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:
On 9/9/2025 9:57 PM, Alan Grayson wrote:
On Tuesday, September 9, 2025 at 5:18:14 PM UTC-6 Brent Meeker wrote:
On 9/8/2025 8:45 PM, Alan Grayson wrote:
On Monday, September 8, 2025 at 9:35:09 PM UTC-6 Brent
Meeker wrote:
On 9/8/2025 11:19 AM, Alan Grayson wrote:
On Monday, September 8, 2025 at 5:06:36 AM UTC-6 John
Clark wrote:
On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson
<[email protected]> wrote:
/> I'm not sure the impossibility of absolute
simultaneity solves the problem,/
*
*
*Watch the video!If you follow what he does
step-by-step you will see that he is right. It's
not difficult. *
*I'll definitely watch it, very soon, but a-priori the
impossibility of absolute simultaneity can't solve the
paradox because it's not its cause. Can you succinctly
state the cause of the paradox? It's the application of
time dilation in SR, under the mistaken assumption that
the twins take symmetric paths; that their situations
are symmetric. This results in the situation that when
they meet and compare clock readings, each concludes
the other is younger. *
No that's wrong. The stay at home twin has a clock that
indicates a longer interval than the traveling twins
clock. They agree that the traveling twin is younger.
Brent
*Can't you understand English? I was stating the paradox and
its cause. With an accurate analysis, the traveling twin is
younger. Also, FWIW, for the traveling twin to return for
the clock comparison, some acceleration is necessary,
although it can be minimized if the comparison is done by
fly-by. a AG *
But notice that the acceleration is entirely incidental, as
illustrated by the case in which Red and Blue each
accelerates the same amount. IT'S JUST GEOMETRY. ONE PATH
IS LONGER THAN THE OTHER.
*In the original statement of the "paradox', the traveling twin
must accelerate to return so the clocks can be compared. Please
explain how this can happen without acceleration. *
I've shown two different ways without acceleration and I've also
shown the paradox with equal accelerations by both twins. Why
can't you just accept that it's geometry; that one path is longer
than the other.
*
*
*If both twins are accelerating, then you've redefined the TP. *
No. But Blue does not accelerate while Red travels out and back.*
*
*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. 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.
Brent
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