On Wed, Mar 5, 2014 at 1:27 PM, Edgar L. Owen <edgaro...@att.net> wrote:
> Jesse, > > Yes, you are right. I phrased it incorrectly. > > What I meant to say was not that each individual view was somehow > weighted, but that all views considered together would tend to cluster > around my results for any distance and motion difference pairs. > Too vague. What does "all views considered together" mean mathematically, if not a weighted average using some specific weighting function? > In other words there would be a lot more views that were close to my > solution, than views that were far from my solution. > How do you count "more" when there are a continuous infinity of frame's views? The only way to "count" different subsets of an infinite set is using some sort of "measure" function (see https://en.wikipedia.org/wiki/Measure_(mathematics) ), which is equivalent to a weighting function--whatever you choose to call it, the idea would be that if you want to compare the "number" or "weight" of frames with velocity between v1 and v2 (the velocities defined relative to some other specific frame, of course) vs. the "number" or "weight" with velocity between v3 and v4, you use a measure/weight function W(v) which gives a value for every specific frame velocity v, and you integrate the function W(v) from v1 to v2, and compare the result to integrating W(v) between v3 and v4 (and if you want to do a "weighted average" of some specific quantity Q(v) that varies from one frame to another, like the amount by which two clocks are out-of-sync, you would integrate Q(v)*W(v) over the frame velocity interval over which you want the weighted average of Q). > And that we can see this because, as you yourself pointed out, as distance > separation and relative motion differences decrease all other frame views > DO tend to converge on my results. > > Thus the aggregate WEIGHT OF ALL VIEWS tends to converge on my solution, > which is what I meant to say. Sort of like a Bell curve distribution with a > point at top representing my solution > > Would you agree to that? > In the case of two clocks at rest and synchronized in a common frame, the only "convergence" I think we agree on is if you consider a series of cases where the distance between the two clocks approaches 0, or where the velocity of the frame whose opinion you're considering relative to the rest frame of the two clocks approaches 0 (which may be what you meant by "as distance separation and relative motion differences decrease all other frame views DO tend to converge on my results"). If you are talking about a FIXED value for the distance between two clocks in their rest frame, and doing a weighted average of larger and larger sets of different frame's opinions about the time difference between the two clocks (eventually including frames with a very large velocity relative to the clocks), then what value this average would "converge" to would depend entirely on the weighting function. As I said a weighting function that looks "uniform" in one frame (equal velocity intervals have equal weight when you integrate over the integral) will look non-uniform in other frames, so I can't see a way to define a weighting function that doesn't privilege one frame at the outset. Jesse -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
