# How much time passed for the passenger traveling with at speed-of-light spaceship? [duplicate]

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?

## marked as duplicate by Danu, Brandon Enright, John Rennie, Qmechanic♦Jun 29 '14 at 6:13

• 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? – ACuriousMind Jun 28 '14 at 22:52
• 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. – albanx Jun 28 '14 at 22:54

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.

• 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... – albanx Jun 28 '14 at 23:11
• This isn't right. "Let's suppose we have a ship with speed of light" doesn't work. See physics.stackexchange.com/questions/110632/… – mmesser314 Jun 29 '14 at 5:05
• @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. – ACuriousMind Jun 29 '14 at 11:48
• That is a misconception. Using the limit creates the impression that you can understand a photon by considering an inertial frame of reference, by thinking that a photon views the world as Lorentz contracted into a flat pancake where no time passes. Keep in mind that a photon has a wavelength and a period. The time and distance a photon travels makes a difference in the state of the photon at arrival. If you try to look at the world from the photon's point of view, you do not see this. The limit does not approach an inertial frame of reference, nor a correct view of a photon. – mmesser314 Jun 29 '14 at 17:36
• @mmesser314: I did not claim that this is what a photon "sees". I claimed that "Physically it just reflects what situation you get ever closer to if you just keep accelerating." You cannot deny that my formula is valid for $v = c - \epsilon, \epsilon > 0$. Taking $\epsilon \rightarrow 0$ and saying that, for $v$ getting closer and closer to $c$ as you like, the situation will become more and more like the limit is certainly not wrong, is it? (I concede that I should have phrased this more carefully in my original answer.) – ACuriousMind Jun 29 '14 at 17:45