Question

In a laboratory experiment, a muon is observed to travel before disintegrating. A graduate student looks up the lifetime of a muon $\left(2\times {10}^{-6}s\right)$and concludes that its speed was

$\mathbf{v}\mathbf{=}\frac{\mathbf{800}\mathbf{m}}{\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{m}\mathbf{/}\mathbf{s}$ .

Faster than light! Identify the student’s error, and find the actual speed of this muon.

Short answer

The student has not considered the time dilation of the muon’s “internal clock,” and the actual speed of the muon will be$2.4\times {10}^8\mathrm{m}/\mathrm{s}$ .

Step-by-step solution

Given Information:

Given data:

The distance traveled by muon observed in a laboratory is d = 800 m .

The actual lifetime of the muon is $t=2\times {10}^{‐6}s$.

Expression for the actual speed of the muon:

Let v be the actual speed of the muon and $t^{\text{'}}$be the lifetime observed in a laboratory. Hence,

$t^{\text{'}}=\gamma t$

${\mathit{t}}^{\mathbf{\text{'}}}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}}$ …… (1)

Here,$t^{\text{'}}=\frac{d}{v}$.

Hence, re-write the equation (1),

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

Squaring on both sides,

$$\begin{gathered} {\left(\frac{d}{v}\right)}^2=\frac{t^2}{1-{\displaystyle \frac{v^2}{c^2}}} \\ \left(\frac{t^2}{d^2}\right)v^2=1-\frac{v^2}{c^2} \\ {v}^2\left[{\left(\frac{t}{d}\right)}^2+\frac{1}{c^2}\right]=1 \\ \frac{v^2}{c^2}=\frac{1}{1+{\left({\displaystyle \frac{tc}{d}}\right)}^2} \end{gathered}$$ …… (2)

Determine the student’s error and the actual speed of the muon:

Solve for the value of$\frac{tc}{d}$.

$$\begin{gathered} \frac{tc}{d}=\frac{\left(2\times {10}^{-6}s\right)\left(3\times {10}^{8}m/s\right)}{800m} \\ \frac{tc}{d}=\frac{6\times {10}^2}{800} \\ \frac{tc}{d}=\frac{6}{8} \\ \frac{tc}{d}=\frac{3}{4} \\ \mathrm{Substitute}\mathrm{the}\mathrm{value}\mathrm{of}\frac{\mathrm{tc}}{\mathrm{d}}\mathrm{in}\mathrm{equation}\left(2\right). \\ \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}=\frac{1}{1+{\left({\displaystyle \frac{3}{4}}\right)}^2} \\ \frac{\mathrm{v}}{\mathrm{c}}=\frac{1}{1+{\displaystyle \frac{9}{16}}} \\ \frac{\mathrm{v}}{\mathrm{c}}=\frac{16}{25} \\ \mathrm{v}=\frac{4}{5}\mathrm{c} \\ \mathrm{On}\mathrm{further}\mathrm{solving}, \\ \mathrm{v}=\frac{4}{5}\times \left(3\times {10}^8\mathrm{m}/\mathrm{s}\right) \\ \mathrm{v}=2.4\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the student has not taken into account the time dilation of the muon’s “internal clock.”

Therefore, the student has not considered the time dilation of the muon’s “internal clock,” and the actual speed of the muon will be $2.4\times {10}^8\mathrm{m}/\mathrm{s}$.