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For constant isothermal sound speed, time-independent shearing-sheet tidal potential ${\Phi }_t$, and inviscid flow, define $E=\Sigma u^2/2+c_s^2\Sigma \ln (\Sigma /{\Sigma }_{\mathrm{ref}})+\Sigma {\Phi }_t$. The barotropic energy density and kinetic energy balances combine to give ${\partial }_{t}E+\nabla \cdot [(E+P)\mathbf{u}]=0$. Coriolis acceleration does no work. Integral conservation requires vanishing net boundary energy flux.

Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.

Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative. For any specific quantity $f$, conservation of mass givesDotting the momentum equation with $\Sigma \mathbf{u}$ produces the kinetic energy balanceThe Coriolis acceleration does no work because $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$. Since the shearing-sheet tidal potential is time-independent, its advected potential-energy density obeysFor the isothermal equation of state, define the barotropic energy density $U=c_s^2\Sigma \ln (\Sigma /{\Sigma }_{\mathrm{ref}})$, with fixed positive reference surface density ${\Sigma }_{\mathrm{ref}}$. The continuity equation impliesAdding these balances combines the two pressure terms into $-\nabla \cdot (P\mathbf{u})$. This establishes isothermal shearing-sheet energy conservation. Thus the conserved energy density and its flux areChoosing the density unit so that ${\Sigma }_{\mathrm{ref}}=1$ reproduces the printed logarithm. Changing the reference adds a multiple of the conserved mass to $E$ and its advective energy flux. The barotropic energy density is the mathematical energy of this fixed-isothermal sound speed closure; it is not the microscopic thermal energy of a thermally isolated gas. Maintaining an isothermal process can require heat exchange.

Let $\mathbf{u}=-(3/2)\Omega x{\mathbf{e}}_y+\mathbf{v}$ and $\rho ={\rho }_0(1+\delta )$, with $\delta$ and $\mathbf{v}$ small. Define the sound speed by $c^2=(dp/d\rho {)}_0$, assumed positive. The background Keplerian shear balances the radial Coriolis force and the shearing-sheet tidal potential. For disturbances preserving axisymmetry ${\partial }_y=0$, its advective derivative ${\mathbf{u}}_0\cdot \nabla$ vanishes, but $\mathbf{v}\cdot \nabla {\mathbf{u}}_0=-(3/2)\Omega v_x{\mathbf{e}}_y$ remains. The linearized Euler equations and continuity equation areThe azimuthal coefficient is $(2-3/2)\Omega =\Omega /2$, not $2\Omega$: it combines the background velocity gradient with the Coriolis force. The uniform background has no buoyancy or vertical stratification in this stated model.

Let $K={\rho }_0|\mathbf{u}{|}^2/2$ be the kinetic energy per unit volume. Dot the momentum equation with ${\rho }_0\mathbf{u}$. The Coriolis acceleration does no work because $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$. Incompressibility turns the advection and pressure terms into divergences, yieldingThus $\mathbf{F}=(K+P)\mathbf{u}$ is the energy flux and $S={\rho }_0{\Omega }^2(3x-2qz)u_x$ is the source in the requested kinetic-energy form.

To decide about conservation of energy, absorb the ordinary radial shearing-sheet tidal potential ${\Phi }_t=-3{\Omega }^2{x}^2/2$ into the density. Its material derivative cancels the $3x$ work term:For $q=0$ this is the usual conservative Jacobi energy in a shearing sheet, subject to vanishing boundary energy flux. For $q\ne 0$, the extra body force has nonzero curl, $\nabla \times [-2{\Omega }^{2}qz{\mathbf{e}}_x]=-2{\Omega }^{2}q{\mathbf{e}}_y$, so no scalar potential can absorb it. A potential $2{\Omega }^{2}qxz$ would necessarily introduce a vertical force $-2{\Omega }^{2}qx{\mathbf{e}}_z$, absent from the model. Therefore this vertical-shear model does not generically conserve the standard total mechanical energy: its maintained shear/irradiation background can supply or remove energy. This is the energy balance of the incompressible vertical-shear model.

Let $\mathbf{\zeta }=\nabla \times \mathbf{u}$ and $\mathbf{\omega }=\mathbf{\zeta }+2\Omega {\mathbf{e}}_z$ be the relative vorticity and absolute vorticity. The nonlinear acceleration identity rewrites the momentum equation asTaking the curl removes the pressure, kinetic energy gradient and shearing-sheet tidal potential. ThusBoth $\nabla \cdot \mathbf{u}$ and $\nabla \cdot \mathbf{\omega }$ vanish. The vorticity equation is consequentlyFor the absolute-vorticity flux tensor $T_{ij}=u_i{\omega }_j-{\omega }_i{u}_j$, use the flux convention $(\nabla \cdot \mathbf{T}{)}_j={\partial }_i{T}_{ij}$, with the first index denoting transport direction. This givesThe scalar $\theta$ is a buoyancy displacement variable, so dimensional consistency gives it units of length; the printed potential-temperature terminology presupposes this normalization. Its curl source directly forces only the $y,z$ components of absolute vorticity. Hence radial vorticity conservation in a shearing sheet takes the local flux formIts volume integral is constant when the net boundary flux vanishes. However $D_t{\omega }_x=\mathbf{\omega }\cdot \nabla u_x$ still permits vortex stretching; the radial component is not generally a materially conserved scalar in three dimensions. Using divergence on the second index of the stated tensor would reverse the transport sign, so the tensor convention matters.

For unit mass, the Lagrangian in cylindrical coordinates isIntroduce shearing sheet coordinates by $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Evaluate all derivatives of $\Phi$ at $(r_0,0)$. The reference circular orbit obeys ${\Phi }_r=r_0{\Omega }_0^2$, while midplane symmetry gives ${\Phi }_z={\Phi }_{rz}=0$. Its Taylor expansion isExpanding the kinetic energy to the same order givesThe terms linear in $x$ cancel by circular-orbit balance. Discard the constant and the term $r_0{\Omega }_0\dot{y}$ by total-time-derivative invariance of a Lagrangian; these do not change the Euler-Lagrange equations. The particle Lagrangian in a shearing sheet is thereforewith shearing-sheet tidal potentialThe second form follows from ${\Phi }_r=r{\Omega }^2$, the orbital shear parameter $q_0=-d\log \Omega /d\log r$ and the vertical epicyclic frequency ${\Omega }_z^2={\Phi }_{zz}$.

In the rotating shearing sheet, write the background orbital shear as ${\mathbf{U}}_0=-q\Omega x{\mathbf{e}}_y$. The rotating momentum equation contains Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{u}$ and the shearing-sheet tidal potential; the background shear balances the radial tidal acceleration.

SetGauss's law for magnetism gives ${\partial }_z{B}_z=0$, and the vertical ideal magnetohydrodynamic induction equation gives ${\partial }_t{B}_z=0$. Horizontal invariance and $v_z=0$ remove the nonlinear horizontal advection. The remaining background-shear term is $(\mathbf{v}\cdot \nabla ){\mathbf{U}}_0=-q\Omega v_x{\mathbf{e}}_y$. The horizontal magnetic tension is $B_z{\partial }_z{\mathbf{B}}_h/{\mu }_0$, while horizontal pressure gradients vanish. Subtracting background balance yieldsFor ideal magnetohydrodynamics, the ideal magnetohydrodynamic induction equation is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}$. Its horizontal components giveThese horizontally invariant magnetized shearing-sheet equations are exact within the stated local, incompressible ansatz, even for finite horizontal amplitudes. The vertical equation determines the pressure needed to balance vertical gravity and magnetic pressure; it does not add another horizontal evolution equation.

For a Keplerian shearing sheet, the shear rate is $S=3\Omega /2$ and the shearing-sheet tidal potential isA uniform steady solution isThe Coriolis acceleration of this linear shear flow balances the radial tidal acceleration. The only nonzero background component of the viscous stress tensor that matters is $T_{xy}=-\nu ({\Sigma }_0){\Sigma }_{0}S$, which is spatially constant. Hence $\nabla \cdot \mathbf{T}=0$: a local uniform patch has no stress gradient or torque divergence to drive an accretion flow.

Use a local Cartesian frame rotating at $\Omega =\sqrt{GM/r_0^3}$ about a circular orbit at $r_0$, with $x=r-r_0\ll r_0$, and neglect curvature beyond leading order. The radial gravitational-plus-centrifugal potential isIts first derivative vanishes at $r_0$ and its second derivative there is $-3{\Omega }^2$. Dropping a constant gives the shearing-sheet tidal potential

A dilute dust component can be approximated as a pressureless fluid with velocity $\mathbf{v}$ and surface density of a disk ${\Sigma }_d$. In a gas velocity $\mathbf{u}$, linear drag force gives acceleration $-(\mathbf{v}-\mathbf{u})/t_s$ for positive aerodynamic stopping time $t_s$. Its equations include Coriolis acceleration, the shearing-sheet tidal potential, and ${\partial }_t{\Sigma }_d+\nabla \cdot ({\Sigma }_d\mathbf{v})=0$. Neglecting gas backreaction is justified only when the prescribed-gas approximation is appropriate. Strong orbit crossing or appreciable dust velocity dispersion can invalidate the single-valued pressureless-fluid description.