What is the figure of 6 years representing in resolving the Twin Paradox?

  • Context: Undergrad 
  • Thread starter Thread starter mrausum
  • Start date Start date
  • Tags Tags
    Paradox Twin paradox
Click For Summary

Discussion Overview

The discussion revolves around the "Twin Paradox" in the context of special relativity, specifically focusing on the time it takes for one twin, Bob, to catch up with the other twin, Ann, after Bob travels at relativistic speeds. Participants explore the calculations involved in determining the time intervals experienced by each twin and the implications of time dilation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant initially questions why it takes Bob 4 years to catch up with Ann at a speed of 15/17c, indicating a lack of clarity in the explanation provided.
  • Another participant claims to have resolved the issue by calculating the time taken for Bob to catch up in his inertial frame and adjusting for time dilation, arriving at a figure of 4 years.
  • A later post discusses the figure of 6 years, seeking clarification on its significance in the context of the problem, while also reiterating the calculation of 4 years based on the distance and relative velocities involved.
  • Participants present calculations involving the distance between the twins and the relative velocities to derive the time taken for Bob to catch up, noting the effects of time dilation on Bob's clock during his return journey.

Areas of Agreement / Disagreement

There is no consensus on the interpretation of the figure of 6 years, as participants express differing views on its relevance and meaning in the context of the Twin Paradox. The discussion includes competing calculations and interpretations without a clear resolution.

Contextual Notes

Participants reference specific velocities and time dilation factors, but the discussion does not resolve the assumptions or definitions that underlie their calculations. The relationship between the figures of 4 years and 6 years remains ambiguous.

mrausum
Messages
44
Reaction score
0
Resolving the "Twin Paradox"

http://mentock.home.mindspring.com/twins.htm

I've been trying to follow this unusual explanation to resolve twin paradox, which uses the lorentz relativistic velocity transformation equation to get the speed bob zooms off after Ann at,15/17 C. I can understand up to that point of the explanation and get the same value, however why does it take him 4 years to catch up with Ann at this speed? It's not explained. Maybe it's obvious? Help!

(please no explanations involving the relativistic dopler shift equation).
 
Last edited by a moderator:
Physics news on Phys.org


Nevermind, i resolved the problem :D. For anyone interested, I worked out the time it took for bob to catch up in the inertial bob frame, then divided this by the new gamma between bobs returning frame and bobs inertial frame (from v = 15/17 C), to get 4 years.
 
Last edited:


mrausum said:
Nevermind, i resolved the problem :D. For anyone interested, I worked out the time it took for bob to catch up in the inertial bob frame, then divided this by the new gamma between bobs returning frame and bobs inertial frame (from v = 15/17 C), to get 6 years.
What is the figure of 6 years supposed to represent? In your original post you were asking how they derived the figure of 4 years for Bob to catch up with Ann again, in the frame where Ann is moving at 3/5c and Bob is moving at 15/17c after Bob's acceleration.

Anyway, here's one way of deriving the 4 year figure. Before Bob accelerates, in this frame Bob is at rest while Ann is moving away at 3/5c, and this lasts for 4 years, so when Bob accelerates the distance between Ann and himself will be (3/5c)*(4 years) = 12/5 light years. Then after he accelerates, he'll be moving towards her at 15/17c while she continues to move away at 3/5c, so the "closing velocity" between them in this frame (the rate at which the distance between them is shrinking in this frame, which is different from the velocity of Ann in Bob's rest frame) will be 15/17c - 3/5c = 75/85c - 51/85c = 24/85c. So, if the distance between them is initially 12/5 light years at the moment Bob accelerates, the time in this frame for Bob to catch up will be (12/5 ly)/(24/85c) = 17/2 years. This is a lot longer than 4 years! But note that Bob's clock is running slow during the return journey, by a time dilation factor of sqrt(1 - (15/17)^2) = sqrt(289/289 - 225/289) = sqrt(64/289) = 8/17. So, during the 17/2 years it takes in this frame for Bob to catch up with Ann after accelerating, Bob's clock only ticks forward by (8/17)*(17/2) = 8/2 = 4 years.
 
Last edited:


JesseM said:
What is the figure of 6 years supposed to represent? In your original post you were asking how they derived the figure of 4 years for Bob to catch up with Ann again, in the frame where Ann is moving at 3/5c and Bob is moving at 15/17c after Bob's acceleration.

Anyway, here's one way of deriving the 4 year figure. Before Bob accelerates, in this frame Bob is at rest while Ann is moving away at 3/5c, and this lasts for 4 years, so when Bob accelerates the distance between Ann and himself will be (3/5c)*(4 years) = 12/5 light years. Then after he accelerates, he'll be moving towards her at 15/17c while she continues to move away at 3/5c, so the "closing velocity" between them in this frame (the rate at which the distance between them is shrinking in this frame, which is different from the velocity of Ann in Bob's rest frame) will be 15/17c - 3/5c = 75/85c - 51/85c = 24/85c. So, if the distance between them is initially 12/5 light years at the moment Bob accelerates, the time in this frame for Bob to catch up will be (12/5 ly)/(24/85c) = 17/2 years. This is a lot longer than 4 years! But note that Bob's clock is running slow during the return journey, by a time dilation factor of sqrt(1 - (15/17)^2) = sqrt(289/289 - 225/289) = sqrt(64/289) = 8/17. So, during the 17/2 years it takes in this frame for Bob to catch up with Ann after accelerating, Bob's clock only ticks forward by (8/17)*(17/2) = 8/2 = 4 years.

Sorry, i meant 4 years for this example. I'm writing this solution to the paradox as a short first year essay/report and so I'm using different values.

Thanks for the explanation, that's exactly what i did, except you explained it a lot clearer. Good to know I'm using the right method.
 
Last edited:

Similar threads

Undergrad Reference systems and apparent counter-explanations on Twin Paradox
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad Understanding Spacetime simultaneity in twin paradox scenarios
  • · Replies 35 ·
2
Replies
35
Views
4K
High School A scenario to help understand the twin paradox in SR
  • · Replies 20 ·
Replies
20
Views
4K
High School My paraphrased explanation of the Twin Paradox
  • · Replies 24 ·
Replies
24
Views
5K
Undergrad What happens when twins synchronize clocks and one teleports?
  • · Replies 9 ·
Replies
9
Views
2K
Undergrad Twin paradox for (accelerated) dummies?
  • · Replies 36 ·
2
Replies
36
Views
6K
Undergrad Twin paradox, virtual clock on ship with Earth time, discontinuity
  • · Replies 115 ·
4
Replies
115
Views
9K
Undergrad How Does the Twin Paradox Apply to Light and Sound Waves?
  • · Replies 48 ·
2
Replies
48
Views
7K
Undergrad Simultaneity and the Twin Paradox
  • · Replies 138 ·
5
Replies
138
Views
11K
High School Twins Paradox Thought Exp: What Happens When Reunited?
  • · Replies 31 ·
2
Replies
31
Views
4K