Length contraction (AQA A-Level Physics): Revision Notes

12.3.4 Length contraction

The proper length $( l_0 )$ of an object is defined as its length when measured by an observer at rest relative to the object. When observed from a frame moving relative to the object, the length $( l )$ appears contracted, and can be calculated using the length contraction formula:

$$l = l_0 \sqrt{1 - \frac{v^2}{c^2}}$$

Where:

• $v$ is the relative speed of the moving observer with respect to the object,
• $c$ is the speed of light in a vacuum. This formula shows that as $v$ approaches $c$, the contracted length $l$ decreases, making the object appear shorter along the direction of its motion.

Important Note: Only the length in the direction of motion is affected by length contraction. The width and height of the object remain unchanged.

Experimental Data:

• Distance between detectors, $d$: 2 km
• Velocity of muons, $v: :highlight[0.996c]$
• Muon half-life (at rest): 1.5 μs
• Count rates:
  • Detector 1: 100 counts per second
  • Detector 2: 80 counts per second

Calculation of Expected Count Rate with Length Contraction

In this scenario:

• The proper length $( l_0 )$ is 2 km, as the detectors are stationary relative to each other.
• We calculate the contracted length $( l )$ as seen by the muons, which are moving relative to the detectors.

1. Calculate the Contracted Length, $( l )$: Using the formula:

$$l = l_0 \sqrt{1 - \frac{v^2}{c^2}}$$ $$l = 2000 \, \text{m} \times \sqrt{1 - \frac{(0.996c)^2}{c^2}} = :highlight[180 m]$$

This contraction means the distance the muons need to travel appears shorter from their own perspective.

2. Calculate the Time Taken to Travel the Contracted Distance: Now, we calculate the time it takes for muons to travel this contracted length:

$$t = \frac{l}{v} = \frac{180 \, \text{m}}{0.996 \times 3 \times 10^8 \, \text{m/s}} = :highlight[6.0 × 10^{-7} s]$$

3. Calculate the Expected Count Rate: Using the half-life formula for decay:

• First, calculate the number of half-lives ( $\frac{t}{\text{half-life}}$):

$$\frac{6.0 \times 10^{-7}}{1.5 \times 10^{-6}} = :highlight[0.4]$$

• Then, calculate the decay factor using:

$$\left(\frac{1}{2}\right)^{0.4} = :highlight[0.76]$$

• Finally, calculate the expected count rate at Detector 2:

$$100 \times 0.76 = :highlight[76 s^{-1}]$$

Since this expected count rate of 76 s^{-1} closely matches the observed count rate of 80 s^{-1}, this experiment provides strong evidence for length contraction.