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For an inviscid barotropic fluid with only potential body forces, the vorticity $\mathbf{\omega }=\nabla \times \mathbf{u}$ obeysThe pressure force is a gradient because ${\rho }^{-1}\nabla p=\nabla {\int }^{\rho }{p}^{\prime }(s)ds/s$. A spatially uniform specific entropy makes a fixed-composition fluid barotropic. Merely having $Ds/Dt=0$ does not exclude baroclinic vorticity generation from spatial entropy gradients.

Consider a circular cylinder of radius $a$ translating at speed $U$ parallel to a stationary wall through a minimum gap $d=ϵa/2$, with $0<ϵ\ll 1$. In the lubrication approximation, the local gap isso the narrow-gap length scale is $\ell \sim a\sqrt{ϵ}$. The leading tangential drag force per unit cylinder length scales aswhere $\mu$ is dynamic viscosity.

For angular velocity $\Omega$, the cylinder surface speed is of order $|\Omega |a$. The associated tangential force scale is $\mu |\Omega |a/\sqrt{ϵ}$, while the rotational torque per unit cylinder length has the additional moment arm $a$:These are distinct physical dimensions: force per length and torque per length.

The scaling follows by integrating a gap shear stress of order $\mu U/h$ over a distance $\ell$: $\mu U\ell /d\sim \mu U/\sqrt{ϵ}$. Translational pressure forces are of the same leading order and must be included for a complete drag calculation. Pressure on the circular surface gives no moment about its centre; the leading rotational resistance comes from tangential stress. The shear and pressure force partition in cylinder lubrication specifies those contributions and explains the parent torque-free cylinder in a lubrication gap result.

The pressure force on a plug-and-film control volume is $\pi a^2\Delta p$. Its opposing wall traction is $2\pi a\mu \int w_y(0)dx$, where $w_y(0)=4U/h-6Q/h^2$. Equating them for a freely moving body proves the displayed formula. Writing $s=2Q/U$, $h=sH$ and $x=\sqrt{s/\kappa }X$ converts it into $H(-\tilde{L})-H(\tilde{L})=C\lambda \int [4/(3H)-1/H^2]dX$, with $C=s/a$ and $\lambda =6\mu U/(\alpha \sqrt{\kappa }{s}^{5/2})$.

Use the source's Gaussian units and neglect the displacement current, as in nonrelativistic magnetohydrodynamics. Write $\Delta p=p_{\perp }-p_{\parallel }$ for the pressure anisotropy. The pressure tensor is $\mathsf{P}=p_{\perp }\mathsf{I}-\Delta p\mathbf{bb}$, so its force density isAmpère's law and the solenoidal magnetic-field constraint give the Lorentz force decompositionAdding these two force densities and using $\mathbf{BB}=B^2\mathbf{bb}$ yieldsThe first term is the gradient of the modified magnetohydrodynamic total pressure; the second is the anisotropic Maxwell stress tensor contribution together with the directional pressure force. Their effective coefficient is the pressure-anisotropy-modified magnetic tension, $\mathcal{T}=B^2/(4\pi )+\Delta p$. At fixed magnetic field, positive pressure anisotropy increases resistance to bending, while negative pressure anisotropy reduces it. When $\mathcal{T}$ becomes negative, bending experiences an amplifying force rather than a restoring force. Gradients of $\Delta p$ can also drive motion, so the curvature interpretation refers specifically to the bending contribution.

A thin spherical shell has outward pressure force $-4\pi r^2(dP/dr)dr$ and inward gravitational force $GM(r)4\pi r^2\rho (r)dr/r^2$. Hydrostatic equilibrium requiresA uniform-density solar model gives $M(r)=M_{\odot }(r/R_{\odot }{)}^3$, $\bar{\rho }=3M_{\odot }/(4\pi R_{\odot }^3)$, and after integration from $R_{\odot }$ to zero,Thus the appropriate orders of magnitude are $10^{14}$–$10^{16}\mathrm{Pa}$ and $10^7\mathrm{K}$. The uniform-density approximation underestimates the actual central pressure because the Sun is centrally concentrated. Radiation pressure is neglected in applying the ideal gas law.

A thin spherical shell has outward pressure force $-4\pi r^2(dP/dr)dr$ and inward gravitational force $GM(r)4\pi r^2\rho (r)dr/r^2$. Hydrostatic equilibrium requiresA uniform-density solar model gives $M(r)=M_{\odot }(r/R_{\odot }{)}^3$, $\bar{\rho }=3M_{\odot }/(4\pi R_{\odot }^3)$, and after integration from $R_{\odot }$ to zero,Thus the appropriate orders of magnitude are $10^{14}$–$10^{16}\mathrm{Pa}$ and $10^7\mathrm{K}$. The uniform-density approximation underestimates the actual central pressure because the Sun is centrally concentrated. Radiation pressure is neglected in applying the ideal gas law.

For a thin spherical shell, outward pressure force is $-4\pi r^2(dP/dr)dr$ and inward weight is $GM(r)\rho (4\pi r^{2}dr)/r^2$. Balance gives $\boxed{dP/dr=-GM(r)\rho /r^2}$. This assumes approximately spherical, slowly evolving structure without important rotational or magnetic support.

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.

In the translating-cylinder frame, a wall moves at $-U$ and the nearby cylinder surface at $W$. With gap $d+x^2/(2a)$ and equal outer pressures, lubrication theory gives the displayed forces with $J_1=\pi \sqrt{2a/d}$. Wall and cylinder shear forces alone do not balance because the sloping cylinder's pressure force is of the same order. Their complete sum is zero. Pressure exerts no moment about the circular cylinder center, while shear supplies the leading singular rotational resistance.