On Wednesday, October 1, 2025 at 11:52:40 PM UTC-6 Alan Grayson wrote:
On Wednesday, October 1, 2025 at 10:41:33 PM UTC-6 Brent Meeker wrote: On 10/1/2025 6:09 PM, Alan Grayson wrote: On Wednesday, October 1, 2025 at 6:11:55 PM UTC-6 Brent Meeker wrote: On 10/1/2025 6:38 AM, Alan Grayson wrote: On Wednesday, October 1, 2025 at 7:20:13 AM UTC-6 John Clark wrote: On Wed, Oct 1, 2025 at 8:29 AM Alan Grayson <[email protected]> wrote: *> Have physicists in the last 120 years claimed that two paths of different lengths in spacetime which start and end at same events, have the same accelerations, except Brent in his diagram? AG* *In a word, yes. Two worldlines between the same events in spacetime can have different lengths even if both involve acceleration. And proper time is the length of your world line. But of course if they have identical acceleration histories then they are in the same worldline, not a different one.* You're writing nonsense. Brent has two worldlines with different lengths, claiming they have identical accelerations. AG And he included diagrams showing the accelerations had the same amplitudes and durations. And that even was redundant. From the diagram it is clear that Red and Blue had the same velocity at the initiation of their accelerations and they turned their velocity thru the same angle in each period of acceleration...hence one can infer mathematically that their (acceleration*duration) products were the same. Brent *That was your intention, and it appears so. But since the longer path goes further out, away from the shorter path, I don't think splitting the necessary accelerations into two parts implies the summed accelerations are the same as the acceleration for the shorter path. As I previously stated, the standard TP problem, with one twin spatially fixed and other traveling, is a limiting case of your diagram, and in that case the accelerations are not equal. AG* The accelerations in my diagram are four equal changes in velocity. I split the turnaround into two burns for Red to make it clear that they were the same delta-v and for Blue's two, as shown at the left. You keep bringing up you "standard TP" problem, but by my examples I've shown that every proposed explanation (e.g. they experience different accelerations is you latest) is nullified by one or more example (like this one). If it can't be due to acceleration in this example, or in the gravitational slingshot example, or in the triplet example, why would you continue to suppose it could be due to acceleration in your "standard" example. *You're trying to show that the difference in clock rates depends only on spacetime length. What I am suspecting, is that in this particular case, the clock rates are different, but your claim that the accelerations are identical, is flawed. I do that by examining two limiting cases, which I acknowledsge isn't a proof, but only a suggestion. AG * *Looking further into this particular scenario, I think you've created a specific geometry which demonstrates the two paths have the same accelerations, and since the elapsed time along each path is different, it can't depend on acceleration. However, there are other paths wherein the accelerations won't be equal, where the endpoints are identical. So, if you want to generally show that the differential elapsed clock time just depends on path length, you can dispense with your diagrams entirely, and instead prove that in spacetime everything moves at light speed, and thus when comparing paths, one need only integrate the velocity along a path, and that will be sufficient to show that when comparing several paths, even those whose endpoints are not juxtaposed, that the longest path produces the shortest elapsed time. If you're so damned smart that you can easily put me down, how come you haven't thought of this solution, which allows you to totally dispense with your diagrams, and do a general solution, to prove that elapsed time depends only path length, and the longer the path length, the shorter the elapsed time, compared to other paths, even those whose endpoints are not juxtaposed? AG* *Concerning your solution using GR, where the turnaround is due to some gravitating body, you claimed that the acceleration is zero since the clock is in free fall. I recall you used the formula F=ma, plugging F=0 and concluding a=0, where F=0 comes from Einstein's "happy thought" that gravity isn't a force. I don't like this line of reasoning because you're conflating GR with Newtonian Mechanics. The mystery I'd like you to address, if you can, is how gravitational acceleration can be non-zero -- which it is, assuming we're not changing our definition of acceleration -- in the absence of any gravitational force? TY, AG* -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/6a25d8fe-0a97-4967-802e-a40290d26d6en%40googlegroups.com.
