Re: [Phys-L] simultaneity at a distance.

[Moses Fayngold <moshfarlan@yahoo.com>]

On Wednesday, August 31, 2016 12:57 PM, "LaMontagne, Bob" <RLAMONT@providence.edu> wrote:



> If it were possible to travel just under the speed of light from Planet A to Planet B, wouldn't the distance travelled >and the time taken both appear to shrink to near zero for the traveler? 
Yes, they would
> And ultimately, aren't all elapsed times zero and all points on top of each other for a photon?

 Yes, they are. The distance AB for a photon goes to zero due to relativistic length contraction. The length of any finite segment AB measured by passing traveler undergoes relativistic length contraction relative to traveler's frame, which takes it to zero at v --->c. The same holds for the corresponding proper time of the traveler. A crude explanation on conceptual level goes like this: imagine a sliding graph of the sinus function representing a monochromatic EM wave propagating from A to B. Let the distance AB measured by us be 1 LY. If we manage to mark some wave crest at the moment of its passing by A, it will pass by B one year later by our time. We will know that, e.g., by counting the corresponding number of similar crests having passed by us by the moment the marked crest reaches B. But an imaginary traveler co-moving with light and sitting on the marked crest will record the same crest when at B, so no time will pass between events A and B. And since AB is sliding past the traveler with finite speed, the distance AB must be also zero. 
<---snip--->
> As such, the t component cannot be zero in any frame.  If it were
> zero, it would mean that the two points are not different.  It would
> mean they are sitting on top of each other.
It would only mean that the two points, just as two moments of time, are merged together in photon's reference frame.  
Moses Fayngold,NJIT

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