Re: Oh my God!

On Wed, 2 Oct 2024 22:26:50 +0000, ProkaryoticCaspaseHomolog wrote:

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On Tue, 1 Oct 2024 22:18:12 +0000, gharnagel wrote:

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On Tue, 1 Oct 2024 19:24:31 +0000, ProkaryoticCaspaseHomolog wrote:

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I have told you several times that I am designating the S' as
stationary, in addition to plainly stating that fact in my drawing.

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And then you "moved" the lab frame and pretended S' was still
stationary, which is contradicted by the fact that the t' and
x' axes are no longer orthogonal.

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Why do you have such difficulty with a ***SIMPLE*** mental switch?
The x' and t' axes are tilted because our S frame is moving.
Our x and t axes are orthogonal from OUR point of view.

Exactly, OUR POV.  That means WE have located ourselves at rest
WRT the S frame.  Previously, we were at rest in the S' frame.
We have switched frames.  Why do you have such difficulty with a
***SIMPLE*** factual switch?

Look, if you want the S frame to be stationary and the S' frame
to be moving, your are perfectly free to do this ***TRIVIAL***
switch.  Physically it makes no difference.

This is exactly what you did, as evidenced by the skewed t' and
x' axes.  Look you show the signal arrow horizontal in all three
panels.  That is wrong.  If the t' and x' axes are skewed, the
signal arrow will be, too, if it's the same signal arrow.  Since
you drew them all horizontal, that means they're not the same
signal.  If you believe that is my thesis then understand it you
do not (Says Yoda).

I only wanted the S' frame to be stationary because I wanted
eventually for you to be able to envision the S' frame to be
surrounded by an INFINITY of moving observers, all of whom
have a different view of the emission and receipt events,
disagreeing on their order.
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Stationary and moving are arbitrary by the first principle, of
course, but the views in panels one and three are from the view
of an observer at rest (stationary) in the lab frame S.  So things
aren't exactly the same:  RoS is not the same.  Do you need proof
of that, or can you figure that out for yourself?

Our frame, the S frame, is moving.

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If it's "our" frame, then we are stationary by the first principle.

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No. We are NOT obligated always to make ourselves the stationary
observer.

By the first principle, we are by necessity,  "We" are always
stationary.  We can have other observers in the lab but "We"
aren't there.  You have put "Us" in three different frames.
As to showing the order of events, all those dots are extreme
overkill.  Properly chosen S and S' frames are sufficient. I
believe it's time for something from by collection of proverbs:
 “Don’t think too much.  You’ll create a problem that wasn’t even
there in the first place.” – Anon.

[Yes, I'm aware of your many contributions to Wiki]
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Anyway, you had so many dots I couldn't tell what was what.  A
simple plot would have been sufficient showing a signal arrow
pointing downward as viewed from a frame, as shown in Figure 2
of DOI: 10.13189/ujpa.2023.170101.  But let's take your version,
completely from the perspective of C and D.
B (at t = vL/c^2 adjacent to D) observes that D sends a signal
at t' = 0 to C at t' = 0.  A, adjacent to C at t = 0, observes
that C receives it.  Does A know that this message came from his
future?  Not until B sends a message telling him so.
So C passes a message to A and A now has a message, purportedly
from his future (but he doesn't know it).  He sends it to B.
If A sends it to B at u = c^2/v to D, D receives the message
immediately as he sent it.  No causality violation.
Could A have sent it to B infinitely fast?  Not from the
perspective of C and D because B is at t = vL/c^2, NOT at
t = 0 in the frame that we started the analysis, in which
we are required to remain!
We could analyze from the perspective of A and B, but still no
causality violation.  My thesis is that tachyon velocity can
be c < u < \infty, but a closed loop cannot violate causality.
Okay, the rest of this is rechewing the fat, so let's skip to
your last comment:

This is ridiculous. AMONG OTHER THINGS, the moving observer is not
attempting to observe the tachyons directly, but is observing the
READOUTS of the tachyon emitters and receivers. At any subluminal v,
there can be no limitation on the observers' ability to monitor their
READOUTS.
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No problem with that, so your claim that "this is ridiculous" is
a bit over the top.  What A observes in the above scenario is that
C obtains a message at t = t' = 0.  A reads the message and
passes it to B.  We're observing from the frame where AB are at
rest, so B waits until D is adjacent ( at t = vl/c^2).  But
the time at C is NOT zero, so D can't send it to C at t' = 0!
RoS is a bitch, baby.