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A linear deep-water gravity wave of surface-displacement amplitude $a$ has equal mean kinetic energy and potential energy, and total mean energy per horizontal area $\mathcal{E}={\rho }_{w}g{a}^2/2$. Its energy transport speed is the group velocity. This amplitude is half the crest-to-trough height.

Neglecting density-gradient kinetic energy in the stationary Gross–Pitaevskii equation gives $n=(\mu -V{)}_{+}/g$ for repulsive interactions. For a radial harmonic trap $V=m{\omega }_r^2{r}^2/2$, this is $n=n_0(1-r^2/R_r^2{)}_{+}$ with $R_r=\sqrt{2g{n}_0/(m{\omega }_r^2)}$. Hard boundaries can truncate this natural support. The approximation fails in vortex cores and boundary healing regions and requires a cloud much larger than its healing length.

For inviscid unforced two-layer QG dynamics, $H_1{F}_1=H_2{F}_2=f_0^2/g\prime$ makes the coupling symmetric in the layer-depth inner product. Multiplication by $H_i{\psi }_i$ gives the positive energy $E=\frac{1}{2}[{\sum }_i{H}_i|\nabla {\psi }_i{|}^2+(f_0^2/g\prime )({\psi }_1-{\psi }_2{)}^2]$. It is conserved after integration when the lateral energy flux vanishes. The interface term is available potential energy; the gradient terms are kinetic energy.

For a twice differentiable position away from the origin, write $\mathbf{r}=r\mathbf{u}$ with radial unit vector $\mathbf{u}$. Its unit-length constraint gives $\mathbf{u}\cdot \dot{\mathbf{u}}=0$ and $\mathbf{u}\cdot \ddot{\mathbf{u}}=-|\dot{\mathbf{u}}{|}^2$, separating radial and directional contributions to speed and kinetic energy. The identity is kinematic and does not require a particular force law.

For a homologous, isolated collapsing sphere, write its Newtonian gravitational potential energy as $U=-aG{M}^2/R$. At spherical-collapse turnaround the coherent kinetic energy vanishes. Conserving total energy and applying the virial theorem to the final equilibrium gives $U_{\mathrm{ta}}=U_{\mathrm{vir}}/2$ and hence $R_{\mathrm{vir}}=R_{\mathrm{ta}}/2$. If the gravitational structure coefficient changes, the ratio is $a_{\mathrm{vir}}/(2a_{\mathrm{ta}})$ instead. Mass loss, surface pressure, external tides and energy exchange also change this estimate.

Here $\Delta \psi =\psi (v^{\prime })+\psi (v_{\ast }^{\prime })-\psi (v)-\psi (v_{\ast })$ is the pairwise collision change, not a Laplace operator. Average over the two particle labels and use collision reversal to obtain the formula. Every collision invariant has $\Delta \psi =0$, giving mass, momentum and kinetic energy conservation. A second symmetrization replaces $f{f}_{\ast }\Delta \psi /2$ by $(f^{\prime }{f}_{\ast }^{\prime }-f{f}_{\ast })(-\Delta \psi )/4$.

The net work on a particle equals its change in kinetic energy: