Time dilation to go a specific distance

[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!

 

[Mentz114]

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

 

[sparks97]

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

Thanks! I figured that out.

 

[tiny-tim]

welcome to pf!

hi sparks97! welcome to pf!

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?

yes

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.

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

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

or just type two #s before and after: $\sqrt{ 1 - \frac{v^2}{c^2}}$

 

[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.

 

[sparks97]

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.

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!

 

[tiny-tim]

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

it goes into the recipient's "MY PF"