Irodov 1.356: No Boost Outruns a Cause – Amit & Deepali Sinha Foundation

Problem Irodov 1.356

Two events happen at two different points of an inertial frame $K$ , the second occurring a time interval $\Delta t$ after the first. In $K$ the two are related as cause and effect, a shot being fired and the bullet reaching its target being one such pair. A frame $K'$ moves uniformly relative to $K$ . Demonstrate that the same two events keep that relationship in $K'$ , and indeed in every inertial frame, so that the cause is never observed to follow its effect.

The two events as the frame K reports them. The shot is fired at the left point and the bullet reaches the target at the right point, a distance Δ x away and a time Δ t later. The arrow is the bullet running between them. — The two events as the frame $K$ reports them. The shot is fired at the left point and the bullet reaches the target at the right point, a distance $\Delta x$ away and a time $\Delta t$ later. The arrow is the bullet running between them.

Given

Two events in frame $K$
$\Delta t = t_2-t_1 > 0$
separation $\Delta x = x_2-x_1$
event 1 causes event 2
$K'$ inertial, moving at $V$ along $x$

Find

Demonstrate that

\Delta t' = t_2'-t_1' > 0

K'

Solution

Step 1Name what the causal link supplies

The phrase “cause and effect” is not decoration here: it asserts that something physical, the bullet, went from the first event to the second. That carrier is a material body or at fastest a light signal, so the speed at which it covers the separation $\Delta x$ in the time $\Delta t$ cannot exceed $c$ . Call that speed $v$ .

v = \frac{\Delta x}{\Delta t}

The hypothesis of the problem is therefore one inequality on $v$ , together with the relation that lets $\Delta x$ be traded for $\Delta t$ in what follows.

\Delta x = v\,\Delta t \qquad\text{with}\qquad |v| \le c

(1)

Step 2Transform the interval into

K'

Let $K'$ move at the constant velocity $V$ along the common $x$ axis, with $|V| < c$ because $K'$ is an inertial frame carrying real clocks and rods. The Lorentz transformation of the time coordinate applied to each event and subtracted gives the interval between the same two events in $K'$ .

\Delta t' = \frac{\Delta t - \dfrac{V\,\Delta x}{c^{2}}} {\sqrt{1-\dfrac{V^{2}}{c^{2}}}}

Substituting the carrier relation (1) removes $\Delta x$ in favor of $\Delta t$ , and $\Delta t$ then comes out of the numerator as a common factor.

\begin{aligned} \Delta t' &= \frac{\Delta t - \dfrac{V v\,\Delta t}{c^{2}}} {\sqrt{1-\dfrac{V^{2}}{c^{2}}}} \\[4pt] &= \frac{\Delta t\left(1-\dfrac{V v}{c^{2}}\right)} {\sqrt{1-\dfrac{V^{2}}{c^{2}}}} \end{aligned} \tag{2}

Step 3Show the one factor that could change sign never does

In (2) the factor $\Delta t$ is positive by hypothesis and the square root is a positive real number for every $|V| < c$ , so the sign of $\Delta t'$ is decided entirely by the bracket. Bounding the product of the two speeds settles it, since neither factor can reach unity.

\left|\frac{V v}{c^{2}}\right| = \frac{|V|}{c}\cdot\frac{|v|}{c} \le \frac{|V|}{c}

\left|\frac{V v}{c^{2}}\right| \le \frac{|V|}{c} < 1

(3)

The bracket therefore lies strictly between $0$ and $2$ , and in particular it is positive. A positive number divided by a positive root and multiplied by a positive $\Delta t$ is positive, which is the proposition asked for.

\begin{gathered} \Delta t' > 0 \quad\text{for every inertial } K', \\ \text{given } \Delta t > 0,\ |v| \le c,\ |V| < c \end{gathered}

(4)

The two events keep their order, so the shot still precedes the hit in $K'$ .

A causal pair in K: the gold world line is the carrier from cause A to effect B, which lies inside the shaded future cone. The dashed x' axis belongs to K' at V = 0.60c, and B sits above it. — A causal pair in $K$ : the gold world line is the carrier from cause $A$ to effect $B$ , which lies inside the shaded future cone. The dashed $x'$ axis belongs to $K'$ at $V = 0.60c$ , and $B$ sits above it.

A pair no carrier could link: D lies outside the light cone of A. The same K' axes now pass above D, putting D earlier than A in K' though later in K. This is the case the hypothesis excludes. — A pair no carrier could link: $D$ lies outside the light cone of $A$ . The same $K'$ axes now pass above $D$ , putting $D$ earlier than $A$ in $K'$ though later in $K$ . This is the case the hypothesis excludes.

Check

Take the fastest possible carrier, a light signal, so that $|v| = c$ and the bound in (3) is saturated. Then $\Delta t' = \Delta t(1-V/c)/\sqrt{1-V^{2}/c^{2}} = \Delta t\sqrt{(1-V/c)/(1+V/c)}$ , which is the relativistic Doppler factor. At $V = 0.60c$ it gives $\Delta t' = 0.50\,\Delta t$ , and as $V \to c$ it shrinks toward zero without ever reaching it, let alone passing through it. The extreme case therefore approaches the boundary of the claim and never crosses it.

Remark

The same algebra says exactly when the order can reverse. The interval $\Delta t'$ measured in $K'$ turns negative once the $K$ quantity $V\Delta x/c^{2}$ exceeds $\Delta t$ , that is once $|V|/c$ exceeds $c\,\Delta t/|\Delta x|$ , and a boost speed below $c$ can do that only if $|\Delta x| > c\,\Delta t$ . That is precisely a spacelike pair, the right-hand figure’s $A$ and $D$ , for which the invariant interval $c^{2}\Delta t^{2}-\Delta x^{2}$ is negative and no carrier exists. So relativity does not protect the time order of events in general, only of events that could have influenced one another; the light cone of the first event, a locus every inertial frame agrees on, is the exact boundary between the two cases.