Predict the growth, then follow its energy

Prerequisites are the rotating-frame coordinates and four real amplitudes from the preceding chapter, exponential growth, and solving a quadratic equation. We first derive a growth rate, then calculate two stresses, and finally compare their work with the energy change. Keep $x$ outward, $y$ along rotation, $u$ multiplying sine, and $b$ multiplying cosine. Magnetic amplitudes retain velocity units.

Replace time derivatives with a growth rate

For a mode whose four amplitudes all have time dependence $e^{st}$, a time derivative multiplies the amplitude by $s$. A positive real $s$ gives exponential growth. The radial induction equation gives $sb_x=Ku_x$. Combining it with radial momentum gives $2\Omega s u_y=(s^2+K^2)u_x$. To eliminate the azimuthal magnetic amplitude, multiply azimuthal momentum by the growth rate and use azimuthal induction. Replace the radial magnetic amplitude with its induction relation. For a nonzero radial mode, the remaining steps are

The second line uses the radial momentum relation above. Expanding the square produces

All terms have units of inverse time to the fourth power. The elimination divides by $s$, so neutral endpoints should also be checked in the original four equations. This polynomial is the unstratified, vertical-wave specialization of Balbus and Hawley's dispersion relation, equations 2.5–2.9. Their frequency convention uses oscillatory time dependence; replacing that frequency with an exponential rate accounts for the changed signs.

Set $y=s^2$. The growing branch is the positive root

For Rayleigh-stable rotation, $q<2$, exponential growth requires $q>0$ and $0<K^2<2q\Omega^2$. At zero coupling, the positive rate goes to zero. At the upper endpoint, tension prevents this exponential branch. Inside the interval, differentiating the root with respect to $K^2$ gives

This maximum varies the wavelength or tension frequency. At a fixed wavelength, weakening the field takes $K$ toward zero and eventually slows growth. A zero exponential rate can describe oscillatory or neutral modes; arbitrary initial disturbances may also have transient or secular responses.

Worked example with an exactly checkable root

Take Keplerian shear $q=3/2$, $\Omega=1/T_0$, and $K=0.6/T_0$. Then $\kappa=1/T_0$. Writing $Y=s^2T_0^2$, the polynomial becomes

The positive root is exactly $Y=0.44$: substituting gives $0.1936+0.7568-0.9504=0$. Thus $s=\sqrt{0.44}/T_0$, approximately $0.663325/T_0$. An amplitude doubles in $\ln2/s$, about $1.04T_0$. Its quadratic energy doubles sooner, in $\ln2/(2s)$, about $0.522T_0$.

The saved real-space calculation uses the same continuum seed on three meshes. Its reported growth estimates approach this reference:

Doubling the mesh reduces this field error by nearly four, consistent with the centered spatial derivative used in these particular calculations. The time step was held fixed in this comparison. A separate temporal refinement gave error ratios near sixteen. These refinement results apply to the saved linear cases.

The fastest Keplerian mode fixes two signs

For $q=3/2$, the fastest tension frequency is $\sqrt{15}\Omega/4$, and the growth rate is $3\Omega/4$. Its four amplitudes can be chosen as

The two velocities have the same sign; the two magnetic amplitudes have opposite signs. Average over a full vertical wavelength. Since the averages of sine squared and cosine squared are both one-half, the outward Reynolds and Maxwell stresses are

Both have units of pressure, or momentum flux. Both are positive for this mode. Substituting the amplitude vector gives $T_{\mathrm{Re}}=\rho A^2/2$, $T_{\mathrm{Max}}=5\rho A^2/6$, and a ratio of $5/3$. This ratio follows from the specific mode. Other initial disturbances can have different stress ratios.

Derive the energy source from the same equations

The vertically averaged perturbation energy density is

Multiply each amplitude equation by its own amplitude and by $\rho/2$, then add. The tension terms cancel between fluid and magnetic energy. The remaining rotation terms give the Reynolds contribution, and winding the radial field gives the Maxwell contribution:

Stress has units of energy per volume; multiplying it by shear rate gives power per volume. Magnetic tension exchanges energy internally, and the maintained background shear supplies the net increase. A growing perturbation energy is therefore compatible with the model's energy equation.

For the fastest vector, $E=4\rho A^2/3$, equal to the total stress. Its energy grows at rate $2s=3\Omega/2$, exactly the rate predicted by the work identity. This is an independent check of the factor of two between amplitude growth and energy growth.

Read an actual energy comparison

For the saved fastest case with initial amplitude $10^{-4}L_0/T_0$, the initial energy is approximately $1.33333\times10^{-8}\rho_0(L_0/T_0)^2$. At $t=3T_0$, it is approximately $1.20023\times10^{-6}\rho_0(L_0/T_0)^2$. The saved independently integrated shear work is approximately $1.18690\times10^{-6}\rho_0(L_0/T_0)^2$, matching the increase.

Here $\rho_0$ is the reference density used for the normalized field record. The unrounded record gives an energy-work residual, normalized by initial energy, of about $2.29\times10^{-11}$. The rounded numbers above cannot reproduce that small residual. The plotted values come from the full saved fields and work record. Work was integrated from stage stresses; it was not defined as the difference of endpoint energies.

Rigid rotation gives a revealing control. Its saved disturbance has a nonzero stress at the final time, but $q=0$, so the shear-work rate is zero. Its total perturbation energy remains constant. Energy extraction depends on the product of stress and shear rate.

Exercises with explained answers

1. Two perturbations have positive amplitude growth rates, but one plot shows energy. Which exponent belongs on that plot? Energy is quadratic in the amplitudes, so its exponential rate is twice the amplitude rate. Check whether the plotted quantity is a component, a norm or an energy before fitting a slope.

2. At fixed vertical wavelength, make the magnetic field arbitrarily weak. Does the fastest possible rate remain attached to that wavelength? No. The fixed wavelength has $K=kv_A$, which tends to zero. In the weak-coupling limit the growing rate approaches $\sqrt{q/(2-q)}K$. The maximum over all allowed wavelengths involves changing the wavelength as the field changes. A finite periodic box supplies only its permitted wavelengths.

3. A program reports a zero-field Keplerian kick with steadily increasing oscillation amplitude. What should you check first? Compare with the epicyclic solution in the preceding chapter. Its invariant is $u_x^2+4u_y^2$. A numerical method can spuriously amplify the Coriolis coupling. The growing curve alone is insufficient evidence for magnetic instability when the magnetic coupling has been set to zero.

4. The saved channel keeps growing. Can you use its last stress to predict a disk's steady accretion rate? This calculation supplies a local linear stress during a prescribed small disturbance. A steady disk rate also needs the global mass and angular-momentum flow and a justified nonlinear transport state. Those ingredients are outside this record.

Sources

Balbus and Hawley (1991), printed pages 216–218, equations 2.2,2.5–2.9 and2.13–2.14b. Primary scan. Stone and Gardiner, arXiv:1006.0139v1, Sections II–III, especially equations 7–13 for shear and epicyclic checks. Pinned version. Numerical examples use the accepted eleven-case periodic linear-field archive and its existing forty-five-gate qualification. Reading and plotting those records adds no evolution step.