Re: Oh my God!

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

>
On Tue, 1 Oct 2024 18:47:16 +0000, gharnagel wrote:

>
And you haven't acknowledged your confusion about what frame
is the "stationary" one in the right and left figures.  Just
proclaiming a frame as stationary doesn't make it so, particularly
when you draw its time axis skewed.

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

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.

Our frame, the S frame, is moving.

If it's "our" frame, then we are stationary by the first principle.
And the skewed t' and x' axes prove it.  Velocity is relative.

To simplify the figures, I have not drawn the S axes, which are
orthogonal. The S' axes are skewed because I am mapping events
in S' to our coordinate system, where our S coordinate system
is moving relative to the S' coordinate system at speeds -0.1c,
0c, and +0.1c.

Velocity is relative.  The left and right panels show that S'
is moving relative to S.

Prok, I have shown that you completely misunderstood my thesis
whereas the reviewer of DOI: 10.13189/ujpa.2023.170101 did not
or he would have rejected it.  Rather than acknowledge your
error and try to understand, you launch another baseless attack
because of your confusion about what v means.  It is the speed
that D must send the signal (Event E1) so it arrives when C and
A are adjacent (E2).  Furthermore, A must send a signal to B
when B is adjacent to D.  Your figures are only half of the full
problem, and they do NOT describe my "proposal."  They are your
imaginings.  If you want to discuss my thesis, then use my
figures (4 and 5, particularly).  Yours are straw men.
>
And you haven't acknowledged your confusion about what frame
is the "stationary" one in the right and left figures.  Just
proclaiming a frame as stationary doesn't make it so, particularly
when you draw its time axis skewed.

>
There is only one person here who is confused, and that is YOU.

Nope.  In fact there is confusion on both sides.

In the S' frame, an infinite speed tachyonic signal is emitted
from (x',t') = (D,0) and is received in zero time at (x',t') = (C,0)
>
That is zero time as measured in the S' frame.

There is no confusion there.

The emission and receipt events are concurrently monitored by
observers in three "S" frames, where the "S" frames are moving
relative to the S' frame at speeds -0.1c, 0c, and +0.1c, and so
forth.

This is certainly possible, but you're three panels represent
perspectives from three different frames, and only the center
panel is from that of S'.

In general, observers in the "S" frames do not consider the signals
as traveling from D to C in zero time.

And there's no confusion there, either; however, your last claim
here is switching from Method II (the "hand-off scenario) to
Method I (direct communication between frames in relative motion.
Let's get some things cleared up here.
(1) We assume that D has a tachyon transmitter and C has a basic
tachyon receiver.
(2) C and D can communicate with each other at any speed
c < u < \infty.  (Infinite speed tachyons have no energy
and can't be detected by a basic tachyon receiver.)
(3) Observers in other frames with a basic tachyon receiver may
or may not be able to "eavesdrop" on their signal, if they don't
have an advanced tachyon receiver.
(4) Having an advanced tachyon receiver means that they have a
receiver with a moving sensor that allows the signal energy to
be greater than zero (the velocity between transmitter and
receiver is closing).  Effectively, this is Method II.
(5) Under these conditions, there are no causality violations
and there is no ripping of spacetime to shreds.
All of this is spelled out in DOI: 10.13189/ujpa.2023.170101.