# Can time dilation affect time taken?

Take an example of a space ship moving at speed comparable with speed of light (like 60% of c) now if spaceship launched from earth to near star does time dilation affects the time taken by space ship to travel from earth to star? I mean does time taken is different for observer on earth and observer on spaceship or it's same?

• What I studied in my class was that by the theory of special relativity the passage of time observed by observer in spaceship is diffrent from the observer on earth so I thought then the time taken by ship should be different for diffrent observer and that confused me like ther is gap in time where observer is at star and for other observer it's going to star. I'm new to this field of study so please tell me where am I wrong.please Jun 15 '20 at 0:15
• Search for "Twin paradox" Jun 15 '20 at 1:23

Yes.

We have an intuition of a 3D isotropic world in which a universal parameter called "time" ticks away uniformly for all observers in all states of motion. Our intuition is wrong, and be glad: that world would have really messy physical laws.

Instead, relativity is correct: we live in a 4D spacetime, and how an observer slices that 4D spacetime into 3 dimensions of Euclidean space and one moment of time (call "now") depends on their motion (which can only be defined relative to other moving observers).

The key results of relativity regarding your question are (1): any observer sees a moving clock tick slower (that is: time ticks slower), and (2): it is not possible to define a unique time interval between distant events.

Putting numbers to it: if your space traveller goes 5 light years at $$v=\frac 3 5 c$$, then according to Earth, it takes him:

$$\Delta t = \frac{5\,{\rm ly}}{\frac 3 5\,{\rm ly/y}}=\,8\frac 1 3\,{\rm y}$$

to arrive. Earth sees the space clock tick off:

$$\Delta t' = \frac 1 {\gamma} \Delta t = \,4\frac 1 6\, {\rm y}$$

because the Lorentz factor $$\gamma=2$$.

In the space traveler's frame, the trip also takes $$\Delta t'$$, as it must. He watched the Earth/Star system zoom by him at $$v=\frac 3 5 c$$ for $$4\frac 1 6\, {\rm y}$$, but the distance between the Earth and Star is Lorentz contracted to 2.5 ly, so everything works out.

The confusing part is as follows: in the space frame, Earth was moving, so the Earth clock was running at half speed in the space frame. For the space traveler, only:

$$\Delta t'' = \frac 1 {\gamma^2} \Delta t = 2\frac 1 {12}\,{\rm y}$$

ticked by on Earth during the trip. Both are correct because they are measuring different time intervals. "Now" depends on motion, so that when the space traveler arrives at the star, "now" on Earth really is only 2.08 years after lift off.

If the spaceship stops, so he is in the Earth inertia frame again, "now" on Earth jumps forward to 8.3333 years after lift off.

If he decides to go home, the moment he accelerates to $$v = -\frac 3 5 c$$, "now" on Earth jumps forward again, and he can return home watching the Earth clock ticking more slowly than his (as it did on the outbound trip), yet when he arrives, the Earthling has aged 16 2/3 years, while the space traveler has aged only 8 1/3 years.

• Maybe you should re-think this statement: "If he decides to go home, the moment he accelerates to v=−(3/5) c, "now" on Earth jumps forward again". Jun 15 '20 at 15:24