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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
6
awarded  Organizer
Apr
6
comment Travel at the speed of light
WetSavannaAnimal: "I categorically did not say the acquired inertia was frame dependent." -- Then the start of your first comment "That of course depends on the reference frame." was a poor choice of wording. To repeat: I object to the phrase ""rest mass" is acquired by a particle with a rest mass of nought when [...] confined" of your answer above; and I ask you again which value of wavelength $\lambda$ "in the box picture" you suppose for a confined photon of energy $E = h \, \nu$.
Apr
6
awarded  Custodian
Apr
6
comment Why does a cup with 100 g water float when placed on another cup with 50 g of water?
@DumpsterDoofus: Since you seem to have the capability to make and include drawings in your answer it might be nicer (and more inviting to practical replications) to draw two equal, slightly conical cups; both similar to this one, together with the appropriate amount of water included. And the corresponding discussion might benefit from mentioning (and even using) "hydrostatic pressure" ...
Apr
6
comment Distance Between Two Photons Calculated in Different Inertial Frames
Nuclear_Wizard: "it's fine for an interval to move at c" -- An interval moving?? A pair of events moving?? Anyways, my main objection was against the OP's question itself speaking of "distance between photons" (I blame the text being studied; not the messenger(s).) Distance is between "[pairs of] points [as pairs of] stationary particles" you considered; but not "even when the points are travelling at the speed of light." Now: "[...] constant in all inertial reference frames, there cannot exist a frame where the speed of a photon is zero" -- Why not some "non-inertial frame"?!?
Apr
6
comment Distance Between Two Photons Calculated in Different Inertial Frames
Nuclear_Wizard: "imagine them as stationary particles" -- +1 for matching Einstein's "space-time propositions amount to the determination of space-time coincidences {such as} encounters between two or more recognizable material points". "photons do not have a rest frame." -- Right. But: "light speed is constant in all reference frames" -- No: in inertial frames. Finally: "even when the points are travelling at the speed of light." -- -1.
Apr
6
comment Travel at the speed of light
WetSavannaAnimal: "[...] depends on the reference frame" -- 1. My comment/question referred to $\nu$ and $\lambda$ "measured by the box walls; incl. equipment at rest wrt. both box walls". (I presumed the box walls remaining at rest to each other.) That seems pretty much implied by your identifying "$E/c^2$" as "the box's inertia increase". So my question stands; "in the box picture". 2. Whatever is frame dependent surely doesn't deserve to be called "some sort of invariant mass". Does your questionable "acquired "rest mass" of the photon" not mean "acquired invariant mass"??
Apr
5
comment Travel at the speed of light
WetSavannaAnimal: "[...] light into a perfectly reflecting box, the box's inertia increases by $E/c^2$, where $E$ is the energy content" -- Fine. ""rest mass" is acquired by a particle with a rest mass of nought when that particle is confined" -- ?!? By Planck's analysis, the frequency of this photon is $\nu = E/h$. Now, what do you suggest is its wavelength: $$\lambda = \frac{c}{\nu} \times \sqrt{1 - \left(\frac{m \, c^2}{E}\right)^2 }$$ for some non-zero "acquired rest mass $m$"? Even for $m = E/c^2$?? p.s. virtual +1 for excellent use of "nucleusses" in a sentence about nuclei.
Apr
5
comment Can we think of gravity as space itself moving?
John Rennie: "The answer is yes [... we can think of gravity as space itself moving]" -- It is answers like these which seem to inspire (follow-up) questions like those: "Can space be accelerated and does it experience length contraction?" (or, indeed, the follow-up question asked by user43783 in the preceding comment).
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