Irodov 3.376: Acceleration Falls as Gamma Cubed – Amit & Deepali Sinha Foundation

Problem Irodov 3.376

A relativistic electron moves along the lines of a uniform electric field of strength $E$ . Find its acceleration at the moment its kinetic energy equals $T$ .

The electron e moving with velocity v along a line of the uniform field E, seen side on. — The electron $e$ moving with velocity $\vect v$ along a line of the uniform field $\vect E$ , seen side on.

Given

$E$
$\vect v$ along the field line
kinetic energy $T$

Find

the acceleration

w

at that moment

Solution

Step 1The law of motion along the line

The force $eE$ is constant and lies along the velocity, so the motion stays on one line. The relativistic second law sets the rate of change of the momentum $p = m\gamma v$ equal to the force, with $\gamma = (1 - v^2/c^2)^{-1/2}$ and $m$ the rest mass:

\frac{d}{dt}\left(m\gamma v\right) = eE \tag{1}

Step 2How the momentum responds to the speed

Differentiating $\gamma v$ with respect to $v$ , using $d\gamma/dv = \gamma^3 v/c^2$ :

\begin{aligned} \frac{d(\gamma v)}{dv} &= \gamma + v\cdot\frac{\gamma^3 v}{c^2} = \gamma^3\left(\frac{1}{\gamma^2} + \frac{v^2}{c^2}\right) \\ &= \gamma^3\left(1 - \frac{v^2}{c^2} + \frac{v^2}{c^2}\right) = \gamma^3 \end{aligned}

So (1) becomes $m\gamma^3\,dv/dt = eE$ , and the acceleration is

w = \frac{dv}{dt} = \frac{eE}{m\gamma^3}

(2)

Step 3The factor

\gamma

from the kinetic energy

The total energy $\gamma mc^2$ is the kinetic energy plus the rest energy, which fixes $\gamma$ :

\gamma mc^2 = T + mc^2 \quad\Longrightarrow\quad \gamma = 1 + \frac{T}{mc^2}

Putting this into (2):

w = \frac{eE}{m\left(1 + T/mc^2\right)^3}

(3)

Check

Limits: for $T \ll mc^2$ the bracket is 1 and $w = eE/m$ , the Newtonian value. At $T = mc^2$ the acceleration is already down to $1/8$ of that, and as $T \to \infty$ it falls to zero, so the speed approaches $c$ without reaching it. Units: $eE/m$ is a force over a mass, and the bracket is a pure number.

Remark

Along its velocity the electron resists a force as if its mass were $m\gamma^3$ . A force component across the velocity only turns $\vect v$ , leaving $\gamma$ unchanged at that instant, and there the same law gives a resistance of only $m\gamma$ . So a relativistic particle’s acceleration is in general not parallel to the force acting on it: the part across $\vect v$ is favoured by the factor $\gamma^2$ .