Let's suppose we have a spaceship with the exact speed of light. If a traveller takes this spaceship to go to proxima centauri (approximately 4 years light away from Earth) and come back, we (as observers on Earth) will see the ship coming back after approximately 8 years.

But how much time would have passed for the traveller on the ship? How can be this calculated with a formula?

  • $\begingroup$ What do you know about relativity (special or general)? What have you tried to solve this yourself? Do you know that is is impossible that matter moves precisely at the speed of light? $\endgroup$
    – ACuriousMind
    Jun 28, 2014 at 22:52
  • $\begingroup$ General relativity to simplify. Yes I know so in this case let say with the same speed of light ( suppose in this case it will be 0 seconds..). I have tried this time ago but right now I need a refresh of the formulas. $\endgroup$
    – albanx
    Jun 28, 2014 at 22:54

1 Answer 1


The quantity that tells you what time an observer travelling along a path $\gamma : [t_0,t_1] \rightarrow \mathbb{R}^4$ experiences is the proper time

$$ \tau = \int_\gamma \sqrt{\mathrm{d}x_\mu\mathrm{d}x^\mu}$$

Assuming flat spacetime, i.e Minkowski metric/special relativity, this reduces to

$$ \tau = \int_\gamma\sqrt{\mathrm{d}t^2 - \frac{1}{c^2}\mathrm{d}x^i\mathrm{d}x_i} =\int_{t_0}^{t_1} \sqrt{1 - \frac{\vec{v}(t)^2}{c^2}}\mathrm{d}t $$

For $\vec{v} = c$, i.e an object travelling with the speed of light, this is $\int_{t_0}^{t_1} 0 \mathrm{d}t = 0$, so anyone hypothetically travelling with the speed of light will indeed not experience any time at all.

By plugging in other values for the travelling speed $\vec{v}$, you are able to calculate the experienced time for arbitary travellers.

  • $\begingroup$ thank you that was what I was supposing. So for the traverller is just an instance jump to the future and despite the distance he could reach any point of the universe at once from his point of view... $\endgroup$
    – albanx
    Jun 28, 2014 at 23:11
  • $\begingroup$ This isn't right. "Let's suppose we have a ship with speed of light" doesn't work. See physics.stackexchange.com/questions/110632/… $\endgroup$
    – mmesser314
    Jun 29, 2014 at 5:05
  • $\begingroup$ @mmesser314: Though you are right that nothing with mass can really travel at the speed of light (as I also said in the comments above), the limit $\mathrm{lim}_{\vec{v}\rightarrow c}$ can nevertheless be examined to simulate that hypothetical situation. Physically it just reflects what situation you get ever closer to if you just keep accelerating. $\endgroup$
    – ACuriousMind
    Jun 29, 2014 at 11:48
  • 1
    $\begingroup$ Again, I make no claims about any photons. I am only claiming that the limit $v \rightarrow c$ represents the situation you get closer to when accelerating. That is certainly correct, since the experienced proper time will shrink the faster you are. Of course, the limit case itself is unattainable, but the traveller will, if his speed is almost $c$, experience almost no time (compared to the frame in which is speed is measured to be almost $c$, of course). I fail to see where that would be faulty reasoning. $\endgroup$
    – ACuriousMind
    Jun 29, 2014 at 18:00
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    $\begingroup$ I have again another question about, this and this time let suppose that the ship travels at 99% of c. When the traveller arrives at proxima centauri (4 years light) should he see the start as it was after 4 years or after 8 years? $\endgroup$
    – albanx
    Jun 30, 2014 at 19:44

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