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Given a flat region of spacetime as set $\mathcal S$ of events together with values of spacetime intervals (up to a common non-zero constant) for each pair of events, $s^2 : \mathcal S \times \mathcal S \rightarrow \mathbb R$,
and considering two (distinct) participants $A$ and $B$ contained in this region(which may or may not have been and remained at rest to each other throughout) where $\mathcal E_A \equiv \{ ... \varepsilon_{AF} ... \varepsilon_{AJ} ... \varepsilon_{AP} ... \} \subset \mathcal S$ denotes the set of (coincidence) events in which $A$ took part, and $\mathcal E_B \equiv \{ ... \varepsilon_{BG} ... \varepsilon_{BK} ... \varepsilon_{BQ} ... \} \subset \mathcal S$ the set of (coincidence) events in which $B$ took part,

can conditions be expressed explicitly in terms of the given spacetime interval values such that
$A$ and $B$ are said to have been and remained "at rest to each other" if and only if all these conditions are satisfied?

user12262
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  • @jld: "Did you invent this notation yourself or did you get it from some other source?" -- The symbol $s^2$ for denoting spacetims intervals is standard (cmp. link in the OP question), as is the notation expressing it being a function from pairs (of elements of a set) into the set of real numbers; cmp. https://en.wikipedia.org/wiki/Distance#General_metric [... continued] – user12262 Apr 14 '16 at 06:06
  • The use of square brackets for enclosing (a comma-separated list of) arguments to a function corresponds to Mathematica (TM) style (StandardForm); see e.g. http://mathematica.stackexchange.com/questions/20166/converting-to-inputform-or-standardform-without-deleting-comments (Btw., this notation style allows to reserve parentheses for the purpose of grouping alone.) [... continued] – user12262 Apr 14 '16 at 06:07
  • Finally (I suppose): I don't know any precedence, other than my own writing, for the notation of events (i.e. the arguments of the spacetime interval funct) I used above (which explicitly lists the participants having been coincident). However: it aims to adhere explicitly to Einstein's prescription that "All our space-time verifications invariably amount to the determination of space-time coincidences {such as} meetings between two or more material points". – user12262 Apr 14 '16 at 06:09

1 Answers1

-1

Conditions for participants $A$ and $B$ having been and remained "at rest to each other" are

(1):
The events in which $A$ took part are straight to each other; and likewise, separately: the events in which $B$ took part are straight to each other; i.e. explicitly $$\forall~\varepsilon_{AF}, \varepsilon_{AJ}, \varepsilon_{AP} \in \mathcal E_A : $$ $$2~s^2[~\varepsilon_{AF}, \varepsilon_{AJ}~]~s^2[~\varepsilon_{AF}, \varepsilon_{AP}~] + 2~s^2[~\varepsilon_{AF}, \varepsilon_{AJ}~]~s^2[~\varepsilon_{AJ}, \varepsilon_{AP}~] + 2~s^2[~\varepsilon_{AF}, \varepsilon_{AP}~]~s^2[~\varepsilon_{AJ}, \varepsilon_{AP}~] = (s^2[~\varepsilon_{AF}, \varepsilon_{AJ}~])^2 + (s^2[~\varepsilon_{AF}, \varepsilon_{AP}~])^2 + (s^2[~\varepsilon_{AJ}, \varepsilon_{AP}~])^2,$$ and likewise $$\forall~\varepsilon_{BG}, \varepsilon_{BK}, \varepsilon_{BQ} \in \mathcal E_B : $$ $$\!2~s^2[~\varepsilon_{BG}, \varepsilon_{BK}~]~s^2[~\varepsilon_{BG}, \varepsilon_{BQ}~] \! + \! 2~s^2[~\varepsilon_{BG}, \varepsilon_{BK}~]~s^2[~\varepsilon_{BK}, \varepsilon_{BQ}~] \! + \! 2~s^2[~\varepsilon_{BG}, \varepsilon_{BQ}~]~s^2[~\varepsilon_{BK}, \varepsilon_{BQ}~] = (s^2[~\varepsilon_{BG}, \varepsilon_{BK}~])^2 + (s^2[~\varepsilon_{BG}, \varepsilon_{BQ}~])^2 + (s^2[~\varepsilon_{BK}, \varepsilon_{BQ}~])^2,$$

(2):
$A$ and $B$ observed pings (signal front roundtrips) between each other throughout; i.e. explicitly $$ \forall~\varepsilon_{AF} \in \mathcal E_A \exists~\varepsilon_{AJ} \in \mathcal E_A, \varepsilon_{BI} \in \mathcal E_B :$$ $$s^2[~\varepsilon_{AF}, \varepsilon_{BI}~] = 0, \qquad s^2[~\varepsilon_{BI}, \varepsilon_{AJ}~] = 0,$$
and likewise $$ \forall~\varepsilon_{BG} \in \mathcal E_B \exists~\varepsilon_{BK} \in \mathcal E_B, \varepsilon_{AH} \in \mathcal E_A :$$ $$s^2[~\varepsilon_{BG}, \varepsilon_{AH}~] = 0, \qquad s^2[~\varepsilon_{AH}, \varepsilon_{BK}~] = 0,$$

(3):
$A$'s ping durations wrt. $B$ are constant; and likewise, separately $B$'s ping durations wrt. $A$ are constant; i.e. explicitly $$ \forall~\varepsilon_{AF}, \varepsilon_{AP} \in \mathcal E_A : \exists~ \varepsilon_{AJ}, \varepsilon_{AU} \in \mathcal E_A, \exists ~\varepsilon_{BI}, \varepsilon_{BT} \in \mathcal E_B |$$ $$s^2[~\varepsilon_{AF}, \varepsilon_{BI}~] = 0, \!\! \qquad \!\! s^2[~\varepsilon_{BI}, \varepsilon_{AJ}~] = 0, \!\! \qquad \!\! s^2[~\varepsilon_{AP}, \varepsilon_{BT}~] = 0, \!\! \qquad \!\! s^2[~\varepsilon_{BT}, \varepsilon_{AU}~] = 0 \implies$$ $$\tau A_F^J = \tau A_P^U,$$ where due to (1) holds $$\frac{\tau A_F^J}{\tau A_P^U} = \sqrt{\frac{s^2[~\varepsilon_{AF}, \varepsilon_{AJ}~]}{s^2[~\varepsilon_{AP}, \varepsilon_{AU}~]}};$$ and likewise $$ \forall~\varepsilon_{BG}, \varepsilon_{BQ} \in \mathcal E_B : \exists~ \varepsilon_{BK}, \varepsilon_{BV} \in \mathcal E_B, \exists~ \varepsilon_{AH}, \varepsilon_{AS} \in \mathcal E_A |$$ $$s^2[~\varepsilon_{BG}, \varepsilon_{AH}~] = 0, \qquad s^2[~\varepsilon_{AH}, \varepsilon_{BK}~] = 0, \qquad s^2[~\varepsilon_{BQ}, \varepsilon_{AS}~] = 0, \qquad s^2[~\varepsilon_{AS}, \varepsilon_{BV}~] = 0 \implies$$ $$\tau B_G^K = \tau B_Q^V,$$ where due to (1) holds $$\frac{\tau B_G^K}{\tau B_Q^V} = \sqrt{\frac{s^2[~\varepsilon_{BG}, \varepsilon_{BK}~]}{s^2[~\varepsilon_{BQ}, \varepsilon_{BV}~]}},$$

(4):
$A$'s constant ping duration wrt. $B$ is equal to $B$'s constant ping durations wrt. $A$; i.e. explicitly $$ \forall~\varepsilon_{AF} \in \mathcal E_A, \varepsilon_{BG} \in \mathcal E_B : \exists~\varepsilon_{AH}, \varepsilon_{AJ} \in \mathcal E_A, \exists~ \varepsilon_{BI}, \varepsilon_{BK} \in \mathcal E_B |$$ $$s^2[~\varepsilon_{AF}, \varepsilon_{BI}~] = 0, \qquad s^2[~\varepsilon_{BI}, \varepsilon_{AJ}~] = 0, \qquad s^2[~\varepsilon_{BG}, \varepsilon_{AH}~] = 0, \qquad s^2[~\varepsilon_{AH}, \varepsilon_{BK}~] = 0 \implies$$ $$\tau A_F^J = \tau B_G^K,$$ where due to (1) holds $$\frac{\tau A_F^J}{\tau B_G^K} = \sqrt{\frac{s^2[~\varepsilon_{AF}, \varepsilon_{AJ}~]}{s^2[~\varepsilon_{BG}, \varepsilon_{BK}~]}}.$$

user12262
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