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.
*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.
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/4a8670f0-e9ab-46b1-91cb-5a78b3188e6f%40gmail.com.
