On 9/29/2025 9:33 PM, Alan Grayson wrote:
On Monday, September 29, 2025 at 10:26:09 PM UTC-6 Brent Meeker wrote: On 9/29/2025 9:11 PM, Alan Grayson wrote:On Monday, September 29, 2025 at 9:50:09 PM UTC-6 Brent Meeker wrote: On 9/29/2025 8:39 PM, Alan Grayson wrote:On Monday, September 29, 2025 at 6:20:57 PM UTC-6 Alan Grayson wrote: On Monday, September 29, 2025 at 4:38:07 PM UTC-6 Alan Grayson wrote: On Monday, September 29, 2025 at 4:21:29 PM UTC-6 Brent Meeker wrote: On 9/29/2025 10:39 AM, Alan Grayson wrote:On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote: No. You're over complicating the problem. It's as simple as the fact that two different thru spacetime are different lengths. Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures). So the rocket, which takes the longer spatial path, experiences less proper time lapse. *Aren't you assuming that the integrated S over both paths is the same? This is the issue I previously flagged. How do we know that both paths when integrated, have identical lengths, S? AG *Certainly not. *The whole point is that they have different length!* The proper time S is what a clock (or age) measures. Brent *If they have different S lengths, how can you conclude the proper time on stationary twin's clock records more time than the traveling twin's clock? Only if the S lengths are identical (which I didn't believe when I posted my question) can you reach that conclusion. AG* *That is, how do you know that a path doesn't exist for the traveling twin, where after you subtract out the spatial squares, the elapsed proper time isn't larger than that measured by the stationary twin when they meet? AG* *To complete your proof, you must assert and prove that in spacetime everything moves at light speed. *No, sorry, I do not. *That's why I don't like proof via diagrams. They're incomplete. AG **Then, if any component on any path is spatially non-zero, the time component of S is reduced, and hence so is proper time, which must be less than that of the stationary twin which has no spatial component. *Yes.*How do we know that in spacetime everything moves at light speed? AG*What's that have to do with anything? The equations are right there on the graphics. Either plug in numbers or do the integral yourself. *Since you're using an arbitrary path for the traveling twin, a complete proof of what I claim, is necessary. AG *I don't know what you're talking about and I don't think you do either. I'm not using an arbitrary path. I've clearly shown a specific path. Necessary for what? And what exactly do you claim? Brent*Your arbitrary path for the traveling twin has its endpoint events juxtaposed with the stationary twin. *
Otherwise how would the compare clocks/ages?
*IMO, it represents any path of traveling twin. *
Everybody's entitled to their opinion.* *
*Using the fact that everything moves at lightspeed in spacetime, solves the problem for any traveling path. AG *
So the problem (whatever it was) is solved. Brent -- 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/1560a950-5586-49aa-bb26-882419de7af5%40gmail.com.
