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In ideal reduced magnetohydrodynamics with constant background sound and Alfvén speeds, perpendicular total-pressure balance gives $\delta p+{\rho }_0{v}_A^2\delta B_{\parallel }/B_0=0$. Put $s=\delta p/p_0-\gamma \delta \rho /{\rho }_0$, $r=\delta \rho /{\rho }_0+s/\gamma$, $c_T=v_A{c}_s/(v_A^2+c_s^2{)}^{1/2}$ and $h=c_s^{2}r/c_T$. Then the parallel momentum, continuity and induction equations yieldThus $f^{\pm }=u_{\parallel }\pm h$ satisfy $D{f}^{\pm }=\mp c_T{\nabla }_{\parallel }{f}^{\pm }$. The two slow-wave populations propagate along bent magnetic field lines and are mixed by the perpendicular Alfvénic turbulence, while the entropy fluctuation is simply advected. They do not force the perpendicular Alfvénic system at this leading order.

For a stationary, $z$-independent ideal magnetohydrodynamics flow with $E_z=0$, nonzero horizontal magnetic field and nonzero magnetohydrodynamic mass loading $k$, the displayed decomposition has $\mathbf{B}\cdot \nabla k=\mathbf{B}\cdot \nabla U=0$. The continuity equation gives the first invariant, and the ideal magnetohydrodynamic induction equation gives the second. Entropy and $V=u_z-B_z/({\mu }_{0}k)$ are also constant on each magnetic field line. Stationarity alone permits constant nonzero $E_z$, so it does not imply this decomposition.