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The following questions (in no particular order) which I had submitted have been "removed from PSE for reasons of moderation":

  1. Which geometric relations obtain between two distinct rest systems?

Consider, as a thought experiment, a set of participants who measure throughout the experiment having been at rest to each other; among them explicitly participants ${\mathbf A}$, ${\mathbf B}$ and ${\mathbf F}$ who determine the ratios of their (chronogeometric) distances between each other as real number values $\frac{{\mathbf A}{\mathbf B}}{{\mathbf A}{\mathbf F}}$, $\frac{{\mathbf B}{\mathbf F}}{{\mathbf A}{\mathbf F}}$, and $\frac{{\mathbf A}{\mathbf B}}{{\mathbf B}{\mathbf F}} = \frac{{\mathbf A}{\mathbf B}}{{\mathbf A}{\mathbf F}} / \frac{{\mathbf B}{\mathbf F}}{{\mathbf A}{\mathbf F}}$.

Further let there be another set of participants (of which neither ${\mathbf A}$, nor ${\mathbf B}$, nor ${\mathbf F}$ are a member) who measure throughout the experiment having been at rest to each other as well; among them ${\mathbf J}$, ${\mathbf K}$ and ${\mathbf Q}$, who determine the ratios of their (chronogeometric) distances between each other as real number values $\frac{{\mathbf J}{\mathbf K}}{{\mathbf J}{\mathbf Q}}$, $\frac{{\mathbf K}{\mathbf Q}}{{\mathbf J}{\mathbf Q}}$, and $\frac{{\mathbf J}{\mathbf K}}{{\mathbf K}{\mathbf Q}} = \frac{{\mathbf J}{\mathbf K}}{{\mathbf J}{\mathbf Q}} / \frac{{\mathbf K}{\mathbf Q}}{{\mathbf J}{\mathbf Q}}$,

such that

  • ${\mathbf J}$ passed ${\mathbf A}$, then passed ${\mathbf B}$,

  • ${\mathbf A}$ passed ${\mathbf J}$, then passed ${\mathbf K}$,

  • ${\mathbf Q}$ passed ${\mathbf F}$, in coincidence with ${\mathbf Q}$ and ${\mathbf F}$ observing ${\mathbf J}$ and ${\mathbf A}$ having passed each other,

  • ${\mathbf B}$ and ${\mathbf F}$ determined that ${\mathbf B}$'s indication of the passage of ${\mathbf J}$ was simultaneous to ${\mathbf F}$'s indication of the passage of ${\mathbf Q}$, and

  • ${\mathbf K}$ and ${\mathbf Q}$ determined that ${\mathbf K}$'s indication of the passage of ${\mathbf A}$ was simultaneous to ${\mathbf Q}$'s indication of the passage of ${\mathbf F}$.

Question:
Is thereby guaranteed that for these distance ratios obtains

(1)
$\frac{{\mathbf A}{\mathbf B}}{{\mathbf A}{\mathbf F}} = \frac{{\mathbf J}{\mathbf K}}{{\mathbf J}{\mathbf Q}}$ ?,

and (moreover)

(2)
$\left( \left(\frac{{\mathbf B}{\mathbf F}}{{\mathbf A}{\mathbf F}}\right)^2 + 1 - \left(\frac{{\mathbf A}{\mathbf B}}{{\mathbf A}{\mathbf F}}\right)^2 \right) \left( \left(\frac{{\mathbf K}{\mathbf Q}}{{\mathbf J}{\mathbf Q}}\right)^2 + 1 - \left(\frac{{\mathbf J}{\mathbf K}}{{\mathbf J}{\mathbf Q}}\right)^2 \right) = 4 \left( 1 - \left( \frac{{\mathbf A}{\mathbf B}}{{\mathbf A}{\mathbf F}} \right) \left( \frac{{\mathbf J}{\mathbf K}}{{\mathbf J}{\mathbf Q}} \right) \right)$ ?

Or otherwise:
What could be concluded if (1) and/or (2) were not found satisfied?


Apr
5
comment Are signal fronts in a beam not at rest to each other?
Evaluating instead the elapsed time (of one front in the beam) between some event (at which some other photon is emitted/passing as well) and just some other event (at which just some other photon is incoming/passing, too) the result is surely: Null. So my argumentation depends on not having anything to be evaluated in the first place. Now, I completely accept that Synge may not have dreamt about applying his "rigidity" definition to signal fronts (photons). (And to boot: It's even stated in the section titled "Born Rigidity"!) Nevertheless, the argument can be made as shown.
Apr
5
comment Are signal fronts in a beam not at rest to each other?
auxsvr: "Synge writes about measuring the elapsed time between sending and receiving a photon" -- To be precise (ABAIU): the elapsed time between sending a photon and (in general separately) receiving not just some other photon(s), but the scattered or reflected one(s), in response to what had been sent. "how do you measure that on a photon?" -- My point is: this questions doesn't even arise; because there is no pair of events (one "sending" and a/the corresponding one "receiving") identifiable to begin with, for which to try to evaluate "elapsed time". Otherwise ...[contd.]
Apr
5
comment Angles between axes after Lorentz transformations
Oops! -- In the last equation I forgot to put the "$(1 - \beta^2_{P}[ Q ])$" in the denominator under the square root; sorry about that. So the "$\sqrt{1 - u^2}$" happens to cancel out, and the whole expression simplifies to: $$\text{Cos}[ \angle_{N}[P, Q]] = \frac{v}{\sqrt{u^2 + v^2 - u^2 v^2} }.$$ For vanishing $v$ this makes (sensibly) $\text{Cos}[ \angle_{N}[P, Q]] = 0$. Correspondingly therefore $$\text{Cos}[ \angle_{Q}[N, P]] = \frac{u}{\sqrt{u^2 + v^2 - u^2 v^2} },$$ which for vanishing $v$ makes $\text{Cos}[ \angle_{Q}[N, P]] = 1$, as surely expected.
Apr
5
comment Are signal fronts in a beam not at rest to each other?
auxsvr: "and a device to send and receive photons" -- For each event in which any one front had a part there's an entire "light cone" of events such that, at least kinematically, photons from the "past half-cone" could've been received by the front, or photons could've been sent from it to the "future half-cone". Certain is only: There's no such "signal round trip" between successive fronts of the same beam. As it happens: that's sufficient to answer my question. "Even Synge says so, the curves must be time-like" -- The "Remark" even quoted it. I just didn't have to use that subclause.
Apr
5
comment Are signal fronts in a beam not at rest to each other?
auxsvr: "You're confused." -- Well, at least one of us seems confused. "it is not that the wave fronts do not exchange photons that makes what you ask not make sense" -- The sheer formal logic of my argumentation seems stringent enough, for starters. (And terminology-wise I rather stick to "signal front".) So your point is(?) 1.: "you'll need at least a clock" -- The possibility to assign parametrizations to a signal front (a.k.a. "null-trajectory") has apparently been considered. And 2.: ... [to be contd.]
Apr
5
comment Query into the cumulative velocity of mounted platforms
Easy, big-Russian-Doll-fella! ;) Glad to help. p.s. To-do list: 1. Repeat my question (you know, that one) about how to derive these especially useful formulas. 2. Think really hard about (possible values of) $$ \angle_{N}[P, Q]] + \angle_{P}[Q, N]] + \angle_{Q}[N, P]] $$ ...
Apr
5
comment Angles between axes after Lorentz transformations
You may find useful that in general, for systems $N$, $P$, $Q$: $$1 - (\beta_{P}[ N ] \, \, \beta_{P}[ Q ] \, \, \text{Cos}[\angle_{P}[N, Q]]) = \sqrt{ \frac{(1 - \beta^2_{P}[ N ]) \, (1 - \beta^2_{P}[ Q ])}{(1 - \beta^2_{N}[ Q ])}}.$$ Setting $\beta_{P}[ N ] = v$, $\beta_{P}[ Q ] = u$ and $\text{Cos}[ \angle_{P}[ N, Q ] ] = 0$ then $$\beta_{N}[ Q ] = \sqrt{u^2 + v^2 - u^2 v^2}$$ and $$\text{Cos}[\angle_{N}[P, Q]] = \frac{\left(1 - (1 - v^2) \sqrt{1 - u^2}\right)}{ v \, \sqrt{u^2 + v^2 - u^2 v^2} }.$$ However, of course that's got nothing to do with coordinates or coordinate transformations...
Apr
5
comment Query into the cumulative velocity of mounted platforms
@DumpsterDoofus: Didn't bother to watch (sorry). At any rate: eventually, I meant something more like that.
Apr
5
revised Query into the cumulative velocity of mounted platforms
(v3.141: cobbling to stick to my lasts.)
Apr
5
comment The scissor paradox: can we pass the information faster than light?
user43796: "When it [the thread] is cut [by the blades], the tip [of the closed scissors] may never tell." -- Since the cut event was supposed to be a sufficiently visible "tell-tale sign", then either the tip can tell, or there was clearly no information passed from thread to tip. "but the information still seems to travel faster than light" -- 1.: Which information might be "passed on by geometrical point $A$ at all? (The color of the yarn?? The outcome, "on which side fell the loose end"?? ...) And 2.: Refer to the main point on top of my answer.
Apr
5
answered Query into the cumulative velocity of mounted platforms
Apr
4
accepted Are signal fronts in a beam not at rest to each other?
Apr
4
comment The scissor paradox: can we pass the information faster than light?
user43796: "Point A and photon travel for a same distance, the photon gets there later." -- ok, let me try again to catch that in my own words: There's a thread placed between/across the still open scissor blades ("close to the start"); the scissors close --> the thread is cut (that's widely visible!). So when (or rather: how) will the scissor tips first learn about that cut event? Well: not necessecarily just because the tips touched each other (i.e. point $A$ had gotten there). Rather: they first learned it when they said they learned it.
Apr
4
comment The scissor paradox: can we pass the information faster than light?
user43796: "[...] Wherever A is, the light will be blocked" -- Allright, I think I got that setup: just e.g. a series (or "bank") of cameras with the scissor blades making a shutter. They're watching some movie. The shutter will close. (According to your setup, i.e. "sequentially rapidly progressing"; or, just as another consideration: the shutter might instead close in front of individual cameras in some arbitrary order.) Now: Is there information being passed from one camera to the other?? How should one camera know what "the final scene" was, that some other camera saw before being shut?
Apr
4
revised The scissor paradox: can we pass the information faster than light?
(v3.14159265: better description of motion in response to signal; which is not perfectly rigid ...)
Apr
4
comment The scissor paradox: can we pass the information faster than light?
user43796: "Point A could exceed the speed of light since it's a geometric object" -- Yes; similar to a "(wave) phase". Therefore its speed is phase speed. Another example is the "(virtual) motion" of a laser pointer dot over a screen. "but then what if I place a detector close enough to where A starts to move?" -- Yes, $A$ can and will arrive at (and pass by) detectors. But: thereby there's no information/signal being passed from fulcrum $O$ to any of those detectors (or likewise from one detector to another).
Apr
4
revised The scissor paradox: can we pass the information faster than light?
(v3.1415926: having proof-read to sloppily ...)
Apr
4
answered The scissor paradox: can we pass the information faster than light?
Apr
4
comment Query into the cumulative velocity of mounted platforms
@DumpsterDoofus: "A railgun with 100,000 stages as drawn would probably end up the size of the solar system, so I think the possibility can be ruled out in the near future, given the current state of space research funding :)" -- Well, projecting the generally expected rate of proposal maturation h_t_a might instead take steps through the long tail of thermal distributions and make his scheme fit snugly in the solar core. p.s. Just noticed jazzwhiz's contribution So: your turn?
Apr
4
comment Query into the cumulative velocity of mounted platforms
@DumpsterDoofus DumpsterDoofus: "That fails once you get to relativistic speeds, but typical rail-gun velocities are between 10000 and 100000 times slower than light, so you can safely ignore relativity" -- Then let's consider the conservative "$\beta_i := 10^{(-5)}$"; or, as hello_there_andy might put it: "v_i := 3000 ms-1". So: What exactly do we tell him about "Russian Dolls" with 100000 or even more such stages ?!? @hello_there_andy -- I remember that drawing (or sth similar?), too. You might find it helpful for now to get rid of any energy considerations and focus on kinematics.