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For inviscid barotropic flow, conservation gives ${\rho }_t+\nabla \cdot (\rho u)=0$ and $\rho (u_t+u\cdot \nabla u)=-\nabla p$. Linearizing about uniform rest ${\rho }_0,p_0$ giveswith the derivative adiabatic for ordinary sound.

Taking curl gives ${\partial }_t(\nabla \times u)=0$. Therefore the acoustic velocity is irrotational if the initial vorticity is zero. Uniform rest as the background does not alone forbid a superposed stationary vortical perturbation: for example $u=\epsilon (\sin y,0,0)$, ${\rho }^{\prime }=p^{\prime }=0$ solves the linearized equations but has nonzero curl. Thus the printed irrotational conclusion requires the usual restriction to the longitudinal acoustic mode. In that mode set $u=\nabla \varphi$ and absorb a time-only gauge into $\varphi$, so $p^{\prime }=-{\rho }_0{\varphi }_t$. Continuity givesThis separates the sound wave from the stationary vorticity mode in linear acoustics.

Define the quadratic acoustic energy density and acoustic energy flux byMultiplying momentum by $u$ and the pressure form of continuity by $p^{\prime }/({\rho }_0{c}_0^2)$ gives

For any progressive plane wave, not necessarily sinusoidal, let $\varphi =F(\xi )$ with $\xi =\hat{k}\cdot x-c_{0}t$. Then $u=F^{\prime }(\xi )\hat{k}$ and $p^{\prime }={\rho }_0{c}_0{F}^{\prime }(\xi )$. Its kinetic and compressional energy densities both equal ${\rho }_0(F^{\prime }{)}^2/2$, soThe energy is transported at the sound speed in the propagation direction. A superposition of oppositely traveling waves can form a standing wave and need not have this pointwise equality; the statement concerns a single progressive wave.

For a spherical outgoing wave, $\varphi =F(r-c_{0}t)/r$. Its radial velocity and pressure areThereforeThey are generally unequal because of the additional near-field velocity term; their difference is ${\rho }_0[F^2/r^4-2F{F}^{\prime }/r^3]/2$. They agree only to leading far-field order, or for special profile values. This is the near-field energy imbalance in a spherical acoustic wave.