A5-4. Special Relativity: Time Dilation and Length ContractionHL ONLY

1. Time Dilation

Moving clocks tick slower. When observing a reference frame that is moving relative to you, you will measure their time as passing more slowly than your own.

Proper Time ($\Delta t_0$): The time interval measured by an observer who is completely at rest relative to the events (the events happen at the exact same spatial location in their frame).

Time Dilation Formula:

$$\Delta t = \gamma \Delta t_0$$

Because $\gamma \ge 1$, the dilated time ($\Delta t$) measured by the stationary outside observer is always greater (longer) than the proper time ($\Delta t_0$).

2. Length Contraction

Moving objects are shorter. When an object moves relative to an observer, the observer measures the object to be contracted along the direction of motion.

Proper Length ($L_0$): The length measured by an observer who is at rest relative to the object.

Length Contraction Formula:

$$L = \dfrac{L_0}{\gamma}$$

Because you divide by $\gamma$, the contracted length ($L$) measured by the outside observer is always shorter than the proper length. (Note: Contraction only occurs parallel to the direction of motion, never perpendicular).

3. The Muon Decay Experiment

The best physical evidence for time dilation and length contraction comes from subatomic particles called muons, created in the upper atmosphere by cosmic rays.

Example 3: Relativistic Muons

Problem: A muon has a proper half-life of $\Delta t_0 = 1.5 \times 10^{-6}\text{ s}$ when at rest. It is created high in the atmosphere and travels toward Earth at $v = 0.98c$ (yielding a $\gamma = 5$). According to classical physics, it shouldn't survive the 4250m trip to the surface. Why does it?

Solution:

  • From the Earth Observer's Frame (Time Dilation):
    The Earth observer sees the muon moving. Therefore, the muon's internal clock ticks slower.
    $\Delta t = \gamma \Delta t_0 = 5 \times (1.5 \times 10^{-6}) = \mathbf{7.5 \times 10^{-6}\text{ s}}$
    Because its lifespan is 5 times longer, it easily has enough time to reach the ground.
  • From the Muon's Frame (Length Contraction):
    In the muon's own frame, it is stationary, and the Earth is rushing up towards it at $0.98c$. Therefore, the atmosphere is length contracted.
    $L = \dfrac{L_0}{\gamma} = \dfrac{4250\text{ m}}{5} = \mathbf{850\text{ m}}$
    Because the distance is 5 times shorter, its short proper lifespan is plenty of time to cross the 850m gap.
  • Conclusion: Both frames successfully conclude the muon hits the ground, preserving the logical consistency of the universe!