Kinetic and dynamic twin effect

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Discussion Overview

The discussion revolves around the "twin paradox" in special relativity (SR), focusing on the roles of time dilation, acceleration, and the implications of the equivalence principle. Participants explore whether the paradox can be resolved purely through kinematic effects or if acceleration introduces additional complexities. The conversation includes theoretical considerations and interpretations of Einstein's work.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants propose that the twin paradox can be understood as a purely kinetic effect, emphasizing time dilation without the need for acceleration.
  • Others argue that acceleration experienced by the returning twin introduces additional time differences that are not accounted for by time dilation alone, suggesting a connection to gravitational effects.
  • A later reply questions the necessity of considering acceleration, asserting that the main resolution lies in how time and distance are measured in the context of the experiment.
  • Some participants highlight that an inertial observer can track an accelerating object using velocity-dependent time dilation, while an accelerating observer must account for changing simultaneity and (pseudo-)gravitational time dilation.
  • There is a discussion about Einstein's explanations in his papers, with some participants noting that his descriptions may have led to confusion regarding the role of acceleration in the twin paradox.
  • One participant suggests that the scenarios of slinging around a planet and bouncing off a spring both involve acceleration, though the implications for the twin paradox may differ.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and implications of acceleration in the twin paradox. While some agree that the paradox can be resolved without considering acceleration, others maintain that it plays a significant role in understanding the time differences involved. The discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Some participants note that the discussion often avoids general-relativistic complications by focusing on scenarios without gravity, which may limit the scope of the arguments presented.

  • #31
Al68 said:
Hi Mike,

Are you looking for an explanation of differential aging for a one way trip, including d? I haven't bothered posting it because I thought we were just discussing how best to explain it to laymen.

The math is easy, for a 10 light year trip at 0.8c, the ship will reach the destination when the ship's clock reads 7.5 yrs, Earth's clock reads 12.5 yrs. Elapsed proper time equals distance/velocity, and the distance in the ship's frame is 6 light yrs, in Earth's frame it's 10 light years. It's as simple as that.

The objection is that in the outbound ship's frame, Earth's clock reads 4.5 yrs when the ship reaches the destination. But that isn't anyone's proper time for the trip. And after the ship comes to rest with earth, we have no clocks or twins in that frame. (But I'll mention a clock left in that frame later.)

Then we can double the results for a round trip. Ship's clock reads 15 yrs. Earth's clock reads 25 yrs.

Some point out that time dilation should be reciprocal. Well it is reciprocal between inertial frames. In either ship inertial frame (outbound or inbound) the elapsed proper time between the ship's departure and return to Earth is 41.66 yrs. So time dilation is reciprocal, we just didn't have a clock in either ship inertial frame for the entire trip. If we left a clock in the ship's outbound frame to keep going past the destination, it would read 41.66 yrs when the ship returned to earth.

I left out a lot due to laziness, and I don't have the best writing skills, but the bottom line is that the differential aging in the twins paradox doesn't actually depend on a turnaround or a reunion. The same effects occur for a one way trip, so we can just "resolve" each one way trip and add them together.

Al

No, Al68, I am not looking for that calculation. I already know how to do it. I am trying to explain that your calculation, and that of others, is not the calculation for a one-way trip. But there is no point in diverting the subject matter of this thread any longer. My apologies to the original poster.
 
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  • #33
MikeLizzi said:
No, Al68, I am not looking for that calculation. I already know how to do it. I am trying to explain that your calculation, and that of others, is not the calculation for a one-way trip. But there is no point in diverting the subject matter of this thread any longer. My apologies to the original poster.
Well, I doubled the one way trip to obtain results for a round trip. For a one way trip only:

For a 10 light year trip at 0.8c, the ship will reach the destination when the ship's clock reads 7.5 yrs, Earth's clock reads 12.5 yrs. Elapsed proper time equals distance/velocity, and the distance in the ship's frame is 6 light yrs, in Earth's frame it's 10 light years. It's as simple as that.

That's a one way trip. Assuming the ship's twin stays at 10 light yrs from earth, at rest with earth, he will be 5 yrs younger than the Earth twin forever.

The spacetime diagram would look like the first half of the twins paradox, then just parallel lines as long as they stay at rest.

Al
 

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