Follow every electron at once in phase space

Mark every electron as a point whose horizontal coordinate is its position and whose vertical coordinate is its velocity. The cloud of points moves like a fluid in this plane. Its local density is the distribution function, and without collisions that density is carried unchanged along each electron’s path.

Sandbox Physics · 2026-10-29

Count electrons in a small box of position and velocity

Call $f(x,v,t)\,dx\,dv$ the number of electrons with position between $x$ and $x+dx$ and velocity between $v$ and $v+dv$ . Integrating over all velocities gives the ordinary density, $n(x,t)=\int f\,dv$ . In this course the ions form a fixed background of density $1$ , so the plasma is neutral when $n = 1$ everywhere.

The Lab draws exactly this plane. The colour shows how far $f$ differs from the smooth, uniform background, so a stripe of warm colour means a few more electrons than average at that position and speed.

Derivation

Carry the density along each path

In plasma units an electron obeys $dx/dt = v$ and $dv/dt = -E$ ; the minus sign is the negative electron charge. Without collisions, electrons are neither created nor removed, and the flow in this plane has zero divergence because $\partial v/\partial x = 0$ and $\partial E/\partial v = 0$ . A small patch of the cloud may stretch and fold, but its area and the number of electrons inside stay fixed, so $f$ is constant along each path.

Writing that statement with the chain rule gives the Vlasov equation. Gauss’s law closes it: the field is produced by the difference between the ion background and the electron density.

\frac{\partial f}{\partial t}+v\frac{\partial f}{\partial x}-E\frac{\partial f}{\partial v}=0,\qquad \frac{\partial E}{\partial x}=1-\int f\,dv

What the equation keeps

Because $f$ is only moved around, every integral of the form $\int\!\!\int G(f)\,dx\,dv$ stays constant. That includes the number of electrons, the sum of $f^2$ and the fine-grained entropy $-\int\!\!\int f\ln f\,dx\,dv$ . Momentum is conserved, and so is the total energy: kinetic energy of the electrons plus the energy stored in the field.

These are the first checks of any solver. In the default run of the phase-space Lab, with a $32\times256$ grid and time step $0.1$ , the relative drift of electron number stays below $10^{-14}$ , the sum of $f^2$ below $10^{-13}$ and the total energy below $10^{-6}$ over 70 time units. The energy error is the largest because splitting a step into a position shift and a velocity shift is accurate only to second order in the time step.

W=\int\!\!\int \frac{v^2}{2}\,f\,dx\,dv+\int \frac{E^2}{2}\,dx=\text{constant}

Worked example

Worked example: a ripple that streams freely

Switch the field off and start from $f_0=(1+\alpha\cos kx)M(v)$ , where $M(v)=e^{-v^2/2}/\sqrt{2\pi}$ is the Maxwellian. Each electron then keeps its velocity, so $f(x,v,t)=f_0(x-vt,v)$ . Integrating over velocity, the ripple in density becomes a Fourier transform of the Maxwellian.

The density ripple therefore shrinks as a Gaussian in time. For $k=0.5$ it falls to $e^{-2}\approx0.135$ of its start by $t=4$ and to $3.4\times10^{-4}$ by $t=8$ . Nothing has been lost: $f$ still carries the full ripple, now tilted into stripes across velocity. The density simply averages over many stripes that cancel.

n(x,t)-1=\alpha\cos(kx)\int M(v)\cos(kvt)\,dv=\alpha\cos(kx)\,e^{-k^2t^2/2}

How the program advances the cloud

The simulation stores $f$ on a grid of positions and velocities. Each step shifts every velocity row in position by $v\,\Delta t$ , computes the new field from the density, and shifts every position column in velocity by $-E\,\Delta t$ . Both shifts are done exactly for the grid with Fourier transforms, and the sequence half-shift, kick, half-shift keeps the error at second order.

This method is accurate while the stripes in velocity are wider than a few grid cells. Chapter 3 shows what the grid does once the stripes become finer than that.

Predict, change, explain

With the field switched off, will the density ripple decay faster or slower than with the field on?

Run the default ripple, then the free-streaming preset. Watch the field strip and the history chart.

Explain the shape of each decay curve on the logarithmic chart: one bends downward ever faster, the other falls along a straight line.

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. Show that the free-streaming solution $f_0(x-vt,v)$ satisfies the Vlasov equation with $E=0$ .

Read an explained answer

Let $s=x-vt$ . Then $\partial f/\partial t = -v\,\partial f_0/\partial s$ and $\partial f/\partial x = \partial f_0/\partial s$ , so $\partial f/\partial t + v\,\partial f/\partial x = 0$ .

2. At what time does a free-streaming ripple with $k=0.3$ fall to $e^{-2}$ of its start?

Read an explained answer

Set $k^2t^2/2=2$ , so $t=2/k\approx6.7$ . Longer waves take longer to mix away because electrons need more time to drift across one wavelength.

3. Why is the conserved entropy $-\int\!\!\int f\ln f$ not a contradiction of the second law?

Read an explained answer

It measures the full, fine-grained distribution, which only moves. Any measurement that averages over fine stripes, such as the density, sees the ripple disappear. Irreversibility enters when collisions or coarse observation erase the stripes, the subject of chapter 7.

Count electrons in a small box of position and velocity

Carry the density along each path

What the equation keeps

Worked example: a ripple that streams freely

How the program advances the cloud

Predict, change, explain

Try explaining it yourself

1. Show that the free-streaming solution f0(x−vt,v)f_0(x-vt,v)f0​(x−vt,v) satisfies the Vlasov equation with E=0E=0E=0.

2. At what time does a free-streaming ripple with k=0.3k=0.3k=0.3 fall to e−2e^{-2}e−2 of its start?

3. Why is the conserved entropy −∫ ⁣ ⁣∫fln⁡f-\int\!\!\int f\ln f−∫∫flnf not a contradiction of the second law?

Sources to read alongside

1. Show that the free-streaming solution $f_0(x-vt,v)$ satisfies the Vlasov equation with $E=0$ .

2. At what time does a free-streaming ripple with $k=0.3$ fall to $e^{-2}$ of its start?

3. Why is the conserved entropy $-\int\!\!\int f\ln f$ not a contradiction of the second law?