A.5.10 (HL)—Time dilation

Use the time-dilation equation

When $\Delta t_0$ is the proper time between two events, an observer for whom the events occur at different positions measures $\Delta t=\gamma\Delta t_0$, where $\gamma=1/\sqrt{1-v^2/c^2}$.

Identify the proper interval first

Find the frame in which the two events occur at the same place; that frame measures $\Delta t_0$. Calculate $\gamma$ using the relative speed, then multiply by $\Delta t_0$ to obtain the longer interval measured in the other inertial frame.

Check the direction of the effect

Because $\gamma\ge1$, the non-proper observer measures a time interval at least as large as the proper interval. At $v=0$, $\gamma=1$ and the two measurements agree.

Worked example from local Question Bank rows 30639–30640

A spacecraft crosses $1.80\times10^{11}\,\mathrm{m}$ at $0.750c$. The station-frame interval is

$$\Delta t=\frac{1.80\times10^{11}}{0.750(3.00\times10^8)}=800\,\mathrm{s}$$

With $\gamma=1/\sqrt{1-0.750^2}=1.51$, the spacecraft clock measures the proper time

$$\Delta t_0=\frac{800}{1.51}=530\,\mathrm{s}$$

The events occur at one place on the spacecraft, so its interval is proper.

Common trap

Do not multiply the proper time by $1/\gamma$ when finding the dilated interval. The inverse is used only when the question gives the larger interval and asks for the proper time.