Chapter 25, Problem 77QAP

To determine

(a)

Does the muon reach Earth's surface within one half-life if Muons have a proper half-life of $1.56\times {10}^{-}{}^{6}s$, suppose a muon is formed at an altitude of 3000 m and travels at a speed of 0.950c straight toward Earth? Complete the problem from both perspectives: the muon's point of view and Earth's reference flame.

Answer to Problem 77QAP

The muon does not make it to Earth

Explanation of Solution

Calculation:

Muons have a proper lifetime of $\Delta t_{proper}=1.56\times {10}^{-}{}^{6}s$. A muon is formed a distance of 3000 m above the surface of Earth and travels straight downward with a speed V = 0.950c. To determine whether or not the muon reaches Earth's surface before it decays, we have to calculate how far the muon travels in $1.56\times {10}^{-}{}^{6}s$relative to both the reference frame of the muon and the reference frame of Earth. The distance the muon travels will be contracted in the muon's reference frame, but the observed time will be longer in Earth's reference frame. In either case, the distance the muon travels must be at least 3000 m if it is to reach Earth's surface before decaying. The minimum necessary speed for the muon to just reach Earth's surface before decaying can be calculated from $V=\frac{L_{rest}}{\Delta t}$ where $\Delta t=\frac{\Delta t_{proper}}{\sqrt{1-\frac{v^2}{c^2}}}$ and $L_{rest}=3000m.$

Length contraction in the muon's reference frame:

  $L=L_{rest}\sqrt{1-\frac{v^2}{c^2}}=3000\times \sqrt{1-\frac{{(0.95c)}^2}{c^2}}=937m$

Distance traveled by the muon in the muon's reference frame:

  $\begin{array}{l}V=\frac{L}{\Delta t_{proper}}\\ L=V\Delta t_{proper}=0.95\times 3\times {10}^8\times 1.56\times {10}^{-}{}^6=444.6m\end{array}$

Because 444.6 m <937 m, the muon does not make it to Earth.

Time dilation in Earth's reference frame:

  $\Delta t=\frac{\Delta t_{proper}}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1.56\times {10}^{-}{}^6}{\sqrt{1-\frac{{(0.95c)}^2}{c^2}}}=5\times {10}^{-}{}^{6}s$

Distance traveled in Earth's reference frame:

  $\begin{array}{l}V=\frac{L_{rest}}{\Delta t}\\ L_{rest}=V\Delta t=0.95\times 3\times {10}^8\times 5\times {10}^{-6}=1425m\end{array}$

Because 1425 m <3000 m, the muon does not make it to Earth.

Conclusion:

The muon does not make it to Earth

To determine

(b)

The minimum speed of the muon so that it just barely reaches Earth's surface after one half-life?

Answer to Problem 77QAP

The minimum speed = 0.988c

Explanation of Solution

Calculation:

  $V=\frac{L_{rest}}{\Delta t}=L_{rest}\frac{\sqrt{1-\frac{v^2}{c^2}}}{\Delta t_{proper}}$

  $\frac{V}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{L_{rest}}{\Delta t_{proper}}$

  ${\left(\frac{V}{c\sqrt{1-\frac{v^2}{c^2}}}\right)}^2={\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^2$

  $\frac{\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}={\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^2$

  $\frac{v^2}{c^2}={\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^{2}1-\frac{v^2}{c^2}$

  $\frac{v^2}{c^2}={\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^2-{\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^2\frac{v^2}{c^2}$

  $\frac{v^2}{c^2}+{\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^2\frac{v^2}{c^2}={\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^2$

  $\begin{array}{l}\frac{v^2}{c^2}\left(1+{\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^2\right)={\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^2\\ \frac{v^2}{c^2}=\frac{{\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^2}{\left(1+{\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^2\right)}\end{array}$

  $\frac{v}{c}=\sqrt{\frac{{\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^2}{\left(1+{\left(\frac{L_{rest}}{c\Delta t_{proper}}\right)}^2\right)}}=\sqrt{\frac{{\left(\frac{3000}{3\times {10}^8\times 1.56\times {10}^{-}{}^6}\right)}^2}{\left(1+{\left(\frac{3000}{3\times {10}^8\times 1.56\times {10}^{-}{}^6}\right)}^2\right)}}=0.988$

  $v=0.988c$

Conclusion:

The minimum speed = 0.988c