Kinetic and dynamic twin effect

[MikeLizzi]

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.

 

[Dale]

Yep, still seems odd.

Thanks for the link DaleSpam. That's pretty much what I suspected.

More on the clock hypothesis from https://www.physicsforums.com/showthread.php?t=292727":

http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html

 

[Al68]

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