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At leading thin-shell order, a spherical balloon with reference radius $a_0$, thickness $h_0$ and current radius $a=\lambda a_0$ has reference elastic energy $4\pi a_0^2{h}_{0}W(\lambda ,\lambda ,{\lambda }^{-2})$. Equating its differential to pressure work $p4\pi a^{2}da$ gives the displayed pressure difference. Its thickness becomes $h_0{\lambda }^{-2}$ by incompressibility. The same result follows from the spherical membrane force balance $p=2h{\sigma }_t/a$, with current tangential Cauchy stress tensor component ${\sigma }_t=\lambda N$ and the equibiaxial nominal tension of an incompressible sheet. Thin-shell validity requires $h/a\ll 1$.

Set ${\partial }_t\mathbf{u}=0$ and $\nu =0$ in the Navier-Stokes equation. Dotting the resulting Euler equation with a displacement $d\mathbf{r}$ tangent to a streamline givesHere mass density $\rho$ is constant, and $B$ is constant along each streamline. The terms are kinetic energy per unit mass, the pressure work per unit mass available from pushing a parcel through the flow, and gravitational potential energy per unit mass. Their sum is conserved mechanical energy per unit mass; $p/\rho$ is not an additional independently conserved parcel energy. Different streamlines can have different $B$ unless, for example, the flow is also irrotational in a connected region.

Choose a smooth radial bump $f$ supported inside a shell where the stellar buoyancy frequency has $N^2<0$. The displayed fluid displacement has $\nabla \cdot \mathbf{\xi }=(\rho g/\gamma p){\xi }_r$, hence its Eulerian pressure perturbation vanishes identically. Its buoyancy contribution to the stellar displacement energy identity is strictly negative and self-gravity is nonpositive. The variational principle therefore proves convective instability with no unsupported assumption that incompressibility alone removes pressure work. Large spherical-harmonic degree also makes perturbed self-gravity small.

Pressure work is mechanical work transferred by pressure forces acting through a moving boundary. For a piston of area $A$ moving through $dx$, $dV=Adx$ and the opposing applied force is $p_{\mathrm{ext}}A$. The work done by the fluid is thereforeThe familiar $\delta W=pdV$ uses the system pressure when it equals the applied boundary pressure, as in a reversible mechanical step. In a free expansion into vacuum, $p_{\mathrm{ext}}=0$ and the boundary work vanishes even if the gas initially has nonzero internal pressure.

In steady incompressible flow, the Bernoulli equation contains $p/\rho$ as flow work per unit mass. This accounts for pressure forces at a control volume's entrances and exits; it is not an independently conserved internal energy of an isolated parcel. The first law of thermodynamics keeps the complete work and energy balance.

For constant mass density $\rho$, kinematic viscosity $\nu >0$, ideal induction, zero boundary velocity and tangent magnetic field, put $E={\int }_D[\rho |\mathbf{v}{|}^2/2+|\mathbf{B}{|}^2/(2{\mu }_0)]dV$. Multiplying the momentum equation by $\mathbf{v}$ and integrating cancels pressure work and advective flux. The magnetic-energy derivative is minus the integral of Lorentz force density dotted with velocity, canceling the same work in the kinetic-energy equation. Integration by parts then gives the displayed identity. For incompressible flow with zero boundary velocity, $\int |\nabla \mathbf{v}{|}^2=\int |\nabla \times \mathbf{v}{|}^2$.