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Suppose I have a space ship that can travel at $0.9c$, and I'm going to a star located at 20 light years or so from the Sun.

From a practical point of view, if I keep pointing the nose of my space ship towards the destination star, am I following the shortest distance path to that star?

The reasoning follows this vague intuition: light itself is telling us the shortest path through spacetime (is that true?), so instead of solving a geodesic based on the shape of the gravitational field (probably nasty because the stars in between move (but maybe not by much for only 20 years)), the path of the light ray is the solution you are looking for? - but there might be a hitch because $0.9c \neq c$ (i.e. all of a sudden, the best path for the light is no longer the best path for the spaceship)? note: this "vague intuition" is begging to be deconstructed and falsified.

[Update: I did some numerical simulations, and it looks like under some circumstances (I was varying the speed of the star, of the spaceship..., it is not possible to not even converge, i.e. I see the spaceship get into a neat trajectory that "follows" the star, without ever getting closer! - of course, it's possible there are bugs in my simulation, but it does raise the question of whether the method even "converges", i.e. reaches the star...]

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  • $\begingroup$ Relative velocities of stars, if I have my numbers right, are in the kilometers per second range (but you can check). That's a good five orders of magnitude slower than your ship. What numbers did you use? $\endgroup$
    – Emilio Pisanty
    Aug 6, 2015 at 2:17
  • $\begingroup$ In the same ballpark for reaching the star, and when I didn't reach it, it was because the star was way too fast - the simulation code was not wrong, but the parameters were definitely wrong. $\endgroup$
    – Frank
    Aug 6, 2015 at 2:25
  • $\begingroup$ No. As in: what velocities, in km/s, did you use for the star? If you put in anything much bigger than 10-100 km/s, your simulation will be unrealistic. If you put the star's velocity in the same ballpark as the ship's then you're wildly out of scope. $\endgroup$
    – Emilio Pisanty
    Aug 6, 2015 at 11:49
  • $\begingroup$ I meant "in the same ballpark as what you are saying", i.e. yes, star at less than 100 km/s. $\endgroup$
    – Frank
    Aug 6, 2015 at 13:37

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Even if your ship traveled at (almost) the speed of light, you would have to follow the trajectory of a ray of light traveling from your point of origin to the star; not from the star to your point of origin.

As described, you keep following the trajectories of many different rays of light (emitted at different times) backwards, putting yourself on a curved trajectory that a) will be longer, and b) will burn a lot of fuel.

In other words, you need to know where the star will be when you encounter it, not where the star was decades earlier when it emitted some light.

Ignoring general relativity and curved spacetime, your trajectory needs to be a straight line in space. Knowing the trajectory of the star, it is an easy exercise in geometry to find the straightline trajectory that happens to intersect the star's just when the star is actually there. If there are multiple solutions, you pick the one with the shortest distance. Relativity theory is not relevant here.

If general relativity cannot be ignored, then you need to know the geodesic structure of spacetime in the neighborhood and find the geodesic that intercepts the target star with the shortest spatial distance. The math may be a lot more complicated, but the principle is the same as in the flat spacetime case.

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  • $\begingroup$ From? Remember - you are the captain of the spaceship and you are expected to give a specific departure direction, at least, so that you can start the journey, right? Which one? $\endgroup$
    – Frank
    Aug 6, 2015 at 1:30
  • $\begingroup$ As I said, knowing the trajectory of the star, you estimate its future position at the time it is intercepted by your spaceship traveling at the given speed. That gives you the departure direction. It is not the direction from which the star's light appears to come. $\endgroup$
    – Viktor Toth
    Aug 6, 2015 at 1:32
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It's often not the quickest because you aren't taking into account how the star was moving when the light was sent.

It is similar to a temperature controller. You can adjust the temperature based on the current temperature (point), the integral of the temperature (integral) and how the temperature is changing (derivative) and using all three (a P.I.D. controller) can make it easier because you don't want to ignore how it is moving, but you also don't want a correction based on how it is moving to make you miss it.

And if missing is your goal and the star keeps getting visually bigger but always in the same visual position that is a very very bad sign.

If hitting it is a goal and you see it straight ahead of you then you still might never reach it. Why?

For instance what if that star was orbiting another (heavier) star and going in a circle around it at 0.9 c then if you are behind it you might chase it forever by running towards it, like a dog chasing his tail. But if you turned around and went the opposite way around the heavier star you might be able to start rushing towards it at faster than 0.9 c. So if you blindly chased it for years without getting closer you should try to find the pattern in the motions if any and exploit that if you can.

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Due to the gravitational pull from other planets and suns it may be the shortest in terms of distance, but not necessarily the quickest. It would really depend on the star you are travelling to.

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    $\begingroup$ Could you expand your second sentence more? $\endgroup$
    – Kyle Kanos
    Aug 6, 2015 at 1:03
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From numerical simulations, I would say that this is method (adjusting the nose to point to the star at all times):

  1. converges (you reach the star) if the star does not go too fast (the spaceship needs to go faster than the star on its orbit),
  2. does not look like it is the shortest path in general, because there is no anticipation of where the star will be, as Viktor pointed out - although, if the speed of the star is small compared to the ship, this method gets very close to the method of pointing the nose towards where the star will be.

In retrospect, following the light emitted several years ago by the destination star "obviously" seems like a bad idea. Much better to anticipate where the star will be, knowing its trajectory around the galaxy center - but again, in 20 years, the star might in fact not move "much".

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