Alfvén waves and their damping

Advanced level

This page uses the equations derived in Resistive MHD: deriving the equations, in particular the energy budget of Step 7. The result is the exact solution the Simulation Lab code is tested against.

A magnetic field line behaves like a string under tension. Pluck it, and the kink travels along the field. This is the Alfvén wave. The tension of a field \(B_0\) is \(B_0^2/\mu_0\) and the mass loading is \(\rho_0\), so, as for a string with speed \(\sqrt{\text{tension}/\text{mass density}}\), the wave moves at the Alfvén speed

\[ v_A = \frac{B_0}{\sqrt{\mu_0\rho_0}} . \]

Question: in resistive MHD, how fast does an Alfvén wave lose its energy, and where does the energy go?

1. Linearizing the resistive MHD equations

The equilibrium. Start from a uniform plasma at rest: density \(\rho_0\), pressure \(p_0\), velocity \(\mathbf{v}_0 = 0\), and a uniform field \(\mathbf{B}_0 = B_0\hat{\mathbf{x}}\). Because everything is uniform, \(\nabla p_0 = 0\) and \(\mu_0\mathbf{J}_0 = \nabla\times\mathbf{B}_0 = 0\): no force acts, so this state stays as it is.

The perturbation. Now disturb it slightly. Every quantity is its equilibrium value plus a small change, marked with a subscript 1:

\[ \rho = \rho_0 + \rho_1, \qquad p = p_0 + p_1, \qquad \mathbf{v} = 0 + \mathbf{v}_1, \qquad \mathbf{B} = \mathbf{B}_0 + \mathbf{B}_1, \qquad \mathbf{J} = 0 + \mathbf{J}_1 , \]

with \(\mu_0\mathbf{J}_1 = \nabla\times\mathbf{B}_1\). "Small" means

\[ \frac{\rho_1}{\rho_0},\ \ \frac{p_1}{p_0},\ \ \frac{|\mathbf{B}_1|}{B_0},\ \ \frac{|\mathbf{v}_1|}{v_A} \;\sim\; \epsilon \ll 1 . \]

Linearizing means substituting these into the equations and keeping only terms of first order in \(\epsilon\). Terms with no subscript-1 quantity are the equilibrium, which already balances; terms with two subscript-1 quantities (such as \(\rho_1\mathbf{v}_1\)) are of order \(\epsilon^2\) and are dropped.

We wish to linearize the following equations, the complete set without gravity and heat conduction:

\[ \begin{aligned} &\frac{\partial\rho}{\partial t} + \nabla\cdot(\rho\mathbf{v}) = 0 \\ &\rho\left(\frac{\partial\mathbf{v}}{\partial t} + \mathbf{v}\cdot\nabla\mathbf{v}\right) = -\nabla p + \mathbf{J}\times\mathbf{B} + \rho\nu\left[\nabla^2\mathbf{v} + \tfrac13\nabla(\nabla\cdot\mathbf{v})\right] \\ &\frac{\partial\mathbf{B}}{\partial t} = \nabla\times(\mathbf{v}\times\mathbf{B}) + \eta\nabla^2\mathbf{B} \\ &\frac{\partial p}{\partial t} + \mathbf{v}\cdot\nabla p + \gamma p\,\nabla\cdot\mathbf{v} = (\gamma-1)\left[\mu_0\eta J^2 + \Phi_\nu\right] \end{aligned} \]

Each one turns into its linear form as follows.

Continuity. Substitute:

\[ \frac{\partial(\rho_0 + \rho_1)}{\partial t} + \nabla\cdot\big[(\rho_0 + \rho_1)\,\mathbf{v}_1\big] = 0 . \]

\(\rho_0\) is constant, so \(\partial_t\rho_0 = 0\) and \(\nabla\cdot(\rho_0\mathbf{v}_1) = \rho_0\nabla\cdot\mathbf{v}_1\). \(\rho_1\mathbf{v}_1\) is second order. This leaves

\[ \frac{\partial\rho_1}{\partial t} + \rho_0\,\nabla\cdot\mathbf{v}_1 = 0 . \]

Momentum. Substitute:

\[ \begin{aligned} (\rho_0 + \rho_1)\left(\frac{\partial\mathbf{v}_1}{\partial t} + \mathbf{v}_1\cdot\nabla\mathbf{v}_1\right) ={}& -\nabla(p_0 + p_1) + \mathbf{J}_1\times(\mathbf{B}_0 + \mathbf{B}_1) \\ &+ (\rho_0 + \rho_1)\,\nu\left[\nabla^2\mathbf{v}_1 + \tfrac13\nabla(\nabla\cdot\mathbf{v}_1)\right] . \end{aligned} \]

Drop \(\rho_1\,\partial_t\mathbf{v}_1\), \(\mathbf{v}_1\cdot\nabla\mathbf{v}_1\), \(\mathbf{J}_1\times\mathbf{B}_1\) and \(\rho_1\nu[\ldots]\) (all second order), and use \(\nabla p_0 = 0\):

\[ \rho_0\frac{\partial\mathbf{v}_1}{\partial t} = -\nabla p_1 + \mathbf{J}_1\times\mathbf{B}_0 + \rho_0\nu\left[\nabla^2\mathbf{v}_1 + \tfrac13\nabla(\nabla\cdot\mathbf{v}_1)\right] . \]

Induction. Substitute:

\[ \frac{\partial(\mathbf{B}_0 + \mathbf{B}_1)}{\partial t} = \nabla\times\big[\mathbf{v}_1\times(\mathbf{B}_0 + \mathbf{B}_1)\big] + \eta\nabla^2(\mathbf{B}_0 + \mathbf{B}_1) . \]

\(\mathbf{B}_0\) is constant and uniform, so \(\partial_t\mathbf{B}_0 = 0\) and \(\nabla^2\mathbf{B}_0 = 0\). \(\mathbf{v}_1\times\mathbf{B}_1\) is second order. This leaves

\[ \frac{\partial\mathbf{B}_1}{\partial t} = \nabla\times(\mathbf{v}_1\times\mathbf{B}_0) + \eta\nabla^2\mathbf{B}_1 . \]

Pressure. Substitute:

\[ \frac{\partial(p_0 + p_1)}{\partial t} + \mathbf{v}_1\cdot\nabla(p_0 + p_1) + \gamma(p_0 + p_1)\,\nabla\cdot\mathbf{v}_1 = (\gamma-1)\left[\mu_0\eta J_1^2 + \Phi_\nu\right] . \]

\(\partial_tp_0 = 0\) and \(\nabla p_0 = 0\). \(\mathbf{v}_1\cdot\nabla p_1\) and \(p_1\nabla\cdot\mathbf{v}_1\) are second order. The heating terms are second order too: \(J_1^2\) is a product of two small quantities, and \(\Phi_\nu\) is quadratic in the velocity gradients. This leaves

\[ \frac{\partial p_1}{\partial t} + \gamma p_0\,\nabla\cdot\mathbf{v}_1 = 0 . \]

So, to first order, the heating drops out of the pressure equation. It does not vanish: it is second order, and section 5 follows it through the energy budget.

2. The linearized equations for an Alfvén wave

Choose a disturbance that is sideways to the field and depends only on \(x\) (along \(\mathbf{B}_0\)) and \(t\):

\[ \mathbf{v}_1 = v(x,t)\,\hat{\mathbf{z}}, \qquad \mathbf{B}_1 = b(x,t)\,\hat{\mathbf{z}} . \]

\(\nabla\cdot\mathbf{B}_1 = \partial b/\partial z = 0\), as it must. Now put this into each linear equation.

Continuity and pressure. \(\nabla\cdot\mathbf{v}_1 = \partial v/\partial z = 0\): the flow does not compress the plasma. The two linear equations become \(\partial_t\rho_1 = 0\) and \(\partial_tp_1 = 0\), so if \(\rho_1\) and \(p_1\) start at zero, they stay zero: \(\rho = \rho_0\) and \(p = p_0\).

Momentum. The current is

\[ \mu_0\mathbf{J}_1 = \nabla\times(b\,\hat{\mathbf{z}}) = \hat{\mathbf{x}}\,\frac{\partial b}{\partial y} - \hat{\mathbf{y}}\,\frac{\partial b}{\partial x} = -\frac{\partial b}{\partial x}\,\hat{\mathbf{y}}, \]

so the Lorentz force is

\[ \mathbf{J}_1\times\mathbf{B}_0 = \left(-\frac{1}{\mu_0}\frac{\partial b}{\partial x}\right)B_0\,(\hat{\mathbf{y}}\times\hat{\mathbf{x}}) = \frac{B_0}{\mu_0}\frac{\partial b}{\partial x}\,\hat{\mathbf{z}} . \]

This is the magnetic tension \((\mathbf{B}_0\cdot\nabla)\mathbf{B}_1/\mu_0\). There is no magnetic-pressure part: that would be \(-\nabla(\mathbf{B}_0\cdot\mathbf{B}_1)/\mu_0\), and \(\mathbf{B}_0\cdot\mathbf{B}_1 = 0\) because the disturbance is perpendicular to \(\mathbf{B}_0\). With \(p_1 = 0\) and \(\nabla\cdot\mathbf{v}_1 = 0\), the viscous term is \(\rho_0\nu\,\partial_x^2v\,\hat{\mathbf{z}}\). The \(z\)-component of the momentum equation is

\[ \rho_0\frac{\partial v}{\partial t} = \frac{B_0}{\mu_0}\frac{\partial b}{\partial x} + \rho_0\nu\frac{\partial^2 v}{\partial x^2} . \tag{1} \]

Induction. \(\mathbf{v}_1\times\mathbf{B}_0 = vB_0\,(\hat{\mathbf{z}}\times\hat{\mathbf{x}}) = vB_0\,\hat{\mathbf{y}}\), and

\[ \nabla\times(vB_0\,\hat{\mathbf{y}}) = -\hat{\mathbf{x}}\,\frac{\partial(vB_0)}{\partial z} + \hat{\mathbf{z}}\,\frac{\partial(vB_0)}{\partial x} = B_0\frac{\partial v}{\partial x}\,\hat{\mathbf{z}} . \]

With \(\eta\nabla^2\mathbf{B}_1 = \eta\,\partial_x^2b\,\hat{\mathbf{z}}\), the \(z\)-component of the induction equation is

\[ \frac{\partial b}{\partial t} = B_0\frac{\partial v}{\partial x} + \eta\frac{\partial^2 b}{\partial x^2} . \tag{2} \]

Equations (1) and (2) are the whole problem. The first term on each right-hand side couples \(v\) and \(b\) (the wave); the second term diffuses each of them (the damping).

3. The ideal wave (η = ν = 0)

Differentiate (1) in time and use (2):

\[ \frac{\partial^2 v}{\partial t^2} = \frac{B_0}{\mu_0\rho_0}\frac{\partial}{\partial x}\frac{\partial b}{\partial t} = \frac{B_0^2}{\mu_0\rho_0}\frac{\partial^2 v}{\partial x^2} = v_A^2\frac{\partial^2 v}{\partial x^2} . \]

This is the wave equation, with speed \(v_A\). A wave travelling in \(+x\), \(v = f(x - v_At)\), gives from (2) \(b = -(B_0/v_A)\,f\), i.e.

\[ b = -\sqrt{\mu_0\rho_0}\;v \qquad (\text{travelling in } +x; \text{ the sign flips for } -x). \]

So \(b^2/2\mu_0 = \rho_0v^2/2\) at every point: the wave carries equal kinetic and magnetic energy.

Writing the field-line displacement as \(\xi(x,t)\), \(v = \partial\xi/\partial t\) and \(b = B_0\,\partial\xi/\partial x\). The tension force per volume is then \(B_0^2\,\partial^2\xi/\partial x^2/\mu_0\): it pulls each bent piece of the line back toward the axis, and the plasma's inertia carries it past, just as for a string.

4. The energy of the wave

The energy density of the disturbance is the kinetic energy plus the magnetic energy:

\[ \frac{\rho_0v^2}{2} + \frac{(B_0\hat{\mathbf{x}} + b\,\hat{\mathbf{z}})^2}{2\mu_0} = \frac{\rho_0v^2}{2} + \frac{B_0^2}{2\mu_0} + \frac{b^2}{2\mu_0} . \]

\(B_0^2/2\mu_0\) is the constant background, and there is no cross term because \(b\perp\mathbf{B}_0\). So the wave energy in a volume \(V\) is

\[ W = \int_V \left(\frac{\rho_0v^2}{2} + \frac{b^2}{2\mu_0}\right)dV = K + M . \]

5. The energy equation in resistive MHD

Multiply (1) by \(v\) and (2) by \(b/\mu_0\), and add:

\[ \frac{\partial}{\partial t}\left(\frac{\rho_0v^2}{2} + \frac{b^2}{2\mu_0}\right) = \frac{B_0}{\mu_0}\left(v\frac{\partial b}{\partial x} + b\frac{\partial v}{\partial x}\right) + \rho_0\nu\,v\frac{\partial^2 v}{\partial x^2} + \frac{\eta}{\mu_0}\,b\frac{\partial^2 b}{\partial x^2} . \]

Each term on the right is rewritten with the product rule:

• \(v\,\partial_x b + b\,\partial_x v = \partial_x(vb)\): a pure flux.
• \(v\,\partial_x^2v = \partial_x(v\,\partial_xv) - (\partial_xv)^2\), and the same for \(b\).

So

\[ \frac{\partial}{\partial t}\left(\frac{\rho_0v^2}{2} + \frac{b^2}{2\mu_0}\right) + \frac{\partial F}{\partial x} = -\underbrace{\rho_0\nu\left(\frac{\partial v}{\partial x}\right)^2}_{\Phi_\nu} - \underbrace{\frac{\eta}{\mu_0}\left(\frac{\partial b}{\partial x}\right)^2}_{\mu_0\eta J^2} , \]

with the energy flux

\[ F = -\frac{B_0vb}{\mu_0} - \rho_0\nu\,v\,\frac{\partial v}{\partial x} - \frac{\eta}{\mu_0}\,b\,\frac{\partial b}{\partial x} . \]

The two sinks are exactly the heating terms of Step 7:

• \(\dfrac{\eta}{\mu_0}(\partial_xb)^2 = \mu_0\eta J^2\), because \(J = -\partial_xb/\mu_0\): ohmic heating.
• \(\rho_0\nu(\partial_xv)^2 = \Phi_\nu\): viscous heating (Problem 4 shows this from the general \(\Phi_\nu\)).

The first part of the flux, \(-B_0vb/\mu_0\), is the Poynting flux. For the travelling wave, \(b = -\sqrt{\mu_0\rho_0}\,v\), it equals \(\rho_0v^2\,v_A\): the wave energy density \(\rho_0v^2\) carried forward at \(v_A\).

In a periodic box (or for a wave far from any boundary), the flux integrates to zero, leaving

\[ \boxed{\;\frac{dW}{dt} = -\int_V\left[\rho_0\nu\left(\frac{\partial v}{\partial x}\right)^2 + \frac{\eta}{\mu_0}\left(\frac{\partial b}{\partial x}\right)^2\right]dV\;} \]

The wave loses energy only to heat, at the rate set by its gradients.

6. A sinusoidal wave: W ∝ exp[−(η+ν)k²t]

Take a single wavelength-\(\lambda\) wave, \(k = 2\pi/\lambda\). For a sinusoid, the gradient squared averages to \(k^2\) times the value squared over a wavelength (\(\sin^2\) and \(\cos^2\) both average to \(\tfrac12\)):

\[ \int(\partial_xv)^2\,dV = k^2\int v^2\,dV = \frac{2k^2}{\rho_0}K, \qquad \int(\partial_xb)^2\,dV = k^2\int b^2\,dV = 2\mu_0k^2M . \]

So

\[ \frac{dW}{dt} = -2k^2\left(\nu K + \eta M\right) . \]

With equal kinetic and magnetic energy, \(K = M = W/2\):

\[ \frac{dW}{dt} = -(\eta + \nu)\,k^2\,W \quad\Longrightarrow\quad \boxed{\;W(t) = W_0\,e^{-(\eta+\nu)k^2t}\;} \]

Does equipartition survive the damping? Look for solutions \(v, b \propto e^{ikx + st}\). Equations (1) and (2) become

\[ (s + \nu k^2)\,v = \frac{ikB_0}{\mu_0\rho_0}\,b, \qquad (s + \eta k^2)\,b = ikB_0\,v . \]

Multiplying them gives the dispersion relation

\[ (s + \nu k^2)(s + \eta k^2) + k^2v_A^2 = 0 , \]

whose roots are

\[ s = -\frac{(\eta+\nu)k^2}{2} \pm i\,\Omega, \qquad \Omega = \sqrt{k^2v_A^2 - \frac{(\eta-\nu)^2k^4}{4}} . \]

• The wave still oscillates (at \(\Omega \approx kv_A\)), and its amplitude decays at \((\eta+\nu)k^2/2\). Energy is amplitude squared, so it decays at twice that, \((\eta+\nu)k^2\): the same result.
• With \(s + \nu k^2 = \tfrac12(\nu-\eta)k^2 + i\Omega\), one finds \(|s + \nu k^2|^2 = k^2v_A^2\), so the first equation gives \(|b| = \sqrt{\mu_0\rho_0}\,|v|\). Each damped mode keeps \(K = M\) exactly, even when \(\eta \ne \nu\).

Only the sum \(\eta + \nu\) matters. Resistivity damps the magnetic half of the energy, viscosity the kinetic half, and the wave keeps passing energy between the two halves, so each dissipates at the average rate.

Answer

A small-amplitude Alfvén wave of wavenumber \(k\) loses energy as \(W = W_0\,e^{-(\eta+\nu)k^2t}\). The energy goes into heat: ohmic heating \(\mu_0\eta J^2\) from the wave's current, and viscous heating \(\Phi_\nu\) from its sheared flow. Shorter waves (larger \(k\)) have steeper gradients and damp faster, as \(k^2\).

7. The code test

In the code's units (\(\rho_0 = B_0 = \mu_0 = 1\), so \(v_A = 1\); Step 8b), the cartesian_2d_alfven setup starts the travelling wave

\[ v_z = a\sin(kx), \qquad b_z = -a\sin(kx), \qquad a = 10^{-3}, \]

i.e. \(b = -v\), the \(+x\) wave of section 3, in a periodic box of length 1 (\(k = 2\pi\)), with \(\eta = \nu = 0.002\). With \(\eta = \nu\), the dispersion relation gives \(\Omega = kv_A\) exactly and \(b = -v\) is exactly one damped mode, so the theory predicts

\[ \frac{W(t)}{W_0} = e^{-(0.004)(2\pi)^2t} = e^{-0.158\,t}, \qquad \frac{W(10)}{W_0} \approx 0.21 . \]

The simulation follows this to 0.3%:

The test checks the tension force, the induction equation, and both dissipation terms at once. If any of them had a wrong sign or factor, the curve would bend away from the dashed line.

Practice problems

Try each problem before opening its solution. Click an option to check a concept question; type a number and press Check for a numerical one (forms like 6.2e7 or 6.2x10^7 both work).

Concept questions

Problem 1

A wave with twice the wavenumber is launched in the same plasma. How does its energy damping rate compare?

• The same
• Twice as fast
• Four times as fast
• Half as fast

The rate is \((\eta+\nu)k^2\). Doubling \(k\) doubles every gradient, and the dissipation goes as gradient squared.

Problem 2

Where does the energy lost by the wave end up?

• It is radiated away as light
• It heats the plasma, raising the pressure
• It accelerates a large-scale flow along \(\mathbf{B}_0\)
• It strengthens the background field \(B_0\)

The sinks are \(\mu_0\eta J^2\) and \(\Phi_\nu\), which appear with a plus sign in the pressure equation. The pressure rise is second order in the amplitude, which is why the linear analysis can ignore it.

Problem 3

A plasma has \(\eta = 0.004\) and \(\nu = 0\). Compared with \(\eta = \nu = 0.002\), its Alfvén waves damp

• at the same rate
• twice as fast
• half as fast
• not at all, because there is no viscosity

Only \(\eta + \nu\) enters. The wave shares its energy equally between flow and field, so it does not matter which half is damped.

Show that…

Problem 4: viscous heating of the wave

Starting from \(\Phi_\nu = 2\rho\nu\left[S_{ij}S_{ij} - \tfrac13(\nabla\cdot\mathbf{v})^2\right]\), with \(S_{ij} = \tfrac12(\partial_iv_j + \partial_jv_i)\), show that for \(\mathbf{v} = v(x,t)\,\hat{\mathbf{z}}\), \(\Phi_\nu = \rho\nu(\partial_xv)^2\).

Hint

Only one derivative of \(\mathbf{v}\) is non-zero. Which components of \(S_{ij}\) does it appear in?

Proof

The only non-zero derivative is \(\partial_xv_z\), so \(\nabla\cdot\mathbf{v} = 0\) and the only non-zero strain components are \(S_{xz} = S_{zx} = \tfrac12\partial_xv\). Then \(S_{ij}S_{ij} = 2\cdot\tfrac14(\partial_xv)^2 = \tfrac12(\partial_xv)^2\), and \(\Phi_\nu = 2\rho\nu\cdot\tfrac12(\partial_xv)^2 = \rho\nu(\partial_xv)^2\).

Problem 5: the Poynting flux of the wave

Show that for the travelling wave \(b = -\sqrt{\mu_0\rho_0}\,v\), the Poynting flux \(-B_0vb/\mu_0\) equals the wave energy density times \(v_A\).

Proof

\(-B_0vb/\mu_0 = B_0\sqrt{\mu_0\rho_0}\,v^2/\mu_0 = \rho_0v^2\,B_0/\sqrt{\mu_0\rho_0} = \rho_0v^2\,v_A\). With equipartition the energy density is \(\rho_0v^2/2 + b^2/2\mu_0 = \rho_0v^2\). So the flux is the energy density moving at \(v_A\).

Numbers

Problem 6

In the code test (\(k = 2\pi\), \(\eta = \nu = 0.002\), code units), after how many Alfvén times has the wave energy fallen by a factor \(e\)?

Solution

\(t = 1/[(\eta+\nu)k^2] = 1/(0.004\times39.48) \approx 6.33\,\tau_A\).