## modphysnoob 2 years ago Frank and Mary are twins. Mary jumps on a spaceship and goes to Alpha-Centauri (4 lightyears away) and returns. She travels at a speed of 0.8c with respect to the Earth and emits a radio signal every week. a) How much time does the trip take according to Frank?

1. modphysnoob

so do we solve it following way 1) distance/speed = time : 1 lightyear = 9.44x1015m (9.44*10^15)*4 / .8 so from this point, do we use time dilation ?

2. Jemurray3

Assuming thats .8 c, then sure.

3. modphysnoob

lol ,"sure"

4. Jemurray3

and take the return trip into account, too.

5. modphysnoob

How much time does the trip take according to Mary? this is one without time dilation

6. Jemurray3

No time dilation, but consider length contraction.

7. modphysnoob

so one frame has lenght contraction while other have time dilation?

8. Jemurray3

yep, but they don't cancel each other -- they make the disagreement even worse, right?

9. modphysnoob

I see, this is quite involved

10. Jemurray3

Hey I was re-browsing through this and I realized I read too quickly and misunderstood what was being asked -- The first thing you did was calculate how long the trip would take for the traveling twin from the perspective of somebody on earth. If you want how much time passes for the twin on earth, then you shouldn't do the time dilation. I thought you were trying to show that both the traveling twin and the twin on earth would agree on the discrepancy, which they would (one using time dilation and the other using length contraction).

11. Jemurray3

12. modphysnoob

a) How much time does the trip take according to Frank? no time dilation or lenght contraction b) How much time does the trip take according to Mary? there would be time dilation

13. Jemurray3

Right. The issue is you could do the calculation from either reference frame and you'd get the same answer. For example: Frank's calculation: "this is the time that would pass on my clock" t = 2d / v "but mary would see a contracted length, so she'd measure" t = 2d/gamma*v on the other hand, we could calculate it from mary's frame: "I would see a contracted length, so my time would be" t = 2d/gamma*v "but frank's clock would be time dilated, so he'd measure" t' = gamma*t = 2d/v See what I mean?

14. modphysnoob

I see, you are measring time in Frank's frame but measuring lenght in mary's fram @Jemurray3

15. Jemurray3

Alternatively you could say that mary will measure the proper time. No matter which way you look at it, it will come out right.

16. modphysnoob

dumb question: why lenght contraction? I know it has something to do with light reaching at different time but I just can't visualize it

17. Jemurray3

When you look at something, what you're really seeing is the light that is coming from that object. Look around whatever room your in. You brain and experience tells you that what you're seeing is the room as it is right this instant, but that's not really true because the light from the nearer parts bounced off the object more recently than the light from the farther parts, so if a desk is closer to you than a chair then you're seeing the desk as it was a very short time ago, and the chair as it was a little bit before that. This time difference is irrelevant unless you're moving extremely close to the speed of light. Imagine you were sprinting towards a barn near the speed of light.... in the finite amount of time it takes the light to get from the back of the barn to the front of the barn, you move an appreciable amount forward. That makes the back of the barn seem closer to the front of the barn.