On Wednesday, September 10, 2025 at 2:30:29 PM UTC-6 Alan Grayson wrote:
On Wednesday, September 10, 2025 at 1:55:07 PM UTC-6 Brent Meeker wrote:
On 9/9/2025 11:52 PM, Alan Grayson wrote:
On Wednesday, September 10, 2025 at 12:08:39 AM UTC-6 Brent Meeker wrote:
On 9/9/2025 10:36 PM, Alan Grayson wrote:
On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:
...
*If both twins are accelerating, then you've redefined the TP. *
No. But Blue does not accelerate while Red travels out and back.
*But earlier you claimed both Red and Blue have the same acceleration, and
that's what your diagram shows. AG*
But not *"while Red travels out and back"*. Which IS what my diagram
shows. You can imagine Blue as decelerating, landing on Earth, and then
accelerating to join up again as Red comes back by. Next time read the
whole sentence, not just up until a question occurs to you.
* Thank you for replying, but please cease adding crap to it. I read it,
and looked at your plot. It sure seemed as if both are accelerating. AG *
*If you have two paths in spacetime, starting at the same point and ending
at the same point, or at a different point, how can you tell which is
longer? AG *
These are paths in *spacetime*. They start and end at the same *event*, a
point in 4-space.
*If the paradox is resolved, then the clocks should read different values
when finally compared. So the end point events are NOT the same. AG*
The clock readings don't define events any more that odometer readings
define distances.
*Correction in black. AG*
*That's what I thought. But when traveling twin returns, it's to a
different event because the time label on the coordinate differs from the
INITIAL event. So the return event is not the same as starting event when
the twins are juxtaposed. AG *
The obvious way to tell which is longer in proper time is to carry an ideal
clock along the two paths and compare the measured intervals. You could
also measure the space distance X along the paths and and compute proper
time S=\sqrt{T^2 - X^2} where T is the coordinate time difference (in the
same reference frame you measured distance).
Brent
*You seem to defying basic physics if this is your claim. I don't deny that
the original problem can be restated in a way which avoids acceleration,
and IMO this is what you've done. *
But I've done more than that. I've done it while maintaining exactly the
same paradox.
*So you admit you're defying the laws of physics? AG *
Why? Because something "seemed to defy basic physics" *to you*? Can you
explain how basic physics is defied?
*I now see you as a magician. In your special model of the problem, your
traveling twin somehow turns around without acceleration. Or maybe you mean
one of the triplets. In physics as I understand it, change in direction is
acceleration and that's how the traveling twin returns. AG *
Brent
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