How do time dilation and length contraction explain a fast particle's journey?

Observers on Earth and travelling with the particle agree about which events occur, but measure different times and positions. In the particle's rest frame, the particle travels zero distance: this is the viewpoint in which it is stationary. It is Earth and its landmarks that move.

Take two events: A is the particle passing a ground marker; B is its decay. Suppose the particle's own clock measures $t_o=4.0\,\mathrm{\mu s}$ between them and Earth measures its constant speed as $v=0.60c$. This is one example interval; unstable particles do not all have identical lifetimes.

The proper time is measured where both events happen at the same position. That is the particle frame. With $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$,

$$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}=1.25,$$ $$\Delta t=\gamma t_o=5.0\,\mathrm{\mu s}.$$

Earth therefore measures a particle displacement

$$\Delta x=v\Delta t=(0.60c)(5.0\,\mathrm{\mu s})=900\,\mathrm{m}.$$

In the particle frame, $\Delta x'=0$. You can check this by using the Lorentz transformations to convert the differences in position and time between the events:

$$\Delta x'=\gamma(\Delta x-v\Delta t)=0,$$ $$\Delta t'=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right)=4.0\,\mathrm{\mu s}.$$

Now imagine a second ground marker at Earth's decay location, $900\,\mathrm{m}$ from the first. Those markers are at rest on Earth, so their proper separation is $L_0=900\,\mathrm{m}$. The particle frame measures their simultaneous separation as

$$L'=\frac{L_0}{\gamma}=720\,\mathrm{m}.$$

The second marker approaches the stationary particle at $0.60c$ and reaches it after $720/(0.60c)=4.0\,\mathrm{\mu s}$. Meanwhile, the first marker has moved $720\,\mathrm{m}$ in the opposite direction. Neither of these distances is the particle's displacement in its own frame (its displacement in its own frame is zero).

The two descriptions agree when each calculation uses measurements from the same frame. Avoid pairing Earth's $900\,\mathrm{m}$ separation with the particle's $4.0\,\mathrm{\mu s}$ interval: that mixes frames. Also keep two tasks separate: locating events A and B, and measuring the distance between ground markers. A length measurement requires the markers' positions at the same time in the frame making the measurement.

Test your frame choices with the cosmic-particle question in the IB Physics space, time and motion questions (Question 64). Before calculating, identify whose clock measures the proper time and which object is moving in each frame.