Irodov 1.377: Both Frames Agree on the Pressure – Amit & Deepali Sinha Foundation

Problem Irodov 1.377

A sphere travels at a relativistic speed $v$ through a gas. In the gas’s own rest frame there are $n$ particles per unit volume, each of mass $m$ , and they move so slowly there that they may be treated as being at rest. The particles scatter elastically off the sphere. Find the pressure $p$ the gas exerts on an element of the sphere’s surface that is perpendicular to the sphere’s velocity, and show that this pressure comes out the same whether it is worked out in the frame fixed to the sphere or in the frame fixed to the gas.

Given

speed $v$
$n$ per unit volume in the gas frame
particle mass $m$
elastic scattering

Find

the pressure

p

on the surface element facing the motion, and that it is the same in both frames

Solution

Step 1The gas is denser in the sphere’s frame

The one scene in two frames. Left, the gas frame: the sphere moves right at v through particles that sit still. Right, the sphere's frame: the sphere at rest, the gas streaming past at v, the particles drawn closer together along the motion. — The one scene in two frames. Left, the gas frame: the sphere moves right at $\vect v$ through particles that sit still. Right, the sphere’s frame: the sphere at rest, the gas streaming past at $\vect v$ , the particles drawn closer together along the motion.

Work first in the frame fixed to the sphere, where the sphere is at rest and the gas blows past it at speed $v$ : there the problem is the familiar one of a stream striking a stationary wall. Take a block of gas holding $N$ particles, at rest in the gas frame and occupying a volume $V_0$ there. In the sphere’s frame that same block is contracted along the motion and unchanged across it, so it holds the same $N$ particles in a smaller volume. With $\beta = v/c$ ,

\begin{aligned} V &= V_0\sqrt{1-\beta^2} \\[2pt] n' &= \frac{N}{V} = \frac{N}{V_0\sqrt{1-\beta^2}} \end{aligned}

and $N/V_0$ is the density $n$ the problem states, which is the density measured in the gas’s own rest frame:

n' = \frac{n}{\sqrt{1-\beta^2}}

(1)

One elastic bounce in the sphere's frame, at the element facing the motion. The hatched face is the sphere's surface, its material to the left. The particle arrives head on with momentum q' and leaves with - q'. — One elastic bounce in the sphere’s frame, at the element facing the motion. The hatched face is the sphere’s surface, its material to the left. The particle arrives head on with momentum $\vect q'$ and leaves with $-\vect q'$ .

Step 2The momentum one particle delivers to the element

In the sphere’s frame a gas particle arrives at the element at speed $v$ , along the normal, carrying the relativistic momentum of a particle of mass $m$ , written $q'$ here so that $p$ stays reserved for the pressure the problem asks for. The scattering is elastic and the sphere is a macroscopic body, so the sphere’s recoil is negligible and the particle leaves at the same speed with its momentum reversed. The momentum handed to the element is therefore twice what one particle carries:

q' = \frac{mv}{\sqrt{1-\beta^2}}

\Delta q' = 2q' = \frac{2mv}{\sqrt{1-\beta^2}}

(2)

Step 3The pressure in the sphere’s frame

Pressure is momentum delivered per unit time per unit area, so it is the flux of particles onto the element times the momentum (2) each one delivers. In the sphere’s frame the stream of density (1) moves at $v$ , so the number crossing unit area of the element per unit time is $n'v$ :

\begin{aligned} p &= \left(n'v\right)\Delta q' \\[2pt] &= \frac{nv}{\sqrt{1-\beta^2}}\cdot\frac{2mv}{\sqrt{1-\beta^2}} \end{aligned}

The two square roots multiply into one factor, and each has done a different job: one crowded the gas, the other loaded each particle with extra momentum.

p = \frac{2nmv^2}{1-\beta^2} = \frac{2nmv^2}{1-v^2/c^2}

(3)

Step 4The same pressure worked out in the gas frame

Now redo the count in the frame fixed to the gas, where the particles are at rest and the element sweeps through them. Nothing may be carried over from the sphere-frame count: the density is $n$ , the flux is set by the element’s own motion, and the momentum each particle ends up with must be transformed from the sphere’s frame. The flux comes first, and it is a matter of counting what the element sweeps up: in a time $dt$ an element of area $dA$ passes through a slab of gas, and the particles standing in that slab are exactly the ones it meets.

\begin{aligned} dV &= v\,dt\,dA \\[2pt] dN &= n\,dV = nv\,dt\,dA \\[3pt] \frac{dN}{dt\,dA} &= nv \end{aligned}

Each of them starts at rest and, after bouncing, moves forward at the speed obtained by composing $v$ with $v$ :

u = \frac{v+v}{1+v^2/c^2} = \frac{2v}{1+\beta^2}

Its Lorentz factor follows from that speed. The numerator is a difference of squares and collapses:

\begin{aligned} 1-\frac{u^2}{c^2} &= \frac{\left(1+\beta^2\right)^2-4\beta^2}{\left(1+\beta^2\right)^2} = \frac{\left(1-\beta^2\right)^2}{\left(1+\beta^2\right)^2} \\[3pt] \frac{1}{\sqrt{1-u^2/c^2}} &= \frac{1+\beta^2}{1-\beta^2} \end{aligned}

So the momentum one particle carries away, and therefore the momentum the element gave it, is

\begin{aligned} \Delta q &= \frac{mu}{\sqrt{1-u^2/c^2}} = m\cdot\frac{2v}{1+\beta^2}\cdot\frac{1+\beta^2}{1-\beta^2} \end{aligned}

\Delta q = \frac{2mv}{1-\beta^2}

(4)

Multiplying the flux $nv$ by (4) gives the pressure in this frame, and it is (3) again:

\left(nv\right)\Delta q = \frac{2nmv^2}{1-\beta^2}

Both frames give the same pressure

p

Step 5Why the agreement was inevitable

The equality is not an accident of the algebra. A pressure is a force divided by an area, and this particular element makes both of them invariant: the element faces the motion, so its area lies entirely across the boost and is unchanged by it, while the force on it lies along the boost, and a longitudinal force on a body at rest in the primed frame takes the same value in both frames.

\Delta A = \Delta A', \qquad F_\parallel = F'_\parallel, \qquad p = \frac{F_\parallel}{\Delta A} = \frac{F'_\parallel}{\Delta A'} = p'

Check

Dimensions first: $nmv^2$ is a mass density times a speed squared, so its units are $\mathrm{kg}/(\mathrm{m}\,\mathrm{s}^2)$ , which is a pascal, and the denominator is a pure number. The Newtonian limit is the second test. For $v\ll c$ the result (3) becomes $p\to2nmv^2$ , which is what the elementary kinetic argument gives for a wall sweeping through stationary particles at speed $v$ : flux $nv$ , and momentum $2mv$ per bounce. The intermediate (4) passes the same test, reducing to the Newtonian $2mv$ that a struck particle carries away when a heavy wall hits it at $v$ .

Remark

The pressure grows without bound as $v\to c$ , even though neither the number of particles the sphere meets per unit of its own path nor the speed of any of them can grow without bound. What diverges is the momentum per particle, and, in the sphere’s frame, the density as well. The result is also a reminder of how narrow the invariance argument above is: it used exactly two properties of this element, that its area lies across the boost and that the force on it lies along the boost, and it is the element’s orientation alone that supplies them. An element tilted away from the motion has neither property, so that argument says nothing about it, and its pressure has to be worked out on its own terms.