Irodov 1.376: Momentum and Energy per Unit Time – Amit & Deepali Sinha Foundation

Problem Irodov 1.376

A steady beam of identical relativistic particles, each of charge $e$ and rest mass $m_0$ and each carrying the same kinetic energy $T$ , runs onto a target that absorbs every particle reaching it. The electric current carried by the beam is $I$ . Find the pressure the beam exerts on the surface it strikes, and the rate at which energy is deposited there.

The beam arrives from the left into the target, the hatched body on the right. Every particle carries the same momentum p along the beam; the gray bracket measures d, the beam's transverse extent, spanning the irradiated target area S. — The beam arrives from the left into the target, the hatched body on the right. Every particle carries the same momentum $\vect p$ along the beam; the gray bracket measures $d$ , the beam’s transverse extent, spanning the irradiated target area $S$ .

Given

$T$ per particle
beam current $I$
charge $e$
rest mass $m_0$
target absorbs every particle

Find

the pressure on the target surface and the power liberated there

Solution

Step 1How many particles arrive each second

The current is the only thing that says how intense the beam is, and it counts charge rather than particles. Dividing it by the charge one particle carries turns it into a rate of arrival $N$ , the number of particles striking the target per unit time.

I = Ne

N = \frac{I}{e}

(1)

The energy-momentum relation as a right triangle: the legs are pc and the rest energy m_0c^2, the hypotenuse the total energy T+m_0c^2. Drawn for the case T = m_0c^2; the shape changes with T, the right angle does not. — The energy-momentum relation as a right triangle: the legs are $pc$ and the rest energy $m_0c^2$ , the hypotenuse the total energy $T+m_0c^2$ . Drawn for the case $T = m_0c^2$ ; the shape changes with $T$ , the right angle does not.

Step 2The momentum one particle brings

A particle’s kinetic energy fixes its total energy, and the energy-momentum relation then fixes its momentum. Writing the total energy as the rest energy plus the kinetic energy and substituting it into that relation clears the rest energy out of the difference:

\begin{aligned} E &= T+m_0c^2 \\[2pt] p^2c^2 &= E^2-m_0^2c^4 \\[2pt] &= \left(T+m_0c^2\right)^2-\left(m_0c^2\right)^2 \\[2pt] &= T^2+2Tm_0c^2+m_0^2c^4-m_0^2c^4 \\[2pt] &= T\left(T+2m_0c^2\right) \end{aligned}

The rest energy terms cancel and the two survivors share the factor $T$ . Taking the square root and dividing by $c$ gives the momentum of one particle:

p = \frac{1}{c}\sqrt{T\left(T+2m_0c^2\right)}

(2)

Step 3The force on the target, and the pressure

What one particle does at the face: it arrives carrying the momentum p of and is absorbed. The gold F is what the target feels, the momentum delivered per unit time by the N arriving particles, directed along the beam. — What one particle does at the face: it arrives carrying the momentum $\vect p$ of (2) and is absorbed. The gold $\vect F$ is what the target feels, the momentum delivered per unit time by the $N$ arriving particles, directed along the beam.

The target absorbs the particles, so each one gives up the whole of its momentum to the target and none is carried back out. The beam falls squarely on the target surface, so the momentum delivered is normal to it. The force is the momentum delivered per unit time, which is the rate (1) times the momentum (2) of one particle:

F = N p

Substituting (1) and (2) into that product leaves the force in the given quantities alone. The pressure, written $\Pi$ here to keep it apart from the momentum $p$ , is that force shared over the area $S$ of target the beam actually covers. The problem does not specify that area, so the data fix the force and the pressure follows from it:

\Pi = \frac{F}{S}, \qquad F = \frac{I}{ec}\sqrt{T\left(T+2m_0c^2\right)}

(3)

Step 4The power liberated at the target

What a stopped particle leaves behind in the target is its kinetic energy: its rest energy stays locked up in its rest mass, which the absorption does not destroy. Multiplying the kinetic energy of one particle by the arrival rate (1) gives the energy liberated per unit time:

P = N T

P = \frac{TI}{e}

(4)

Check

Both results have the right dimensions and the right Newtonian limit. In $F$ , the current over the charge is a number per unit time and the root is an energy, so $F$ carries energy per unit time divided by a speed, which is a force; in $P$ , energy times number per unit time is a power. For $T\ll m_0c^2$ the root becomes $\sqrt{2m_0c^2T}$ , so $F \to \left(I/e\right)\sqrt{2m_0T} = N m_0 v$ , the Newtonian momentum flux of a beam of particles of speed $v = \sqrt{2T/m_0}$ . In the opposite limit $T\gg m_0c^2$ the root becomes $T$ itself and $F \to TI/(ec) = P/c$ , the familiar relation between the force and the power absorbed from a beam moving at practically the speed of light.

Remark

The answers (3) and (4) are not independent: dividing one by the other gives $F/P = p/T$ , the momentum a particle carries per unit of kinetic energy it delivers, and that single ratio is what interpolates between the two limits checked above. It equals $2/v$ for a slow particle, since then $p = m_0v$ and $T = m_0v^2/2$ , and it falls to $1/c$ when $T$ swamps the rest energy. So the same beam current pushes hardest per watt when its particles are slow, and a beam that delivers a given power presses on an absorber at least as hard as light of the same power would, with equality only in the ultrarelativistic limit. All of this assumes what the problem states: one kinetic energy $T$ shared by every particle, and a target that stops each of them.