A spacecraft starts from being at rest at the origin and accelerates at a constant rate \(g\), as seen from Earth, taken to be an inertial frame, until it reaches a speed of \(c / 2\). (a) Show that the increment of proper time is related to the elapsed time in Earth's frame by: \(d \tau=\sqrt{1-v^{2} / c^{2}} d t\) (b) Find an expression for the elapsed time to reach speed C/2 as seen in Earth's frame. (c) Use the relationship in (a) to obtain a similar expression for the elapsed proper time to reach \(c / 2\) as seen in the spacecraft, and determine the ratio of the time seen from Earth with that on the spacecraft to reach the final speed.

To derive the increment of proper time relation, we'll start with the definition of proper time, which is the time measured by an observer who is at rest relative to the accelerated object (the spacecraft). Proper time \(\tau\) is related to the time measured in Earth's frame, 't', and the spacecraft's velocity (relative to Earth) 'v', by the time dilation formula in special relativity as follows: \[d \tau=\sqrt{1-\frac{v^{2}}{c^{2}}} \, dt\] Here, \(c\) is the speed of light. This relation explains part (a) of the problem.

Now let's find an expression for the elapsed time to reach a speed of c/2. Since the spacecraft accelerates at a constant rate 'g', the velocity 'v' as a function of time can be expressed as: \[v=gt\] Among the given conditions, the final speed of the spacecraft is \(c/2\), which can be written as: \[\frac{c}{2} = g t_f\] Solving for the elapsed time \(t_f\), we get: \[t_f = \frac{c}{2g}\] This expression represents part (b) of the problem.

To find the elapsed proper time (\(\tau_f\)) to reach a speed of c/2 as seen in the spacecraft, we will integrate the time dilation relation derived in Step 1, with respect to 't' (elapsed time on Earth), from 0 to \(t_f\): \[\tau_f = \int_0^{t_f} \sqrt{1 - \frac{v^2}{c^2}} \, dt\] The velocity 'v' can be replaced using the relation \(v = gt\), and then we can integrate: \[\tau_f = \int_0^{\frac{c}{2g}} \sqrt{1-\frac{(gt)^2}{c^2}} dt \] To simplify the integral, we can make a substitution for the variable 't': \[u = \frac{gt}{c}\] \[du = \frac{g}{c} dt\] The integral becomes: \[\tau_f = \frac{c}{g} \int_0^{\frac{1}{2}} \sqrt{1-u^2} du\] To find the numerical value for the proper time, you can integrate this expression. Next, we find the ratio of elapsed times \(R\), as seen from Earth and onboard the spacecraft. The expression derived for Earth's frame time is \(t_f = \frac{c}{2g}\): \[R = \frac{t_f}{\tau_f} = \frac{\frac{c}{2g}}{\frac{c}{g} \int_0^{\frac{1}{2}} \sqrt{1-u^2} du} = \frac{1}{2 \int_0^{\frac{1}{2}} \sqrt{1-u^2} du}\] This ratio demonstrates the difference between the perception of time on the spacecraft and the time perceived on Earth, as required for part (c) of the problem.