In Example 2.5, we noted that Anna could go wherever she wished in as little time as desired by going fast enough to length-contract the distance to an arbitrarily small value. This overlooks a physiological limitation. Accelerations greater than about $\mathbf{30}\mathit{g}$are fatal, and there are serious concerns about the effects of prolonged accelerations greater than $\mathbf{1}\mathit{g}\mathbf{.}$ Here we see how far a person could go under a constant acceleration of 1g, producing a comfortable artificial gravity.

(a) Though traveller Anna accelerates, Bob, being on near-inertial Earth, is a reliable observer and will see less time go by on Anna's clock $\mathbf{(}\mathit{d}\mathit{t}\mathbf{\text{'}}\mathbf{)}\mathbf{}$than on his own $\mathbf{\left(}\mathit{d}\mathit{t}\mathbf{\right)}\mathbf{.}$Thus, $\mathit{d}{\mathit{t}}^{\mathbf{\text{'}}}\mathbf{\hspace{0.33em}}\mathbf{=}\mathbf{\hspace{0.33em}}\mathbf{(}\mathbf{1}\mathbf{/}\mathit{\gamma }\mathbf{)}\mathit{d}\mathit{t}$, where u is Anna's instantaneous speed relative to Bob. Using the result of Exercise $\mathbf{117}\mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{,}$with $\mathit{g}$ replacing F/m, substitute for $\mathit{u},$then integrate to show that

$\mathit{t}\mathbf{=}\frac{\mathbf{c}}{\mathbf{g}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\frac{\mathbf{g}{\mathbf{t}}^{\mathbf{\text{'}}}}{\mathbf{c}}$

(b) How much time goes by for observers on Earth as they “see” Anna age 20 years?

(c) Using the result of Exercise 119, show that when Anna has aged a time t’, she is a distance from Earth (according to Earth observers) of

$\mathit{x}\mathbf{=}\frac{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{g}}\left(cosh\frac{g{t}^{\text{'}}}{c}-1\right)$

(d) If Anna accelerates away from Earth while aging 20 years and then slows to a stop while aging another 20. How far away from Earth will she end up and how much time will have passed on Earth?

Anna is accelerating at with respect to Bob who is stationary on earth. The time on Anna’s clock$d{t}^{\text{'}}$as observed by Bob with respect to his clock will be given by

$d{t}^{\text{'}}=\sqrt{1-u^2/c^2}dt$

The relativistic velocity is given by the following.

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

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

$$\begin{gathered} d{t}^{\text{'}}=\left[1-\frac{(gt{)}^2}{c^2\left(1+(gt/c{)}^2\right)}\right]dt \\ =\left[1-\frac{(gt/{)}^2}{\left(1+(gt/{)}^2\right)}\right]dt \\ =\left[\frac{1}{\left(1+(gt/{)}^2\right)}\right]dt \end{gathered}$$

Integrating the above expression,

$$\begin{gathered} t^{\text{'}}=\int \frac{dt}{\left(1+(gt/c{)}^2\right)} \\ =\frac{arcsinh\left(\frac{gt}{c}\right)}{g/c}+\mathrm{C} \\ =\frac{\mathrm{c}}{\mathrm{g}}arcsinh\left(\frac{\mathrm{g}}{\mathrm{c}}\mathrm{t}\right)+\mathrm{C} \end{gathered}$$

As the clocks of Anna and Bob were synchronized at $t\hspace{0.33em}=\hspace{0.33em}0$, the integration constant $C=\hspace{0.33em}0$. Therefore,

$$\begin{gathered} t^{\text{'}}=\frac{c}{g}arcsinh\left(\frac{gt}{c}\right) \\ or \\ t=\frac{c}{g}\sinh \left(\frac{g}{c}{t}^{\text{'}}\right) \end{gathered}$$

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’s frame isas follows.

$d{t}^{\text{'}}\hspace{0.33em}=\frac{dt}{\gamma }$

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

$$\begin{gathered} t=\frac{c}{g}\sinh \left(\frac{g}{c}{t}^{\text{'}}\right) \\ =\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{9.8\mathrm{m}/{\mathrm{s}}^2}\sinh \left(\frac{9.8\mathrm{m}/{\mathrm{s}}^2\times 20\times 3.154\times {10}^7\mathrm{s}}{3\times {10}^8\mathrm{m}/\mathrm{s}}\right) \\ =0.306\times {10}^8\sinh (20.61) \\ =1.36\times {10}^{16}s \end{gathered}$$

Hence,the time in years would be

$t\hspace{0.33em}=\hspace{0.33em}4.3\hspace{0.33em}\times \hspace{0.33em}{10}^8\hspace{0.33em}yrs$

So, 430 million years would go by as observers on earth as they see Anna age 20 years. For Anna’s 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 $F/m$with and using the expression of in $t$terms of $t\text{'}$which we derived in the previous part, i.e., $t=\frac{c}{g}\sinh \left(\frac{g{t}^{\text{'}}}{c}\right)$,we get,

$$\begin{gathered} x=\frac{c^2}{g}\left[\sqrt{1+{\left(\frac{g}{c}\left(\frac{c}{g}\sinh \left(\frac{g{t}^{\text{'}}}{c}\right)\right)\right)}^2}-1\right] \\ x=\frac{c^2}{g}\left[\sqrt{1+{\sinh }^2\left(\frac{g{t}^2}{c}\right)}-1\right] \end{gathered}$$

Using the identity of Hyperbolic functions, that is, $cos{h}^{2}x\hspace{0.33em}-sin{h}^{2}x\hspace{0.33em}=\hspace{0.33em}1$, yields as

Let’s consider the first half of the question and determine the distance traveled by Anna as she ages 20 yearsasmeasured by Bob.

$$\begin{gathered} x=\frac{c^2}{g}\left(\cosh \frac{g{t}^{\text{'}}}{c}-1\right) \\ =\frac{{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}{9.8\mathrm{m}/{\mathrm{s}}^2}\left(\cosh \left(\frac{9.8\mathrm{m}/{\mathrm{s}}^2\times 20\times 3.154\times {10}^7\mathrm{s}}{3\times {10}^8\mathrm{m}/\mathrm{s}}\right)-1\right) \\ =0.918\times {10}^{16}\left(\cosh \right(20.61)-1) \\ =4.1\times {10}^{24}m \end{gathered}$$

Hence, the distance in light-years is

$x=4.1\times {10}^{8}ly$

This value is the same as the value in part(b) because Bob sees Anna traveling at almost the speed of light. From Anna’s 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 $8.4\times {10}^{8}ly$. As a result, we can say that during Anna’s journey the observers on Earth would have aged$8.4\times {10}^{8}yrs.$