5.3 Time dilation  (Page 4/8)

Calculating $\text{Δ}t$ For a relativistic event

Suppose a cosmic ray colliding with a nucleus in Earth’s upper atmosphere produces a muon that has a velocity $v=0.950c.$ The muon then travels at constant velocity and lives 2.20 μs as measured in the muon’s frame of reference. (You can imagine this as the muon’s internal clock.) How long does the muon live as measured by an earthbound observer ( [link] )?

As we will discuss later, in the muon’s reference frame, it travels a shorter distance than measured in Earth’s reference frame.

Strategy

A clock moving with the muon measures the proper time of its decay process, so the time we are given is $\text{Δ}\tau =2.20\mu \text{s}.$ The earthbound observer measures $\text{Δ}t$ as given by the equation $\text{Δ}t=\gamma \text{Δ}\tau .$ Because the velocity is given, we can calculate the time in Earth’s frame of reference.

Solution

1. Identify the knowns: $v=0.950c,\text{Δ}\tau =2.20\mu \text{s.}$
2. Identify the unknown: $\text{Δ}t.$
3. Express the answer as an equation. Use:
   
   $\text{Δ}t=\gamma \text{Δ}\tau$
   
   with
   
   $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.$
4. Do the calculation. Use the expression for $\gamma$ to determine $\text{Δ}t$ from $\text{Δ}\tau$ :
   
   $\begin{array}{cc}\hfill \text{Δ}t& =\gamma \text{Δ}\tau \hfill \\ & =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\text{Δ}\tau \hfill \\ & =\frac{2.20\mu \text{s}}{\sqrt{1-{(0.950)}^2}}\hfill \\ & =7.05\mu \text{s.}\hfill \end{array}$
   
   Remember to keep extra significant figures until the final answer.

Significance

One implication of this example is that because $\gamma =3.20$ at 95.0% of the speed of light $\left(v=0.950c\right),$ the relativistic effects are significant. The two time intervals differ by a factor of 3.20, when classically they would be the same. Something moving at 0.950 c is said to be highly relativistic.

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