simultaneity

………. and a bag of $latex \text{ c }$s

EMROPTCE-3: Let’s calculate! — “Constraints” vs. “Degrees of freedom”

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Keeping in mind the caveats sketched in the preceding part 2 of this blog post series let’s now focus on the specific basic “construction \frak C” being proposed for expressing mutual rest of participants through coincidence events; namely the ping relations between four “corners” (A, B, F, G), and six “half-way-betweens” (J, K, P, Q, U, W):

Require:

– for any signal event in which A had taken part (and had thus stated a signal indication), A observed in coincidence that

* B and F and G had observed this signal event, and

* J had observed that A had observed that J had observed this signal event, and

* likewise for K and P;

– for any signal event in which B had taken part, B observed in coincidence that

* A and F and G had observed this signal event, and

* J had observed that B had observed that J had observed this signal event, and

* likewise for Q and U;

– etc. for F and for G;

– for any signal event in which J had taken part, J observed in coincidence that

* A and B and K and P and Q and U had observed this signal event;

– etc. for K, P, Q, U and for W.

(At this point some may ask for some additional guidance being provided through a sketch. But I’m hesitant, for fear of a sketch being even more “leading” than referring to A, B, F, G as “corners”, and to J, K, P, Q, U, W as “half-way-betweens”. Let’s say I’ll add a sketch here if and when WordPress supports(1) drawing with PSTricks commands.)

Perhaps the best and decisive illustration is by noticing and comprehending that the set of these requirements provides constraints on the geometric relations between these 10 participants (A through W, a.k.a. “set \mathcal T_{\frak C}“), and by spelling out concretely how many actual contraints are thereby obtained (which can be expressed as actual algebraic “constraint equations”, as they will appear later in this series of blog posts):

for any one signal event in which participant A had taken part (and had thus stated a signal indication) A is supposed to determine one coincidence event in which A’s observations of ping echos regarding six distinct other participants (namely concerning B, F, G, J, K, and P) were made strictly together (and not in any way “sequentially”). This gives 5 “actual constraints” provided by participant A, for any one signal event in which participant A had taken part. The exact same count applies to the constraints provided by B, F, and G, respectively, since the “requirements on” B, F, and G are exactly similar to those on A.

The “requirements on” J, K, P, Q, U, and W are likewise exactly similar to each other (but somewhat different from those on A, B, F, and G). However, incidentally, J (for instance) is also supposed to determine one coincidence event in which J’s observations of ping echos (from any one signal event in which J had taken part) regarding six distinct other participants (namely concerning A, B, K, P, Q, and U) were made strictly together (and not in any way “sequentially”). Therefore J provides 5 “actual constraints” for any one signal event, too; and K, P, Q, U, and W likewise.

Now (“Let’s calculate!”) try to compare the number of these constraints to the number of “degrees of freedom” of these 10 participants (set \mathcal T_{\frak C}) under consideration (which (optimistically!) are provided and indeed limited by their characterizations with respect to any suitable “tentative inertial frame \mathcal F_{\frak C*}“):

Considering 10 signal events (which must be distinct events, because –oh!, seems I forgot to mention — the 10 participants of set \mathcal T_{\frak C} ar of course supposed to never meet each other) that’s (optimistically!?) \approx 40 initial degrees of freedom. (Strictly, it might actually some fewer; “effectively” perhaps rather \approx 40 - 4 - 3 = \approx 33; or even only \approx 33 - 6 = \approx 27, considering that the initial signal events in which J, K, P, Q, U, and W took part, respectively, are not independent of the initial signal events pertaining to A, B, F, or G).

Since any subsequent events are not completely independent of the initial signal events under consideration, but related by (mutual) observation and (even) pings (a.k.a. “signal round-trips”), the number of “degrees of freedom” increases by 3 (rather than by 4) for each participant at each subsequent events.

For the explicit (basic) set of “requirements” as quoted above, the count of “degrees of freedom vs. constraints” therefore stands (roughly, but optimistically) at

27 + 2*3*4 + 2*3*6 - 50 = 27 + 24 + 36 - 50 = 37,

where the number 2*3*4 = 24 refers to the “degrees of freedom” of A, B, F, and G in two events each (following the initial signal events), and the number 2*3*6 = 36 refers to the “degrees of freedom” of J, K, P, Q, U, and W, in two subsequent events each.
Clearly, that’s “too few constraints” to draw some distinctive conclusion such as the proposed “mutual rest”.

However: as noted earlier already, these “requirements” (i.e. the “contruction \frak C“) can be “extended or refined (iterated, combined)“; it can be repeated over and over because any event in which one of the ten participants took part may in turn serve as initial signal event for the next/appended iteration of the “construction”. The important point is: the number of “constraints” obtained by appending successive “construction” steps grows more rapidly than the corresponding number of “degrees of freedom”; specificly as

27 + 2*3*4 + 3*3*6 - 50 - 5*6 = 37 + 18 - 30 = 25,

27 + 3*3*4 + 3*3*6 - 50 - 5*6 - 5*4 = 25 + 12 - 20 = 17,

27 + 3*3*4 + 4*3*6 - 50 - 2*5*6 - 5*4 = 17 + 18 - 30 = 5,

27 + 4*3*4 + 4*3*6 - 50 - 2*5*6 - 2*5*4 = 5 + 12 - 20 = -3

There are already (optimistically) more “constraints” than “degrees of freedom”; meaning more practically “more equations than unknowns”, which indicates that there is some “solution” to be evaluated, as I propose and hope.

A “practical difficulty” is of course immediately obvious, too: any desired concrete “solution” would seemingly have to be obtained from a system of about one hundred or more “constraint equations”. (That’s certainly a bit too much of a problem for an old laptop running an almost as old version of Mathematica (TM) …)

[To be continued.]

(1: Instant PSTricks test: \begin{pspicture}(5,5)  %% Triangle in red:  \psline[linecolor=red](1,1)(5,1)(1,4)(1,1)  %% Bezier curve in green:  \pscurve[linecolor=green,linewidth=2pt,%  showpoints=true](5,5)(3,2)(4,4)(2,3)  %% Circle in blue with radius 1:  \pscircle[linecolor=blue,linestyle=dashed](3,2.5){1}  \end{pspicture}
)

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