In the frame in which they are at rest, the number of muons at time t is given by

$\mathbf{N}\mathbf{=}{\mathbf{N}}_{\mathbf{0}}{\mathbf{e}}^{\mathbf{-}\mathbf{l}\mathbf{/}\mathbf{T}}$

Where ${\mathbf{N}}_{\mathbf{0}}$ is the number at t = 0 and T is the mean life-time 2.2 $\mathit{\mu }\mathit{s}$. (a) If muons are produced at a height of 4.0 km , heading toward the ground at 0.93 c, what fraction will survive to reach the ground? (b) What fraction would reach the ground if classical mechanics were valid?

Consider a lifetime of muons at rest is $2.2\mu s$.

Consider a distance at a height of 4000 m.

Consider a speed of muons is 0.93 c.

Write the formula of fraction of muons that would survive given relativistic effects.

${\mathbf{N}}_{\mathbf{a}}\mathbf{=}{\mathbf{N}}_{\mathbf{0}\mathbf{,}\mathbf{a}}{\mathbf{e}}^{\mathbf{-}\frac{{\mathbf{t}}_{\mathbf{a}}}{\mathbf{T}}}$ …… (1)

Here, ${\mathbf{e}}^{\mathbf{-}\frac{{\mathbf{t}}_{\mathbf{a}}}{\mathbf{T}}}$are exponentially decay muons.

Write the formula of fraction of muons that would survive disregarding relativistic effects.

${\mathbf{N}}_{\mathbf{a}}\mathbf{=}{\mathbf{N}}_{\mathbf{0}\mathbf{,}\mathbf{b}}{\mathbf{e}}^{\mathbf{-}\frac{{\mathbf{t}}_{\mathbf{a}}}{\mathbf{T}}}$ …… (2)

Here, ${\mathbf{e}}^{\mathbf{-}\frac{{\mathbf{t}}_{\mathbf{a}}}{\mathbf{T}}}$ are exponentially decay muons.

The following equation may be used to show how a relativistic effect causes an object's length to contract:

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

According to the following connection, the quantity of muons decreases exponentially with time:

$N=N_0{e}^{-\frac{t}{T}}$

Relativistic effects have an impact on the distance from a height to the ground because muons move very quickly. Consequently, the separation would decrease. Eq. (1) may be used to compute the distance:

$$\begin{gathered} l=l_0\sqrt{1-\frac{v_{ave}^2}{c^2}} \\ =\left(4000\right)\sqrt{1-\frac{{\left(0.93c\right)}^2}{c^2}} \\ =4000\sqrt{1-{\left(0.93\right)}^2} \\ =1470\mathrm{m} \end{gathered}$$

Hence, it would take a distance of 1470 m for the muons to travel the distance.

Next, since we are aware of the link between speed, time, and distance, we write $v=\frac{d}{t}$ . If we set d = t, we can calculate how long it takes the muons to traverse 1470 m by using the following formula:

$$\begin{gathered} v=\frac{d}{t_a} \\ {t}_a=\frac{d}{v} \\ =\frac{1470}{0.93c} \\ =\frac{1470}{0.93.3.{10}^8} \end{gathered}$$

Solve further as:

$t_a=5.27\mu s$

Using Eq (2), we can get the percentage of muons that would endure if the quantity of muons decayed exponentially with time since we were able to compute the duration for the muons to travel:

Determine themuons would survive upon reaching the ground at a fraction.

Substitute 5.27 for $t_a$ and 2.2 for T into equation (1).

$$\begin{gathered} \frac{N_a}{N_{0,a}}=e^{\frac{-5.27}{2.2}} \\ \approx 0.091 \end{gathered}$$

Therefore, the value of muons would survive upon reaching the ground at a fraction of 0.091 .

We set $d=l_0$when we simply take into account classical mechanics and ignore relativistic effects. Using the connection between speed, time, and distance, we can determine how long it takes the muons to travel a distance:

$$\begin{gathered} v=\frac{d}{t_b} \\ {t}_b=\frac{d}{v} \\ =\frac{4000}{0.93c} \\ =\frac{4000}{0.93.3.{10}^8} \end{gathered}$$

Solve further as:

$t_b=14.3\mu s$

We were able to determine the muons' journey duration; therefore we used Eq. to get the percentage of muons that would survive if the quantity of muons decayed exponentially with time (2):

Determine the muons would survive upon reaching the ground at a fraction.

Substitute 14.3 for $t_b$ and 2.2 for T into equation (2).

$$\begin{gathered} \frac{N_b}{N_{0,b}}=e^{-\frac{t_b}{2.2}} \\ =e^{\frac{-14.3}{2.2}} \\ \approx 0.0015 \end{gathered}$$

Therefore, the value of muons would survive upon reaching the ground at a fraction of 0.0015.