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Relativity and the twin paradox

[SOLVED] Relativity and the twin paradox

1. Homework Statement
A friend of yours who is the same age as you travels to the star Alph Centauri, which is 4 light years away, and returns immediately. He claims that the entire trip took just 5 years. How fast did he travel?


2. Homework Equations
Time dialation: [tex]\Delta[/tex]t = [tex]\gamma[/tex][tex]\Delta[/tex]t'
Length contraction: L = Lp/[tex]\gamma[/tex]
where Lp is the proper length and gamma is:
[tex]\gamma[/tex] = 1/[tex]\sqrt{1-(v^2/c^2)}[/tex]


3. The Attempt at a Solution
I know that the twin paradox utilizes both time dialation and length contraction, but I'm stuck as to how to account for both.

Thanks in advance! =)
 
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Answers and Replies

tiny-tim
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Welcome to PF!

1. Homework Statement
A friend of yours who is the same age as you travels to the star Alph Centauri, which is 4 light years away, and returns immediately. He claims that the entire trip took just 5 years. How fast did he travel?
Hi dakillerfishy! Welcome to PF! :smile:

(btw, if you type alt-j and alt-v, it prints ∆ and √ )

Let's rewrite the essential part of the question in letters that you're familiar with:

He goes 8 light-years (in your measurements), and take 5 years (in his measuements).

(It doesn't matter that he turned round - only the 8/5 ratio matters!)

So ∆x = 8. And ∆t´ = 5.

Does that help? :smile:
 
Hi tiny-tim! Thanks for the help.

So from your post, I gather that the proper length, Lp = 8 lightyears and [tex]\Delta[/tex]t' = 5.

I was thinking that since the Lorentz transformation ([tex]\gamma[/tex]) was the same in both cases, that I could use it to substitute into the time dialation it looks somthing like this:

[tex]\Delta[/tex]t = [tex]\frac{L}{Lp}[/tex] [tex]\Delta[/tex]t'

However, I'm not sure where to go from there.
 
mgb_phys
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Not quite, the question says he returns immediately! (do you get airmiles on a stargate?)
So he travels 4light years and took 5years to do it.

The length contraction is irrelevant - it doesn't mean a contraction of the distance - it means that he (and his ship) were shorter when he was traveling.
 
Aha that makes much more sense. :smile:

So, in that case only the time dialation equation is relevant. So, am I correct in assuming that the proper time, [tex]\Delta[/tex]t, will be 8?
 
mgb_phys
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Proper time for the traveller was 5years, you don't know proper time for the person at home
 
I'm sorry, I think I mis-phrased that. I meant the [tex]\Delta[/tex]t which is the time it would've taken without time dialation.
 
Well, I've been playing with numbers and I got this...

[tex]\Delta[/tex]t = [tex]\gamma[/tex][tex]\Delta[/tex]t'
[tex]\Delta[/tex]t = [tex]\frac{1}{\sqrt{1-v^{2}/c^{2}}}[/tex][tex]\Delta[/tex]t'

Then plugging in numbers...
5/8 = [tex]\sqrt{1-v^{2}/c^{2}}[/tex]
v = .6124 c

Which is apparently not right... what am I missing?
 
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mgb_phys
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He travelled 4 light years and it took 5 years from his point of view.

It's always worth having an idea of the answer before doing the maths - so just try a few numbers.
If he went 0.6c then it would have taken 5/0.6=6.7years (for a home observer), but from his pov it would take 6.7 * sqrt(1-0.36) = 5.3y, so he must have gone faster than 0.6c
At 0.65c it wowuld have taken 6.5years, but to him only 6.5*sqrt(1-0.42) = 4.9years

So the answer is slightly less than 0.65c
 
Dick
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.6124=sqrt(1-5/8). You want sqrt(1-(5/8)^2). Right? Do you see why?
 
mgb_phys
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It depends what returns 'immediately' means in the question.
If it means that he spend no time at the Alpha cen then dick is correct and it's 8light years distance for 5years (traveller pov) time.

If it means he returned taking no time ( presumably star gates are available from the same suppliers of frictionless planes and massless springs ) then it's 4light years distance and 5 years travel time.
 
Dick
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He travelled 4 light years and it took 5 years from his point of view.

It's always worth having an idea of the answer before doing the maths - so just try a few numbers.
If he went 0.6c then it would have taken 5/0.6=6.7years (for a home observer), but from his pov it would take 6.7 * sqrt(1-0.36) = 5.3y, so he must have gone faster than 0.6c
At 0.65c it wowuld have taken 6.5years, but to him only 6.5*sqrt(1-0.42) = 4.9years

So the answer is slightly less than 0.65c
I don't get that at all. Be careful. This isn't called the twin "paradox" for nothing. Just analyze everything from the viewpoint of the home observer. He stays in an inertial frame and you just have to worry about the time dilation factor. If you try to look at things from the viewpoint of the travelling twin, it's really easy to get confused.
 
mgb_phys
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I was - but I misunderstood ( or differently interpreted) the question, I took 'immediately' to mean that he took no time to return home.
If you take 'immediately' in the question to simply mean that he took no time to turn around the question is differnet and 5/8 is the answer.
 
Dick
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.6124=sqrt(1-5/8). You want sqrt(1-(5/8)^2). Right? Do you see why?
Sorry. I'm not correct either. I just jumped in and corrected the math without thinking the problem through. The round trip travel time from the home twins perspective isn't 8yrs. It's (8yrs)/(v/c). Can you correct your work to allow for that? Forget about wormholes etc. He can't return 'immediately' there's no such concept as absolute simultaneity in SR to allow you to define 'immediately'.
 
Oh I understand now. So you need to take the spacetime interval and rearrange it, so it reads:

([tex]\Delta[/tex]t)[tex]^{2}[/tex] = ([tex]\Delta[/tex]t')[tex]^{2}[/tex] - ([tex]\Delta[/tex]x/c)[tex]^{2}[/tex]

Except you need the time dialation equation to solve for [tex]\Delta[/tex]t, so you plug that into the above equation, right?
 
mgb_phys
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He can't return 'immediately' there's no such concept as absolute simultaneity in SR to allow you to define 'immediately'.
It's a hypothetical exam question - there is no such thing in relativity as massless springs or dimensionless points but you can't refuse to answer any question involving them!
 
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The problem is that the trip doesn't take 8 years for the observer on earth. Because the speed of the traveler is less than c it takes longer than 8 years.

I think the problem is easier done with lenght contraction after all.

I'll express the speed in lightyears/year since all distances and times are given in
lightyears and years, so [tex]c = 1[/tex]
For the traveler, the distance to Alpha centauri is shortened by [tex] \sqrt{1-v^{2}} [/tex].
Now you can just use distance = speed * time for the traveler

[tex] 8 \sqrt{1-v^{2}} = 5 v[/tex]

and solve for v.
 
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I think in this example, we're supposed to assume that he does return immediately, just for simplicity's sake. We also ignore the fact that he has to accelerate/decellerate, which will ultimately change the time dialation.
 
tiny-tim
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… use the full Lorentz equations! …

Well, I've been playing with numbers and I got this...

[tex]\Delta[/tex]t = [tex]\gamma[/tex][tex]\Delta[/tex]t'
[tex]\Delta[/tex]t = [tex]\frac{1}{\sqrt{1-v^{2}/c^{2}}}[/tex][tex]\Delta[/tex]t'

Then plugging in numbers...
v = .6124 c

Which is apparently not right... what am I missing?
Hey dakillerfishy - stop talking about time dilation! :frown:

"Time dilation" is for explaining it to non-physicists.

It's a short-cut, and it usually doesn't work.

You must use the Lorentz equations! :smile:

t = 8/v, x = 8, t´ = 5.

So t´ = (t - vx)/√(1 - v^2),
or t´^2(1 - v^2) = (t - vx)^2.

Putting in the numbers: 5 = 8(1/v - v)/√(1 - v^2)
= 8√(1 - v^2)/v.

So 25v^2 = 64(1 - v^2), or v = 8/√89 = 0.848

Doesn't look pretty, does it? Where have I gone wrong? :cry:
 
Dick
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It's a hypothetical exam question - there is no such thing in relativity as massless springs or dimensionless points but you can't refuse to answer any question involving them!
You have to refuse to answer them in relativity questions. Or you get nonsense.
 
Got it. :smile:

Thanks everyone! You've all been really helpful. :smile:
 
Dick
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Oh I understand now. So you need to take the spacetime interval and rearrange it, so it reads:

([tex]\Delta[/tex]t)[tex]^{2}[/tex] = ([tex]\Delta[/tex]t')[tex]^{2}[/tex] - ([tex]\Delta[/tex]x/c)[tex]^{2}[/tex]

Except you need the time dialation equation to solve for [tex]\Delta[/tex]t, so you plug that into the above equation, right?
Nah. It's not that hard. Just go back to post 8 with 5/8=sqrt(1-v^2/c^2) and replace '8' with the correct value of 8/(v/c). Now solve for v.
 
Ohhhh... I understand now. It goes back to what tiny-tim was saying about Lorentz transformations. :smile:
 

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