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From: hitlong@yahoo.com (gharnagel)
Newsgroups: sci.physics.relativity
Subject: Re: Oh my God!
Date: Tue, 1 Oct 2024 13:51:55 +0000
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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?
>
> ======================================================================
> 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).
>
> ======================================================================
> 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.
>
> ======================================================================
> 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:

Please look at my Figure 4.  A and B are stationary (the t and x
axes are orthogonal, call this the S frame).  Points along
horizontal lines are simultaneous for A and B.  In this figure,
A launches the tachyonic signal to B, but it could be any of the
other players doing the launch as well.  But let's go with A to B
at t = vL/c^2 and B passes the message to D at that time.  The two
arrows with "?" presents the question: where (or when, actually)
can D send a signal to C?  To C at t = 0 or C at t = vL/c^2?

From the perspective of C and D in S', D should be able to send
it to C at t' = t = 0 because that is along a horizontal line
(the x' axis), but from the perspective of A and B, C is NOT at
t = 0: C is at t = vL/c^2.  It is only possible to claim that D
can send it infinitely fast from the S' frame.  It cannot be done
from the S frame.  THIS is why my arrow is horizontal in S.  This
is relativity of simultaneity in the raw.

This is one reason why Morin, Taylor, Wheeler and Recami say to
stay in one frame to solve the problem.  Staying in S in Figure 4,
D cannot send the signal to C at t = 0 because C is at t = vL/c^2.

Now, look at Figure 5.  A similar ambiguity appears when A sends
the signal to B.  Note, however, that A at t = t' = 0 can send
the signal to B at t' = 0, but B's time is t = vL/c^2.  Therefore,
according to C and D, A can send it no faster than c^2/v.

Actually, A could send it infinitely fast to B, but B isn't
adjacent to D, so B would have to wait until t' = 0.

Spacetime diagrams tend to desensitize one's faculties about RoS.