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Ignoring the small radiative back-reaction, conservation of energy gives $\dot{x}=\sqrt{2(E-f(x))/m}$ and Newton's equation gives $\ddot{x}=-f^{\prime }(x)/m$. Integrating the Larmor formula with $dt=dx/\dot{x}$ yields, for $E>V_0$,If $E<V_0$, the particle turns at $x_{\ast }$ defined by $f(x_{\ast })=E$ and traverses the accelerated region twice, so the same formula holds with twice the integral from $0$ to $x_{\ast }$.

For a particle in the scalar potential energy $U(x)$,The canonical commutation relation $[x,p]=i\hslash$ givesThe Ehrenfest theorem with $O=x$ therefore givesSince $[p^2,p]=0$ and direct action on a wavefunction gives $[U(x),p]=i\hslash U^{\prime }(x)$,Finally $H$ has no explicit time dependence and $[H,H]=0$, so

The first two equations are the expectation-value counterparts of momentum equals mass times velocity and Newton's second law; the last is conservation of energy. They become the classical equations directly when the force varies negligibly across the wave packet, so that $⟨U^{\prime }(x)⟩\simeq U^{\prime }(⟨x⟩)$. These conclusions are summarized by the classical equations from Ehrenfest theorem.

The force is central force, so its torque about the star vanishes. The specific angular momentum is therefore constant:Taking the scalar product of $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$ with $\dot{\mathbf{r}}$ givesThus conservation of energy gives the second constant

Conservation of energy and Fourier's law giveAt steady state $k{T}_z=F_c$ is constant. Since $k=a/T$,The two temperature boundary conditions therefore giveAt the basal phase boundary, the Stefan condition isThe first term removes latent heat released by freezing, while the oceanic flux $F_w$ supplies heat to the interface. A steady shell has $\dot{h}=0$, hence $F_c=F_w$ andHere $T_s$ must simultaneously satisfy the surface balance from part a with $F_c=F_w$.

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.

For fixed mass $M$, a uniform sphere of physical radius $R$ has $W_G=-3G{M}^2/(5R)$ and $W_{\Lambda }=-\Lambda M{R}^2/10$, using the convention in which its outward acceleration is $\Lambda r/3$. The virial theorem reads $2T+W_G=2W_{\Lambda }$. Defining $x=R_f/R_{\mathrm{ta}}$ and $\eta =\Lambda /(4\pi G{\rho }_{\mathrm{ta}})$, conservation of energy gives $2\eta x^3-(2+\eta )x+1=0$. The collapsing branch has $x=1/2-\eta /8+O({\eta }^2)$. For small positive cosmological constant, it is more compact at fixed turnaround data, even though the outward force inhibits turnaround and collapse.