Irodov 3.381: When the Cyclotron Clock Starts Slowing

Problem Irodov 3.381

An electron and a proton each revolve in a uniform magnetic field. At high speeds the period of revolution becomes longer than its value for non-relativistic speeds. For each particle, find the kinetic energy at which the period exceeds the non-relativistic period by the fraction $\eta = 1.0\%$ .

Given

$\eta = 1.0\%$
an electron and a proton in uniform $\vect B$

Find

the kinetic energy

T

for each particle at which

\tau

exceeds

\tau_0

\eta

Solution

Step 1The period at any speed

The force $q\,\vect v\times\vect B$ is perpendicular to $\vect v$ , so the speed $v$ and the factor $\gamma = (1 - v^2/c^2)^{-1/2}$ stay constant. The relativistic equation of motion then loses its $d\gamma/dt$ term:

\frac{d}{dt}\left(\gamma m\vect v\right) = \gamma m\,\frac{d\vect v}{dt} = q\,\vect v\times\vect B

This is the non-relativistic equation with $m$ replaced by $\gamma m$ . For motion across $\vect B$ the path is a circle of radius $r$ , on which the force is centripetal:

A positive charge on its circular orbit in a uniform B out of the page: the velocity v, the force F = q v· B toward the centre, and the radius r. — A positive charge on its circular orbit in a uniform $\vect B$ out of the page: the velocity $\vect v$ , the force $\vect F = q\vect v\times\vect B$ toward the centre, and the radius $r$ .

\begin{aligned} \frac{\gamma m v^2}{r} &= q v B \quad\Longrightarrow\quad r = \frac{\gamma m v}{qB}, \\ \tau &= \frac{2\pi r}{v} \end{aligned}

\tau = \frac{2\pi\gamma m}{qB}

(1)

At non-relativistic speeds $\gamma \to 1$ , so the period there is $\tau_0 = 2\pi m/(qB)$ , independent of the speed.

Step 2The condition on the energy

The ratio of the two periods is $\gamma$ itself. The period exceeds $\tau_0$ by the fraction $\eta$ when

\frac{\tau - \tau_0}{\tau_0} = \gamma - 1 = \eta

The kinetic energy of a relativistic particle is its total energy less its rest energy, and $\gamma - 1$ is just the fraction $\eta$ :

T = (\gamma - 1)\, m c^2

Below the energy this gives, the period exceeds $\tau_0$ by less than $\eta$ .

Step 3The two particles

Putting in the rest energies, $m_ec^2 = 0.511$ MeV for the electron and $m_pc^2 = 938$ MeV for the proton, with $\gamma - 1 = \eta = 0.010$ :

\begin{aligned} T_e &= 0.010 \times 0.511\text{ MeV}, \\ T_p &= 0.010 \times 938\text{ MeV} \end{aligned}

\begin{gathered} T = \eta\, m c^2 \\ T_e \approx 5.1\text{ keV}, \\ T_p \approx 9.4\text{ MeV} \end{gathered}

(2)

Check

The classical limit: at low speed $\gamma - 1 \approx v^2/(2c^2)$ and the kinetic energy is $m v^2/2$ , so the fractional lengthening of the period is about $T/(mc^2)$ , the boxed result read the other way. For the electron at $5.1$ keV, $v^2 = 2T/m$ gives $v \approx 0.14\,c$ and $v^2/(2c^2) \approx 0.010$ , equal to $\eta$ at this accuracy, as it should be for so small a correction. The proton needs the same speed, so its energy is larger by the mass ratio $m_p/m_e = 1836$ , which turns $5.1$ keV into $9.4$ MeV.

Remark

The threshold depends on $\gamma$ alone, so it is the same speed, about $0.14\,c$ , for every particle; only the energy scale differs. This is why a classical cyclotron, whose fixed-frequency field relies on $\tau$ being independent of the energy, works for protons up to some MeV but fails for electrons already at keV energies, and why the synchrocyclotron lowers its frequency as $\gamma$ grows.