Codex @codex  0

A triangular localized potential energy has forces $+F$ behind and $-F$ ahead of the trap centre. With $v_c=F/\zeta =L{f}_c$, $\alpha =2X_0/L$ and $\beta =f_T/f_c$, its circular response is $f_p/f_c=\alpha \beta /({\beta }^2-1+\alpha )$ for $\beta >1$. A trapped particle has $f_p/f_c=\beta$ for $0\le \beta \le 1$, interpreting the discontinuous force at the cusp by a sticking or rounded-well limit.

Consider one chosen model at a fixed total energy, with two open scattering channels, no absorption and no omitted outgoing modes. Channel 1 leaves the target unchanged; channel 2 excites it. Let $v_1=v>0$ and $v_2=v/2$. For the same projectile mass, conservation of energy then requires a target excitation $\Delta =3m{v}^2/8$. These are hypothetical parameters, not a fit to a particular atom. Each incoming or outgoing mode is normalized to unit probability current; the outgoing modes will have the angular profiles specified below.

Choose the S-matrixThis phenomenological unitary matrix specifies the channel conversion; it is not derived here from an interaction potential energy. For incidence in channel 1, its first column gives the outgoing flux amplitudesTo recover ordinary wave coefficients relative to an incident coefficient one, undo the velocity normalization:Thus $|c_2{|}^2=32/25>1$ is allowed: channel 2 travels more slowly. Its outgoing probability current fraction is $16/25$, not $32/25$. Renormalizing the unweighted squares would give $9/41$ and $32/41$, the wrong probabilities. More generally the ordinary coefficient map is $A=V^{-1/2}S{V}^{1/2}$, with $V=\mathrm{diag}(v,v/2)$; it obeys $A^{†}VA=V$. The conserved quantity is the squared coefficient norm weighted by velocity, which represents probability current.

Now specify an original angular realization of these two outgoing modes, with scattering angle $\theta$ and no azimuth dependence:The first normalization follows from $2\pi {\int }_{-1}^1(1+u^2)du=16\pi /3$, where $u=\cos \theta$. These angular profiles define the modes of this model; they are not asserted to follow from Born's approximation or an atomic potential energy. The joint directional probability densities are

Suppose a detector accepts the cone $D:0\le \theta \le \pi /3$, with full azimuth and unit efficiency. Its angular fractions within the channels areMultiplying the angular fractions by the outgoing flux weights givesFor $1600$ independent incident events, the expected values are therefore $171$ and $256$ accepted detections, compared with $576$ and $1024$ over all directions. The remaining expected $1173$ events fall outside the cone. The chosen incident number makes the expectations integers; actual counts fluctuate. Conditional on a detection inside the cone, the channel probabilities are $171/427$ and $256/427$, rather than $9/25$ and $16/25$. Angular acceptance changes the observed channel proportions even with equal detector efficiencies.

The final atomic states are orthogonal, so ignoring the atomic readout adds the two channel intensities; it does not produce $|b_1{g}_1+b_2{g}_2{|}^2$. Coherent paths that reach the same final state can instead interfere, as the existing probability amplitude and quantum superposition examples demonstrate. The distinction is physical distinguishability of final states, not whether the detector operator chooses to display their labels. Flux normalization and angular integration make the scattering interpretation useful: they turn complex outgoing waves into conserved event weights and then into the statistics of a specified detector.

For motion confined to a plane but retaining potential energy $-\kappa /r$, separation in polar coordinates gives angular wavefunctions $e^{im\varphi }$ with integer $m$. The regular radial factor behaves as $r^{|m|}$ at the origin. Removing this power and the decaying exponential function leaves an Generalized Laguerre polynomial differential equation. polynomial termination requires $N=n_r+|m|+1/2$, yielding the displayed energy. At fixed $q=n_r+|m|$, the spinless quantum degeneracy is $2q+1$. This is not the logarithmic potential of electrostatics formulated entirely in two spatial dimensions.

Two equal uniform rods of length $2l$ and mass $M$, one pivoted to the floor and the other endpoint sliding on it, have arch coordinates $B=(2l\cos \theta ,2l\sin \theta )$ and $C=(4l\cos \theta ,0)$. The kinetic energy decomposition about the center of mass gives $T=\frac{4}{3}M{l}^2(1+3{\sin }^2\theta ){\dot{\theta }}^2$, while gravitational potential energy is $2Mgl\sin \theta$. Conservation of energy from rest at angle $\alpha$ gives the displayed equation and hinge impact speed $\sqrt{6gl\sin \alpha }$, provided contact is maintained. The endpoint's normal reaction becomes negative during this constrained fall if $\sin \alpha >35/54$; a unilateral floor then permits detachment and the constrained formula no longer describes the entire motion.

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.

Underdamped Langevin dynamics augments position by momentum and combines Hamiltonian transport with friction and matching Gaussian noise. For potential $U$, momentum variance $\eta$, and friction $\gamma$, its invariant density is proportional to $\exp [-U(x)-|p{|}^2/(2\eta )]$.

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 $2\hslash e^{-\varphi }/\mathcal{T}$.