It has already been asked here how fast a probe would have to travel to reach Alpha Centauri within 60 years. NASA has done some research into a probe that would take 100 years to make the trip. But I am interested in this question from a slightly different perspective. I would like to know how long the trip from our solar system to Proxima Centauri would take for both a ship and for an observer on Earth while accounting for the effects of general relativity on a spaceship under constant acceleration.
The following assumptions may be made:
- The ship starts and ends at 0 km/s relative to the Earth observer.
- The ship experiences a constant force of acceleration of 20 g until the half-way point and then a constant force of deceleration of 20 g until the destination.
- The path traveled is such that accleration is in only one spatial direction.
- The distance traveled is 4.25 light years.
Calculate (accounting for general relativistic effects):
- Time of trip for Earth-based observer.
- Time of trip for ship-based observer.
- Maximum speed.
You don't need to consider mass, or life-support systems, or method of propulsion; I am really only interested in the effects of relativity on the perceived time of the trip.
Best Answer
To perform such a calculation, we will use the flat Minkowski space-time of special relativity. Assuming that the traveller does not come sufficiently close to any massive body during the trip.
Now, in order to perform this calculation relativistically (assuming you want to include the effects of a changing Lorentz factor and associated acceleration) we must first obtain/derive an expression for how the Lorentz factor of a moving object transforms. This will allow us to derive the required relativistic expression for the objects proper-acceleration.
So, let us consider two inertial frames S (the 'stay-at-home observer) and S' (the traveller) in 'standard configuration' (that is, assuming that S' is moving in the positive x-direction with speed $v$). Let $\mathbf{u} = (u_{1}, u_{2}, u_{3})^{\mathsf{T}}$ be the instantaneous vector of velocity in S of the traveller. We now wish to find the velocity and $\mathbf{u}' = (u'_{1}, u'_{2}, u'_{3})^{\mathsf{T}}$ of the traveler in frame S'. We can define
$$\mathbf{u} = (\mathrm{d}x/\mathrm{d}t, \mathrm{d}y/\mathrm{d}t, \mathrm{d}z/\mathrm{d}t))^{\mathsf{T}},$$ $$\mathbf{u}' = (\mathrm{d}x'/\mathrm{d}t', \mathrm{d}y'/\mathrm{d}t', \mathrm{d}z'/\mathrm{d}t'))^{\mathsf{T}}.$$
From this definition and the fact that the two frames are in the 'standard configuration', we can immediately write the well know velocity transformation formulae (without derivation):
$$u'_{1} = \frac{u_{1} - v}{1 - u_{1}v/c^{2}}, \; u'_{2} = \frac{u_{2}}{\gamma(1 - u_{1}v/c^{2})}, \; u'_{3} = \frac{u_{3}}{\gamma(1 - u_{1}v/c^{2})}. $$
No assumptions about uniformity were made here and these formulae apply equally to the instantaneous velocity in a non-uniform motion.
Let us now write $u = (u_{1}^{2} + u_{2}^{2} + u_{3}^{2})^{\frac{1}{2}}$ and $u' = ({u'}_{1}^{2} + {u'}_{2}^{2} + {u'}_{3}^{2})^{\frac{1}{2}}$ for the magnitudes of the corresponding velocities in S and S'. Now let us choose the signature of our metric tensor $g_{\mu \nu}$ of our Minkowski space-time so that for our two inertial frames we can write
$$c^{2}\mathrm{d}t'^{2} - \mathrm{d}x'^{2} - \mathrm{d}y'^{2} - \mathrm{d}z'^{2} = c^{2}\mathrm{d}t^{2} - \mathrm{d}x^{2} - \mathrm{d}y^{2} - \mathrm{d}z^{2}.\;(\mathrm{A})$$
In our 'standard configuration' the Lorentz transformations for our coordinates are given by
$$\mathrm{d}x' = \gamma(\mathrm{d}x - v\mathrm{d}t), \; \mathrm{d}y' = \mathrm{d}y, \; \mathrm{d}z' = \mathrm{d}z, \; \mathrm{d}t' = \gamma(\mathrm{d}t - v\mathrm{d}x/c^{2}).\;(\mathrm{B})$$
Now, factoring out $\mathrm{d}t'^{2}$ and $\mathrm{d}t^{2}$ form the LHS and RHS of (A), respectivley and using (B) we can write
$$\mathrm{d}t^{2} (c^{2} - u^{2}) = \mathrm{d}t'^{2}(c^{2} - u'^{2}) = \mathrm{d}t^{2}\gamma^{2}(v)(1 - u_{1}v/c^{2})^{2}(c^{2} - u'^{2}).\;(\mathrm{C})$$
Now, cancelling $\mathrm{d}t^{2}$ from the above we can now obtain the following transformation for $u^{2}$, the squared magnitude of our traveller's velocity:
$$c^{2} - u'^{2} = \frac{c^{2}(c^{2} - u^{2})(c^{2} - v^{2})}{(c^{2} - u_{1}v)^{2}}.$$
Note here $u_{1}v = \mathbf{u}.\mathbf{v}$ so that the RHS is actually symmetric in $\mathbf{u}$ and $\mathbf{v}$ - meaning that this holds for any two subliminal 3-velocities. Now, rewriting the above interms of $\gamma(u)$ and $\gamma(u')$, with some work we get the following useful relations
$$\frac{\gamma(u')}{\gamma(u)} = \gamma(v)(1 - \frac{u_{1}v}{c^{2}})$$
This expression shows how the Lorentz factor of a moving object transform (for +ive $v$). We can now use this to get an expression for our proper-acceleration.
Now using the rapidity function we can simplify the following derivation (in a big way!). The rapidity function $\phi(u)$ can be written as
$$\phi(u) = \tanh^{-1} (\frac{u}{c}), $$
which allows the velocity addition formula to be rewritten in the remarkably simple form
$$\phi(u) = \phi(v) + \phi(u'),$$
now differentiating with respect to (WRT) $t$ yields
$$\frac{\mathrm{d}}{\mathrm{d}t}\phi(u) = \frac{\mathrm{d}}{\mathrm{d}t'}\phi(u')\frac{\mathrm{d}t'}{\mathrm{d}t}.\;(\mathrm{D})$$
Which can be written as
$$\frac{\mathrm{d}}{\mathrm{d}t}\phi(u) = \frac{1}{c}\gamma^{2}(u)\frac{\mathrm{d}u}{\mathrm{d}t},$$
and from (C) above we can write $\mathrm{d}t'/\mathrm{d}t = \gamma(u')/\gamma(u)$. Substituting this and the equation above (and its primed version) into the above expression for $\phi(u)$ we can write the desired acceleration transformation formula (hoping I have made no mistakes! :])
$$\gamma^{3}(u')\frac{\mathrm{d}u'}{\mathrm{d}t'} = \gamma^{3}(u)\frac{\mathrm{d}u}{\mathrm{d}t}.$$
Now if we define the proper acceleration $\alpha$ (say), as that which is measured in our travellers rest-frame S', we find on setting $u' = 0$ and $\mathrm{d}u'/\mathrm{d}t' = \alpha$, using our acceleration transformation equation we get
$$\alpha = \gamma^{3}(u)\frac{\mathrm{d}u}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}[\gamma(u)u].$$
This proper acceleration $\alpha$ is exactly the push we feel in an accelerating rocket. Now finally, in our case of interest, that of rectilinear motion with constant proper acceleration $\alpha$. We can integrate the above equation once, choosing $t = 0$ when $u = 0$
$$\alpha t = \gamma(u) u. \; (\mathrm{E})$$
Square this, solve for $u$ and integrate again with the same initial conditions give us the following equation of motion
$$x^{2} - c^{2}t^{2} = c^{4}/\alpha^{2}.$$
hence why motion with constant proper accelleration is called hyperbolic!
We can now solve your question. It is likely that at 20g for half the distance we will be well beyond the speed of light. Let us take the version of the derived relativistic equation of motion above for the frame S. Now, setting the distance as 2.125 light years with an acceleration of 20g, we can work out the time taken to reach the halfway point (using the relativistic equation of motion above), from the home observers reference frame which turns out to be 1531 days (or 4.19 years). This motion will be symmetric so the time taken for the entire trip in frame S (taking into account full relativistic motion!) will be 3062 days (or 8.39 years).
Now for the time taken as measured in frame S'... I will let you work that out! It is not as simple as using a Lorentz transform on the total time taken in this case; as we have seen that the Lorentz factor will change for and accelerating body.
As for the maximum speed I will also leave this as an exercise - I have purposely missed out the step where we derive equation for $u$. You can get $u$ from (E), and work out the max speed accordingly.
You will also notice that in the Newtonian calculation, the time taken to get to the half-way point is 166 days. This is because the speed of light is reached in 17.69 days at a distance of 212 light minutes; giving a speed at the half-way mark of a whopping 9.38c! The relativist calculation reflects the limit of c in the calculation.
I hope you enjoy reading this as much as I did going through it. You can tell I am off work!
All the best.