Irodov 1.379: Hyperbolic Motion Hides a Constant Force

Problem Irodov 1.379

In a reference frame $K$ a particle of rest mass $m_0$ travels along the $x$ axis, its coordinate at the time $t$ being

x = \sqrt{a^2+c^2t^2},

where $a$ is a constant length and $c$ is the speed of light. Find the force acting on the particle, as measured in this same frame $K$ .

The stated law of motion as a position-time graph, in units with a = c = 1. The gold curve x(t) leaves x = a with zero slope and closes on the royal light line ct from above without touching it. — The stated law of motion as a position-time graph, in units with $a = c = 1$ . The gold curve $x(t)$ leaves $x = a$ with zero slope and closes on the royal light line $ct$ from above without touching it.

Given

rest mass $m_0$
$x=\sqrt{a^2+c^2t^2}$
$a$ constant
motion along $x$ in $K$

Find

the force

F

on the particle in the frame

K

Solution

Step 1The velocity the stated law implies

The force in relativity acts on the momentum, and the momentum needs the velocity first, so the law of motion is differentiated once. The derivative of the square root brings its own argument down into the denominator:

\begin{aligned} v &= \frac{dx}{dt} = \frac{d}{dt}\sqrt{a^2+c^2t^2} \\[3pt] &= \frac{2c^2t}{2\sqrt{a^2+c^2t^2}} \end{aligned}

v = \frac{c^2t}{\sqrt{a^2+c^2t^2}}

(1)

Step 2The Lorentz factor of this motion

The momentum needs $\sqrt{1-v^2/c^2}$ , so evaluate it from (1) before assembling anything. Squaring the velocity and dividing by $c^2$ leaves a single fraction, and the $c^2t^2$ terms cancel:

\begin{aligned} \frac{v^2}{c^2} &= \frac{c^2t^2}{a^2+c^2t^2} \\[3pt] 1-\frac{v^2}{c^2} &= \frac{a^2+c^2t^2-c^2t^2}{a^2+c^2t^2} = \frac{a^2}{a^2+c^2t^2} \end{aligned}

Taking the square root of that leaves the constant $a$ over the same root as before:

\sqrt{1-\frac{v^2}{c^2}} = \frac{a}{\sqrt{a^2+c^2t^2}}

(2)

Step 3The momentum grows linearly with time

Assembling the momentum from (1) and (2) is the whole of the problem, because the awkward root appears once above and once below and cancels:

\begin{aligned} p &= \frac{m_0v}{\sqrt{1-v^2/c^2}} \\[3pt] &= m_0\cdot\frac{c^2t}{\sqrt{a^2+c^2t^2}} \cdot\frac{\sqrt{a^2+c^2t^2}}{a} \end{aligned}

p = \frac{m_0c^2t}{a}

(3)

Step 4The force is the slope of that line

The relativistic equation of motion says the force is the rate of change of the momentum, and (3) is linear in $t$ with a constant coefficient, so differentiating it returns that coefficient:

F = \frac{dp}{dt} = \frac{d}{dt}\left(\frac{m_0c^2}{a}\,t\right)

F = \frac{m_0c^2}{a}

(4)

Momentum against time, with p in units of m_0c and t in units of a/c. The gold line p(t) is straight; its constant slope is the force. The marked point is t = a/c, where p = m_0c. — Momentum against time, with $p$ in units of $m_0c$ and $t$ in units of $a/c$ . The gold line $p(t)$ is straight; its constant slope is the force. The marked point is $t = a/c$ , where $p = m_0c$ .

Check

The combination $m_0c^2/a$ is a rest energy divided by a length, which is a force, as the answer must be. The early-time behavior is the second test: expanding the law of motion for $ct \ll a$ gives $x \approx a+c^2t^2/2a$ , a Newtonian uniform acceleration $c^2/a$ starting from rest at $x = a$ , and Newton’s second law then reads $F = m_0c^2/a$ , the boxed value. The late-time behavior is the third: the momentum (3) grows without bound under this constant force, while (1) keeps the speed below $c$ forever, which is exactly how a relativistic particle absorbs an impulse that never stops.

Remark

Squaring the law of motion gives $x^2-c^2t^2 = a^2$ , so the particle’s worldline is a hyperbola with the light lines through the origin as its asymptotes; a constant force acting along the line of motion always produces a hyperbola of this kind, which is relativity’s stand-in for uniformly accelerated motion. Two constant quantities go with it, both belonging to the particle rather than to the frame $K$ : its proper acceleration, the acceleration read in its own instantaneous rest frame, which is $c^2/a$ ; and the rate at which its rapidity advances in its own proper time. The rapidity itself is not one of them. Writing $\eta$ for it, defined by $v/c = \tanh\eta$ , the motion satisfies $\sinh\eta = ct/a$ , which with $d\tau = dt\sqrt{1-v^2/c^2}$ integrates to $\eta = c\tau/a$ : $\eta$ climbs without bound, and it is the derivative $d\eta/d\tau = c/a$ that holds fixed. So it is the rapidity, not the speed, that grows uniformly in the particle’s own proper time $\tau$ . The stated law also runs backwards through $t<0$ , where it describes the particle arriving from large $x$ , decelerating, and coming momentarily to rest at $x = a$ at $t = 0$ : the force is the same constant throughout, and only its own sense relative to the motion changes.