Problem — Spacecraft and the beeps

Problem

A spacecraft moves away from Earth at $v = 0{,}60\,c$. On the spacecraft, a clock reads that $1{,}0$ s passes between one beep and the next. How much time passes between two beeps as measured from Earth?

Solution

Proper time. The interval between two beeps is measured by a single onboard clock, at the same point: it is the proper time $\Delta\tau = 1{,}0$ s.

Lorentz factor. With $v = 0{,}60\,c$, $\dfrac{v^2}{c^2} = 0{,}36$:

$\gamma = \frac{1}{\sqrt{1 - 0{,}36}} = \frac{1}{\sqrt{0{,}64}} = \frac{1}{0{,}8} = 1{,}25$

Time dilation. The observer on Earth measures a dilated interval:

$\Delta t' = \gamma\,\Delta\tau = 1{,}25 \times 1{,}0\;\text{s}$

$\ev{\Delta t' \approx 1{,}25\;\text{s}}$

For Earth, the beeps of the moving clock arrive more spaced out: the spacecraft’s clock “ticks more slowly”.

Links

Topics: Relatività ristretta Concepts: Dilatazione dei tempi Skills: Impostazione simbolica