Time dilation (AQA A-Level Physics): Revision Notes

12.3.3 Time dilation

Time Dilation

• Definition: Time dilation is an effect predicted by Einstein's theory of special relativity. It states that time will run at different rates for observers in different frames of reference, particularly when they are moving at high velocities relative to one another.
• Key Principle: Time dilation only occurs in inertial frames (frames moving at a constant velocity) and causes time to pass slower for an observer in motion relative to a stationary observer.

Key Concepts:

• Stationary Observer vs External Observer:
  • The stationary observer is considered to be at rest relative to the event being measured.
  • The external observer is moving relative to the stationary observer's frame of reference.
  • Proper Time ($t₀$): Time experienced by the stationary observer.
  • Dilated Time ($t$): Time experienced by the external observer (observed to be longer than proper time).
• Time Dilation Formula:

$$t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

Where:

• $t$ = time observed by the external observer
• $t_0$ = proper time observed by the stationary observer
• $v$ = relative velocity between the observers
• $c$ = speed of light
• Important Note: The proper time will always be less than the dilated time $t$ observed by the external observer.

Experimental Evidence: Muon Decay

• Muon Decay provides practical evidence of time dilation. Muons are high-speed particles that enter Earth's atmosphere from space and decay very quickly.
• Experiment Setup: Detectors are placed at two altitudes to measure muon count rate, allowing for calculation of the time dilation effect on their decay.

Experimental Setup for Muon Decay Observation

1. Detector Placement:

• A detector $1$ is placed at a high altitude, and a detector $2$ is set up further down, closer to the Earth's surface.
• The distance $d$ between the two detectors is 2 km.

2. Observations:

• Count rate at detector $1: 100 muons per second.
• Count rate at detector $2: 80 muons per second.

3. Known Values:

• Muon speed $( v )$: 0.996 times the speed of light $( c )$.
• Muon half-life (at rest): 1.5 μs (or $1.5 \times 10^{-6}$ seconds.

Calculating Expected Count Rate at Detector 2

This experiment involves two calculations:

4. Without considering special relativity (assuming no time dilation).
5. Considering special relativity (accounting for time dilation).

1. Expected Count Rate at Detector $2$ (Ignoring Time Dilation)

• Step 1: Calculate the time $( t )$ for muons to travel between the detectors:

$$t = \frac{d}{v} = \frac{2 \times 10^3}{0.996 \times 3 \times 10^8} = 6.69 \times 10^{-6} \, \text{s}$$

• Step 2: Calculate the number of half-lives $( n )$ expected during this time:

$$n = \frac{t}{\text{half-life}} = \frac{6.69 \times 10^{-6}}{1.5 \times 10^{-6}} = 4.46$$

• Step 3: Determine the expected count rate at detector $2$:
  • Using the formula for exponential decay:

$$\text{Expected count rate} = \text{Initial count} \times \left(\frac{1}{2}\right)^n = 100 \times \left(\frac{1}{2}\right)^{4.46} \approx 4.5 \, \text{s}^{-1}$$

• Conclusion: This predicted count rate of 4.5 s⁻¹ is significantly lower than the observed 80 s⁻¹, suggesting that ignoring special relativity leads to an incorrect result.

2. Expected Count Rate at Detector 2 (Considering Time Dilation)

• Step 1: Calculate the time $( t )$ from the laboratory frame (external observer's perspective):

$$t = \frac{d}{v} = 6.69 \times 10^{-6} \, \text{s}$$

• Step 2: Calculate the proper time $( t_0 )$, which is the time experienced by the muons (moving frame):

$$t_0 = t \sqrt{1 - \frac{v^2}{c^2}}$$

• Substitute values:

$$t_0 = 6.69 \times 10^{-6} \times \sqrt{1 - (0.996)^2} = 6.0 \times 10^{-7} \, \text{s}$$

• Step 3: Calculate the number of half-lives that occur during the muon's proper time:

$$n = \frac{t_0}{\text{half-life}} = \frac{6.0 \times 10^{-7}}{1.5 \times 10^{-6}} = 0.4$$

• Step 4: Determine the expected count rate at detector $2$:
  • Using exponential decay:

$$\text{Expected count rate} = 100 \times \left(\frac{1}{2}\right)^{0.4} \approx 76 \, \text{s}^{-1}$$

• Conclusion: This value (76 s⁻¹) is much closer to the observed value of 80 s⁻¹, which supports the existence of time dilation as predicted by special relativity.