.90^\circ}$ , which is $2\arctan(2.294) - 90^\circ$ exactly. Nearly half a right angle between where you push and where the proton goes — at a speed a modest cyclotron reaches. Since $\Delta_{\max} \to 90^\circ$ as $\gamma \to \infty$ , at LHC energies a suitably angled force produces an acceleration essentially perpendicular to itself. The Newtonian picture of a body obediently accelerating along the push is not a poor approximation here — it is the wrong shape of statement.

Physics of the portrait

Newton's second law is the most famous equation in physics that most people misquote. The Principia does not say $\vec F = m\vec a$ . It says that the change of motion is proportional to the impressed force — and by "motion" (motus) Newton meant what we call momentum. Written his way, the law needed no amendment when relativity arrived.

Origins · 1 minNewton's second law and how this was discovered

Newton's second law

1687–present · Principia, Book I · Planck's 1906 relativistic recasting · every accelerator on Earth

The second law appears in the Philosophiæ Naturalis Principia Mathematica (1687) as Lex II: "Mutationem motus proportionalem esse vi motrici impressae" — the change of motion is proportional to the motive force impressed. Newton's quantity of motion is mass times velocity, so his statement is about $\mathrm{d}\vec p/\mathrm{d}t$ , not $m\vec a$ ; the two coincide only because Newtonian mass is constant, and generations of textbooks quietly took the derived form for the primitive one. Relativity punished that substitution and vindicated the original. When Planck founded relativistic dynamics in 1906, his decisive move was simply to keep Newton's form and let the momentum become $\gamma m\vec v$ — from which the transverse coefficient came out as $\gamma m$ , correcting the $\gamma^2 m$ Einstein had obtained in 1905 from a different force definition. The law in Newton's own words has survived a change in the structure of spacetime without alteration — a claim almost nothing else in the Principia can make.

History

1687

— Newton's Principia states the second law (Lex II) as a proportionality between impressed force and the change in the quantity of motion — that is, as a statement about momentum, not about acceleration.

1904

— Lorentz's electron theory carries velocity-dependent longitudinal and transverse electromagnetic masses, with the values γ³m and γm.

1905

— Einstein, in the final section of "Zur Elektrodynamik bewegter Körper", obtains a longitudinal mass γ³m but a transverse mass γ²m — a consequence of the force definition he adopts, not an error of physics.

1906

— Planck founds relativistic dynamics by keeping Newton's F = dp/dt with p = γmv; the transverse coefficient comes out as γm, and this is the value used ever since.

1908

— Bucherer's beta-ray deflection experiments are reported as favouring the Lorentz-Einstein velocity dependence over Abraham's rigid-sphere electron; later analysis showed the data were not precise enough to discriminate.

1989

— Okun's "The concept of mass" (Physics Today 42(6), 31) argues for retiring "longitudinal mass", "transverse mass" and "relativistic mass" together: m is the invariant rest mass, and the gamma factors belong to the response, not the particle.

Going further

Texts. Rindler, Introduction to Special Relativity, 2nd ed., §37 · Griffiths, Introduction to Electrodynamics, 4th ed., §12.2.3 · Morin, Introduction to Classical Mechanics, ch. 12 · Okun, "The concept of mass", Physics Today 42(6), 31 (1989).

Provenance. Signature $(+,-,-,-)$ ; $m$ is the invariant rest mass throughout, and no relativistic mass appears anywhere — the "longitudinal/transverse mass" names are quoted only to deprecate them. Everything above is recomputed in verify_aq-sr-relativistic-force.py, which parses this article's own printed digits: $\mathrm{d}\gamma/\mathrm{d}t = \gamma^3(\vec v\cdot\vec a)/c^2$ ; equation (2) and equation (3) as its exact 3D inverse; the identity $1+\gamma^2\beta^2=\gamma^2$ ; the coefficients $\gamma^3m$ , $\gamma m$ ; the work–energy theorem; the owned $\tan\theta_a = \gamma^2\tan\theta_F$ and $\Delta_{\max} = 2\arctan\gamma - 90^\circ$ (against brute-force numerical maximisation at six speeds); and the printed $\gamma = 2.294$ , $\gamma^2 = 5.263$ , $\gamma^3 = 12.07$ , $\theta_a = 79.24^\circ$ , $\Delta = 34.24^\circ$ , $\theta_F^{\max} = 23.55^\circ$ , $\Delta_{\max} = 42.90^\circ$ , and the LHC rigidity $0.299792458 \times 8.33 \times 2803.95 = 7002\ \mathrm{GeV}/c$ .

Scope. The four-force $K^\mu = \mathrm{d}p^\mu/\mathrm{d}\tau$ — the covariant version of all this — is not derived here; it belongs to The Four-Force, which owns the orthogonality $K\cdot U = 0$ and the invariant $K\cdot K$ . Hyperbolic motion under a constant proper force belongs to Relativistic Uniform Acceleration; the electromagnetic four-force $K^\mu = (q/c)F^{\mu\nu}U_\nu$ to the relativistic-electrodynamics band. This article owns the three-force law and the angle between $\vec F$ and $\vec a$ .

Builds on. Relativistic Momentum · Relativistic Energy · Four-Velocity · Invariant Mass. Leads to. The Four-Force · Four-Acceleration · Conservation of Four-Momentum · Relativistic Uniform Acceleration.

Is Newton's second law wrong in relativity?

No — the version Newton wrote is exactly right; the version we teach is wrong. The Principia states the law as a proportionality between impressed force and the change in the "quantity of motion" (momentum), i.e. $\vec F = \mathrm{d}\vec p/\mathrm{d}t$ , and that survives relativity untouched. What fails is $\vec F = m\vec a$ — a derived form valid only when $m$ is constant and $\vec p = m\vec v$ . General lesson: when a theory is extended, the primitive statement survives and the convenient specialisation does not.

If gamma-cubed and gamma are not masses, what are they?

Coefficients in a linear response. "How much force does a unit of acceleration cost, in this direction, at this speed?" — $\gamma^3 m$ along the motion, $\gamma m$ across it. That is a property of the force law at a particular velocity, not of the particle. The cleanest proof they are not masses: they are two numbers for one body at one instant, and at a general angle neither works — no scalar relates $\vec F$ to $\vec a$ , because the two are not parallel. A quantity that fails to exist in the general case was never a mass.

So does a particle get harder to accelerate as it speeds up, or not?

Yes — but "harder" is direction-dependent, which is exactly what the word "mass" obscures. Along the motion the cost per unit acceleration grows as $\gamma^3$ ; across it, only as $\gamma$ . At $\beta = 0.9$ those are $12.07\,m$ and $2.294\,m$ . Saying "the mass increases" collapses two numbers into one and cannot explain why the particle accelerates in a different direction from the push; saying "the response coefficients are $\gamma^3$ and $\gamma$ " keeps all the information and predicts the $34.24^\circ$ misalignment.

Can the force ever be perpendicular to the acceleration?

Not exactly perpendicular, but arbitrarily close. $\Delta = \arctan(\gamma^2\tan\theta_F) - \theta_F$ is maximised at $\tan\theta_F = 1/\gamma$ , giving $\Delta_{\max} = 2\arctan\gamma - 90^\circ$ — which tends to $90^\circ$ as $\gamma\to\infty$ but never reaches it. At $\beta = 0.9$ it is $42.90^\circ$ ; at $\gamma = 100$ , $88.9^\circ$ . For an ultra-relativistic particle the acceleration is very nearly perpendicular to a suitably angled force — one way of saying such a particle will do almost anything except go faster.

Why is a magnetic force so useful in an accelerator if it does no work?

Because doing no work is precisely the job. An accelerator has two problems: keep the beam on a closed path, and raise its energy. A magnetic force solves the first without touching the second — $\vec F\cdot\vec v = 0$ gives $\mathrm{d}E/\mathrm{d}t = 0$ exactly, so you can bend a 7 TeV proton through $360^\circ$ and hand it back with the same 7 TeV. Energy is added separately, by RF cavities acting longitudinally. That clean separation exists only because a transverse force is guaranteed energy-neutral — which is equation (5).

Why does the speed of light come out as a limit here without being assumed?

Because the longitudinal response coefficient diverges. Apply a constant force along the motion: $a = F/(\gamma^3 m) \to 0$ as $\gamma\to\infty$ . The particle's willingness to gain speed evaporates faster than the force can supply it, and $v$ asymptotes to $c$ in infinite time. Note what is not happening: the energy is not saturating (it grows without bound), the power is not vanishing ( $\vec F\cdot\vec v \to Fc$ ), and nothing has been forbidden. The work is still being done — it goes into momentum, which is unbounded, rather than into speed, which is not. The limit is an output of the dynamics, not an input.