The Energy Function and Jacobi Integral

Between the Earth and the Moon there is a gate. Not a physical structure — an energetic one: a narrow neck in an invisible surface that a spacecraft simply cannot cross unless a single number, computed once at engine cutoff, is small enough. Fire your trans-lunar burn a hundred metres per second too gently and the gate stays sealed; the Moon remains on the far side of a forbidden region no amount of clever steering can enter. That number is the Jacobi integral, and the strangest thing about it is what it is not: it is not the energy. In the rotating frame where the Earth and Moon stand still, the mechanical energy $T+V$ is not conserved at all — yet this other quantity, built from the Lagrangian by a fixed recipe, sits perfectly constant for the entire mission. The recipe is the energy function $h = \sum_i \dot q_i\,\partial L/\partial\dot q_i - L$ , and this article is about the two separate questions it forces you to keep apart. First: is $h$ conserved? That depends only on whether the Lagrangian carries an explicit clock — $\mathrm{d}h/\mathrm{d}t = -\partial L/\partial t$ . Second: is $h$ the energy $T+V$ ? That depends on something entirely different — whether the coordinates connect to inertial space without time creeping into the transformation. The two switches flip independently, and every famous "paradox" about energy in rotating frames, cranked wires, and moving supports is someone conflating them. Master the distinction and you own the cleanest conservation law in Lagrangian mechanics — the one that guides real spacecraft today.

Key takeaways

The energy function is $h(q,\dot q,t) = \sum_i \dot q_i\,\partial L/\partial\dot q_i - L$ — a fixed recipe applied to any Lagrangian, engineered so the Euler–Lagrange equations cancel every implicit time dependence.
Its evolution law is $\mathrm{d}h/\mathrm{d}t = -\partial L/\partial t$ : $h$ is conserved exactly when the Lagrangian has no explicit time dependence.
Whether $h$ equals the mechanical energy $T+V$ is a separate question: it requires the kinetic energy to be purely quadratic in the velocities ( $T = T_2$ ), which holds when the coordinate transformation to inertial space is time-independent (scleronomic).
With moving constraints or rotating frames, $T = T_2 + T_1 + T_0$ and the general result is $h = T_2 - T_0 + V$ — conserved, but not the energy.
The classic payoff is the Jacobi integral of the circular restricted three-body problem: conserved in the rotating frame while the inertial energy is not, it bounds all possible motion through zero-velocity surfaces.
Numerically $h$ equals the Hamiltonian; the Legendre transform to $H(q,p,t)$ changes the variables, not the value.
Key idea: $h=\sum_i\dot q_i\,p_i - L$ is conserved iff $L$ has no explicit time dependence; it equals $T+V$ only when, in addition, the kinetic energy is homogeneous quadratic in the velocities (time-independent constraints).

Where this lives today

The Jacobi integral is working infrastructure. Every libration-point mission — JWST holding station around Sun–Earth $L_2$ , the Artemis-era Gateway riding its near-rectilinear halo orbit around Earth–Moon $L_2$ — is designed on CR3BP energy surfaces first, with the Jacobi constant selecting which families of halo and Lyapunov orbits exist and how much station-keeping they cost. Low-energy transfers thread the $L_1$ / $L_2$ necks at Jacobi constants barely below critical, trading months of flight time for hundreds of metres per second of fuel; trajectory designers speak of "opening the gate" in exactly the sense of worked example 2, and mission-design software exposes $C_J$ as a first-class plotting variable next to $\Delta v$ . The same integral organizes structure far beyond spacecraft. In galactic dynamics, a star cluster orbiting a galaxy is a restricted three-body problem writ large: the cluster's tidal (Jacobi) radius is set by the $L_1$ / $L_2$ zero-velocity surfaces, and stars that leak across the gates form the leading and trailing tidal streams that the Gaia catalogue now maps in exquisite detail across the Milky Way halo. Fitting those streams — today a standard technique for weighing the Galaxy's dark matter — begins with the same equation (9) you used above, with a galaxy where the Earth was and a globular cluster where the spacecraft sat.

The physical picture

Every system answers two independent questions, and the entire content of this article is that they are different questions. Question one: does the Lagrangian carry an explicit clock — does $t$ appear anywhere other than inside $q(t)$ and $\dot q(t)$ ? If not, the number $h$ built by the recipe above cannot change: nothing in the dynamics distinguishes this instant from the next, so the quantity that tracks "the ticking of the clock" has nothing to track. Question two: do the generalized coordinates connect to inertial space through a frozen map $\mathbf r_a(q)$ , or does the map itself move — a crank turning, a support oscillating, a frame rotating? A frozen map keeps the kinetic energy purely quadratic in the velocities and $h$ collapses to the familiar $T+V$ . A moving map injects linear and velocity-independent pieces into $T$ , and $h$ becomes something else — still perfectly lawful, still possibly conserved, but no longer the number a lab-frame calorimeter would report.

Figure

Two independent switches decide what the energy function is and does

The picture to hold onto for the moving-map case is a bead threaded on a straight horizontal wire that a motor cranks at fixed angular velocity $\omega$ . As the bead drifts outward the motor must torque harder and harder to keep the wire's rotation rate fixed, pumping energy into the bead: the lab-frame energy $E$ climbs relentlessly. And yet the Lagrangian, written in the natural coordinate $r$ along the wire, contains no explicit $t$ at all — so something is conserved. That something is $h = \tfrac12 m\dot r^2 - \tfrac12 m\omega^2 r^2$ : kinetic-looking term minus the centrifugal term. The conserved quantity has a sign flip exactly where the naive energy has a plus, and the gap between the two is precisely the work the motor feeds in.

Figure

The bead on a cranked wire — lab energy climbs while the energy function stays flat

Finally, keep one bookkeeping fact in view: $h$ is numerically the same object as the Hamiltonian. It is the Hamiltonian wearing velocity clothes — expressed in $(q, \dot q)$ rather than $(q, p)$ . The Legendre transform that produces $H(q,p,t)$ , treated fully in the Hamiltonian formalism, changes which variables are independent, never the value. Everything proved about $h$ here transfers to $H$ wholesale.

The recipe and its master equation

Start from any Lagrangian $L(q, \dot q, t)$ with generalized momenta $p_i = \partial L/\partial\dot q_i$ (their anatomy is the business of generalized forces and momenta). The energy function is the combination

h(q,\dot q,t) \;=\; \sum_i \dot q_i\,\frac{\partial L}{\partial \dot q_i} \;-\; L \qquad (1)

where: $h$ — the energy function (called the Jacobi integral when it is conserved); the sum runs over all $n$ generalized velocities. The combination looks arbitrary until you watch it under differentiation: it is engineered so that every way $h$ can change through the motion itself cancels by the equations of motion, leaving only the explicit ticking of the clock. For the simplest case $L = \tfrac12 m\dot x^2 - V(x)$ the recipe gives $h = m\dot x^2 - (\tfrac12 m\dot x^2 - V) = \tfrac12 m\dot x^2 + V$ — the energy, with the sign of $V$ flipped back to plus by the subtraction.

Derivation AThe Master Equation for the Energy Function

We are deriving the single most useful identity of this article: along any actual trajectory, $\mathrm{d}h/\mathrm{d}t = -\partial L/\partial t$ — the energy function changes only through the explicit time dependence of the Lagrangian. The strategy is direct: differentiate the definition (1) with the product rule, expand $\mathrm{d}L/\mathrm{d}t$ with the chain rule, and watch two separate cancellations do the work — the $\ddot q$ terms cancel identically (pure algebra, no physics), while the $\dot q$ terms cancel only on-shell, by the Euler–Lagrange equations. That second cancellation is where the physics enters, so keep the two kinds of cancellation distinct: the identity is a statement about solutions, not about arbitrary curves.

Step 1. Differentiate the definition of $h$ totally with respect to time, using the product rule on each $\dot q_i\,\partial L/\partial\dot q_i$ :

\frac{\mathrm{d}h}{\mathrm{d}t} = \sum_i\left(\ddot q_i\,\frac{\partial L}{\partial\dot q_i} + \dot q_i\,\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial\dot q_i}\right) - \frac{\mathrm{d}L}{\mathrm{d}t}

Step 2. Expand the last term with the chain rule — $L$ changes because $q$ , $\dot q$ , and possibly the clock all advance:

\frac{\mathrm{d}L}{\mathrm{d}t} = \sum_i\left(\frac{\partial L}{\partial q_i}\,\dot q_i + \frac{\partial L}{\partial\dot q_i}\,\ddot q_i\right) + \frac{\partial L}{\partial t}

Why the distinction matters: $\mathrm{d}L/\mathrm{d}t$ is the change of $L$ along the trajectory; $\partial L/\partial t$ is the change of $L$ at frozen $(q,\dot q)$ — the part contributed by the clock alone. Conflating them is the single most common error in this derivation.

Step 3. Subtract. The $\ddot q_i\,\partial L/\partial\dot q_i$ terms cancel identically — no equations of motion needed — leaving

\frac{\mathrm{d}h}{\mathrm{d}t} = \sum_i \dot q_i\left(\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial\dot q_i} - \frac{\partial L}{\partial q_i}\right) - \frac{\partial L}{\partial t}

Step 4. Now impose that the trajectory is a solution. Each bracket is exactly the left side of an Euler–Lagrange equation and vanishes on-shell:

\frac{\mathrm{d}h}{\mathrm{d}t} = -\,\frac{\partial L}{\partial t} \qquad (2)

where: the right side is evaluated holding $q$ and $\dot q$ fixed. Why this is a conservation theorem: if $t$ never appears explicitly in $L$ , the right side is identically zero and $h$ is a constant of the motion — no further conditions, no assumptions about the form of $T$ , no requirement that constraints be time-independent.

Step 5. Read equation (2) physically even when it is not zero: $-\partial L/\partial t$ is the power flowing into the system through whatever makes the Lagrangian time-dependent — a driven support, a ramped field, a scheduled crank. The energy function leaks at exactly the rate the external schedule injects. As a quick check: for $L = \tfrac12 m\dot x^2 - V(x,t)$ equation (2) reduces to $\mathrm{d}E/\mathrm{d}t = \partial V/\partial t$ — the elementary driven-potential result.

Equation (2) settles the conservation question completely. It says nothing yet about the identity question — whether the conserved number is the mechanical energy. For that we need to look inside the kinetic energy.

When is h the energy? The anatomy of T

Suppose the particle positions connect to the generalized coordinates through a map that may carry explicit time, $\mathbf r_a = \mathbf r_a(q_1,\dots,q_n,t)$ — time appears when a constraint moves on a schedule (a rotating wire, an oscillating support), which is what rheonomic means (see the classification of constraints). Differentiating the map gives each particle's velocity, and squaring it sorts the kinetic energy by powers of $\dot q$ :

T \;=\; T_2 + T_1 + T_0, \qquad T_n \ \text{homogeneous of degree } n \text{ in the } \dot q_i \qquad (3)

where: $T_2 = \tfrac12\sum_{jk} a_{jk}(q,t)\,\dot q_j\dot q_k$ is the familiar quadratic form; $T_1 = \sum_j b_j(q,t)\,\dot q_j$ and $T_0(q,t)$ are cross terms between the $\dot q$ -motion and the map's own drift velocity $\partial\mathbf r_a/\partial t$ . If the map is frozen — scleronomic constraints — then $\partial\mathbf r_a/\partial t = 0$ and $T_1$ , $T_0$ vanish identically: the kinetic energy is purely quadratic. A moving map is the only way $T_1$ and $T_0$ get in.

Go deeper · Derivation B · 2 minWhat Survives the Recipe: h Equals T2 Minus T0 Plus V

Derivation BWhat Survives the Recipe: h Equals T2 Minus T0 Plus V

The goal is the exact identity $h = T_2 - T_0 + V$ for any Lagrangian $L = T - V$ with velocity-independent potential — the formula that answers "is $h$ the energy?" in complete generality. The tool is Euler's theorem on homogeneous functions: a function satisfying $f(\lambda x) = \lambda^n f(x)$ obeys $\sum_i x_i\,\partial f/\partial x_i = n f$ . The strategy is to apply the theorem degree by degree to the decomposition (3) and feed the result into the definition (1). Watch two things: the potential contributes nothing to the momentum sum (it has no $\dot q$ in it), and the linear piece $T_1$ cancels out of $h$ entirely — a small algebraic miracle with physical consequences, because $T_1$ is exactly where Coriolis-type terms live.

Step 1. Apply Euler's theorem to each homogeneous piece of $T$ . Degree 2 returns twice itself, degree 1 returns itself once, degree 0 returns nothing:

\sum_i \dot q_i\,\frac{\partial T}{\partial\dot q_i} \;=\; 2T_2 + T_1 \qquad (4)

Why you can trust this without expanding any matrices: differentiating a quadratic form in $\dot q$ and re-contracting with $\dot q$ doubles it; doing the same to a linear form reproduces it; $T_0$ has no $\dot q$ to differentiate.

Step 2. Insert $L = T - V$ into the definition of $h$ , using $\partial V/\partial\dot q_i = 0$ :

h = \sum_i \dot q_i\,\frac{\partial T}{\partial\dot q_i} - T + V = (2T_2 + T_1) - (T_2 + T_1 + T_0) + V

Step 3. Collect:

h \;=\; T_2 - T_0 + V \qquad (5)

where: $T_1$ has cancelled between the momentum sum and $-T$ . Why the cancellation matters physically: in a uniformly rotating frame $T_1$ is the Coriolis coupling $m\,\Omega\,(x\dot y - y\dot x)$ — the term that bends trajectories without doing work. It bends the motion but leaves no fingerprint on the conserved number.

Step 4. Specialize to a frozen map. Scleronomic constraints give $T_1 = T_0 = 0$ , hence

h = T + V = E \qquad (6)

where: $E$ is the ordinary mechanical energy. Why both conditions were needed: equation (5) already assumed $V$ velocity-independent; equation (6) additionally needs $T = T_2$ . Drop either and $h$ detaches from $E$ .

Step 5. Note the sign structure of (5): $T_0$ enters with a minus sign, while in the energy $E = T_2 + T_1 + T_0 + V$ it enters with a plus. Whenever a moving map generates a $T_0$ , the conserved $h$ and the lab energy differ by $2T_0 + T_1$ — a computable, generally time-varying gap that the constraint's driver supplies. The bead example below realizes the sign flip of $T_0$ explicitly, and the rotating-frame derivation recovers the Jacobi constant, including the conventional factor $-2$ .

Which statement answers which question

you are asked	the test	the tool
is anything conserved here?	does $t$ appear explicitly in $L$ ?	master equation (2)
is the conserved thing the energy?	is $T$ purely quadratic ( $T_1 = T_0 = 0$ ) and $V$ velocity-free?	anatomy result (5), (6)
what exactly is conserved in a rotating frame?	compute $T_2 - T_0 + V$ in that frame	Jacobi integral (8)
how fast does energy leak in a driven system?	evaluate $-\partial L/\partial t$	master equation (2) as a power balance
what is $h$ in momentum variables?	Legendre transform, same value	the Hamiltonian, deferred to the canonical formalism

The Jacobi integral of the rotating frame

The reward for all this care is the celebrated conserved quantity of the circular restricted three-body problem (CR3BP): a test particle — spacecraft, asteroid, ballistic Moon probe — moving in the gravity of two primaries (Earth and Moon, say) that circle their common barycentre at fixed angular velocity $\Omega$ . In the inertial frame the potential rotates, energy is not conserved, and no first integral is available. The masterstroke is to ride the rotation.

Go deeper · Derivation C · 2 minThe Jacobi Integral of the Circular Restricted Three-Body Problem

Derivation CThe Jacobi Integral of the Circular Restricted Three-Body Problem

We derive the one conserved quantity the restricted three-body problem possesses: the Jacobi integral. The strategy is a frame change followed by pattern-matching against the general theory above. In the frame co-rotating with the primaries at constant $\Omega\hat z$ , both primaries stand still, so the gravitational potential $U(x,y)$ loses its clock — the Lagrangian becomes time-independent and the master equation (2) hands us a conserved $h$ for free. The real work is identifying $T_2$ , $T_1$ , $T_0$ in the rotating coordinates and applying $h = T_2 - T_0 + V$ ; the watch-out is that the answer is not the rotating-frame kinetic-plus-potential energy — the centrifugal term enters with the opposite sign, exactly as the general theory demands.

Step 1. Relate velocities. A particle at rotating-frame position $(x,y)$ with rotating-frame velocity $(\dot x, \dot y)$ has inertial velocity $\mathbf v_{\mathrm{in}} = \dot{\mathbf r} + \boldsymbol\Omega\times\mathbf r = (\dot x - \Omega y,\ \dot y + \Omega x)$ . The Lagrangian is the inertial kinetic energy minus the gravitational potential, expressed in rotating coordinates:

L = \tfrac12 m\big[(\dot x - \Omega y)^2 + (\dot y + \Omega x)^2\big] - m\,U(x,y)

Step 2. Expand and sort by powers of the rotating-frame velocities:

T = \underbrace{\tfrac12 m(\dot x^2 + \dot y^2)}_{T_2} \;+\; \underbrace{m\,\Omega\,(x\dot y - y\dot x)}_{T_1} \;+\; \underbrace{\tfrac12 m\,\Omega^2(x^2 + y^2)}_{T_0} \qquad (7)

Why each piece is what it is: $T_2$ is the kinetic energy the rotating observer would naively write; $T_1$ is the Coriolis coupling, linear in velocity; $T_0$ is the sweep energy of the frame itself at the particle's position — the term that will become the centrifugal potential.

Step 3. Check the clock. $\Omega$ is constant and the primaries are pinned at fixed rotating-frame positions, so $U = U(x,y)$ with no explicit $t$ : $\partial L/\partial t = 0$ , and by the master equation (2) $h$ is conserved.

Step 4. Apply the anatomy result (5), $h = T_2 - T_0 + V$ , remembering the sign flip on $T_0$ :

h = \tfrac12 m(\dot x^2 + \dot y^2) - \tfrac12 m\,\Omega^2(x^2 + y^2) + m\,U(x,y) \qquad (8)

where: the middle term is the centrifugal potential — the $T_0$ of the frame, subtracted. The Coriolis piece $T_1$ has vanished from the conserved quantity entirely, though it very much remains in the equations of motion. Astronomers quote the same content as the Jacobi constant $C_J = -2h/m$ , so that

C_J = \Omega^2(x^2+y^2) - 2U(x,y) - v^2 \qquad (9)

where: $v^2 = \dot x^2 + \dot y^2$ is the rotating-frame speed squared. Why the repackaging is useful: with $C_J$ fixed by initial conditions, demanding $v^2 \ge 0$ carves configuration space into allowed and forbidden regions bounded by the zero-velocity curves $v = 0$ — motion is trapped without solving any equation of motion.

Step 5. Extract the punchline. Setting $v = 0$ in (9) defines the zero-velocity surfaces for a given $C_J$ . For large $C_J$ (gentle launches) the allowed region near the Earth is a closed pocket: the spacecraft physically cannot leave it. As $C_J$ drops past the critical value at the interior Lagrange point $L_1$ , the pocket around the Earth and the pocket around the Moon connect through a neck — the gate opens. Note the two structural features of the result: the conventional factor $-2$ in the definition of $C_J$ , and the complete vanishing of the Coriolis term $T_1$ from the conserved integral — present in the equations of motion, absent from the constant.

Figure

Zero-velocity curves in the Earth-Moon rotating frame — the Jacobi constant decides which regions exist for you

The same construction with a bead and a motor instead of a Moon and a month is worked numerically below — the physics is identical, only the hardware budget differs.

Limiting cases

case	result	meaning
scleronomic constraints, $V(q)$	$h = T + V$ , conserved	both switches favourable: the textbook conservation law, equation (6)
moving constraint, no explicit $t$ in $L$	$h = T_2 - T_0 + V = \mathrm{const} \neq E$	Jacobi-integral regime: cranked wire, CR3BP — conserved but not the energy
explicit $t$ in $L$	$\mathrm{d}h/\mathrm{d}t = -\partial L/\partial t \neq 0$	driven systems: $h$ leaks at exactly the drive's power
$\Omega \to 0$ in the rotating frame	$T_1, T_0 \to 0$ , $h \to T + V$	the Jacobi integral degrades gracefully into the ordinary energy
charged particle, static fields	$h = T + q\phi$	the magnetic term, linear in $\dot q$ , cancels from $h$ like any $T_1$
relativistic free particle	$h = \gamma m c^2$	the recipe (1) applied to $L = -mc^2/\gamma$ returns the total relativistic energy

Common trapConserved means it is the energy — and if the energy is not conserved, something is wrong

Both halves of this reflex are false, and the counterexamples are not exotic. The bead on a uniformly cranked wire has a conserved $h$ that is not the energy — the lab energy grows without bound while $h$ sits exactly still. A particle in a slowly ramped trap $V(x,t)$ has $h = T + V$ equal to the energy at every instant — and draining, per $\mathrm{d}h/\mathrm{d}t = -\partial L/\partial t$ . "Is it conserved?" interrogates the clock in $L$ ; "is it the energy?" interrogates the time dependence of the coordinate map. Two switches, four combinations, all four realized in ordinary mechanics. And when the lab energy of a constrained system does change, energy conservation is not violated — the constraint's driver (motor, crank, support) is doing work through the constraint force, and a full accounting that includes the driver balances to zero. What $h$ offers is a number that stays constant without having to audit the driver.

Worked examples

1 · The bead on a cranked wire, with numbers (confidence builder)

A bead of mass $m = 0.20\ \mathrm{kg}$ slides without friction on a straight horizontal wire that a motor spins at constant $\omega = 10\ \mathrm{rad/s}$ . At $t = 0$ the bead is at $r_0 = 0.10\ \mathrm{m}$ from the axis, at rest along the wire. Find the conserved quantity, the motion, and the energy the motor has supplied by $t = 0.20\ \mathrm{s}$ .

Step 1. Build the Lagrangian. With the wire's angle enslaved to the clock, $\phi = \omega t$ , the bead's only freedom is $r$ : $T = \tfrac12 m(\dot r^2 + r^2\omega^2)$ and $V = 0$ , so $L = \tfrac12 m\dot r^2 + \tfrac12 m\omega^2 r^2$ . No explicit $t$ appears — $\omega t$ was absorbed into the coordinate choice — so $h$ is conserved.

Step 2. Sort the kinetic energy: $T_2 = \tfrac12 m\dot r^2$ , $T_1 = 0$ , $T_0 = \tfrac12 m\omega^2 r^2$ . The rotation converted what looks like kinetic energy into a velocity-independent $T_0$ — the rheonomic signature. By equation (5), $h = T_2 - T_0 = \tfrac12 m\dot r^2 - \tfrac12 m\omega^2 r^2$ . At $t=0$ : $h = 0 - \tfrac12(0.20)(10)^2(0.10)^2 = \boxed{-0.100\ \mathrm{J}}$ , and it stays there forever.

Step 3. Solve the motion. The Euler–Lagrange equation is $\ddot r = \omega^2 r$ — centrifugal runaway — with solution $r(t) = r_0\cosh\omega t$ for this release. At $t = 0.20\ \mathrm{s}$ : $\omega t = 2$ , $\cosh 2 = 3.762$ , $\sinh 2 = 3.627$ , so $r = 0.376\ \mathrm{m}$ and $\dot r = r_0\,\omega\sinh 2 = 3.63\ \mathrm{m/s}$ .

Step 4. Audit both meters. $T_2 = \tfrac12(0.20)(3.63)^2 = 1.316\ \mathrm{J}$ and $T_0 = \tfrac12(0.20)(100)(0.376)^2 = 1.415\ \mathrm{J}$ . The conserved quantity checks: $h = 1.316 - 1.415 = -0.100\ \mathrm{J}$ ✓. The lab energy is $E = T_2 + T_0 = 2.73\ \mathrm{J}$ — it started at $0.100\ \mathrm{J}$ and has grown by a factor of 27. The motor supplied $E - E_0 = \boxed{2.63\ \mathrm{J}}$ in a fifth of a second, and its instantaneous power at that moment, $P = 2m\omega^2 r\dot r = 54.6\ \mathrm{W}$ , is climbing exponentially. Holding "constant angular velocity" is expensive; the conserved $h$ knew that without ever consulting the motor.

2 · Opening the lunar gate (exam level)

A spacecraft sits in a circular parking orbit at 200 km altitude ( $r = 6578\ \mathrm{km}$ , inertial speed $7.784\ \mathrm{km/s}$ ). Using the planar Earth-Moon CR3BP with $\mu = m_M/(m_E + m_M) = 0.01215$ , mean separation $a = 384{,}400\ \mathrm{km}$ , and frame angular rate $n = 2.662\times10^{-6}\ \mathrm{rad/s}$ : (a) find its Jacobi constant; (b) decide whether the Moon is reachable; (c) find the perigee speed at which the $L_1$ gate opens, given $C(L_1) = 3.1883$ .

Step 1. Set up units. Lengths in units of $a$ , speeds in units of $V = n\,a = 1.023\ \mathrm{km/s}$ . The Earth sits at $x = -\mu$ , so the parking orbit is at $x = -\mu + 6578/384400 = 0.00496$ , i.e. $r_1 = 0.01711$ from the Earth. The rotating-frame speed subtracts the frame sweep $n r = 0.0175\ \mathrm{km/s}$ : $v_{\mathrm{rot}} = 7.784 - 0.018 = 7.767\ \mathrm{km/s} = 7.591$ units.

Step 2. Evaluate the Jacobi constant (9) in nondimensional form, $C = (x^2 + y^2) + 2(1-\mu)/r_1 + 2\mu/r_2 - v^2$ . The terms: $x^2 + y^2 = 2.5\times10^{-5}$ (negligible this deep in the well), $2(1-\mu)/r_1 = 115.45$ , $2\mu/r_2 = 0.025$ , $v^2 = 57.62$ . So $C = \boxed{57.86}$ .

Step 3. Compare with the gate. $C = 57.9 \gg C(L_1) = 3.1883$ : the zero-velocity surface is a small closed shell around the Earth, and the Moon lies deep inside the forbidden region. No trajectory, however cleverly steered, reaches the Moon on this budget — the verdict required no trajectory integration at all, which is exactly the power of the integral.

Step 4. Open the gate. Increasing perigee speed $v_p$ lowers $C$ (speed enters with a minus sign). Setting $C(v_p) = C(L_1)$ with all position terms fixed: $v_p^2$ must absorb $115.47 - 3.188 = 112.28$ units², giving $v_{\mathrm{rot}} = 10.596$ units $= 10.842\ \mathrm{km/s}$ , i.e. an inertial perigee speed $v_p = \boxed{10.86\ \mathrm{km/s}}$ .

Step 5. Sanity-check against known mission numbers. Circular speed there is $7.78\ \mathrm{km/s}$ and parabolic escape is $11.01\ \mathrm{km/s}$ : the Moon gate opens just $150\ \mathrm{m/s}$ shy of escaping Earth entirely, which is why every trans-lunar injection in history is a near-escape burn. A real TLI targeting $10.92\ \mathrm{km/s}$ gives $C = 1.93 < C(L_2) = 3.172$ — both gates wide open, with margin for the three-dimensional, Sun-perturbed reality.

Origins · 1 minCarl Gustav Jacob Jacobi and how this was discovered

Carl Gustav Jacob Jacobi

1804–1851 · Königsberg and Berlin

Jacobi was the most formidable pure calculator of nineteenth-century mechanics — a Berlin-born prodigy who taught at Königsberg for seventeen years and made elliptic functions, determinants, and partial differential equations into precision instruments. His 1836 note to the Paris Academy dropped a bombshell into celestial mechanics: the restricted three-body problem, hopeless in the inertial frame, admits an exact integral in the rotating frame — the quantity now bearing his name — even though the energy itself is not conserved there. Characteristically, he found it by trusting the formalism rather than physical intuition; asked what mathematics was for, he famously answered "the honour of the human spirit." The integral seeded G. W. Hill's proof that the Moon cannot wander off, Poincaré's discovery that it is essentially the only integral the problem has, and — a century and a half later — the energy bookkeeping of every low-energy lunar mission. His broader legacy in this course is everywhere: the Jacobi identity, Jacobians, and the Hamilton–Jacobi equation that closes the classical story.

History

1788

— Lagrange's Mécanique analytique builds mechanics from the Lagrangian; the energy function is implicit in his formalism but not yet named.

1836

— Jacobi announces the integral of the circular restricted three-body problem in the rotating frame, in a note to the Paris Academy.

1878

— G. W. Hill uses the integral's zero-velocity curves to prove the Moon's orbit is permanently bounded — the birth of Hill regions.

1892

— Poincaré's Méthodes nouvelles shows the Jacobi constant is essentially the only integral of the restricted problem, opening the door to chaos.

1991

— Belbruno's ballistic-capture rescue of Japan's Hiten probe turns Jacobi-constant reasoning into an operational low-energy transfer technique.

Going further

References. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §2.7 (energy function) and §3.3 (Jacobi integral) · Landau & Lifshitz, Mechanics, 3rd ed., §6 (conservation of energy) · Morin, Introduction to Classical Mechanics, Ch. 6 (the Lagrangian method) · Taylor, Classical Mechanics, §7.8 (conservation of energy in Lagrangian mechanics) The key symbolic results are: the master equation $\mathrm{d}h/\mathrm{d}t=-\partial L/\partial t$ derived from the Euler–Lagrange equations; Euler's homogeneous-function theorem giving $\sum\dot q_i\partial T/\partial\dot q_i=2T_2+T_1$ ; and the rotating-frame Lagrangian yielding $C_J=\tfrac12 v^2-\tfrac12\Omega^2(x^2+y^2)+U$ . Worked-example numbers on this page (bead: $h = -0.100$ J, $E = 2.73$ J at $t = 0.2$ s, motor input 2.63 J; lunar gate: $C_{\mathrm{LEO}} = 57.9$ , $C(L_1) = 3.1883$ , gate-opening perigee speed 10.86 km/s vs escape 11.01 km/s) are illustrative arithmetic recomputed from the boxed formulas with $\mu = 0.01215$ , $a = 384{,}400$ km, $V = na = 1.023$ km/s.

Builds on: The Lagrangian Function · The Euler-Lagrange Equations · Generalized Forces and Generalized Momenta · Generalized Coordinates and Configuration Space · Constraints and Their Classification Leads to: Symmetries and Continuous Transformations · Cyclic Coordinates and Conservation of Conjugate Momenta · Hamilton's Principle of Stationary Action

Is h just the Hamiltonian with a different name?

Numerically yes, conceptually almost. The energy function $h(q,\dot q,t)$ and the Hamiltonian $H(q,p,t)$ have the same value on any state; they differ in which variables are treated as independent. $H$ is obtained from $h$ by the Legendre transform — invert $p_i = \partial L/\partial\dot q_i$ for the velocities and substitute — which requires the Hessian $\partial^2 L/\partial\dot q_i\partial\dot q_j$ to be invertible. The distinction earns its keep later: the whole canonical formalism (Poisson brackets, phase flow, canonical transformations) needs $H$ as a function of $(q,p)$ , while everything in this article works directly in velocity variables. Goldstein keeps the separate symbol $h$ precisely to flag the difference in functional dependence.

The bead's energy grew by a factor of 27 in a fifth of a second. Who paid, and is energy conservation violated?

The motor paid, through the constraint force, and global energy conservation is intact. The wire must push the bead sideways (perpendicular to the wire) to keep it co-rotating; that push is a genuine force doing genuine work at rate $P = 2m\omega^2 r\dot r$ , which you can check equals $\mathrm{d}E/\mathrm{d}t$ exactly. What is remarkable is that the Lagrangian analysis never needed to model any of this: choosing the coordinate $r$ along the wire absorbed the drive into the coordinate map, the clock disappeared from $L$ , and a conserved $h$ emerged that silently accounts for the motor. If the motor instead struggled and $\omega$ sagged, $\phi(t)$ would no longer be $\omega t$ , explicit time would enter $L$ , and neither $E$ nor $h$ would be constant.

Why does T0 enter h with a minus sign when it enters the energy with a plus sign?

Trace it through the recipe. The momentum sum $\sum\dot q_i\,\partial T/\partial\dot q_i$ weights each homogeneous piece by its degree — $2T_2 + 1\cdot T_1 + 0\cdot T_0$ — so $T_0$ contributes nothing there. Then the $-L$ subtracts all of $T$ once, including $T_0$ . Net: $+2T_2 - T_2 = +T_2$ , $+T_1 - T_1 = 0$ , $0 - T_0 = -T_0$ . The physical reading of the flip: $T_0 = \tfrac12 m\Omega^2\rho^2$ is the frame's own sweep energy at your position, and in the conserved bookkeeping it plays the role of a potential well (the centrifugal potential $-\tfrac12 m\Omega^2\rho^2$ ), not a kinetic asset. That is why zero-velocity curves exist at all: $T_0$ shapes the landscape rather than adding to the budget.

Can h and E both be conserved and still be different numbers?

Yes, and the free particle in a rotating frame is the cleanest case. Describe an interaction-free particle from a turntable rotating at constant $\Omega$ : the Lagrangian has no explicit time, so $h$ is conserved; meanwhile the inertial energy $E$ is trivially conserved too. But $h = E - \Omega L_z$ , where $L_z$ is the (also conserved) angular momentum about the rotation axis — three constants, two of them independent, no contradiction. This identity $h = E - \boldsymbol\Omega\cdot\mathbf L$ holds generally for uniformly rotating frames and is the fastest way to convert between the Jacobi integral and inertial-frame quantities; astronomers use exactly this to relate $C_J$ to a comet's inertial orbital elements in the Tisserand criterion.

Why do astronomers write C = −2h instead of just using h?

Convention with a purpose. Multiplying by $-2$ (and dividing by mass) turns the zero-velocity condition into the tidy $C_J = \Omega^2(x^2+y^2) + 2(1-\mu)/r_1 + 2\mu/r_2$ evaluated at $v = 0$ — all terms positive, larger $C_J$ meaning less energetic and more confined, which matches the intuition of "climbing out of a well." It also makes landmark values easy to tabulate: for the Earth–Moon system $C(L_1) = 3.1883$ and $C(L_2) = 3.1722$ , so a glance at a spacecraft's $C_J$ against the row of critical values tells you instantly which gates are open. The physics content is identical to $h$ ; only the packaging differs, and you should be fluent in both directions of the conversion.

Does a magnetic field break the equality h = T + V, since its potential depends on velocity?

It threatens to, and then doesn't — for a reason you can now diagnose in one line. The electromagnetic Lagrangian $L = \tfrac12 mv^2 - q\phi + q\mathbf A\cdot\mathbf v$ has a velocity-linear interaction term, which behaves exactly like a $T_1$ : the momentum sum reproduces it once, the $-L$ removes it once, and it cancels from $h$ entirely. The result is $h = \tfrac12 mv^2 + q\phi$ — kinetic plus electric potential energy only. The magnetic field steers the motion (it is very much present in the equations of motion) but leaves no trace in the conserved number, the same disappearing act the Coriolis term performs in the rotating frame. For static fields $\partial L/\partial t = 0$ , so this $h$ is conserved — the familiar statement that magnetic forces do no work, upgraded to a Lagrangian theorem.

If L has explicit time dependence, is the energy function useless?

Not at all — equation (2) is then a precision instrument rather than a conservation law. $\mathrm{d}h/\mathrm{d}t = -\partial L/\partial t$ tells you the exact instantaneous power the time dependence injects, which is the starting point for driven-system energetics, parametric amplification, and perturbation theory. Two special regimes are worth knowing. If the drive is periodic, $h$ is not conserved but its change over one period can average to zero or pump steadily, which is how parametric resonance is diagnosed. If the drive is slow compared with the internal motion, $h$ changes but a subtler quantity — the action integral $\oint p\,\mathrm{d}q$ — is nearly frozen; those adiabatic invariants are treated in their own article later in the track and inherit everything from equation (2).

Is h always the energy in some cleverly chosen frame?

Often, not always — and the counterexamples matter. For a uniformly rotating frame, $h = E - \Omega L_z$ is a linear combination of inertial conserved quantities, and you may regard it as "the energy in the co-rotating sense." But the theorem $h = T_2 - T_0 + V$ is coordinate-fact, not frame-fact: choose deliberately weird generalized coordinates (say, $q = x - \alpha t$ for a system with no physical drift) and you manufacture $T_1$ , $T_0$ pieces and a corresponding $h$ that matches no observer's calorimeter, while remaining exactly conserved if $L$ lost its clock. The honest statement: $h$ is the Noether charge of time-translation invariance in your chosen description — the article on Noether's theorem makes that construction general — and its physical costume depends on how the description was cut.

The Energy Function and Jacobi Integral

Where this lives today

The physical picture

The recipe and its master equation

When is h the energy? The anatomy of T

Which statement answers which question

The Jacobi integral of the rotating frame

Limiting cases

Worked examples

1 · The bead on a cranked wire, with numbers (confidence builder)

2 · Opening the lunar gate (exam level)

Going further

Before this

Where this leads