Re: Oh my God!

On Thu, 3 Oct 2024 12:18:55 +0000, ProkaryoticCaspaseHomolog wrote:

On Thu, 3 Oct 2024 9:17:40 +0000, gharnagel wrote:
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On Thu, 3 Oct 2024 7:57:10 +0000, ProkaryoticCaspaseHomolog wrote:

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You are stuck with some sort of language difficulty.

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It is more than a "language" difficulty.  The frame in which time
and space axes are orthogonal IS the frame in which "WE" are at
rest.  In the center panel of the top trio, "WE" are at rest in
the S frame.  In the left panel, "WE" are at rest in the S' frame.
In the right panel "WE" are at rest in the S'' frame.  In order
to do that, "WE" had to undergo significant acceleration:  "WE"
switched frames.

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NO NO NO NO NO!!!!
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See Figure 4

WHAT is "Figure 4"?  Is that your new attachment?

Let us change the description of the diagram slightly to one that
you might agree with better.
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The emission of a signal from D and its receipt by C in the S'
frame is being concurrently monitored by observers 1, 2, and 3
in frames S1, S2, and S3.
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Three "outside" observers of the same events. No frame jumping.
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Any objections to that?

Just as long as when an analysis is performed, it is done from
one frame, it doesn't matter which one.
"An extremely important strategy in solving relativity problems
is to plant yourself in a frame and stay there. The only thoughts
running through your head should be what you observe. That is,
don’t try to use reasoning along the lines of, 'Well, the person
I’m looking at in this other frame sees such-and-such.'  This will
almost certainly cause an error somewhere along the way, because
you will inevitably end up writing down an equation that combines
quantities that are measured in different frames, which is a no-no.
 -- David Morin, "Introduction to Classical Mechanics," p. 522.
"one should never mix together the descriptions of one phenomenon
yielded by different observers, otherwise--even in ordinary
physics-- one would immediately meet contradictions"
 -- E. Recami, "Classical Tachyons and Possible Applications,"
Rivista Del Nuovo Cimento, 9:6 (1986), p. 66.

The emission event occurs at (x',t') = (D,0)
The receipt event occurs at (x',t') = (C,0)
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1, 2, and 3 are all stationary within their own frames.
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If you understand the relabeled diagram, please look at my other
discussions with this improved understanding.

I have no problem with the attached figure except that it labels
the panels S', S' and S' whereas in the above text you define
observers 1, 2 and 3 in frames S1, S2 and S3.  I agree with your
labeling in the text.
I do have a problem that you seem to believe that WE can switch
views from one panel to another with impunity.  When WE look at
the first panel, WE have adopted the perspective of observer 1
in S1, etc.  So stay there and solve the problem.  In panel 1,
the arrow is observed by observer 1 as moving at u1 = -c^2/v.
In panel 2, it is observed by observer 2 as u2 = - \infty.  In
panel 3, it is not observed by observer 3 if he only has a basic
tachyon receiver.  This is because tachyon energy is frame
dependent and
E = mc^2/sqrt(u^2/c^2 - 1) in S2, where u = -\infty.  Therefore
E = 0.  (We're assuming this as a limit for analysis purposes).
In S1, D is moving toward the observer's receiver, so the signal
relative to S1 is E > 0.  Essentially it's a Doppler effect.
In S3, however, D is moving AWAY from observer 3, the energy goes
down. Is it negative?  The 4-momentum formalism says yes, but it
is mathematically incorrect in the situation.  However, one can
assume, in a sense, that it is negative energy, which can't activate
the receiver at rest in S3.
As you say, the observer in S3 knows that C received a signal.  If
there were two observers in S3, A and B (B adjacent to D when D
launched the signal and A adjacent to C when C received the signal),
then A would know that C received a signal at tA = 0 (tC' = 0)
and B would know that D sent a signal at tB = vL/c^2 (tB' = 0).
They don't know that it's the same signal.  So let's say there
was a message in the signal that D sent to C, and C passed that
message to A, and A sent it to B at u = \infty.  So B gets it at
tB = 0, but B isn't adjacent to D, so B must wait until tB = vL/c^2
to pass it to D.  D receives it just as D sent it (or a bit after
because of energy considerations - receivers must have SOME energy
to work on, therefore u < \infty).  No causality problem.