---
title: "Predict the growth, then follow its energy"
description: "Derive the channel growth rate, separate velocity and magnetic stresses, and compare energy gain with shear work in one saved field."
canonical_url: "https://sandboxphysics.com/stories/magnetorotational-instability/growth-stress-and-energy/"
language: "en"
page_type: "story"
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  zh-CN: "https://sandboxphysics.com/zh/stories/magnetorotational-instability/growth-stress-and-energy/"
last_modified: "2026-11-12"
---

L2 · Derivation and worked examples

# Predict the growth, then follow its energy

Sandbox Physics · 2026-11-12

The four channel equations predict a growing disturbance when outward-decreasing rotation overcomes magnetic tension at the chosen wavelength. The same equations predict the direction of angular-momentum transport and the energy drawn from shear. These predictions can be checked against one set of saved velocity and magnetic fields.

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

$$
\begin{gathered}
s(s^2+K^2)u_y=\Omega[-(2-q)s^2+qK^2]u_x,\\
(s^2+K^2)^2+\kappa^2s^2-2q\Omega^2K^2=0.
\end{gathered}
$$

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

$$
\begin{gathered}
s^4+(\kappa^2+2K^2)s^2+K^2(K^2-2q\Omega^2)=0,\\
\kappa^2=2(2-q)\Omega^2.
\end{gathered}
$$

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

$$
s^2=\frac{\sqrt{\kappa^4+16\Omega^2K^2}
-(\kappa^2+2K^2)}{2}.
$$

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

$$
K_{\mathrm{fast}}^2=\frac{q(4-q)}4\Omega^2,
\qquad s_{\max}=\frac q2\Omega.
$$

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

$$
Y^2+1.72Y-0.9504=0.
$$

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:

| Vertical cells | Growth estimate in inverse local time | Final relative field error |
| --- | --- | --- |
| 64 | 0.662936 | 0.00128772 |
| 128 | 0.663228 | 0.000321942 |
| 256 | 0.663301 | 0.0000804864 |

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

$$
(u_x,u_y,b_x,b_y)
=A\left(1,1,\sqrt{5/3},-\sqrt{5/3}\right).
$$

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

$$
T_{\mathrm{Re}}=\frac\rho2u_xu_y,
\qquad T_{\mathrm{Max}}=-\frac\rho2b_xb_y.
$$

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

$$
E=\frac\rho4(u_x^2+u_y^2+b_x^2+b_y^2).
$$

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:

$$
\frac{dE}{dt}=q\Omega
\left(T_{\mathrm{Re}}+T_{\mathrm{Max}}\right).
$$

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.

![Saved radial amplitude and continuum exponential reference, then saved energy increase and integrated shear work](https://sandboxphysics.com/_astro/03-growth-work.en.desktop.BLZY0NLX.svg)

Top: radial amplitude from the saved spatial fields and a separate continuum exponential reference. Bottom: the energy increase and accumulated shear work from the same saved calculation, normalized by initial energy. The lines nearly coincide. The plotted perturbations stay small throughout; saturation requires nonlinear evolution.

## 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](https://articles.adsabs.harvard.edu/pdf/1991ApJ...376..214B). Stone and Gardiner, arXiv:1006.0139v1, Sections II–III, especially equations 7–13 for shear and epicyclic checks. [Pinned version](https://arxiv.org/html/1006.0139v1#S2). 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.

## Try the prediction

### [Compare growth, energy and shear work](https://sandboxphysics.com/labs/magnetorotational-instability/)

1. Predict growth for the reference, fastest and strong-tension couplings in Keplerian shear.
2. For each case, choose the preset or coupling, Pause if playing, then Save the current channel. Presets can restart playback; use common time values from the exported history for equal-time comparisons.
3. In those matched history rows, compare state and energy growth; compare integratedShearWork with the energy increase since the first history row.
4. Explain the factor of two in energy growth and why a local linear stress gives no steady global accretion rate.

## Keep reading

Magnetic tension and the growth of a disk disturbance · Reading path

### Magnetic tension and the growth of a disk disturbance

1. [Article **Who carries angular momentum through an accretion disk?**](https://sandboxphysics.com/stories/magnetorotational-instability/who-carries-angular-momentum/)
2. [Article **Four real fields in a rotating patch of a disk**](https://sandboxphysics.com/stories/magnetorotational-instability/four-fields-in-a-rotating-patch/)
3. [You are here **Predict the growth, then follow its energy**](https://sandboxphysics.com/stories/magnetorotational-instability/growth-stress-and-energy/)
4. [Article **Why a vertical mesh changes a linear channel**](https://sandboxphysics.com/stories/magnetorotational-instability/why-a-vertical-mesh-changes-a-channel/)
5. [Article **Checking growth, transport and shear work**](https://sandboxphysics.com/stories/magnetorotational-instability/checking-growth-and-shear-work/)
6. [Article **Check an MRI calculation from its saved fields**](https://sandboxphysics.com/stories/magnetorotational-instability/check-saved-mri-fields/)
7. [Article **Read beyond linear MRI growth**](https://sandboxphysics.com/stories/magnetorotational-instability/read-beyond-linear-mri/)