Astronaut heading to Proxima Centauri

Problem

An astronaut sets off for Proxima Centauri ($\sim 4$ light years) travelling at $v = 0{,}80\,c$. What does their clock measure for the outbound trip? What do the clocks on Earth measure? How does this square with the idea that “time is time”?

Solution

Lorentz factor. With $\beta = 0{,}80$:

$\gamma = \frac{1}{\sqrt{1-0{,}80^2}} = \frac{1}{\sqrt{1-0{,}64}} = \frac{1}{\sqrt{0{,}36}} = \frac{1}{0{,}6} = \frac{5}{3} \approx 1{,}67.$

Time measured by Earth clocks. In Earth’s frame the distance is $d = 4$ light years and the speed is $0{,}80\,c$, so

$\Delta t_{\text{Terra}} = \frac{d}{v} = \frac{4\ \text{a.l.}}{0{,}80\,c} = 5\ \text{years}.$

Proper time on board. The astronaut’s clock reads the proper time, reduced by the factor $\gamma$:

$\Delta\tau = \frac{\Delta t_{\text{Terra}}}{\gamma} = 5 \times 0{,}6 = 3\ \text{years}.$

Check from the astronaut’s point of view. For them, Earth and Proxima are moving, and the distance is contracted: $d' = d/\gamma = 4 \times 0{,}6 = 2{,}4$ l.y. The travel time is then $d'/v = 2{,}4/0{,}80 = 3$ years, in agreement.

$\ev{\Delta\tau_{\text{bordo}} = 3\ \text{years}, \qquad \Delta t_{\text{Terra}} = 5\ \text{years}}$

“Time is time”. There is no contradiction: there is no absolute time equal for everyone. Every clock correctly measures time along its own world line. The astronaut has travelled a different spacetime path from the one who stays on Earth, so they accumulate less proper time. On return, they will truly be younger than their peers who stayed behind.

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Topics: Relatività ristretta Concepts: Dilatazione dei tempi · Contrazione delle lunghezze Skills: Cambio di sistema di riferimento