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. In other words there 
would be a lot more views that were close to my solution, than views that 
were far from my solution. 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 

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?


On Wednesday, March 5, 2014 11:00:19 AM UTC-5, jessem wrote:
> On Wed, Mar 5, 2014 at 8:38 AM, Edgar L. Owen <<javascript:>
> > wrote:
>> Jesse,
>> Here's another point for you to ponder:
>> You claim that all frame views are equally valid. What would you say the 
>> weighted mean of all frame views is?
> Weighted how? I can't see any "weighing" that doesn't itself depend on 
> privileging one frame over others. For example, suppose I label frames 
> using velocity relative to my rest frame, and use a uniform distribution on 
> velocity values as my weight function, which implies that the collection of 
> frames with velocities between 0.1c and 0.1c + dV will have the same total 
> weight as the collection of frames with velocities between 0.9c and 0.9c + 
> dV, since these are equal-sized velocity intervals (for example, if 
> dV=0.05c then we are looking at the frames from 0.1c to 0.15c, and the 
> frames from 0.9c to 0.95c). But if we look at all the frames in these two 
> intervals, and translate from their velocities relative to ME to their 
> velocities relative to another frame B that is moving at say 0.8c relative 
> to me, then these two bunches of frames do NOT occupy equal-sized velocity 
> intervals when we look at their velocities relative to frame B (an interval 
> from 0.1c to 0.15c in my frame translates to the interval from -0.761c to 
> -0.739c in B's frame, while an interval of 0.9c to 0.95c in my frame 
> translates to an interval from 0.357c to 0.625c in B's frame). So if we 
> "weigh" them equally using MY velocity labels, that would translate to an 
> unequal weighing relative to B's velocity labels, so we are privileging my 
> frame's definitions over the definitions of other frames like B.
>> I would suspect that it converges towards my solution. It is clear from 
>> your own analysis that it does converge to my solution as separation and 
>> relative motion diminishes, so I strongly suspect it converges towards my 
>> solution in all cases.
>> Correct? And if so I would argue that this also tends to validate my 
>> solution as the actual correct 1:1 correlation of proper ages, even though 
>> I agree completely that all observers cannot direct observe this 
>> correlation...
>> In fact this is tantalizingly similar to the notion of a wavefunction 
>> representing the probabilities of all possible locations of a particle. If 
>> we take all possible frame views as a continuous 'wavefunction' of the 
>> actual age correlation can we begin to assign probabilities based on their 
>> weighted mean, and if so isn't that going to be my solution?
> This doesn't really help your case unless you can find a "weight" function 
> for the continuous infinity of different possible frames that doesn't 
> itself privilege one frame's definitions from the start.
> Jesse

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