Pioneering the Applications of Interphasal Resonances 
http://tech.groups.yahoo.com/group/teslafy/

     On Wednesday, December 20, 2017 5:26 PM, Quora <mention-nore...@quora.com> 
wrote:
 

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| Guido Wuyts mentioned you in an answer to "How would you apply the time 
dilation equation when both objects are moving in opposite directions of each 
other at >0.5 speed of light?" |
| (May 28, 2017)As a complement to Bryan Pawlina’s answer, and to illustrate 
that time dilation is a mutual (or reciprocal) feature between any two 
observers, take a look at this picture:(from my SRT page MySRT and TwinParadox 
)Thanks to relativistic simultaneity, both the moving observer and the one “at 
rest” observe reciprocally dilation of each other’s time.Now to come to your 
case of two observers moving in opposite directions. Let the observer at rest 
get to move at velocity -v. His (x, ct) white axes will transform to the mirror 
image of the moving observer’s (x’, ct’) green axes, say, they’ll become 
(x”,ct”) axes orientated toward the left. But the reciprocal time dilation 
‘scissor’ lines will remain reciprocal! (they’ll have become symmetric wrt the 
ct axis).Edit Dec. 2017…I just got a comment from Harvey D Norris . I don’t 
understand how such comments are generated when I don’t see them right here as 
a comment to my answer (Quora has its secrets for me…). But I’ll answer here 
some of his remarks and add new pictures showing the case graphically.The 
reciprocity indeed does not mean the statement that
If A = B minus 20%, then B = A plus 20%,
which is plainly wrong.
But this is not because Lorentz reciprocity is a special case, this is a 
general feature of percentages. That’s why rather than using percentages, it is 
less tricky to use ratios:
If A = B times ratio R, then B = A divided by R, or B = A times ratio 1/R.But 
the confusing aspect of Lorentz reciprocity is, I think, that people come to 
asking the question:
How can it be, given that in the rest system
L’(moving system) = L(rest system) / gamma(v)
[ so that according to the previous one would think L = L’ * gamma],
(so, how can it be…) that Lorentz states also that in the moving system
L(rest system) = L’(moving system) / gamma(-v)???The fallacy is, of course, in 
failing to notice that the events determining the lengths for the rest system, 
are different from the events determining the lengths for the moving system! 
Hence the ‘scissors’-like reciprocities in the picture I put above. Let me add 
some more pictures showing this important distinction, allowing to fully grasp 
the Lorentz reciprocities.(‘Rest’ system = cyan, ‘moving’ system = blue, 
rightward)(‘Rest’ system = blue, ‘moving’ system = cyan, leftward)An animation 
showing Minkowsky space, shifting its reference axes continuously between 
‘cyan’ and ‘blue’ systems:
vt-media-tumblr-com/tumblr_p1a41...(Pictures available on my Tumblr page Wugi, 
another medium I only understand poorly…)
 

>From what I see here when the author notes that that L' noting a compressed 
>Length  L is related or equal to that rest fame L value divided by a value. 
>(It has been compressed with respect to the original) The authour also notes 
>that the equivalence between two magnitudes A and B is related by a ratio 
>whereby one quantity will see the other quantity in equivalence by either a 
>multiplication or division of the (gamma) ratio. So initially he is stating;  
>L’(moving system) = L(rest system) / gamma(v). The division indicates a 
>reduction in magnitude of L' with respect to L (rest). Since the amount of 
>contraction depends on v or velocity that relationship is included. Now 
>according to my reasoning the inverse of all of the above would imply that as 
>a reciprocal relationship that exists between the two where the ratio showing 
>equivalency would now be viewed in reverse or instead of a division by the 
>ratio it would be a multiplication by the ratio resulting in an enlargement 
>with respect to the second observers point of reference. Thus the author then 
>states; [ so that according to the previous one would think L = L’ * gamma],( 
>I am a bit confused to where the v in the relationship went unless this is a 
>typographical error but at least this mathematical language seem clarified a 
>bit) he then concludes; (so, how can it be…) that Lorentz states also that in 
>the moving system
L(rest system) = L’(moving system) / gamma(-v)???
This where I am a bit lost unless he is referring to the opposite observer on 
the quora question?
He then concludes;  The fallacy is, of course, in failing to notice that the 
events determining the lengths for the rest system, are different from the 
events determining the lengths for the moving system! 
Now compare what I wrote on quora in my reply to him; 
"THE 200,000 cycles/sec emitted in a single compressed second does not have the 
same amount of cycles received as reception when compared to stationary cycles 
per second for the simple fact that we have now proposed to use the compressed 
second as the measuring stick, instead of the stationary one made on first 
enquiree." 
At half the speed of light the time dilation noted by T(motion) = 
T(stationary)/[sq rt(1-(v/c)^2)] therefore T(motion) = T (stationary)/sq 
rt(3/4) or a 15.5% time dilation in movement compared to the stationary one. If 
two radio stations @ 200,000 hz are made to separate in space at velocity C/2 
the non relativistic doppler effect will show each station now receiving 
100,000 hz. If we view one of the radio stations as being the stationary one 
and apply the time dilation to the moving stations reception the 100,000 hz 
would appear 15.5% greater at 115,500 hz. (100,000 cycles in .866 sec = 115,473 
hz. ) Consider if the moving radio station broadcasts it's 200,000 hz signal 
using the compressed time reference to the stationary one. THE 200,000 
cycles/sec emitted in a single compressed second does not have the same amount 
of cycles received as reception when compared to stationary cycles per second 
for the simple fact that we have now proposed to use the compressed second as 
the measuring stick, instead of the stationary one made on first enquiree. 
Using the original 200,000 oscillations as a single pulse of one compressed 
second we find that these same oscillations are now spread across 1.15 seconds 
of stationary reception, BEFORE the Doppler effects themselves half that value 
making it 100,000 cycles per sec. We find that the stationary system in one 
second of IT'S time frame will receive only 1/1.1547 or 86.6% of the total 
waves emitted from the compressed time source or 86,600 hz. Note ALSO that 
these GAIN (115,500 hz/LOSS 86,660 hz) ratios are non symmetrical from the 
median 100,000 hz predicted by conventional Doppler effect. 
 In any case I still haven't seen the light at the end of tunnel from these 
diagrams he has shown.
Sincerely HDN

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