91影视

Anna is accelerating at $g$with respect to Bob who is stationary on earth. The time on Anna鈥檚 clock $dt\text{'}$as observed by Bob with respect to his clock $dt$will be given by $dt\text{'}=\sqrt{\raisebox{1ex}{$1-u^2$}\!\left/ \!\raisebox{-1ex}{$c^{2}dt.$}\right.}$

The relativistic velocity is given by the following.

$$\begin{gathered} u=\frac{1}{\sqrt{1+{\left({\displaystyle \frac{Ft}{mc}}\right)}^2}}\frac{Ft}{m} \\ =\frac{gt}{\sqrt{1+{\left({\displaystyle \frac{gt}{c}}\right)}^2}} \end{gathered}$$

Putting the expression for $u$in $dt\text{'}$.

$$\begin{gathered} dt\text{'}=\left[1-\frac{{\left(gt\right)}^2}{c^2\left(1+{\left({\displaystyle \raisebox{1ex}{$gt$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2\right)}\right]dt \\ =\left[1-\frac{{\left({\displaystyle \raisebox{1ex}{$gt$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2}{\left(1+{\left({\displaystyle \raisebox{1ex}{$gt$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2\right)}\right]dt \\ =\left[\frac{1}{\left(1+{\left({\displaystyle \raisebox{1ex}{$gt$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2\right)}\right]dt \end{gathered}$$

Integrating the above expression,

$$\begin{gathered} t\text{'}=鈭�\frac{dt}{\left(1+{\left({\displaystyle \raisebox{1ex}{$gt$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2\right)} \\ =\frac{arc\sin h\left({\displaystyle \frac{gt}{c}}\right)}{{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}}+c \\ =\frac{c}{g}arc\sin h\left(\frac{g}{c}t\right)+c \end{gathered}$$

As the clocks of Anna and Bob were synchronized at $t=0$, the integration constant $c=0$Therefore,

$$\begin{gathered} t\text{'}=\frac{c}{g}arc\sin h\left(\frac{gt}{c}\right) \\ or \\ t\text{'}=\frac{c}{g}\sin h\left(\frac{g}{c}t\text{'}\right) \end{gathered}$$

Suppose Anna was traveling at a constant velocity comparable to the speed of light, the time required for Anna to cover some distance with respect to Bob鈥檚 frame is as follows.

$dt\text{'}=\frac{dt}{y}$

In this situation, Anna is accelerating at under constant force with respect to stationary Bob. The expression for time passed on Earth鈥檚 in terms of time passed on Anna鈥檚 frame as derived in the previous part is as follows.

$$\begin{gathered} t=\frac{c}{g}\sin h\left(\frac{g}{c}t\text{'}\right) \\ =\frac{3脳{10}^8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{9.8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}}\sin h\left(\frac{9.8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2脳20脳3.154脳{10}^{7}s$}\right.}}{3脳{10}^8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}\right) \\ =0.306脳{10}^8\sin h\left(20.61\right) \\ =1.36脳{10}^{16}s \end{gathered}$$

Hence, the time in years would be $t=4.3脳{10}^8$yrs.

So, 430 million years would go by as observers on earth as they see Anna age 20 years. For Anna鈥檚 clock to run so slowly, the velocity must have been very close to the speed of light. As you might recall from Exercise 118, it takes 6.8 years with respect to an observer on Earth for Anna to reach the velocity of 0.99c from stationary.

In Exercise 119, we derived an expression for the relativistic position of an object under constant force accelerating at $g$is given below.

$x=\frac{m{c}^2}{F}\left(\sqrt{1+{\left(\frac{Ft}{mc}\right)}^2}-1\right)$

Replacing $\raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$with $g$and using the expression of $t$in terms of $t\text{'}$which we derived in the previous part, i.e., $t=\frac{c}{g}\sin h\left(\frac{gt\text{'}}{c}\right)$, we get,

$$\begin{gathered} x=\frac{c^2}{g}\left[\sqrt{1+{\left(\frac{g}{c}\left(\frac{c}{g}\sin h\left(\frac{gt\text{'}}{c}\right)\right)\right)}^2}-1\right] \\ x=\frac{c^2}{g}\left[\sqrt{1+\sin h^2\left(\frac{gt\text{'}}{c}\right)}-1\right] \end{gathered}$$

Using the identity of Hyperbolic functions, that is, $\cos h^{2}x-\sin h^{2}x=1$, yields as

$x=\frac{c^2}{g}\left[\cos h\left(\frac{gt\text{'}}{c}\right)-1\right]$

Let鈥檚 consider the first half of the question and determine the distance traveled by Anna as she ages 20 years as measured by Bob.

$$\begin{gathered} x=\frac{c^2}{g}\left(\cos h\frac{gt\text{'}}{c}-1\right) \\ =\frac{{\left(3脳{10}^8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}\right)}^2}{9.8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}}\left(\cos h\left(\frac{9.8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2脳20脳3.154脳{10}^{7}s$}\right.}}{3脳{10}^8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}\right)-1\right) \\ =0.918脳{10}^{16}\left(\cos h\left(20.61\right)-1\right) \\ =4.1脳{10}^{24}m \end{gathered}$$

Hence, the distance in light-years is $x=4.2脳{10}^8$ly.

This value is the same as the value in part(b) because Bob sees Anna traveling at almost the speed of light. From Anna鈥檚 frame, it takes Anna 20 years to stop. Bob will see Anna move the same distance as before. Therefore, Bob will see Anna travel a total distance of $8.4脳{10}^8$ly . As a result, we can say that during Anna鈥檚 journey the observers on Earth would have age$8.4脳{10}^8$yrs.