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Define the friction velocity by $u_{\ast }=(|{\tau }_w|/\rho {)}^{1/2}$, where ${\tau }_w$ is the total tangential wall stress. It is a stress-derived velocity scale, not the local mean velocity. In the overlap region, $y$ is small relative to the channel half-width, so the total stress is approximately its wall value. Molecular stress is small there, giving $-\overline{u^{\prime }{v}^{\prime }}\simeq u_{\ast }^2$.

Combining this Reynolds stress with the logarithmic shear gives $\mathcal{P}\simeq u_{\ast }^3/(\kappa y)$. For locally equilibrated wall turbulence, energy transfer to small scales occurs over the local eddy turnover time and is balanced by molecular dissipation; energy storage and the net energy-transport divergence are small in the leading overlap-layer balance. ThusThis is the local turbulent kinetic energy balance. It is a modeling approximation rather than a general identity for all wall positions. The constant-energy solution derived next is consistent with it: its modeled $k$ diffusion term is exactly zero, while the $ϵ$ diffusion term remains essential.

The constant Reynolds stress and the logarithmic shear imply the eddy viscosity ${\nu }_T=u_{\ast }\kappa y$. On the other hand the K-epsilon model gives ${\nu }_T=C_{\mu }{k}^2/ϵ$. Insert $ϵ=u_{\ast }^3/(\kappa y)$:Canceling the positive factors yields $C_{\mu }{k}^2=u_{\ast }^4$. Since energy is nonnegative, the log-layer solution of the k-epsilon model hasindependent of $y$ to this leading overlap-layer accuracy. In particular $k_y=0$, and the $k$ diffusion term vanishes, confirming the local production–dissipation balance in the energy equation.

Let $E(k)$ be the one-dimensional turbulent energy spectrum, normalized so that $\frac{1}{2}⟨|{\mathbf{u}}^{\prime }{|}^2⟩={\int }_0^{\infty }E(k)dk$. Large energy-containing eddies receive energy from the forcing. In the inertial range, nonlinear transfer passes this energy towards smaller scales at a mean rate $ϵ$ per unit mass. Dimensional analysis of Kolmogorov 1941 theory givesThe dissipation range begins near the inverse of the Kolmogorov length scale, where kinematic viscosity removes the transferred kinetic energy. The large-scale spectrum depends on the forcing; the rising low-$k$ branch in this sketch is illustrative, rather than a universal power law.

The tube has neither homogeneous nor isotropic large-scale turbulence: gravity selects a vertical direction, its walls constrain the motion, and its density gradient varies with height. The width $W$ bounds the transverse size of the energy-containing eddies. At sufficiently high Reynolds number, much smaller eddies may become approximately locally isotropic and exhibit an inertial range, even though the full tube flow does not satisfy the homogeneous-isotropic assumptions.

To obtain the Reynolds stress, write $u_i=U_i+u_i^{\prime }$ and $p=P+p^{\prime }$, with $⟨u_i^{\prime }⟩=0$. Averaging the incompressible Navier-Stokes equations uses $⟨u_i{u}_j⟩=U_i{U}_j+⟨u_i^{\prime }{u}_j^{\prime }⟩$ and gives the Reynolds-averaged momentum equationHere the final term is the mean buoyancy force, with the hydrostatic reference removed. Thus averaging the quadratic momentum flux introduces the covariance $⟨u_i^{\prime }{u}_j^{\prime }⟩$; multiplied by ${\rho }_0$, it has the dimensions of a stress. Its divergence transfers momentum between the mean flow and the fluctuations.

An eddy viscosity model represents this Reynolds stress byThe isotropic term can be absorbed into the mean pressure, while the deviatoric term gives a diffusion of mean momentum with turbulent viscosity ${\nu }_T\sim u_{\ast }\ell$, where $u_{\ast }$ and $\ell$ are representative turbulent speed and mixing length. Unlike kinematic viscosity, this coefficient depends on the flow. The closure assumes a local relation between stress and mean strain; anisotropic, buoyancy-driven large eddies may require a more detailed Reynolds stress model.

The magnetorotational instability extracts energy from differential rotation through magnetic tension. It can destabilize a disk whose specific angular momentum increases outward, even though such a disk satisfies Rayleigh's circulation criterion. The distinction is that an unmagnetized displaced parcel approximately conserves its own angular momentum, whereas a field line transfers angular momentum between parcels. If angular velocity decreases outward, an outward-displaced parcel lags its inner partner. Magnetic tension transfers angular momentum to the outer parcel and removes it from the inner one, making their radial separation increase. A sufficiently stiff magnetic connection suppresses the separation instead. The instability therefore requires a finite range of field-line-bending strengths.

Make this mechanism quantitative in the local shearing sheet of a cylindrical rotation profile $\Omega (r)$. Put $q=-d\log \Omega /d\log r$, and take $0<q<2$ so the unmagnetized flow is stable. Its radial epicyclic frequency is ${\kappa }^2=2(2-q){\Omega }^2$. Let the uniform field be vertical, $B_z{\mathbf{e}}_z$, with constant mass density, and use an axisymmetric horizontal disturbance proportional to $e^{st+ikz}$. Write $\mathbf{h}=\mathbf{b}/\sqrt{{\mu }_0\rho }$ and ${\omega }_A=k{B}_z/\sqrt{{\mu }_0\rho }$. The linearized ideal magnetohydrodynamics equations areThe first pair contains Coriolis acceleration and the epicyclic response; the last equation is the shear-winding of a radial field perturbation into an azimuthal one. Omitting that winding term would remove the driving mechanism.

Eliminating $h_r,h_{\varphi }$ gives $(s^2+{\omega }_A^2)u_r=2\Omega s{u}_{\varphi }$ and $s(s^2+{\omega }_A^2)u_{\varphi }+\Omega [(2-q)s^2-q{\omega }_A^2]u_r=0$. Their determinant is the ideal magnetorotational dispersion relationIts discriminant as a quadratic in $s^2$ is ${\kappa }^4+16{\Omega }^2{\omega }_A^2>0$. Since $2{\omega }_A^2+{\kappa }^2>0$, a positive $s^2$ exists precisely when its constant term is negative:Thus angular velocity, rather than specific angular momentum, must decrease outward. A Keplerian disk has $q=3/2$, ${\kappa }^2={\Omega }^2$ and is hydrodynamically stable, yet magnetic modes with $0<k^2{v}_A^2<3{\Omega }^2$ grow.

The growing root is $s^2=-{\omega }_A^2-{\kappa }^2/2+\frac{1}{2}\sqrt{{\kappa }^4+16{\Omega }^2{\omega }_A^2}$. Differentiation with respect to ${\omega }_A^2$ gives the maximum growth rate of ideal magnetorotational instability:For Keplerian shear these are $3\Omega /4$ and $15{\Omega }^2/16$. The ideal maximum is set by shear; reducing field strength reduces the fastest wavelength instead of reducing this maximum growth rate. This is why the exactly zero-field problem differs from a sequence of weak-field problems with progressively larger wave number. At fixed wave number, growth does tend to zero as the field vanishes. The wavelength must fit the actual disk, and diffusion eventually matters at sufficiently short scales.

The instability has the required angular-momentum-transfer sign. In a growing mode, $u_{\varphi }/u_r=(s^2+{\omega }_A^2)/(2\Omega s)>0$. At the fastest mode this ratio is one, while induction gives $h_{\varphi }=-h_r$ after allowing for their common Fourier phase. Thus the averaged Reynolds stress minus Maxwell stress tensor, $\rho ⟨u_r{u}_{\varphi }-h_r{h}_{\varphi }⟩$, is positive: angular momentum moves outward. Its product with $q\Omega$ is the energy supplied by the background shear. These local mechanisms and coefficient conventions are also checked in the primary MRI stability calculation.

More general fields require separating field-line bending from field gradients and curvature. At leading local order, for a stationary compatible field in an incompressible flow with axisymmetric wavevector $(k_r,0,k_z)$, the bending frequency isPressure elimination projects the radial force by $f=k_z^2/k^2$. The radial momentum equation becomes $s{u}_r-2f\Omega u_{\varphi }=i{\omega }_A{h}_r$; the azimuthal and induction equations retain the form above. Eliminating the amplitudes gives the oblique axisymmetric magnetorotational dispersion relationHere $k_z\ne 0$. The unstable range is $0<{\omega }_A^2<-fd{\Omega }^2/d\log r$. The poloidal magnetic field therefore determines whether an axisymmetric perturbation bends a field line. A locally uniform toroidal magnetic field does not enter $\mathbf{k}\cdot \mathbf{B}$ for axisymmetry, and can contribute magnetic pressure that is projected out of this incompressible leading-order problem. Thus a toroidal component does not suppress the weak-field MRI merely by being present. Conversely a purely toroidal field has no ordinary axisymmetric local bending MRI in this approximation. This does not imply that every globally curved toroidal equilibrium is stable.

A steady general poloidal magnetic field must also be compatible with the equilibrium rotation. The azimuthal ideal magnetohydrodynamic induction equation contains ${\partial }_t{B}_{\varphi }=r{\mathbf{B}}_p\cdot \nabla \Omega$. A stationary axisymmetric equilibrium with no meridional flow consequently requires ${\mathbf{B}}_p\cdot \nabla \Omega =0$, Ferraro's law of isorotation. A radial field crossing cylindrical differential-rotation surfaces would otherwise wind secularly; it cannot simply be substituted into a stationary normal-mode calculation. Boundary conditions and the along-field eigenvalue problem replace a freely chosen local $k$ in nonuniform or bounded equilibria. Strong enough field-line tension can then stabilize all admissible wavelengths.

An explicit axisymmetric effect of nonuniform field strength and curvature is obtained with a purely toroidal field $B_{\varphi }(r){\mathbf{e}}_{\varphi }$, constant mass density and cylindrical rotation. A radial fluid displacement changes the field byThis follows directly from $\mathbf{b}=\nabla \times (\mathbf{\xi }\times \mathbf{B})$ and incompressibility. Its radial curvature force, after absorbing the magnetic-pressure perturbation, is $-2B_{\varphi }{b}_{\varphi }/({\mu }_0\rho r)$. The axisymmetric azimuthal momentum equation is $s{u}_{\varphi }+(2\Omega +r{\Omega }^{\prime })u_r=0$. Combining it with the radial force and then projecting out pressure givesThis is the local Michael criterion for axisymmetric toroidal-field interchange. The new term can either stabilize or destabilize: for $B_{\varphi }\propto r^p$, $\mathcal{D}={\kappa }^2+2(1-p)v_{A\varphi }^2/r^2$. A current-free $B_{\varphi }\propto r^{-1}$ is stabilizing; $B_{\varphi }\propto r$ leaves this axisymmetric interchange coefficient unchanged; a sufficiently steep increase with $p>1$ can destabilize even a centrifugally stable rotation law. That last instability can persist without differential rotation, so its energy source is the magnetic-current equilibrium rather than the ordinary MRI. Nonaxisymmetric current-driven modes are outside the requested essay scope.

For mixed poloidal and toroidal fields, the full induction perturbation contains $(\mathbf{B}\cdot \nabla )\mathbf{\xi }-(\mathbf{\xi }\cdot \nabla )\mathbf{B}-\mathbf{B}\nabla \cdot \mathbf{\xi }$, and the linear Lorentz force contains both $(\nabla \times \mathbf{b})\times \mathbf{B}$ and $(\nabla \times \mathbf{B})\times \mathbf{b}$. These terms explain why gradients and curvature cannot be replaced everywhere by a scalar $k^2{v}_A^2$: they can draw on equilibrium-current energy and couple the radial and vertical motions. Compressibility introduces additional magnetic pressure and magnetosonic responses; a strong toroidal field can alter the dispersion even though it disappeared from the leading incompressible weak-field result.

Finally, a vertically or radially stratified general field also supports part of the equilibrium weight through magnetic pressure. A displaced flux tube carries its frozen-in mass-to-flux relation, so its mass density at the surrounding total pressure can differ from its new environment. This introduces magnetic buoyancy instability and competition with ordinary buoyancy. Axisymmetric interchange displacements can access this energy without an azimuthal wave number, while poloidal tension resists variation along the field. Growth that survives when $d\Omega /dr=0$ must not be attributed to shear-driven MRI. The general-field stability problem therefore combines shear, field-line tension, current/curvature forces, compressibility, buoyancy and boundary constraints; the uniform-field result isolates the basic angular-momentum feedback mechanism.

Let the Eulerian time average commute with space and time derivatives and let the statistically steady mean have no time derivative. Define $U=\overline{u}$, $V=\overline{v}$, $\hat{u}=u-U$ and $\hat{v}=v-V$, so $\overline{\hat{u}}=\overline{\hat{v}}=0$. Averaging instantaneous incompressibility givesand subtracting gives ${\hat{u}}_x+{\hat{v}}_y=0$ for the fluctuations. Start from the instantaneous streamwise Navier-Stokes equationUse incompressibility to write $u{u}_x+v{u}_y={\partial }_x(u^2)+{\partial }_y(uv)$. The averaged products are $\overline{u^2}=U^2+\overline{{\hat{u}}^2}$ and $\overline{uv}=UV+\overline{\hat{u}\hat{v}}$. The mean conservative terms reduce back to $U{U}_x+V{U}_y$ because $U_x+V_y=0$. The Reynolds-averaged momentum equation is consequentlyThe signs use the positive covariance convention for the Reynolds stress. In particular, both quadratic averages involve the fluctuations, not products of the mean $U$ with a fluctuation; the converted TeX loses the overbar/hat structure here.

Let $L$ be the streamwise evolution scale, $\delta$ the jet width, and $U_c$ its velocity scale. A slender jet has $\epsilon =\delta /L\ll 1$; mean incompressibility gives $V=O(\epsilon U_c)$. Both mean advection terms are then $O(U_c^2/L)$. For turbulent parcels crossing a transverse distance $l$, retention of their original streamwise momentum gives a fluctuation $\hat{u}\sim -l{U}_y$; a comparable transverse fluctuating speed gives the down-gradient mixing length closureThe mixing length is an effective distance of momentum retention, with statistical coefficients absorbed into $l$; it is not the net transverse displacement of a mean streamline. This closure models the Reynolds stress rather than deriving it exactly.

In a slender equilibrium jet, the fluctuation magnitude inferred from momentum balance is $u^{\prime 2}\sim U_c^2\delta /L=\epsilon U_c^2$. The transverse Reynolds stress gradient then has size $u^{\prime 2}/\delta \sim U_c^2/L$, whereas the streamwise normal-stress gradient is $u^{\prime 2}/L\sim \epsilon U_c^2/L$. At sufficiently high Reynolds number, molecular transverse diffusion is smaller providedStreamwise molecular diffusion is smaller still by ${\epsilon }^2$. A free jet has ambient mean pressure constant to leading order; transverse momentum corrections imply only a higher-order mean streamwise pressure force. Thus high speed must be accompanied by the slender-jet ordering, not used alone to discard every other term.

For a symmetric jet decreasing away from the centre, $U_y<0$ on $y>0$ and $U_y>0$ on $y<0$. Accordingly $\overline{\hat{u}\hat{v}}=\mathrm{sgn}(y)l^2(U_y{)}^2$ off the centreline, and the leading equation isAt the centreline the signed form requires the stress to tend continuously to zero; otherwise differentiation would add a distributional term. The undifferentiated Reynolds stress form is the appropriate global equation.

Use mean incompressibility to convert advection into momentum-flux divergence:Therefore the reduced Reynolds-averaged momentum equation givesFor a localized free jet, $UV$ and the Reynolds stress tend to zero in the quiescent ambient as $y\to \pm \infty$. Assume sufficient decay to differentiate its momentum integral. Integrating across the jet givesThus $M_0=\int U^{2}dy$ is constant. It is a specific momentum flux per unit span: multiplication by density gives the momentum flux. A clipped profile with nonzero stress just inside its artificial edge would fail this boundary-flux assumption unless an additional external momentum source were supplied.

First derive the formal turbulent plane jet similarity equations, then check the printed edge conditions. Define a streamfunction by $U={\psi }_y$, $V=-{\psi }_x$, so mean incompressibility is automatic. Momentum conservation gives $U_c^2\delta \sim M_0$. With $l=C_1|y|$, the closure balance $U_c^2/x\sim C_1^2{U}_c^2/\delta$ gives $\delta \propto x$. Hence $U_c\propto M_0^{1/2}{x}^{-1/2}$ and the consistent self-similar ansatz isAn odd $f$ gives symmetric $U$ and antisymmetric $V$. Direct differentiation givesThe advection and Reynolds stress terms becomeThus the two required consistency equations, the first for $\eta >0$ and the second for a symmetric finite-width profile, areThe second is momentum normalization; it is the equation referenced as (2) in the PDF, although its display in part (d) has no printed number. Use an infinite upper limit for a decaying profile of infinite support. Symmetry sets $f(0)=0$. A finite centreline velocity and continuous zero centreline stress give the integrated first equationThese equations establish the similarity reduction. They also give a direct finite-edge stress condition for a mixing-length jet. At a finite edge ${\eta }_w>0$, $U=0$ requires $f^{\prime }({\eta }_w)=0$. For finite $f({\eta }_w)$, the integrated equation forces $f^{\prime \prime }({\eta }_w)=0$. A nontrivial decreasing velocity profile has nonzero shear somewhere inside, so zero edge shear cannot be its maximum. Allowing a one-sided nonzero edge shear instead produces nonzero edge stress and an unbalanced jump in the conservative momentum equation.

For a concrete diagnostic, take $f^{\prime }=1-{\eta }^2$ on $0\le \eta \le 1$, so $f=\eta -{\eta }^3/3$. It has zero edge velocity and its largest one-sided shear magnitude at the edge, matching that descriptive condition. Yet at $\eta =1$ the integrated equation has left-hand side zero and right-hand side $8C_1^2>0$: such a profile is not a solution of the stated momentum closure.

The centreline is also problematic for a smooth interpretation. If $f^{\prime }(0)=a>0$, then $f\sim a\eta$, and the decreasing branch of the integrated equation givesA continuous formal profile can have this cusp, but it is not a differentiable symmetric velocity field, and molecular diffusion cannot be neglected uniformly in that core. Therefore there is no nontrivial regular similarity solution satisfying all the printed edge requirements. The boxed consistency equations are valid; the additional coincidence of a zero-velocity edge with maximum shear is not. A consistent jet model must relax that edge definition and regularize the core or use a different mixing length.

An Eulerian time average of an incompressible constant-density Navier-Stokes equation gives the displayed mean momentum equation. Decompose velocity into its mean and a zero-mean fluctuation. In conservative form, the averaged momentum flux is $\overline{u_i{u}_j}=U_i{U}_j+\overline{u_i^{\prime }{u}_j^{\prime }}$; mean incompressibility then turns its first term into mean advection. The fluctuation covariance is the Reynolds stress divided by density, using the positive covariance convention. Closing this extra unknown requires a turbulence model.

In zonally averaged barotropic quasi-geostrophic approximation flow with zero mean meridional transport, the Taylor identity for quasi-geostrophic flux relates the meridional potential vorticity flux to the convergence of Reynolds stress. A local downgradient closure $\overline{v^{\prime }{q}^{\prime }}=-\kappa {\overline{Q}}_y$, with $\kappa \ge 0$, therefore gives westward mean acceleration where the background gradient is positive. Rossby wave breaking can stir PV into fine filaments; coarse-graining or weak diffusion produces the inferred mixing. Globally the momentum-flux convergence integrates to boundary stresses, so localized westward changes need compensating transport or momentum changes elsewhere in an isolated system.

For complex velocity amplitudes $v_x=v_y=V$ and magnetic amplitudes $b_x=-b_y=B$ in Alfvén-speed units, phase averaging gives the displayed stress. For fixed horizontal velocity energy, equal radial and azimuthal components maximize their positive Reynolds stress. Opposite equal magnetic components maximize the outward tension contribution of the Maxwell stress tensor, since $-B_x{B}_y$ is then positive. These conclusions follow from $2|ab|\le |a{|}^2+|b{|}^2$. The vertical channel transports angular momentum outward and extracts energy from decreasing angular velocity.