🥋 The Move
Identify the proper time \(\Delta\tau\) (measured by the clock that’s co-located with both events), compute \(\gamma=\dfrac{1}{\sqrt{1-\beta^2}}\) with \(\beta=v/c\), and use
\[
\boxed{\Delta t=\gamma\,\Delta\tau}
\]
to get the dilated time in any frame where the clock is moving. Proper time is always the shortest time between the two events.
📘 Canonical Problem
A spaceship moves at speed \(v\) relative to Earth. A clock on the ship measures a proper time interval \(\Delta\tau\) between two ticks. What time interval \(\Delta t\) does an Earth observer measure between those ticks?
- Given: \(v,\ \Delta\tau,\ c\)
- Find: \(\Delta t\) as seen from Earth
- Assume: inertial frames; events occur at the same place in the ship frame (so \(\Delta\tau\) is proper).
Sketch notes (no figure):
- Define \(\beta=v/c\), \(\gamma=(1-\beta^2)^{-1/2}\).
- Invariant interval: \(c^2\Delta\tau^2=c^2\Delta t^2-\Delta x^2\).
- For the ship’s clock seen from Earth: \(\Delta x=v\,\Delta t\Rightarrow \Delta t=\gamma\Delta\tau\).
🔎 Setup (Preview)
- Pick the proper frame. In the ship frame, the two ticks occur at one place \(\Rightarrow \Delta x’=0\) and \(\Delta t’=\Delta\tau\).
- Relate frames. In Earth frame, the clock moves: \(\Delta x=v\,\Delta t\). Plug into \(c^2\Delta\tau^2=c^2\Delta t^2-\Delta x^2\) to get \(\Delta t=\gamma\Delta\tau\).
Full derivation, exam-speed forms, and traps at Black Belt.