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Dec 9, 2014 at 5:04 answer added Mark H timeline score: 0
Nov 7, 2014 at 3:28 answer added Sofia timeline score: 0
Nov 6, 2014 at 22:32 answer added BowlOfRed timeline score: 2
Nov 6, 2014 at 21:38 comment added user1318216 I envsioned it as sum(x^k-x^(k+1)) for every other number from 1 to n. Where x is the percent imparted momentum (.1 above), k is the number of times it's been reflected and n is the number of times it is reflected before it is expelled. As for where the photons come from, they could be generated on or outside of the vehicle, it wouldn't matter in the above example.
Nov 6, 2014 at 21:33 comment added njzk2 if you consider the sum of exchanged momentum on one side minus on the other, you get something like (1 - k) where k is the amount of conserved energy (very very close to 1). Also, how do you give the initial energy and momentum to the photon?
Nov 6, 2014 at 20:38 comment added user1318216 Just added some clarification on how the two are related.
Nov 6, 2014 at 20:36 history edited user1318216 CC BY-SA 3.0
Added further clarification.
Nov 6, 2014 at 20:26 comment added Floris The first part of your question is quite long and confusing. The second part ("Put a different way...") is clear - and the answer is "yes". Since the photon lost energy on the first bounce, it has less energy for the second bounce. The difference might be measurable in the wavelength shifting of the photon - but I can guarantee you cannot measure it on your "mirror". But maybe if you repeated this billions of times. Of course once a mirror has started moving after the first impact, the situation changes... I wonder if you can tighten up the question a bit.
Nov 6, 2014 at 20:08 review First posts
Nov 6, 2014 at 20:19
Nov 6, 2014 at 20:06 history asked user1318216 CC BY-SA 3.0