On 9/9/2025 11:52 PM, Alan Grayson wrote:
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.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*
The clock readings don't define events any more that odometer readings define distances.* **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*
Why? Because something "seemed to defy basic physics" /to you/? Can you explain how basic physics is defied?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 *
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