Let's say I have a rocket that can provide constant (proper) acceleration (let's say of 1 gee) for an infinte amount of time. Let's say I want to use this rocket as a means of "suspended animation," shooting off to relativistic speeds and returning back after a set number of centuries or millennia, which I will have experienced as a mere number of years or decades. Assuming my goal is to mimimize the proper time (time that elapsed from my reference frame), what is my ideal trajectory?

I'm hoping I can do better than: travel to a distant point, accelerating half the time, braking the other half, then reversing the journey.

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    $\begingroup$ Welcome to physics stackexchange! I think this question is a bad fit for the website at the moment, because it shows insufficient research effort. Take the equation at the end of joshphysics's post here physics.stackexchange.com/questions/75487/… and piece together three such paths, and then integrate to get the proper time, then use calculus to minimize/maximize whatever parameters you want. Start doing that, and once you get stuck at a conceptual difficulty, you can ask that question here. $\endgroup$ – user12029 Sep 1 '15 at 19:09
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    $\begingroup$ Indeed - there have been many questions on the twin paradox, and very few good ones. It seems to be something that people grab on to and think they have a new and novel solution (much like diffraction fringes). Sigh. You actually have to slog through the physics and math to get a real answer. $\endgroup$ – Jon Custer Sep 1 '15 at 19:20
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    $\begingroup$ Personally I think this a good question. OP isn't suggesting a new or novel solution in the slightest, but simply asking which trajectory will result in one having aged the least, for a given amount of time elapsed back on Earth. Should you just travel in a straight line, accelerating half the way there and decelerating the rest, or could you move in some kind of circle? It's interesting. $\endgroup$ – gj255 Sep 1 '15 at 19:52
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    $\begingroup$ Of some possible interest here: Elisha Huggins' article "Gravity, Time, and Lagrangians" in The Physics Teacher vol 48. $\endgroup$ – dmckee Sep 2 '15 at 3:32
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    $\begingroup$ Why is this a bad question? It seems an excellent question and quite a hard one for the non-relativity nerd. See my answer - it takes a lot of explaining. How many of the site members could have answered it? Not many would be my guess. $\endgroup$ – John Rennie Sep 2 '15 at 14:59

The time measured by the clock you carry with you on your journey is called the proper time, and the proper time is just the length of your world line (give or take a factor of $c$). So if we calculate the length of your world line as you accelerate away then back, that gives us your elapsed time. Better still, the proper time is an invariant i.e. all observers in all coordinate systems will calculate the same value for it, so we can use whatever coordinate system is most convenient to do the calculation.

I'm going to do the calculation in my inertial frame here on Earth. We use the usual spatial coordinates $x$, $y$ and $z$ and the time coordinate $t$. The geometry of our (flat) spacetime is described by the Minkowski metric:

$$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$

For convenience we'll assume that you're moving only along the $x$ axis so $dy = dz = 0$ and the metric simplifies to;

$$ c^2d\tau^2 = c^2dt^2 - dx^2 \tag{1} $$

What equation (1) is telling us is that if I observe you to move a distance $dx$ in a time $dt$ then the time change on your clock will be $d\tau$. To see this, suppose you're moving at a constant velocity $v$ that is:

$$ \frac{dx}{dt} = v $$


$$ dx = vdt $$

We can substitute this value for $dx$ in equation (1) and we get:

$$ c^2d\tau^2 = c^2dt^2 - v^2dt^2 $$

and with a minor bit of rearranging we get:

$$ d\tau = dt \sqrt{1 - \frac{v^2}{c^2}} = \frac{dt}{\gamma} \tag{2} $$

which you should immediately recognise as the usual expression for time dilation at constant velocity. So far so good.

The problem here is that you are accelerating so your velocity isn't constant. In that case we write the velocity as a function of time, $v(t)$, in equation (2), and we get the proper time $\tau$ by integrating:

$$ \tau = \int_0^T \sqrt{1 - \frac{v^2(t)}{c^2}} dt \tag{3} $$

It's worth clarifying exactly what equation (3) is telling us. If I observe you to depart Earth at time $t = 0$ and return at time $t = T$, and during that time I observe your velocity to be some function of time $v(t)$, then using equation (3) will calculate the time $\tau$ measured on the clock you are carrying. What you're asking is how to choose the function for the velocity $v(t)$ to minimise the elapsed time $\tau$ subject to the constraint that the maximum acceleration of your rocket is 1g (or whatever).

The rigorous way to do this is using the calculus of variations and vary the function $v(t)$ to find the stationary value for $\tau$. But this is a hard calculation, and in any case we don't need to go to all that effort because it's obvious that we minimise $\tau$ by making $v(t)$ as big as possible. In other words you have to accelerate as hard as possible.

So I'm afraid the answer is precisely the boring one that you were hoping to avoid. You minimise your proper time by accelerating continuously at the highest possible acceleration.

Some of the comments have mentioned joshphysics' answer to Derivation of hyperbolic motion in Special Relativity. While equation (3) looks at first glance an easy way to calculate the elapsed time $\tau$ the problem is that constant acceleration in your frame (constant proper acceleration) is not constant acceleration in my frame. Obviously not, since constant acceleration $a$ as observed in my frame would result in you moving faster than light after a finite time $t = c/a$, and this isn't possible. What Josh calculates is the equation for your motion as observed in my frame when you are using constant acceleration. The function, $v(t)$, for your motion at constant acceleration turns out to be:

$$ v(t) = \frac{at}{\sqrt{1 + \frac{a^2t^2}{c^2}}} $$

To use equation (3) you'd have to use this equation for the velocity and patch together the acceleration and deceleration phases of the journey.


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