Irodov 1.380: Two Inertias, One Rest Mass – Amit & Deepali Sinha Foundation

Problem Irodov 1.380

A particle of rest mass $m_0$ moves with velocity $\vect v$ in an inertial reference frame, and a force $\vect F$ acts on it. Its acceleration in that same frame is $\vect w = d\vect v/dt$ . Take as the starting point the fundamental equation of relativistic dynamics, which equates the rate of change of the particle’s momentum $\vect p = m_0\vect v/\sqrt{1-v^2/c^2}$ to the force $\vect F$ . Throughout, $\beta$ denotes the speed ratio $v/c$ . Find:

the circumstances under which the acceleration $\vect w$ points in the same direction as the force $\vect F$
the proportionality factor $k$ in $\vect F = k\vect w$ , for the case in which $\vect F$ is perpendicular to $\vect v$ and for the case in which $\vect F$ lies along the line of $\vect v$

The particle at the instant considered, moving with velocity v while the force F acts at an angle θ to that velocity. The acceleration is deliberately absent from this drawing: where it points is the first thing the problem asks. — The particle at the instant considered, moving with velocity $\vect v$ while the force $\vect F$ acts at an angle $\theta$ to that velocity. The acceleration is deliberately absent from this drawing: where it points is the first thing the problem asks.

Given

$m_0$
$\vect v$
$\vect F$
$\vect w = d\vect v/dt$
$\beta = v/c$

Find

(a) the circumstances in which

\vect w

is along

\vect F

, (b) the factor

k

\vect F = k\vect w

for

\vect F \perp \vect v

and for

\vect F \parallel \vect v

Solution

Step 1The law that governs the motion

Everything is measured in one inertial frame, the frame in which the particle has the velocity $\vect v$ . The law that survives unchanged in relativity is written for the momentum, and the whole velocity dependence of the inertia sits inside that momentum.

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

Step 2Differentiating the momentum splits the force in two

The product $\gamma\vect v$ has to be differentiated as a product, because $\gamma$ depends on the time through the speed. The chain rule applied to $\gamma = (1-\vect v\cdot\vect v/c^2)^{-1/2}$ gives its rate of change in terms of the acceleration.

\begin{aligned} \frac{d\gamma}{dt} &= -\frac{1}{2}\left(1-\frac{v^2}{c^2}\right)^{-3/2} \left(-\frac{2\,\vect v\cdot\vect w}{c^2}\right) \\[3pt] &= \frac{\gamma^3(\vect v\cdot\vect w)}{c^2} \end{aligned}

Feeding that into the derivative of the momentum gives the force in terms of the acceleration.

\begin{aligned} \vect F &= m_0\frac{d}{dt}\bigl(\gamma\vect v\bigr) \\[3pt] &= m_0\gamma\vect w + m_0\vect v\,\frac{d\gamma}{dt} \end{aligned}

\vect F = m_0\gamma\vect w + \frac{m_0\gamma^3(\vect v\cdot\vect w)}{c^2}\,\vect v

(1)

The second term points along $\vect v$ , not along $\vect w$ , which is already the reason the two vectors need not agree.

Step 3Inverting the law to get the acceleration

Equation (1) delivers $\vect F$ once $\vect w$ is known, and both questions need it the other way round. The only obstacle is the scalar $\vect v\cdot\vect w$ buried in the second term, so the first move is to take the scalar product of (1) with $\vect v$ and solve for it.

\begin{aligned} \vect F\cdot\vect v &= m_0\gamma(\vect v\cdot\vect w) + \frac{m_0\gamma^3(\vect v\cdot\vect w)\,v^2}{c^2} \\[3pt] &= m_0\gamma(\vect v\cdot\vect w)\bigl(1+\gamma^2\beta^2\bigr) \end{aligned}

The bracket collapses, because $\gamma^2$ and $\gamma^2\beta^2$ differ by exactly one.

1+\gamma^2\beta^2 = \frac{(1-\beta^2)+\beta^2}{1-\beta^2} = \gamma^2

so the scalar product of the velocity with the acceleration is fixed by the scalar product of the velocity with the force.

\vect F\cdot\vect v = m_0\gamma^3(\vect v\cdot\vect w) \qquad\Longrightarrow\qquad \vect v\cdot\vect w = \frac{\vect F\cdot\vect v}{m_0\gamma^3}

Returning that to (1), the factor $m_0\gamma^3$ cancels against the one already standing in the second term.

\begin{aligned} m_0\gamma\vect w &= \vect F - \frac{m_0\gamma^3(\vect v\cdot\vect w)}{c^2}\,\vect v \\[3pt] &= \vect F - \frac{m_0\gamma^3}{c^2} \cdot\frac{\vect F\cdot\vect v}{m_0\gamma^3}\,\vect v \end{aligned}

\vect w = \frac{1}{m_0\gamma} \left[\vect F - \frac{(\vect F\cdot\vect v)}{c^2}\,\vect v\right]

(2)

The same particle with the acceleration drawn in; θ is smaller here than in the setup figure, so the wedge can be marked. The acceleration w sits at the larger angle θ' from v, and the marked wedge is θ'-θ. — The same particle with the acceleration drawn in; $\theta$ is smaller here than in the setup figure, so the wedge can be marked. The acceleration $\vect w$ sits at the larger angle $\theta'$ from $\vect v$ , and the marked wedge is $\theta'-\theta$ .

Step 4Resolving the response along and across the velocity

Take components of (2) along $\vect v$ and perpendicular to it, writing $F_\parallel$ and $F_\perp$ for the two parts of the force and $w_\parallel$ and $w_\perp$ for the two parts of the acceleration. The subtracted vector lies entirely along $\vect v$ , with component $F_\parallel\beta^2$ there, so it touches the parallel line only.

\begin{aligned} m_0\gamma\,w_\parallel &= F_\parallel - F_\parallel\beta^2 = F_\parallel(1-\beta^2) \\[3pt] m_0\gamma\,w_\perp &= F_\perp \end{aligned}

Dividing each line by $m_0\gamma$ and using $(1-\beta^2)/\gamma = \gamma^{-3}$ on the first gives the two independent responses.

w_\parallel = \frac{F_\parallel}{m_0\gamma^3}, \qquad w_\perp = \frac{F_\perp}{m_0\gamma}

(3)

The two divisors are different, and that single fact answers both parts of the problem.

Step 5When the acceleration follows the force

Read (2) as it stands. The acceleration is the force minus a vector directed along $\vect v$ , so it can point along $\vect F$ only if that subtracted vector is itself along $\vect F$ or is absent altogether.

\vect w \parallel \vect F \quad\Longleftrightarrow\quad (\vect F\cdot\vect v)\,\vect v \parallel \vect F \quad\Longleftrightarrow\quad \vect F\cdot\vect v = 0 \ \ \text{or}\ \ \vect v \parallel \vect F

The same two cases show up in the angles of the figure above. Dividing the two lines of (3) into each other turns the tangent of the acceleration’s angle into the tangent of the force’s.

\tan\theta' = \frac{w_\perp}{w_\parallel} = \gamma^2\,\frac{F_\perp}{F_\parallel} = \gamma^2\tan\theta

Since $\gamma^2>1$ at every $v>0$ , the two angles agree only where the tangent is unmoved by a factor, namely at $\tan\theta = 0$ and at $\theta = 90^\circ$ . Both divisors in (3) are positive, so in those two cases the acceleration takes not merely the line of the force but its sense as well.

Only for

\vect F\perp\vect v

or for

\vect F

along

\vect v

Step 6The transverse factor

For the first case put $F_\parallel = 0$ in (3). The whole acceleration is then the perpendicular part, so $w = w_\perp$ , and the second line of (3) inverts into the factor asked for.

w = \frac{F_\perp}{m_0\gamma}

F_\perp = m_0\gamma w = \frac{m_0w}{\sqrt{1-\beta^2}}

(4)

Step 7The longitudinal factor

For the second case put $F_\perp = 0$ instead. Now the acceleration lies along $\vect v$ , so $w = w_\parallel$ , and the first line of (3) inverts with the cube of $\gamma$ in place of $\gamma$ .

w = \frac{F_\parallel}{m_0\gamma^3}

F_\parallel = m_0\gamma^3 w = \frac{m_0w}{(1-\beta^2)^{3/2}}

(5)

The factors and in units of m_0, over 0 <= β <= 0.8. The gold curve is the longitudinal factor, the royal one the transverse; at β = 0.8 they read 4.63 m_0 and 1.67 m_0. — The factors (4) and (5) in units of $m_0$ , over $0 \le \beta \le 0.8$ . The gold curve is the longitudinal factor, the royal one the transverse; at $\beta = 0.8$ they read $4.63\,m_0$ and $1.67\,m_0$ .

Check

Putting (2) back into (1) must return $\vect F$ , and it does: dotting (2) with $\vect v$ gives $\vect v\cdot\vect w = (\vect F\cdot\vect v)(1-\beta^2)/(m_0\gamma) = (\vect F\cdot\vect v)/(m_0\gamma^3)$ , so the second term of (1) is exactly $(\vect F\cdot\vect v)\vect v/c^2$ , which is precisely what the first term $m_0\gamma\vect w$ falls short of $\vect F$ by. At $\beta = 0$ both (4) and (5) reduce to $m_0$ and (1) becomes $\vect F = m_0\vect w$ , the Newtonian law. Their ratio is $F_\parallel/F_\perp = 1/(1-\beta^2) = \gamma^2$ , greater than $1$ at every $v>0$ , so producing a given acceleration always costs more force along the motion than across it, as the plotted curves show.

Remark

The factors (4) and (5) are the quantities once called the longitudinal mass $m_0\gamma^3$ and the transverse mass $m_0\gamma$ . Neither is a property of the particle: both belong to the particle’s state of motion in the frame doing the measuring, and in the particle’s own instantaneous rest frame $\beta = 0$ and the two collapse onto the single invariant $m_0$ . A body that resists a push differently depending on the push’s direction is exactly why a single velocity-dependent mass was given up as a useful idea, leaving the invariant $m_0$ and the momentum form of the law to carry the dynamics. The tilt has a vivid ultrarelativistic reading, still in the frame where the particle’s velocity is $\vect v$ : with $\tan\theta' = \gamma^2\tan\theta$ , a force held at some fixed small angle to the motion produces, as $\gamma$ grows, an acceleration turned ever closer to the perpendicular. Both responses die away there, the transverse one as $1/\gamma$ and the longitudinal one as $1/\gamma^3$ ; what the tilt records is the comparison, the turning of the particle’s direction outlasting the change in its speed by two powers of $\gamma$ .