Irodov 1.357: The Interval Answers Both Questions – Amit & Deepali Sinha Foundation

Problem Irodov 1.357

Three events $A$ , $B$ and $C$ take place on the $x$ axis of an inertial frame $K$ . Their positions and times are read off a space-time chart that plots $ct$ against $x$ with both coordinates in meters, and the pairs $(x, ct)$ come out as $A\,(2, 1)$ , $B\,(5, 6)$ and $C\,(7, 4)$ . Find:

the time interval between $A$ and $B$ measured in the inertial frame in which those two events happen at one and the same point;
the distance between the points at which $A$ and $C$ happen, measured in the inertial frame in which those two events are simultaneous.

The three events A, B and C on the chart of frame K. Both coordinates are lengths in meters on a common scale, so a light signal runs at 45°. — The three events $A$ , $B$ and $C$ on the chart of frame $K$ . Both coordinates are lengths in meters on a common scale, so a light signal runs at $45^\circ$ .

Given

$A\,(x,ct) = (2,1)$ m
$B\,(x,ct) = (5,6)$ m
$C\,(x,ct) = (7,4)$ m
measured in frame $K$
$c = 3.00\times 10^{8}$ m/s

Find

(a)

\Delta t'

for

A

B

where they coincide in space; (b)

\Delta x'

for

A

C

where they coincide in time

Solution

Step 1Set down the two separations in

K

Only differences of coordinates matter, so subtract $A$ ’s pair from $B$ ’s and from $C$ ’s. Both entries of each pair are lengths, which is the whole convenience of plotting $ct$ rather than $t$ .

\begin{aligned} \Delta x_{AB} &= 5-2 = 3\ \mathrm{m}, & c\,\Delta t_{AB} &= 6-1 = 5\ \mathrm{m},\\ \Delta x_{AC} &= 7-2 = 5\ \mathrm{m}, & c\,\Delta t_{AC} &= 4-1 = 3\ \mathrm{m}. \end{aligned} \tag{1}

Step 2Evaluate the one quantity every frame agrees on

Lorentz transformations mix $x$ and $ct$ into one another, but they leave the combination $c^{2}\Delta t^{2}-\Delta x^{2}$ untouched, exactly as a rotation leaves $x^{2}+y^{2}$ untouched. That combination is the squared interval, and it is the only thing here that can be carried from $K$ into the frames the problem asks about.

s^{2} = c^{2}\Delta t^{2}-\Delta x^{2}

Putting the numbers of (1) into it for each pair,

\begin{aligned} s_{AB}^{2} &= 5^{2}-3^{2} = 25-9, \\ s_{AC}^{2} &= 3^{2}-5^{2} = 9-25 . \end{aligned}

s_{AB}^{2} = +16\ \mathrm{m^{2}}, \qquad s_{AC}^{2} = -16\ \mathrm{m^{2}}

(2)

Step 3Part (a): the frame in which

A

and

B

coincide in space

The timelike pair, A at the origin. The ct' axis of the frame at V = 0.60c runs through B. The dashed hyperbola of interval 4 m cuts ct at 4 m and crosses ct' at B. — The timelike pair, $A$ at the origin. The $ct'$ axis of the frame at $V = 0.60c$ runs through $B$ . The dashed hyperbola of interval $4$ m cuts $ct$ at $4$ m and crosses $ct'$ at $B$ .

The frame asked for is the one in which the two events share a point, so in it $\Delta x' = 0$ and the interval collapses onto the time term alone. Writing the invariance of $s^{2}$ between $K$ and that frame,

s_{AB}^{2} = c^{2}\Delta t'^{2}-\underbrace{\Delta x'^{2}}_{=\,0} = c^{2}\Delta t'^{2}

so that $c\,\Delta t'$ is the square root of the interval found in (2).

c\,\Delta t' = \sqrt{s_{AB}^{2}} = \sqrt{16\ \mathrm{m^{2}}} = 4.0\ \mathrm{m}

Dividing by the speed of light turns that length into a time. Numerically,

\Delta t' = \frac{4.0\ \mathrm{m}}{3.00\times 10^{8}\ \mathrm{m/s}}

\Delta t' = \frac{\sqrt{s_{AB}^{2}}}{c} = 1.3\times 10^{-8}\ \mathrm{s} = 13\ \mathrm{ns}

(3)

Step 4Part (b): the frame in which

A

and

C

coincide in time

The spacelike pair, A at the origin. The x' axis of the frame at V = 0.60c runs through C: that frame calls A and C simultaneous. The sideways hyperbola cuts x at 4 m and crosses x' at C. — The spacelike pair, $A$ at the origin. The $x'$ axis of the frame at $V = 0.60c$ runs through $C$ : that frame calls $A$ and $C$ simultaneous. The sideways hyperbola cuts $x$ at $4$ m and crosses $x'$ at $C$ .

This time the frame asked for is the one in which the two events share a time, so $\Delta t' = 0$ and the interval collapses onto the space term alone.

s_{AC}^{2} = \underbrace{c^{2}\Delta t'^{2}}_{=\,0}-\Delta x'^{2} = -\Delta x'^{2}

The interval for this pair is negative, so $-s_{AC}^{2}$ is a positive number and its square root is a real length.

\Delta x'^{2} = -s_{AC}^{2} = 16\ \mathrm{m^{2}}

\Delta x' = \sqrt{-s_{AC}^{2}} = 4.0\ \mathrm{m}

(4)

Check

Both frames can be exhibited explicitly, which also confirms that they exist. Take $V = 0.60c$ along $+x$ , so that $\beta = 0.60$ and $\gamma = 1/\sqrt{1-0.36} = 1.25$ . For the pair $A$ , $B$ the transformation gives $\Delta x' = \gamma(\Delta x-\beta\,c\Delta t) = 1.25(3-0.60\times 5) = 0$ and $c\,\Delta t' = \gamma(c\Delta t-\beta\,\Delta x) = 1.25(5-0.60\times 3) = 4.0$ m, reproducing part (a). For the pair $A$ , $C$ the same boost gives $c\,\Delta t' = 1.25(3-0.60\times 5) = 0$ and $\Delta x' = 1.25(5-0.60\times 3) = 4.0$ m, reproducing part (b).

Remark

One boost served both parts above, and that is not a general theorem but an arithmetic accident of these particular coordinates: the frame that co-locates a timelike pair must move at $\beta = \Delta x/(c\Delta t)$ , here $3/5$ , while the frame that synchronizes a spacelike pair must move at $\beta = c\Delta t/\Delta x$ , here also $3/5$ , because the two pairs happen to carry the same two numbers $3$ and $5$ swapped over. What is general is the reciprocal form of those two conditions, and with it the reason each has a solution below $c$ exactly when its own $s^{2}$ has the right sign. Note also that the single frame moving at $0.60c$ is one in which $A$ and $B$ occupy one point while $A$ and $C$ occupy one instant, so a single observer there could report both answers from a single laboratory notebook.