Calculating how fast a plasma wave fades

A small wave in a Maxwellian plasma fades at a rate fixed by one number: the imaginary part of the complex frequency at which the plasma’s dielectric function is zero. For $k=0.5$ that root is $\omega=1.4157-0.1534i$ . The simulation’s fitted rate is $-0.1536$ .

Sandbox Physics · 2026-10-29

Derivation

Linearise around a uniform plasma

Write $f=f_0(v)+f_1$ , where $f_0$ is the uniform distribution and $f_1$ is small. Keep only terms of first order in $f_1$ and the field. The term $E\,\partial f_1/\partial v$ is second order and drops out, leaving a linear equation that is driven by the field acting on the slope of the background.

Look for a mode $f_1,E\propto e^{i(kx-\omega t)}$ . The Vlasov equation gives $f_1$ in terms of $E$ , and Poisson’s equation turns the density of $f_1$ back into $E$ . A nonzero field needs the dielectric function to vanish.

f_1=\frac{i\,E\,f_0'(v)}{\omega-kv},\qquad \varepsilon(k,\omega)=1-\frac{1}{k^2}\int\frac{f_0'(v)}{v-\omega/k}\,dv=0

The pole at the wave speed

The integrand has a pole where an electron moves at the wave speed, $v=\omega/k$ . Landau showed that the initial-value problem fixes how to pass it: continue the integral from the upper half of the complex $\omega$ plane, so the contour runs below the pole. This choice adds a term proportional to the slope $f_0^{\prime}$ at the wave speed.

For a Maxwellian the integral is written with the plasma dispersion function $Z$ . The program evaluates $Z$ through the Faddeeva function and finds the root with Newton’s method. In this formula $\zeta=\omega/(\sqrt2\,k)$ , and the root lies in the lower half plane: the wave decays.

\varepsilon(k,\omega)=1-\frac{1}{2k^2}\,Z'\!\left(\zeta\right),\qquad \zeta=\frac{\omega}{\sqrt2\,k},\qquad Z'(\zeta)=-2\left[1+\zeta Z(\zeta)\right]

Worked example

Worked example: three wave numbers

Solving $\varepsilon=0$ gives $\omega=1.1598-0.0126i$ for $k=0.3$ , $1.4157-0.1534i$ for $k=0.5$ and $1.7999-0.5346i$ for $k=0.8$ . Short waves damp much faster because their phase speed is closer to the thermal speed, where many more electrons can ride along.

The Lab measures the same rates from the field history alone. It fits a straight line to the logarithm of the peaks of the field, between $t=4$ and the time when nonlinear trapping or a grid recurrence could start. The fits give $-0.0127$ , $-0.1536$ and $-0.5358$ . The real parts also agree to within $0.2\%$ .

For long waves a short formula captures most of this. With $k=0.5$ it gives $\gamma\approx-0.151$ , close to the root. Its error grows quickly at other wave numbers, so the comparison below uses the exact root and keeps the formula for quick estimates.

\gamma\approx-\sqrt{\frac{\pi}{8}}\;\frac{1}{k^3}\,\exp\!\left(-\frac{1}{2k^2}-\frac32\right)

What the rate does and does not say

The damping rate describes the field and the density, the quantities that average over velocity. It does not mean the distribution relaxes: the energy leaves the field and stays in the electrons as finer and finer velocity stripes. The rate also assumes the wave stays small. Chapter 5 shows what changes when a strong wave holds on to the electrons it is trying to hand energy to.

An experiment to think through

Predict, change, explain

Use the three roots above to predict how many time units it takes the $k=0.8$ field to fall by a factor of $e^{5}$ .
Move the wave-number slider in the phase-space Lab from 0.3 to 0.8 and read the measured and root rates side by side.
Explain why the agreement stays good when you raise the ripple depth a little, and fails when you raise it a lot.

Watch a ripple fold into velocity stripes Follow the energy to resonant electrons

The three phase-space Labs solve a one-dimensional electron model in plasma units. Each Lab exports its parameters, model version and field history.

Try explaining it yourself

1. For $k=0.8$ , how long does the field take to fall by $e^{5}$ ?

Read an explained answer

The decay rate is $0.5346$ , so the time is $5/0.5346\approx9.4$ time units, a little more than two oscillation periods of $2\pi/1.80\approx3.5$ .

2. Why does the linear calculation lose the term $E\,\partial f_1/\partial v$ , and when is that unsafe?

Read an explained answer

It is a product of two small quantities. It becomes important when the field is strong enough to bend electron orbits within the damping time; chapter 5 measures that with the bounce time.

3. A distribution has a positive slope at the wave speed. What sign do you expect for the imaginary part of the root?

Read an explained answer

Positive: the slope term reverses sign, so the wave grows instead of decaying. Chapter 5 checks this with a bump on the tail.

Sources to read alongside

R. Fitzpatrick, Plasma Physics lecture notes, University of Texas at Austin: Landau damping, resonant particles and the plasma dispersion function
Three linked pages derive the linear response of a Maxwellian plasma, the role of electrons near the wave speed, and the plasma dispersion function used for our dielectric roots.
J. H. Malmberg and C. B. Wharton, Collisionless Damping of Electrostatic Plasma Waves, Physical Review Letters 13, 184 (1964)
A laboratory measurement of collisionless wave damping. Only the bibliographic record was checked; none of its data are used here, and the rates in chapter 4 come from our dielectric roots and simulations.

Linearise around a uniform plasma

The pole at the wave speed

Worked example: three wave numbers

What the rate does and does not say

Predict, change, explain

Try explaining it yourself

1. For k=0.8k=0.8k=0.8, how long does the field take to fall by e5e^{5}e5?

2. Why does the linear calculation lose the term E ∂f1/∂vE\,\partial f_1/\partial vE∂f1​/∂v, and when is that unsafe?

3. A distribution has a positive slope at the wave speed. What sign do you expect for the imaginary part of the root?

Sources to read alongside

1. For $k=0.8$ , how long does the field take to fall by $e^{5}$ ?

2. Why does the linear calculation lose the term $E\,\partial f_1/\partial v$ , and when is that unsafe?