Post by gharnagelPost by gharnagelI only discussed HALF of the necessary analysis. In order to prove
that a solution is consistent (i.e, a closed loop solution is found
which does not violate RoS), it must be so from the perspective of
A and B as well as from the perspective of C and D.
Your schema does so with infinite speed tachyons in both frames.
Other arrangements are not so forgiving. Some are consistent in
one frame, but not in the other when obeying RoS. In that case,
a closed loop is not a valid solution. There are several examples
of this type.
A further addendum: The analysis of panel 3, concluding that the
closed loop elapsed time was zero was based on the conventional
approach, which violates relativity of simultaneity. The question
we must ask ourselves is, is it possible to violate RoS?
The attached figure is panel 3 with the lab frame info added in green.
In that frame the horizontal lines represent the points of simultaneity.
That is, when A is at t = 0, B is at t = 0.
However, C is also at t = 0 (t' = 0) AND D is ALSO at t = 0 (t' does
MOT equal zero). RoS is baked into spacetime diagrams because they
are representations of the LT equations.
The blue arrow in the attachment shows the line of simultaneity for
the CD frame (pontifically called the S frame). That arrow violates
RoS in the lab frame. Prok seems to be under the mistaken impression
that I am claiming that the blue arrow should look like a horizontal
arrow in the lab frame. I am NOT! I am claiming that as far as the
lab frame is concerned that the arrow in the lab frame is horizontal
because the arrow in the S frame is going upward (violet arrow).
I hope this puts to rest the hyper-ventilation expressed as "ripping
spacetime to shreds!" It does, of course, deserves some cogitation
about why D can't send an infinitely-fast signal to C.
Post by gharnagelI only discussed HALF of the necessary analysis. In order to prove
that a solution is consistent (i.e, a closed loop solution is found
which does not violate RoS), it must be so from the perspective of
A and B as well as from the perspective of C and D.
Your schema does so with infinite speed tachyons in both frames.
Other arrangements are not so forgiving. Some are consistent in
one frame, but not in the other when obeying RoS. In that case,
a closed loop is not a valid solution. There are several examples
of this type.
A further addendum: The analysis of panel 3, concluding that the
closed loop elapsed time was zero was based on the conventional
approach, which violates relativity of simultaneity. The question
we must ask ourselves is, is it possible to violate RoS?
The attached figure is panel 3 with the lab frame info added in green.
In that frame the horizontal lines represent the points of simultaneity.
That is, when A is at t = 0, B is at t = 0.
However, C is also at t = 0 (t' = 0) AND D is ALSO at t = 0 (t' does
MOT equal zero). RoS is baked into spacetime diagrams because they
are representations of the LT equations.
The blue arrow in the attachment shows the line of simultaneity for
the CD frame (pontifically called the S frame). That arrow violates
RoS in the lab frame. Prok seems to be under the mistaken impression
that I am claiming that the blue arrow should look like a horizontal
arrow in the lab frame. I am NOT! I am claiming that as far as the
lab frame is concerned that the arrow in the lab frame is horizontal
because the arrow in the S frame is going upward (violet arrow).
I hope this puts to rest the hyper-ventilation expressed as "ripping
spacetime to shreds!" It does, of course, deserves some cogitation
about why D can't send an infinitely-fast signal to C.
Going through your Universal J. of Physics and Application paper, I am
VERY much reminded of David Seppala, who constantly returned to these
forums presenting the same two or three basic scenarios with a few new
added complications that he hoped would show some inconsistency with
special relativity.
YOUR tactic is to add additional observers to basic, very well-known
demonstrations of causality violation associated with superluminal
signaling and to argue about what these additional observers are or
aren't capable of seeing.
First of all, you do not appear to comprehend the modern usage of the
word "observer". Otherwise you would not have written such absurdities
as "Thus a signal cannot be sent round-trip in this configuration
since A isnât adjacent to C at t = vL/c2."
Taylor and Wheeler discussed the modern concept of "observer" in
Spacetime Physics. The classical usage of the word "observer" very
often led to the same sort of reasoning difficulties that you exhibit,
so they championed a revised definition.
See Figure 2-6 from their textbook.
https://www.eftaylor.com/spacetimephysics/02_chapter2.pdf
For valid pedagogical reasons, their description is a bit verbose.
In Wikipedia I explained the concept with somewhat fewer words as
follows: (probably at least 95% my original wording.)
======================================================================
| "Imagine that the frame under consideration is equipped with a
| dense lattice of clocks, synchronized within this reference frame,
| that extends indefinitely throughout the three dimensions of
| space. Any specific location within the lattice is not important.
| The latticework of clocks is used to determine the time and
| position of events taking place within the whole frame. The term
| observer refers to the whole ensemble of clocks associated with
| one inertial frame of reference.
| "In this idealized case, every point in space has a clock
| associated with it, and thus the clocks register each event
| instantly, with no time delay between an event and its recording.
| A real observer, will see a delay between the emission of a signal
| and its detection due to the speed of light. To synchronize the
| clocks, in the data reduction following an experiment, the time
| when a signal is received will be corrected to reflect its actual
| time were it to have been recorded by an idealized lattice of
| clocks."
======================================================================
In the attached figure, I modified Taylor and Wheeler's diagram by
placing a blue laptop computer in the lower righthand corner to which
all of the clocks in the infinite lattice of clocks report the events
that they have detected.
In a two-dimensional Minkowski diagram, I would diagram this concept
with a dense line of black clocks, with one blue dot representing the
laptop compouter.
In your Figure 4, your propose that D should not be able to send the
signal faster than uâ² = âc^2/v.
*** THIS IS STUPID ***
Frame S' is surrounded by an INFINITE number of other frames traveling
at an INFINITY of different velocities v with respect to S'. It is
IMPOSSIBLE that the speed at which D can send a signal to C should in
any way be dictated by relative speed v of any other frame because
there are an infinite number of different v's to choose from.
If you think otherwise, you are nuts.
In my re-drawing of your Figure 4, D sends an infinite speed signal
to C as measured in the S' frame. The signal includes a GUID, the
globally unique identifier 87f01be4-0a75-4428-b296-409ca23312c4
The RECEIPT of the signal by C is detected up by the infinite array
of clocks, and the time of this event and the GUID are transmitted
to the blue laptop computer which I have drawn off to the side.
Several seconds later, the TRANSMISSION of the signal by D is
detected by the same infinite array of clocks, and the time of this
event and the GUID are transmitted the the blue laptop computer.
An analysis program on the laptop computer notes that the transmission
of the uniquely labeled event occurred after its receipt.
*** THERE ARE NO WORRIES ABOUT WHO IS ADJACENT TO WHOM ***
