91影视

When the two events occur at the same point in a particular frame of reference, the time interval between the two events in the frame is known as the proper time interval.

If the frame moves with constant velocity relative to second frame, the time interval between the events is observe in the second frame is longer than$鈭�t_0$ .

$$\begin{gathered} 鈭�t=\frac{鈭�t}{\sqrt{1-u^2/c^2}} \\ =纬鈭�t_0 \\ 纬=\frac{1}{\sqrt{1-u^2/c^2}} \end{gathered}$$

Given that,

$$\begin{gathered} h_0=55km \\ 鈭�t_0=2.20渭s \end{gathered}$$

The initial height in the muon鈥檚 rest frame is:

$$\begin{gathered} h=h_0\sqrt{1-\frac{u^2}{c^2}} \\ =\left(55\right)\sqrt{1-{\left(0.9860\right)}^2} \\ =9.17km \end{gathered}$$

Hence, in the muon鈥檚 frame, its initial height above the surface of the earth is 9.17 km .

The change in the muon鈥檚 initial height in S' is:

$$\begin{gathered} 鈭�h_s=u鈭�t_0 \\ =0.9860\left(3脳{10}^8\right)\left(2.20脳{10}^{-6}\right) \\ =0.651km \end{gathered}$$

The faction of the moun鈥檚 original height is:

$$\begin{gathered} \frac{鈭�h_s}{h}=\frac{0.651}{9.81} \\ =7.1\% \end{gathered}$$

Hence, the fraction of the muon鈥檚 original height, as measured in the muon鈥檚 frame is 7.1% .

The muon鈥檚 lifetime in the earth frame S is:

$$\begin{gathered} 鈭�t=\frac{鈭�t_0}{\sqrt{1-u^2/c^2}} \\ =\frac{2.20}{\sqrt{1-{\left(0.9860\right)}^2}} \\ =13.2渭s \end{gathered}$$

The change in the muon鈥檚 initial height in S is:

$$\begin{gathered} 鈭�h_s=u鈭�t \\ =0.9860\left(3脳{10}^8\right)\left(13.2脳{10}^{-6}\right) \\ =3.90km \end{gathered}$$

The faction of the moun鈥檚 original height h is:

$$\begin{gathered} \frac{鈭�h_s}{h_0}=\frac{3.90}{55} \\ =7.1\% \end{gathered}$$

Hence, the fraction of the muon鈥檚 original height in the earth鈥檚 frame is 7.1% .