Re: Oh my God!

On Tue, 1 Oct 2024 13:51:55 +0000, gharnagel wrote:

On Mon, 30 Sep 2024 23:48:27 +0000, ProkaryoticCaspaseHomolog wrote:

>
On Sun, 29 Sep 2024 20:53:41 +0000, gharnagel wrote:

>
Your initial statement is that the S' frame is stationary and you
are going to move the lab frame.  Then you place two observers, C
and D, orthogonal because they are stationary in S'.  Then you
send a signal from D to C infinitely fast.
>
Now in the triptych, you move the lab frame S.  That's fine, but
S' is the STATIONARY frame: C and D should still be vertical.
So I must assume that you are actually switching to the S frame
and making IT stationary, right?

>
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I stated very clearly that I am using the inverse Lorentz transform
to map the events of STATIONARY frame S' to my MOVING lab frame. To
keep the diagrams uncluttered, I did not draw the coordinates of my
lab frame. My coordinates remain orthogonal.
======================================================================

>
Prok, Prok, Prok!  You're smarter than that, and so am I! t' and x'
are NOT orthogonal in the left and right figures, so you HAVE switched
frames.  You left out the t and x axes, which hides the fact that
they ARE orthogonal, which would prove that the viewpoint in those
figures is from the S frame.
>

So you have done exactly what Morin, Taylor, Wheeler and Recami
say NOT to do.  You have switched horses (er, frames) in the
middle of the stream (er, problem setup).

>
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No, I am not switching back and forth between frames.
======================================================================

>
Yes, you did.  You have confused yourself, not me.
>

======================================================================
For the purpose of any one diagram, my moving frame is moving at a
constant rate relative to the stationary frame, and I stick to that
one frame without jumping around.
======================================================================

>
No, you didn't, as the non-orthogonal t' and x' axes prove.
>

So let's look at the figure on the right with v = 0.1c.  As viewed
from S, t1 = \gamma (0 + 0.1L) = 0.1gL.  At that time in S, the time
in C is not at t = 0.  This is relativity of simultaneity (RoS).
Prok, YOU are the one trying to rip SR to shreds by pretending you
can ignore RoS.  Figures 4 and 5 in DOI: 10.13189/ujpa.2023.170101
obey RoS, your figures do not.

>
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There is no problem with my triptych diagrams. As I have drawn them,
the event associated with receipt of the signal is at t'=0 in all
three scenarios. It is the diagram below the triptych labeled "Gary's
proposal" which is totally INSANE. In order to prevent the arrow from
moving backwards in time as observed in my moving frame, you would
have the context of the receiving event SHIFT as the result of my
movement.
======================================================================

>
No, Prok, you misunderstand.  I'm saying that the signal should
arrive at t' = vL/c^2. not t' = 0 to save RoS.  There is no "ripping
spacetime to shreds" :-).  To help you understand:

Claiming that t' = vL/c^2 DESTROYS RoS, does not rescue it.
There can be any number of observers of the same events. Suppose you
are in the lab frame moving to the left at 0.1 c. You claim that to
save RoS, the S' observer must receive the signal at t' = 0.1*L/c^2
I am concurrently observing the same events from a frame moving at v=0.
To save RoS, the S' observer must receive the signal at t'=0.
My wife is concurrently observing the same events from a frame moving
at v=-0.1c. To save RoS, the S' observer must receive the signal at
t' = -0.1*L/c^2.
Which is it, Gary? Does the S' observer receive the signal at
t' = 0.1*L/c^2, t' = 0, or t' = -0.1*L/c^2 ?
Does reality change as a result of the motion of external observers?
Does reality split into an infinity of worlds? Is your "theory" a
"many worlds" interpretation of special relativity?