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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?


Dec
7
revised Consistent answers in special relativity
(v3.141592: minor copy-editing.)
Dec
7
answered Consistent answers in special relativity
Dec
6
comment Proof in physics
George Smyridis: If you follow the suggestion by @ACuriousMind of reading the Wikipedia article on the Scientific method you may note especially that 1: there is (presently) no (explicit) mentioning of "proving theories" or "falsifying theories" or "testing theories", but rather: hypotheses. But 2: It's still claimed there quite explicitly that definitions (as part of "characterizations"), too, are subject of experimental tests.
Dec
6
comment Proof in physics
anna v: "premises [...] for example in classical physics all variables identified with measurable quantities are continuous" -- The definitions of (how to measure) "quantities", i.e. the main contents and objective of any physics-related theory, include the definitions of their range from the outset. Any "new datum/observation" about some defined quantity cannot falsify its defined range. (The reason to reject/replace "classical physics" is rather its, in hindsight, glaring lack of any careful definitions of "how to measure").
Dec
5
comment Formal definition of an observer?
John O: "two ideas that need to be cleanly separated [...] a grid of sensors, and a single sensor" -- Correct: these are two separate notions. Now, in order to separate them cleanly suitable distinct verbiage (terminology) must be available and used consistently. The key phrase here is "grid of" (or "system") which implies and refers to several distinct elements, possibly in some particular relation(s) between each other. Consequently there must be verbiage/terminology to denote such elements, namely as "sensors", or "material points", or: "observers".
Dec
5
comment Is operator $\hat{O}_{\alpha}:|\phi,\psi\rangle\mapsto |e^{i~\alpha[\phi,\psi]}~\phi,e^{-i~\alpha[\phi,\psi]}~\psi\rangle$ unitary?
ACuriousMind: "The operator, as you try to write it, is not well-defined on the whole of the space [...]" -- Thanks, but I really can't quite follow. But I have to admit that what I had been writing (in the question title) apparently does not quite express some definite operator. So: notation is at least one of my problems ... I've been mislead by the resemblance with $$| \phi \rangle = |v\uparrow + ~w\downarrow\rangle \mapsto \hat O_1 | \phi \rangle = | e^{i \alpha[v, w]} v \uparrow + ~e^{-i \alpha[v, w]} w \downarrow\rangle$$ (where $\hat O_1$ is clearly not unitary, in general).
Dec
5
comment Is operator $\hat{O}_{\alpha}:|\phi,\psi\rangle\mapsto |e^{i~\alpha[\phi,\psi]}~\phi,e^{-i~\alpha[\phi,\psi]}~\psi\rangle$ unitary?
ACuriousMind: Here's an argument by which I've meanwhile convinced myself that, shall we say, "my question, as it stands, is a dead end": namely by failing to construct an explicit example as $$| \phi,\psi \rangle = | a\uparrow\uparrow +b\uparrow\downarrow +f\downarrow\uparrow +g\downarrow\downarrow \rangle \mapsto \hat O_X ~| \phi,\psi \rangle = | j\uparrow\uparrow +k\uparrow\downarrow +p\downarrow\uparrow +q\downarrow\downarrow \rangle.$$ Now, for your argument: [to be continued]
Dec
4
comment Proof in physics
anna v: I appreciate your being responsive. Now: "If [...] a new datum/observation falsifies not only the model but a basic premise/postulate of the theory on which the model is based, [...]" -- Yes, this clarifies your answer: it's now more clearly wrong! Because: "a premise is an assumption that something is true.". Therefore any premise belongs to a particular (falsifiable) model, while the applicable theory is (only) concerned with defining the applicable "something" in the first place, regardless of whether that's then held "true", or "false".
Dec
3
comment Proof in physics
anna v: Your answer carefully describes and distinguishes the notions "theory" vs. "model"; therefore I find it largely very agreeable. But there's one important exception (i.e. one inconsistency in your otherwise great answer): "If the observations and experiments are correctly described by a model based on a theory," -- ... yes ... "and a new datum/observation falsifies the theory, [...]" -- No: if new/additional observations or (derived) measured values are not correctly described by some particular model then they falsify that model; without affecting the underlying theory.
Dec
3
comment Is operator $\hat{O}_{\alpha}:|\phi,\psi\rangle\mapsto |e^{i~\alpha[\phi,\psi]}~\phi,e^{-i~\alpha[\phi,\psi]}~\psi\rangle$ unitary?
ACuriousMind: "If the complex values are non-constant [...] it is not linear" -- 1: Fair enough; but: Is it unitary?. 2: It'd sure be helpful to make more explicit in your answer how non-constant $a, b$ (and/or non-constant $\alpha[ \phi, \psi ]$) lead to operator $\hat O_{\alpha}$ being different from the/any "do-nothing operator". "we really want a map on the projective Hilbert spaces, where each point is a state" -- What's stopping us from considering e.g. $|\phi,\psi\rangle$ and $|e^{i/5}\phi,e^{-i/5}\psi\rangle$ as different "points in projective Hilbert space"?
Dec
3
comment Is operator $\hat{O}_{\alpha}:|\phi,\psi\rangle\mapsto |e^{i~\alpha[\phi,\psi]}~\phi,e^{-i~\alpha[\phi,\psi]}~\psi\rangle$ unitary?
@Sofia: "[...] unitary if its action can be reversed" -- Well, then: does the suggested operator $\hat O_{\alpha}$ satisfy that? For arbitrary "phase functions $\alpha[ \phi, \psi ]$"? Or only for some? Also: note that the operator is defined for a composite state, as I've tried to describe ... (It'd be nice if you could demonstrate and discuss this as an answer. My own focus had rather been on the property of unitary operators to "preserve the inner product".)
Dec
3
comment Is operator $\hat{O}_{\alpha}:|\phi,\psi\rangle\mapsto |e^{i~\alpha[\phi,\psi]}~\phi,e^{-i~\alpha[\phi,\psi]}~\psi\rangle$ unitary?
ACuriousMind: Does your answer really account for the possibility of the complex values $a, b$ being non-constant functions of the individual states (ranging over either of the Hilbert spaces)? Certainly I had not been aiming to construct a "do-nothing operator"; but rather, if you like, a "do-something,-but-gentle-enough" operator. I had been expecting that the suggested "operator $\hat O_{\alpha}$" would be quite different, in general, from the/any "do nothing operator" when it comes to considerations of "linearity" ...
Dec
3
comment Is operator $\hat{O}_{\alpha}:|\phi,\psi\rangle\mapsto |e^{i~\alpha[\phi,\psi]}~\phi,e^{-i~\alpha[\phi,\psi]}~\psi\rangle$ unitary?
ACuriousMind: "What is a "state with two explicit components"?" -- Well, my question is motivated by (and aiming to dispute) this answer; especially its EDIT. Correspondingly, the two components would be Bob's positron state (say $|\psi\rangle$) together with some suitable separate "blank" state (say $|\phi\rangle$, for instance the state of some particular proton). Together, formally: $|\phi,\psi\rangle$. "Do you mean an element of the tensor product?" -- You tell me, please ...
Dec
3
asked Is operator $\hat{O}_{\alpha}:|\phi,\psi\rangle\mapsto |e^{i~\alpha[\phi,\psi]}~\phi,e^{-i~\alpha[\phi,\psi]}~\psi\rangle$ unitary?
Dec
2
comment Question about the no-clone theorem
Martin: "I'm afraid your map is just not linear" -- That's right. (The equations of my previous comment should be helpful to recognize that.) I still need to convince myself that the suggested operator/map is (therefore) not unitary either; referring in particular to this Wikipedia derivation. But anyways, that's a wrinkle I had not considered. So: thanks for bringing this to my attention.
Dec
1
comment Question about the no-clone theorem
Martin: "what your operator is supposed to do" -- Bob would use $\hat B$ (which is not "cloning") to estimate "his (positron) state $\psi_p$". Note: $$\hat B |\text{blank},\text{exp[} i\alpha\pi\text{]}\text{up}\rangle := |\text{up},\text{exp[}i\alpha\pi\text{]}\text{up}\rangle $$,$$\hat B |\text{blank},\text{exp[} i\beta\pi\text{]}\text{down}\rangle := |\text{down},\text{exp[}i\beta\pi\text{]}\text{down}\rangle $$,$$ \hat B |\text{blank},\text{exp[}i\gamma\pi\text{]}(\text{up}\pm\text{down})\rangle :=|(\text{up}+\text{down}),\text{exp[} i\gamma\pi\text{]}(\text{up}\pm\text{down})\rangle $$.
Dec
1
comment Question about the no-clone theorem
Martin: "a version of this instantaneous communication protocol [...] Alice and Bob share a maximally entangled [...] pair and we assume that they are totally anti-correlated (i.e. choosing a basis, if the outcome is up for Alice, it will be down for Bob [...]) Now Alice chooses one of the bases and measures the electron [...] If Bob could clone the particle [then ...]" -- I wonder whether Bob is allowed to use operator $\hat B$ such that $$ \hat B|\text{blank},~\psi_p\rangle=\hat B|\text{blank},(p_u~\text{up}+p_d~\text{down})\rangle :=|(|p_u|~\text{up}+|p_d|~\text{down}),\psi_p\rangle$$ ...
Nov
28
awarded  Inquisitive
Nov
28
awarded  Curious
Nov
27
answered Confusion about Length Contraction (ex in Muon decay)