# Time dilation to go a specific distance

1. Dec 17, 2013

### sparks97

Traveling on a spaceship to Alpha Centauri A, 4.37 light years away, at a velocity of 0.92 c,
if you just calculate the time by dividing the distance by the velocity:
$t = \frac{d}{v} = 1.496 x 10^8 sec = 4.74 years$

Is this the time from the Earth reference frame?
Then the time on the spaceship is:

$t' = t \sqrt{ 1 - \frac{v^2}{c^2}} = 1.86 years$
This is less than the time it takes light to reach Earth. Using length contraction on the spaceship, then determining the time to travel the shorter distance results in this same time.

OR, the time of 4.74 years is the time onboard the spaceship and the time on Earth is 12.09 years.

I'm confused!

Last edited: Dec 17, 2013
2. Dec 17, 2013

### Mentz114

Hi, change the backslash in the \tex to a forward slash and it will display properly.

3. Dec 17, 2013

### sparks97

Thanks! I figured that out.

4. Dec 17, 2013

### tiny-tim

welcome to pf!

hi sparks97! welcome to pf!
yes
the spaceship regards both alpha centauri and earth as moving, so the length is contracted

since the spaceship clock reading is also lower, by the same amount, that means the spaceship measures the speed of the earth as the same, v
or just type two #s before and after: $\sqrt{ 1 - \frac{v^2}{c^2}}$

5. Dec 17, 2013

### ghwellsjr

There's an ongoing thread with almost the same scenario. The astronaut is coming back from a star almost 4 light-years away, taking 2 years of his time to be here. But the principles are the same and might help you understand.

6. Dec 17, 2013

### sparks97

I posted a Thanks to Tiny Tim. I'm not sure where it went.

Thanks for posting this link to the other thread. Great forum here!

7. Dec 17, 2013

### tiny-tim

it goes into the recipient's "MY PF"