Irodov 3.377: Sideways Push Slows the Forward Speed – Amit & Deepali Sinha Foundation

Problem Irodov 3.377

A relativistic proton enters, at $t = 0$ , a region of uniform electric field $\vect E$ , its velocity $\vect v_0$ at entry being perpendicular to $\vect E$ . As functions of time, find:

the angle $\theta$ between the proton’s velocity $\vect v$ and its initial direction of motion;
the component $v_x$ of $\vect v$ along that initial direction.

The proton p arriving with velocity v_0 at the region of uniform field E, which is perpendicular to v_0; the field is seen side on. — The proton $p$ arriving with velocity $\vect v_0$ at the region of uniform field $\vect E$ , which is perpendicular to $\vect v_0$ ; the field is seen side on.

Given

$\vect v_0 \perp \vect E$
$E$ uniform
entry at $t = 0$

Find

(a)

\theta(t)

· (b)

v_x(t)

Solution

Step 1The two components of the momentum

Take $x$ along $\vect v_0$ and $y$ along $\vect E$ , with $\beta_0 = v_0/c$ and $m$ the proton’s rest mass. There is no force along $x$ , so $p_x$ keeps its entry value $p_0$ ; along $y$ the constant force $eE$ builds the momentum up from zero:

p_x = p_0 = \frac{m v_0}{\sqrt{1 - \beta_0^2}}, \qquad p_y = eEt \tag{1}

The proton's momentum p at time t: the constant component p_0 along the initial direction x and the growing component eEt along the field. The velocity has the direction of p, at the angle θ to x. — The proton’s momentum $\vect p$ at time $t$ : the constant component $p_0$ along the initial direction $x$ and the growing component $eEt$ along the field. The velocity has the direction of $\vect p$ , at the angle $\theta$ to $x$ .

Step 2(a) The angle of the velocity

The velocity $\vect v = \vect p c^2/W$ , with $W$ the total energy, is parallel to the momentum, so $\theta$ is the angle of $\vect p$ . Its tangent is the ratio of the components (1):

\tan\theta = \frac{p_y}{p_x} = \frac{eEt}{m v_0/\sqrt{1 - \beta_0^2}}

\tan\theta = \frac{eEt}{m v_0}\sqrt{1 - \beta_0^2}

(2)

Step 3The total energy as it grows

The total energy follows from the momentum through $W^2 = (pc)^2 + (mc^2)^2$ . At entry it is $W_0 = mc^2/\sqrt{1 - \beta_0^2}$ , and the growing $p_y$ adds to it:

W^2 = (p_0c)^2 + (mc^2)^2 + (eEct)^2 = W_0^2 + (eEct)^2 \tag{3}

Step 4(b) The forward component of the velocity

The forward component is $v_x = p_xc^2/W$ . At entry $p_0c^2 = v_0W_0$ ; with (3) for $W$ , then dividing through by $W_0$ :

\begin{aligned} v_x &= \frac{p_0c^2}{W} = \frac{v_0W_0}{\sqrt{W_0^2 + (eEct)^2}} = \frac{v_0}{\sqrt{1 + (eEct/W_0)^2}}, \\ \frac{eEct}{W_0} &= \frac{eEt}{mc}\sqrt{1 - \beta_0^2} \end{aligned}

v_x = \frac{v_0}{\sqrt{1 + (1 - \beta_0^2)\,(eEt/mc)^2}}

(4)

Check

Units: $eEt$ is an impulse, so $eEt/(mc)$ and $eEt/(mv_0)$ are pure numbers. Newtonian limit $c \to \infty$ : $\beta_0 \to 0$ and $eEt/(mc) \to 0$ , so $\tan\theta \to eEt/(mv_0)$ and $v_x \to v_0$ , the parabola of a slow particle. At $t = 0$ , $\theta = 0$ and $v_x = v_0$ . At late times $v_x$ falls off as $1/t$ while $v_y = p_yc^2/W \to c$ , so $\theta \to 90^\circ$ and the speed stays below $c$ .

Remark

No force acts along $x$ , yet $v_x$ falls. The momentum $p_x = W v_x/c^2$ is what stays constant; the field keeps pumping energy $W$ into the proton, so the forward velocity must drop to keep the product fixed. In Newtonian mechanics the two motions are independent; in relativity the transverse push drags on the forward motion through the shared energy.