On 9/10/2025 1:30 PM, Alan Grayson wrote:
You don't understand what an event is. It's a point in spacetime. It's independent of what coordinate labeling exists, just as a point on Earth is independent of what map you use. It's a physical thing. Have you never read a book on relativity?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.*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 event. So the return event is not the same as starting event when the twins are juxtaposed. AG *
The point of the slingshot turnaround is that is shows the effect on the clock has nothing to do with acceleration. The whole trip is in free-fall. An accelerometer would read zero the whole time.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 *
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