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There are clocks (with observers) at every theoretically possible point in the “original observer’s” inertial frame of reference.

And all of these clocks, in the inertial frame of reference of the original observer, are synchronized.

(Perhaps we can call him the “OG observer.”)

All of the “clock/observers” are requested to report the “event” of the part of the square that is right next to him at one specified time (such as “12:00:00”).

And so, when 12:00:00 rolls around on all of the synchronized clocks, one of the clock/observers will say the top right corner of the square is next to him and another will say that at 12:00:00 the lower left corner is next to him and so on (and not just for the four corners but for every point along each side there will be a clock/observer at 12:00:00 who can report what “event” is next to him at this time).

And this way we can find out the “height” of the length contracted square in the inertial frame of reference of the “original observer.”

(If the square is moving at a constant relative speed this this same “height” will be found no matter what agreed upon time we use (11:59:59 or whatever).)

And the same thing is done for the “height” (the y displacement) of the vertically moving rectangle.

This is why I don’t think “Relativity of Simultaniety” is the solution here. (I’ve tried but I still don’t see it.) The clocks that are relevant to the measurements here of the square and rectangle are in the rest frame and so are not un-synchronized. (But it’s possible that when we switch to the square’s rest frame or to the rectangle’s rest frame this “relativity of simultaneity” with the square or the rectangle then pops up from this switch and saves the day (paradox-wise) … but I can’t see it yet.)

However …

After reading your link Mike-from-the-Bronx … I may be able to start seeing this other way to the solution. I’m definitely not there yet … but I think I’m beginning to understand the basics of the kind of switching (double-switching) you are talking about.

I’m still thinking … I can start to see something. Thank you.

And thank you all!

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[Thales et al #26] Thanks once again for the detailed info. I am fairly well up on that sort of stuff, but my memory lets me down.

Re my #24 & Mike-from-the-Bronx's #23 & #25, if Mike is correct then it means that the angle of the frame of reference of the stationary observer at 0,0 makes a difference to the paradox. Surely this cannot be correct.

In effect the scenario in my #24 is that i have turned the FOR 45dg so that the diamond-diamond is now moving parallel to the xx axis, & i said that there is contraction in the xx direction & zero contraction in the yy direction (compared to stationary). How can i possibly be wrong ??. How can changing the xy angle of the FOR possibly make a difference ??(except ease of calculation). I don't remember Alby mentioning any such requirement in any of his postulates principles laws suppositions presumptions assumptions.

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In a frame of reference described by x and y co-ordinates is it reasonable to describe a diagonal motion (what axis is this on?). Does this 'diagonal' motion not have x and y components of speed and therefore will appear to have contraction in both x and y directions?

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In a frame of reference described by x and y co-ordinates is it reasonable to describe a diagonal motion (what axis is this on?). Does this 'diagonal' motion not have x and y components of speed and therefore will appear to have contraction in both x and y directions?

If I understand you correctly, your conclusion would be that a Square moving at a 45 degree angle would just be a smaller square. Relativistic contraction does not work that way. (Edit: added "When....")When transforming reference frames in relativity velocities don't combine by the rules of vectors and neither does contraction.

Example: If you take an object that is initially a square at rest with respect to you and transform to a reference frame that is moving northwest, the northwest and southeast corners of the square are contracted (closer to the center). The northeast and southwest corners are not effected at all. Connect the corners to determine the shape and it will be a symmetric diamond.

Edited by Mike-from-the-Bronx

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