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Two identical uniform rigid bodies of rod shape, each of length $2l$ and mass $M$, form a symmetric arch with a free common hinge and both outer ends sliding without friction on a horizontal floor. At angle $\theta$, each center has speed $l|\dot{\theta }|$, giving total kinetic energy $4M{l}^2{\dot{\theta }}^2/3$. The total gravitational potential energy is $2Mgl\sin \theta$. Conservation of energy from rest at $\alpha$ gives the displayed hinge impact speed. Unlike fixing one outer endpoint, allowing both endpoints to slide keeps the arch center horizontally fixed. The floor reaction on each rod is $Mg[1+9{\sin }^2\theta -6\sin \alpha \sin \theta ]/4$, nonnegative throughout this fall.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.

In a uniform U-tube containing incompressible liquid of density $\rho$, cross section $A$ and total moving length $l$, let one surface fall by $y$ while the other rises by $y$. The height difference is $2y$, and potential energy relative to equilibrium is $U=\rho gA{y}^2$. The whole moving mass is $\rho Al$, so $K=\rho Al{\dot{y}}^2/2$. Newton's second law or energy conservation gives $\ddot{y}+(2g/l)y=0$ and ${\omega }_0=\sqrt{2g/l}$. This assumes small excursions that keep the free surfaces in the uniform vertical limbs.

A homogeneous canonical scalar field with positive scalar potential has energy density ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and pressure $P_{\varphi }={\dot{\varphi }}^2/2-V$. When its potential energy dominates its kinetic energy, the equation-of-state parameter approaches $-1$. If other components and spatial curvature are negligible, the Friedmann acceleration equation gives accelerating expansion. Small initial kinetic energy is not enough to ensure long cosmic inflation; the field must also remain on a sufficiently flat part of the scalar potential, as measured by the potential slow-roll parameters.

The interaction potential energy associated with a Van der Waals force.

A horizontal bar of mass $M$ hangs from a spring of incremental stiffness ${\kappa }_0>0$, and two equal masses $m$ hang separately from the bar through springs of stiffness ${\kappa }_1>0$. The bar remains horizontal and all motion is vertical. For physical displacements $q=(q_1,q_2,q_3)$ from the loaded equilibrium, the kinetic energy and quadratic potential energy areGravity determines the static extensions; its linear terms cancel at equilibrium. Exchange of the lower masses splits the normal modes into the antisymmetric mode $(0,1,-1)$ with ${\omega }^2={\kappa }_1/m$, and two symmetric modes $(A,B,B)$. The latter obey $({\kappa }_0+2{\kappa }_1-M{\omega }^2)A-2{\kappa }_{1}B=0$ and $-{\kappa }_{1}A+({\kappa }_1-m{\omega }^2)B=0$. Their determinant gives the title quadratic in ${\omega }^2$, and $B/A={\kappa }_1/({\kappa }_1-m{\omega }^2)$. For $M=2,m=1,{\kappa }_0=4,{\kappa }_1=1$, their squared angular frequencies are $2\pm \sqrt{2}$, with displacement ratios $(1,1\mp \sqrt{2},1\mp \sqrt{2})$ respectively. The lower symmetric mode moves all masses in phase; the higher one moves the lower pair against the bar.

The virial theorem relates the average value of a function along a trajectory to the forces that bind it. For an isolated system of particles interacting through Newtonian gravitational potential energy, it gives $2⟨K⟩+⟨U⟩=0$: the average kinetic energy is half the magnitude of the average negative potential energy. An eccentric Kepler orbit makes the distinction vivid: kinetic energy and potential energy vary throughout the orbit, although their sum is constant. The relation concerns their time averages, rather than their values at every instant. It underlies the virial mass estimate and the negative heat capacity of a virialized gravitational system.

Use constant positive particle masses $m_i$, positions ${\mathbf{r}}_i$, mechanical momenta ${\mathbf{p}}_i=m_i{\dot{\mathbf{r}}}_i$, and total kinetic energy $K={\sum }_i{\mathbf{p}}_i^2/(2m_i)$. The auxiliary quantity $G(t)$ is a sum of dot products, not the gravitational constant, which we denote here by $G_N$:This finite-$T$ identity is exact wherever Newton's second law holds along a regular trajectory; ${\mathbf{F}}_i$ must include every force. The endpoint term is what a finite observation window leaves behind. For the infinite-time theorem, suppose the averages on the left exist and $(G(T)-G(0))/T\to 0$. A sufficient condition is uniformly bounded positions and velocities for finitely many particles. More generally $G(T)=o(T)$ suffices, even without those individual bounds. Bounded positions alone do not supply a bound on momenta, especially near a singular potential energy; nor does a small endpoint term prove thermodynamic equilibrium. Periodic motion with $G(T)=G(0)$ gives the identity without an endpoint term at that finite $T$. For an isolated system, use an inertial centre of mass frame so translational drift does not contribute bulk kinetic energy.^[ref]

For a differentiable potential energy that is a homogeneous function of degree $q$, choose its additive constant so that the scaling law really holds. With conservative forces, the Euler theorem for homogeneous functions givesThus a harmonic oscillator has $q=2$ and $⟨K⟩=⟨U⟩$. An isolated system with Newtonian gravitational potential energy hasEach interacting pair occurs once, and $U=0$ at infinite separation. Here the total energy $E=K+U$ is conserved. An external field or a mixture of different scaling powers requires its own force term; one cannot assign the whole potential energy a single degree $q$. The virial theorem for a homogeneous pair potential states the corresponding general energy relation.

For a differentiable total pair potential energy homogeneous of degree $q$ in all particle positions, the Euler theorem for homogeneous functions gives ${\sum }_i{\mathbf{r}}_i\cdot {\nabla }_{i}W=qW$. Since ${\mathbf{F}}_i=-{\nabla }_{i}W$, the force virial is $-qW$. The bounded-motion time-averaged virial theorem consequently gives the displayed equation. For $q\ne 0$, total energy is $E=(q+2)⟨K⟩/q$. A positive kinetic-temperature coefficient makes its heat capacity negative exactly when $-2<q<0$. For a pair potential proportional to $r^{p+1}$ this means $-3<p<-1$.

Consider the exact two-body problem with masses $m_1,m_2$. In the inertial centre of mass frame, its internal kinetic energy is $K=m|\dot{\mathbf{r}}{|}^2/2$, where $\mathbf{r}$ is their relative position, $m=m_1{m}_2/(m_1+m_2)$ is the reduced mass, and $U=-\kappa /r$ with $\kappa =G_N{m}_1{m}_2$. Take a noncollision elliptic Kepler orbit with semi-major axis $a$ and orbital eccentricity $0<e<1$. Write $\eta$ for eccentric anomaly, to avoid confusing it with total energy $E$. With mean motion $n=\sqrt{\kappa /(m{a}^3)}$ and orbital period $P=2\pi /n$, Kepler's equation and the orbit geometry giveThe time origin is periapsis. Since $dt=(1-e\cos \eta )d\eta /n$, the time average of $1/r$ over one orbital period is exactly $1/a$. Therefore $⟨K{⟩}_P=\kappa /(2a)$ and $⟨U{⟩}_P=-\kappa /a$, independently of orbital eccentricity. Equal steps in eccentric anomaly are not equal time steps; unweighted averaging of a drawing of the ellipse gives the wrong answer.

The original dimensionless example in both diagrams has $m=a=\kappa =1$ and $e=0.6$. At periapsis, $r=0.4$, $K=2$, $U=-2.5$; at apoapsis, $r=1.6$, $K=0.125$, $U=-0.625$. Both have $E=-0.5$, but $2K+U$ is respectively $1.5$ and $-0.375$. Only the orbital period averages are $⟨K{⟩}_P=0.5$ and $⟨U{⟩}_P=-1$. The instantaneous zero of $2K+U$ occurs at $r=a$, not at every point of the orbit.

The endpoint term can also be evaluated exactly: $G=m\mathbf{r}\cdot \dot{\mathbf{r}}=mn{a}^{2}e\sin \eta$. Starting at periapsis, a window ending at eccentric anomaly $\eta$ hasAt $T=P/4$, Kepler's equation gives $\eta \simeq 2.09133$ and a positive residual $\simeq 0.3314\kappa /a$. At $T=P/2$ or $P$, $G$ returns to zero and the residual vanishes exactly. The lower panel shows the running residual and its endpoint expression coinciding; the shaded quarter-orbital period window explains why a short observation need not satisfy the infinite-time theorem.

The volt is the SI unit of electric potential and electric potential difference: $1\mathrm{V}=1J/C$, one joule per coulomb. A positive electric charge of $2\mathrm{C}$ moved through a potential increase of $3\mathrm{V}$ gains $\Delta U=q\Delta \Phi =6\mathrm{J}$ of potential energy; the electric field does $-6\mathrm{J}$ of work. ^[ref].

For a finite jump in a one-dimensional potential energy, with constant mass and no delta potential, both the wavefunction and its first spatial derivative are continuous. Integrating the time-independent Schrödinger equation across a vanishing interval makes the jump in ${\psi }^{\prime }$ vanish, since the integral of $(V-E)\psi$ vanishes. A jump in $\psi$ itself would produce a derivative of a Dirac delta function in ${\psi }^{\prime \prime }$, with no matching term in the equation. The second derivative can jump. This matching allows transmission and quantum tunnelling; a finite step is not a hard wall.

For a smooth symmetric confining potential energy with four separated simple classical turning points $-x_2<-x_1<x_1<x_2$, defineThe Airy turning-point connection formula applied to a decaying exterior wavefunction gives a central WKB approximation proportional to ${\chi }^{-1/2}[2\cos \theta e^K+\sin \theta e^{-K}]$, where $\chi =\sqrt{2m(V-E)}$ and $K(x)={\hslash }^{-1}{\int }_x^{x_1}\chi dx$. Since ${\chi }^{\prime }(0)=0$ and $2K(0)=\varphi$, the even Neumann boundary condition gives $\tan \theta =2e^{\varphi }$, while the odd Dirichlet boundary condition gives $\tan \theta =-2e^{\varphi }$. For $\varphi \gg 1$, $\theta =(n+\frac{1}{2})\pi \mp \frac{1}{2}{e}^{-\varphi }+O(e^{-3\varphi })$, with the minus correction for even parity. Since ${\theta }^{\prime }=\mathcal{T}/(2\hslash )$ for the single-well classical oscillation period $\mathcal{T}$, the leading double-well tunneling splitting is $2\hslash e^{-\varphi }/\mathcal{T}$.

For a point mass of constant mass $m$ observed in an inertial frame, the net work along its actual path equals its change in kinetic energy:The middle expression is a line integral; the dot product selects the component of force along the displacement. Positive net work increases speed, negative net work decreases it, and zero net work leaves the endpoint speeds equal. Equal endpoint speeds do not imply equal velocities: a perpendicular force can continually change direction while doing zero work. ^[ref]

The bridge to Newton's second law is short: take the dot product of $m\dot{\mathbf{v}}={\mathbf{F}}_{\mathrm{net}}$ with velocity. The chain rule gives $d(m|\mathbf{v}{|}^2/2)/dt={\mathbf{F}}_{\mathrm{net}}\cdot \mathbf{v}$, equating the rate of change of kinetic energy to instantaneous power. Integrating gives the theorem. No conservative force assumption entered this argument; friction also contributes to the net work.

For the harmonic oscillator with spring stiffness $k=2N/m$, release a mass $m=0.5\mathrm{kg}$ from rest at $x=0.3\mathrm{m}$. The spring exerts $F=-kx$. Its work before the first passage through $x=0$ isThus $m{v}^2/2=0.09\mathrm{J}$ gives speed $0.6m/s$ there. The velocity is negative on this passage because the position is decreasing. The energy calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent potential energy $V$, the conservative force contributes work $-\Delta V$. Separating all other work as $W_{\mathrm{other}}$ gives $\Delta (T+V)=W_{\mathrm{other}}$. Consequently conservation of mechanical energy requires zero net other work; the work-energy theorem itself does not require that conservation. With damping, mechanical energy decreases even though the theorem remains valid. ^[ref]

In 1935, Hideki Yukawa proposed a new massive exchange quantum to explain nucleon binding. His paper began with the inadequacy of an electron–neutrino exchange description of the binding strength. It then required a potential energy falling faster with distance than a Coulomb potential, introduced an exponentially decaying field, and related its inverse range to the exchange particle's rest mass. This was a theoretical proposal, not a detection with an apparatus of his own.^[ref]

The crucial idea is that a finite rest mass supplies a length scale even in empty space. In a scalar model, the static limit of a Klein-Gordon equation produces the modified Helmholtz equation. Its decaying Green's function gives the Yukawa potential. The derivation below isolates this range mechanism; one-pion exchange also carries spin and isospin structure.

The nuclear emulsion track experiment used developed photographic emulsions exposed to cosmic rays and inspected microscopically. Stopping pions gave daughter muon tracks with closely grouped ranges. Comparing penetrating muons with pion-induced nuclear breakup distinguished their interactions; this was not a direct static force measurement. The muon is a lepton, not Yukawa's strongly interacting meson.^[ref]

For potential energy homogeneous of degree $-2$, Newton's second law gives $d^2{r}^2/d{t}^2=4E/m$. Indeed, Euler's identity implies $\mathbf{r}\cdot \mathbf{F}=2U$, so the second derivative equals $2v^2+4U/m=4(K+U)/m$. Thus a trajectory with zero total energy and initially zero radial velocity remains on a sphere of constant radius. A positive electric charge released from rest on the equator of a fixed ideal electric dipole has these properties; in its meridional plane it oscillates on the semicircle on the negative-pole side. A finite two-charge electric dipole does not have exactly homogeneous potential energy and perturbs this result.

A zero-rest-length spring has quadratic potential energy in its actual length $s$, with no subtracted natural length. It is a useful idealization, not an ordinary unstretched spring of arbitrary length. For an end-to-end spring on a two-link pendulum, changing the natural length changes the quadratic bending stiffness about the straight configuration.