Irodov 1.378: Momentum Grows Forever, Speed Saturates

Problem Irodov 1.378

A particle of rest mass $m_0$ is at rest until the moment $t = 0$ . From that moment on a constant force $\vect F$ , unchanging in both magnitude and direction, acts on it. Working in the reference frame in which the particle was initially at rest, find how the particle’s speed $v$ and the distance $l$ it has covered depend on the elapsed time $t$ .

The particle starts from rest at t = 0 and is pushed by the constant force F; l is the distance it has covered by the time t. The force keeps its direction throughout, so the whole motion runs along that one line. — The particle starts from rest at $t = 0$ and is pushed by the constant force $\vect F$ ; $l$ is the distance it has covered by the time $t$ . The force keeps its direction throughout, so the whole motion runs along that one line.

Given

$m_0$
$\vect F = \mathrm{constant}$
$v = 0$ at $t = 0$

Find

the speed

v(t)

and the distance covered

l(t)

Solution

Step 1The impulse fixes the momentum, not the velocity

The law that survives unchanged in relativity is the one written for momentum: the rate of change of momentum equals the applied force. The particle starts at rest and the force never turns, so the momentum stays along $\vect F$ and the problem is one dimensional from the outset.

\frac{d\vect p}{dt} = \vect F, \qquad \vect p = \frac{m_0\vect v}{\sqrt{1-v^2/c^2}}

Integrating the constant force from the moment the motion starts gives the momentum at time $t$ .

p(t) = \int_0^t F\,dt' = Ft

Writing the left side out in terms of the speed turns that into an equation for $v$ alone:

\frac{m_0 v}{\sqrt{1-v^2/c^2}} = Ft

(1)

Step 2Unpacking the speed

Equation (1) is implicit in $v$ , so it has to be squared and solved. Squaring both sides clears the root and leaves a linear equation in $v^2$ .

\begin{aligned} m_0^2v^2 &= F^2t^2\left(1-\frac{v^2}{c^2}\right) \\[2pt] v^2\left(m_0^2+\frac{F^2t^2}{c^2}\right) &= F^2t^2 \\[2pt] v^2 &= \frac{F^2t^2}{m_0^2+F^2t^2/c^2} \end{aligned}

Taking the positive root and multiplying numerator and denominator by $c$ gives the speed at any time.

v = \frac{Ft}{\sqrt{m_0^2+F^2t^2/c^2}} = \frac{Fct}{\sqrt{m_0^2c^2+F^2t^2}}

(2)

The result carries its own time scale, the instant $\tau = m_0c/F$ at which the accumulated impulse $Ft$ reaches $m_0c$ .

Speed against time, with τ = m_0c/F. The gold curve is , the royal line the classical Ft/m_0 it tracks while t is small. The dotted gold line is c, approached but never reached. At t = τ the speed is 0.71c. — Speed against time, with $\tau = m_0c/F$ . The gold curve is (2), the royal line the classical $Ft/m_0$ it tracks while $t$ is small. The dotted gold line is $c$ , approached but never reached. At $t = \tau$ the speed is $0.71c$ .

Step 3The distance covered

The distance is the time integral of the speed from the start of the motion. The integrand has the form of a derivative of a square root, so the substitution $u = m_0^2c^2+F^2t'^2$ , for which $du = 2F^2t'\,dt'$ , carries it through with the limits transformed along with it.

\begin{aligned} l &= \int_0^t v(t')\,dt' = \int_0^t \frac{Fct'\,dt'}{\sqrt{m_0^2c^2+F^2t'^2}} \\[3pt] &= \frac{c}{2F}\int_{m_0^2c^2}^{m_0^2c^2+F^2t^2}\frac{du}{\sqrt{u}} \\[3pt] &= \frac{c}{F}\Bigl[\sqrt{u}\,\Bigr]_{m_0^2c^2}^{m_0^2c^2+F^2t^2} \end{aligned}

Evaluating the bracket and taking the constant $c/F$ inside both square roots gives the distance covered.

l = \sqrt{\left(\frac{m_0c^2}{F}\right)^{\!2}+c^2t^2} \; - \; \frac{m_0c^2}{F}

(3)

Distance against time. The gold curve leaves the origin tangent to the t axis, then straightens into a line of slope c running parallel to the royal light line ct, approaching a fixed m_0c^2/F behind it, the value ticked on the vertical axis. — Distance against time. The gold curve leaves the origin tangent to the $t$ axis, then straightens into a line of slope $c$ running parallel to the royal light line $ct$ , approaching a fixed $m_0c^2/F$ behind it, the value ticked on the vertical axis.

Step 4What the two results say at early and late times

Both results, (2) and (3), are governed by the single ratio $t/\tau$ . For $t \ll \tau$ the impulse is still small compared with $m_0c$ , and both expressions reduce to the Newtonian ones.

\begin{aligned} v &= \frac{Ft}{m_0\sqrt{1+t^2/\tau^2}} \approx \frac{Ft}{m_0} \\[3pt] l &= \frac{m_0c^2}{F}\left(\sqrt{1+\frac{t^2}{\tau^2}}-1\right) \approx \frac{m_0c^2}{F}\cdot\frac{t^2}{2\tau^2} = \frac{Ft^2}{2m_0} \end{aligned}

For $t \gg \tau$ the rest mass term under each root is negligible beside the impulse term, and the motion becomes the ultrarelativistic one.

v \to c, \qquad l \to ct-\frac{m_0c^2}{F}

Check

Differentiating (3) must give back (2), and it does. Writing $R$ for the square root that stands in $l$ , the derivative is $dl/dt = c^2t/R$ , and multiplying that above and below by $F/c$ turns it into the speed (2) exactly. The combination $m_0c^2/F$ is an energy divided by a force and so is a length, as (3) needs it to be. Finally $v/c = t/\sqrt{\tau^2+t^2}$ is smaller than $1$ at every finite time, so the speed stays under $c$ however long the force acts.

Remark

The distance (3) rearranges into $\left(l+m_0c^2/F\right)^2-c^2t^2 = \left(m_0c^2/F\right)^2$ , a hyperbola whose asymptotes are the light lines through $l = -m_0c^2/F$ , the motion known as hyperbolic motion. Written with the momentum as $p = m_0c\sinh\eta$ , so that $\eta$ is the rapidity, (1) reads $\sinh\eta = t/\tau$ , and $d\eta/d\tau_{\text{pr}} = F/\left(m_0c\right)$ , a constant. In this straight-line case, with the force parallel to the velocity, a force held constant in the laboratory is also the constant push felt in the particle’s instantaneous rest frame, and it is the rapidity, not the speed, that grows uniformly in the particle’s own proper time.