Velocity-Dependent and Generalized Potentials

Every tokamak and stellarator simulation running in 2026 leans on this page's central object. Gyrokinetic codes — the workhorses of fusion design at ITER and the private fusion ventures — follow billions of charged-particle guiding centers through magnetic geometries for millions of timesteps, and the quantity they monitor to certify the physics hasn't drifted is the conserved canonical momentum: in an axisymmetric tokamak the canonical angular momentum $p_\varphi = mRv_\varphi + qRA_\varphi$ is an exact invariant, and its numerical conservation is both a confinement theorem (it caps how far particles can wander radially) and the standard regression test of the integrator. Modern "structure-preserving" integrators are built variationally — discretizing the Lagrangian (4) itself rather than the force law — precisely so that the discrete dynamics inherits a conserved canonical momentum by construction. The deeper legacy is minimal coupling. The prescription $\mathbf{p} \to \mathbf{p} - q\mathbf{A}$ that equation (8) reads off classically is, essentially unmodified, how the Standard Model couples every charged field to the photon — promoted to a local gauge principle, the derivative-replacement rule of QED. And the once-philosophical question this page raises — is $\mathbf{A}$ real, or mere bookkeeping? — was settled experimentally: electrons diffracting around a solenoid acquire the Aharonov–Bohm phase $q\Phi/\hbar$ through regions where $\mathbf{B} = 0$ but $\mathbf{A} \neq 0$, a phase shift measured to exquisite precision in electron holography and now engineered deliberately in mesoscopic rings and flux qubits. The velocity slot Lagrange left open turned out to be where gauge theory lives.

Watch two forces act on a moving particle and you will see two different characters at work. A gravitational or spring force is a landscape force: it points downhill in a fixed terrain $V(\mathbf{r})$, does work as the particle descends, and cares nothing about the particle's velocity. A magnetic force is something else entirely. It is always exactly perpendicular to the velocity, so it can bend a trajectory into a circle forever without doing a single joule of work — a proton in a cyclotron travels kilometers of arc length and its speed never changes. Stop the proton and the force evaporates. No landscape behaves like this: a gradient force at a point is one fixed vector, while the magnetic force at that same point is a whole rule, a linear map that takes your velocity in and hands a sideways push back.

So the taxonomy of forces splits three ways, and the split is the intellectual spine of this article. Landscape forces come from $V(q)$: conservative, work-doing, the textbook case. Gyroscopic forces are velocity-dependent but work-free — magnetic forces, and their mechanical twin the Coriolis force. These, it turns out, still fit inside Hamilton's principle, provided we let the potential borrow the velocity slot. Dissipative forces — friction, drag — are velocity-dependent and energy-draining, and they are permanently outside: no function $U$ of any kind generates them, as we will prove. The skill this page teaches is telling the second family from the third, because they look superficially alike ("force depends on velocity") and behave nothing alike. Why should a velocity-dependent potential be possible at all? Look at the operator the Euler–Lagrange equations already apply to every term of $L$: $\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial}{\partial\dot q_j} - \frac{\partial}{\partial q_j}$. Feed it a position-only $-V(q)$ and the first slot returns zero — you recover $-\partial V/\partial q$, the old rule, as a special case. But nothing stops you from feeding it a $U$ that lives in both slots. The operator itself defines the widest class of forces the Lagrangian formalism can host; the definition of a generalized potential is simply the demand that the force match what this operator produces. Mechanics did not have to be modified to admit magnetism — the door was always this wide, and electromagnetism was the first force to walk through it.

Start from where d'Alembert's principle left the equations of motion: $\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial T}{\partial\dot q_j} - \frac{\partial T}{\partial q_j} = Q_j$, with the generalized force $Q_j$ still completely general. The question is which $Q_j$ can be absorbed into a Lagrangian. The answer is the defining equation of this article:

where: $Q_j$ — the generalized force conjugate to $q_j$; $U(q,\dot q,t)$ — the generalized (velocity-dependent) potential. Read the structure: the second term is the familiar "downhill" pull, while the first term is new — it lets whatever $U$ stores in its velocity slot re-emerge as force through the time derivative. If $U$ happens not to depend on $\dot q$, the first term dies and (1) collapses to $Q_j = -\partial U/\partial q_j$: the ordinary conservative rule is the zero-velocity-dependence corner of this larger space. Forces expressible as (1) are called monogenic — generated by a single function — and they are exactly the forces Hamilton's principle can digest. Absorbing (1) into the equations of motion takes one line. Substitute it into d'Alembert's form and move both $U$ terms across:

where: the combination $T - U$ is the Lagrangian, $L = T - U$. This is why the recipe "$L = T - V$" survives verbatim with $V$ promoted to $U$: the Euler–Lagrange operator is linear, and (1) was designed so that the $U$ terms slot in beside the $T$ terms with the right signs. Every consequence of the Lagrangian formalism — cyclic coordinates and their conserved momenta, the energy function, the passage to Hamiltonian mechanics — now applies to velocity-dependent forces at no extra charge. The only casualties are our Newtonian instincts about what "momentum" and "energy" mean, and auditing those instincts is the real work of the rest of this page.

Everything abstract above becomes concrete in one physical system: a point charge $q$ moving through electromagnetic fields described by the scalar potential $\phi(\mathbf{r},t)$ and the vector potential $\mathbf{A}(\mathbf{r},t)$, with $\mathbf{E} = -\nabla\phi - \partial\mathbf{A}/\partial t$ and $\mathbf{B} = \nabla\times\mathbf{A}$. The claim — and it is the single most important example in this corner of mechanics — is that the generalized potential

where: $\mathbf{v} = \dot{\mathbf{r}}$ is the particle velocity, generates the entire Lorentz force through rule (1). The first term is ordinary electrostatic potential energy. The second is the novelty: a coupling linear in velocity, projecting $\mathbf{v}$ onto the local vector potential. Linearity is not an accident — a term linear in $\dot q$ is precisely the kind that produces work-free, direction-bending forces, because (as the energy function will confirm) it contributes force without contributing energy. The Lagrangian is then

where: $m$ is the particle's mass and $v^2 = \dot{\mathbf{r}}\cdot\dot{\mathbf{r}}$. Note what this Lagrangian is not: it is not "kinetic minus potential energy," because $-q\mathbf{v}\cdot\mathbf{A}$ is not an energy stored anywhere — it is a coupling term whose job is to encode $\mathbf{B}$ through the curl of $\mathbf{A}$. The target it must reproduce is the force Lorentz distilled in 1892 from the electron theory of matter:

where: the electric term accelerates and decelerates, while the magnetic term only steers. That a single scalar function (3) contains the full vector structure of (5) — including the cross product — is the small miracle the next derivation verifies line by line.

Step 1 of the derivation quietly delivered the most consequential formula on this page. The momentum conjugate to position — the object the Lagrangian formalism actually tracks — is

where: $m\mathbf{v}$ is the kinetic momentum (what a radar gun measures) and $q\mathbf{A}$ is the field contribution, present purely because $U$ depends on velocity. The sum is the canonical momentum. It is the canonical, not the kinetic, momentum that obeys the conservation theorems: if $\phi$ and $\mathbf{A}$ do not depend on $x$, then $x$ is cyclic and $p_x = m\dot x + qA_x$ is a constant of the motion, while $m\dot x$ alone oscillates freely. A charged particle can trade momentum with the field and back, and only the combined ledger (6) balances. Worked example 2 turns this single conservation law into a complete solution of a penetration problem that would otherwise need the full trajectory. There is a satisfying geometric way to see how different the two momenta are. Put a proton on a circular cyclotron orbit in a uniform $\mathbf{B}$ and choose the symmetric gauge $\mathbf{A} = \tfrac12\mathbf{B}\times\mathbf{r}$ centered on the orbit. Then $q\mathbf{A}$ is everywhere antiparallel to $m\mathbf{v}$ with exactly half its magnitude, so the canonical momentum $\mathbf{p} = m\mathbf{v} + q\mathbf{A}$ points along the velocity but with half the kinetic value — a vector you would never guess by staring at the motion, because half of it lives in the field bookkeeping.

Once the canonical momentum exists, the whole Hamiltonian edifice follows by expressing energy in terms of it. Inverting (6) for the velocity, $m\mathbf{v} = \mathbf{p} - q\mathbf{A}$, and performing the Legendre transform gives

where: the recipe visible here — take the free Hamiltonian and replace $\mathbf{p} \to \mathbf{p} - q\mathbf{A}$, then add $q\phi$ — is minimal coupling, the exact prescription by which electromagnetic interactions enter quantum mechanics, where $\mathbf{p}$ becomes the operator $-i\hbar\nabla$. Landau levels, the Aharonov–Bohm phase, flux quantization in superconductors: each is equation (8) doing quantum work. Classical mechanics wrote the coupling; quantum mechanics inherited it unchanged. One more ledger needs auditing: energy. The energy function $h = \sum_j \dot q_j\,\partial L/\partial\dot q_j - L$ evaluates, for the Lagrangian (4), to

where: the $q\mathbf{v}\cdot\mathbf{A}$ coupling has cancelled out entirely — a term linear in velocity contributes $\dot q\,\partial_{\dot q}$ of itself exactly once and appears in $L$ exactly once, and the two occurrences subtract. This is the formal fingerprint of a gyroscopic force: it steers the dynamics (through the equations of motion) while contributing nothing to the conserved energy (the magnetic field does no work). Only the scalar potential stores energy.

Which object answers which question

The potentials $\phi$ and $\mathbf{A}$ are not measurable; the fields $\mathbf{E}$ and $\mathbf{B}$ are. For any smooth function $\Lambda(\mathbf{r},t)$, the replacement

where: $\Lambda$ is called the gauge function — leaves $\mathbf{E}$ and $\mathbf{B}$ untouched, because the curl of a gradient and the combination $-\nabla(-\partial_t\Lambda) - \partial_t(\nabla\Lambda)$ both vanish. The Lagrangian (4) is not invariant under (7) — but it changes in the one way that costs nothing, as the next derivation shows. This is the physicist's first encounter with a theme that dominates modern field theory: the description carries redundancy, and physical content lives in what the redundancy cannot touch. The general theory of when two Lagrangians describe identical dynamics has its own article; here we verify the electromagnetic instance explicitly.

The generalized potential admits magnetism; does it admit friction? Both are velocity-dependent forces, and a beginner's reasonable guess is that a clever enough $U$ handles drag too. The guess is provably wrong, and the proof is short enough to be beautiful.

1 · A proton's two momenta (confidence builder)

A proton ($m = 1.673\times10^{-27}\ \mathrm{kg}$, $q = 1.602\times10^{-19}\ \mathrm{C}$) moves at $v = 2.0\times10^{5}\ \mathrm{m/s}$ perpendicular to a uniform field $B = 0.50\ \mathrm{T}$. Find the cyclotron frequency, the orbit radius, and — in the symmetric gauge centered on the orbit — the kinetic, field, and canonical momenta.

Step 1. Cyclotron frequency. The magnetic force supplies the centripetal acceleration, $qvB = mv^2/r$, giving the speed-independent angular frequency

equivalently $f = \omega_c/2\pi = 7.6\ \mathrm{MHz}$ — the dial setting of an early cyclotron's oscillator.

Step 2. Orbit radius, from $r = v/\omega_c$:

a tabletop circle, traversed once every $T = 2\pi/\omega_c = 131\ \mathrm{ns}$.

Step 3. Kinetic momentum: $mv = (1.673\times10^{-27})(2.0\times10^{5}) = \mathbf{3.35\times10^{-22}\ kg\,m/s}$, tangent to the orbit.

Step 4. Field momentum. In the symmetric gauge $\mathbf{A} = \tfrac12\mathbf{B}\times\mathbf{r}$ with origin at the orbit center, $|\mathbf{A}| = \tfrac12 Br$ on the orbit, so

— exactly half the kinetic momentum, and directed against it (check the geometry: for a positive charge circling in the sense the magnetic force enforces, $\tfrac12\mathbf{B}\times\mathbf{r}$ is antiparallel to $\mathbf{v}$).

Step 5. Canonical momentum: $p = mv - \tfrac12 mv = \mathbf{1.67\times10^{-22}\ kg\,m/s}$, tangent to the orbit with half the kinetic magnitude — the configuration drawn in the second figure. The moral is not the specific factor $\tfrac12$ (it is gauge- and origin-dependent) but the vivid demonstration that canonical and kinetic momentum are different physical ledgers that happen to coincide only when $\mathbf{A} = 0$.

2 · How deep does a proton punch into a magnetic wall? (exam level)

A proton travelling at $v_0 = 3.0\times10^{5}\ \mathrm{m/s}$ in the $+x$ direction crosses $x=0$ into a region of uniform field $\mathbf{B} = 0.20\ \mathrm{T}\,\hat{\mathbf{z}}$ filling $x>0$. Using canonical momentum conservation — no trajectory integration — find the maximum penetration depth and the time spent inside the field region before the turning point.

Step 1. Choose a gauge adapted to the symmetry. With $\mathbf{A} = (0,\ Bx,\ 0)$ we get $\nabla\times\mathbf{A} = B\hat{\mathbf{z}}$ ✓, and nothing in the Lagrangian depends on $y$: the coordinate $y$ is cyclic.

Step 2. Write the conserved canonical momentum, eq. (6), for the cyclic coordinate:

At entry ($x=0$, $\dot y = 0$): $p_y = 0$. So everywhere inside, $m\dot y = -qBx$ — the sideways kinetic momentum is enslaved to the penetration depth. (Note $m\dot y$ changes freely; $p_y$ stays exactly zero. That is the warning box, live.)

Step 3. Add the work-free property. The magnetic force never changes speed, so $\dot x^2 + \dot y^2 = v_0^2$ at all times. The turning point of penetration is where $\dot x = 0$, hence $|\dot y| = v_0$ there.

Step 4. Combine: at the turning point, $m v_0 = qB\,x_{\max}$, so

— which is precisely the Larmor radius, as it must be: the orbit inside the field is a circular arc of radius $mv_0/qB$, entering perpendicular to the boundary.

Step 5. Time inside. From entry to maximum depth the velocity direction rotates by a quarter turn (from $+x$ to $-y$ for this charge sign), at the uniform rate $\omega_c = qB/m = 1.92\times10^{7}\ \mathrm{rad/s}$:

Step 6. Audit. The two conservation laws (canonical $p_y$, kinetic energy) replaced the entire differential-equation solve; the geometry check ($x_{\max}$ = Larmor radius, quarter-period timing) confirms both. This two-line style — cyclic coordinate, conserved canonical momentum, work-free constraint — is the standard opening move for charged-particle problems in plasma physics.

References. Goldstein, Classical Mechanics, 3rd ed., §1.5 (velocity-dependent potentials) and §1.6 · Landau & Lifshitz, Mechanics (Vol. 1), §17; and Classical Theory of Fields, §16 · Taylor, Classical Mechanics, §7.9 · Morin, Introduction to Classical Mechanics, Ch. 6

Provenance. Equations and the Lorentz-force derivation follow Goldstein, Classical Mechanics, 3rd ed., §1.5; energy function and gauge structure cross-checked against Landau–Lifshitz Mechanics §17 and Classical Theory of Fields §16. A verifier should recompute: (1) the cancellation of $\dot x\,\partial_x A_x$ in the EL equation; (2) the curl identification $(\mathbf{v}\times\mathbf{B})_x=\dot y B_z-\dot z B_y$; (3) that $\delta L=q\,\mathrm{d}\Lambda/\mathrm{d}t$ under a gauge transformation; (4) that $h=\tfrac12 mv^2+q\phi$ with the $\mathbf{v}\cdot\mathbf{A}$ term absent. Cyclotron frequency $\omega_c=qB/m$ is standard. Worked-example numbers on this page ($\omega_c = 4.79\times10^{7}\ \mathrm{rad/s}$, $r = 4.2\ \mathrm{mm}$, $qA = \tfrac12 mv = 1.67\times10^{-22}\ \mathrm{kg\,m/s}$; $x_{\max} = 1.57\ \mathrm{cm}$, $t = 82\ \mathrm{ns}$) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: The Lagrangian Function · The Euler-Lagrange Equations · Generalized Forces and Generalized Momenta · d'Alembert's Principle. Leads to: Cyclic Coordinates and Conservation of Conjugate Momenta · The Energy Function and Jacobi Integral · Generalized Forces and Generalized Momenta.