Special Relativity: Twin Paradox Question

In summary, Joe's trip to the distant planet takes less than 7 years in his reference frame because space is contracted.
  • #1
leonmate
84
1
I've already completed most of the question, it's an add on at the end that has stumped me.

I've calculated using time dilation the difference in ages between the two twins, Joe traveled at v = 24/25 (c=1 units) to a planet for 7 years in HIS reference frame and returned at v = 12/25

The question at the end is this:

In Ed's frame (Ed stayed at home, whilst his twin Joe ventured through the galaxy) the distance between Earth and the distant planet is greater than 7 light years. So how does Joe explain that the outbound trip took less than seven years in his frame?

I guessed this is something to do with length contraction, in Ed's frame the ship would contract. In Joe's frame he's standing still and space is wooshing past him, so would space contract? Doesn't seem to make sense to me, if we use the length contraction equation:

L = L' / γ

Where γ = (1 - v^2)^-1/2

Applied to Joe's frame, I get 1.96 lyr (using L' = 7 lyr)
But then traveling at v = 24/25 he would get the in 2-3 years, not 7?
 
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  • #2
You are correct that the distance between the Earth and the distant planet is contracted in Joe's frame. But you made a mistake in assuming L' is 7 Lyr.
 
  • #3
Ok, if that's the case then can I assume that L is greater than 7 so:

7.1 = L'/γ
7.1 * γ = L'
L' = 25.36 lyr

Surely that's not right either... is it because I'm using c = 1 units that I'm not getting the answer I expected?

When i stick in γ = (1-v^2/c^2)^-0.5 I get out 7.1 again :S
 
  • #4
How did you get L = 7.1 for Ed?

You should be able to deduce L' by using the information given for Joe. You can imagine a large stick stretching from the Earth to the distant planet. From Joe's perspective that stick is of length L' and the stick moves past him in a time of 7 years. Knowing how fast the stick is moving past him, Joe can deduce L'
 
  • #5
Ok,

Joe travels at v = 24/25 for 7 years so,
L = 6.72 lyrs

From Joe's reference frame the 'stick' travels at v = 24/25

So Joe sees the stick at:

L = L'γ
L' = 6.72/(1-(24/25)^2)^-1/2
L' = 1.88 lyrs

Too small? Or is that right?
Also if he has a speed of v = 24/25 then how does it take 7 years to travel that distance? Brain hurts.
 
  • #6
leonmate said:
Ok,Joe travels at v = 24/25 for 7 years so,
L = 6.72 lyrs

Which length are you trying to get here? The 7 years is the time as measured by Joe. In Joe's frame, he sees himself at rest (v = 0) and the stick is moving (v = 24/25 c). If the stick moves for 7 years (according to Joe) at a speed of v = 24/25 c (according to Joe), how far does the stick travel according to Joe? Would this give you L or L'?
 
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  • #7
I think I'm getting it (hopefully),

From Joe's frame,
The stick is flying past at v = 24/25 for 7 years. Thus Joe measures it at 6.72lyr
So, L = 6.72lyr in Joe's frame

L = L'/γ
Lγ = L' = 24 lyrs

So at rest the stick is 24 lyrs long. This would be the measurement Ed takes also.

But how can Joe get there faster than a pulse of light that would take 24 years?
 
  • #8
leonmate said:
I think I'm getting it (hopefully),

From Joe's frame,
The stick is flying past at v = 24/25 for 7 years. Thus Joe measures it at 6.72lyr
So, L = 6.72lyr in Joe's frame

L = L'/γ
Lγ = L' = 24 lyrs

Yes, for Joe the distance between the Earth and the planet is 6.72 lyrs.

So at rest the stick is 24 lyrs long. This would be the measurement Ed takes also.

But how can Joe get there faster than a pulse of light that would take 24 years?

Good. For Ed, the distance of the trip is 24 lyrs. How long does the trip take according to Ed?
 
Last edited:
  • #9
TSny said:
Yes, for Joe the distance between the Earth and the planet is 6.72 lyrs.
Good. For Ed, the distance of the trip is 24 lyrs. How long does the trip take according to Ed?

According to Ed it takes 25 years!
Ok, it's coming together now. Not as intuitively as I'd hoped, thanks a lot for your help though!
 
  • #10
OK, good work.
 

Related to Special Relativity: Twin Paradox Question

1. What is the Twin Paradox in special relativity?

The Twin Paradox is a thought experiment that illustrates the effects of time dilation in special relativity. It involves two identical twins, one who stays on Earth and the other who travels at high speeds near the speed of light. When the traveling twin returns to Earth, they will have aged less than the twin who stayed on Earth, due to the effects of time dilation.

2. How does the Twin Paradox challenge our understanding of time?

The Twin Paradox challenges our understanding of time because it shows that time is not absolute and can be affected by motion. The traveling twin experiences time passing more slowly due to their high speed, while the stationary twin experiences time at a normal rate. This contradicts our everyday experience of time being the same for everyone.

3. Is the Twin Paradox a real phenomenon or just a thought experiment?

The Twin Paradox is a real phenomenon that has been demonstrated through experiments and observations. The effects of time dilation have been observed in high-speed particles and in atomic clocks on airplanes. However, the Twin Paradox is often used as a thought experiment to help understand the principles of special relativity.

4. Can the Twin Paradox be explained by the theory of general relativity?

While the Twin Paradox is often associated with special relativity, it can also be explained by the theory of general relativity. General relativity takes into account the effects of gravity on time, which can also cause time dilation. The Twin Paradox can be seen as a combination of both special and general relativity.

5. Are there any real-life applications of the Twin Paradox?

The Twin Paradox has implications for space travel, as astronauts traveling at high speeds will experience time dilation. This must be taken into account when planning long-distance space missions. Additionally, the principles of the Twin Paradox are used in technologies such as GPS, where precise time measurements are necessary for accurate location tracking.

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