
#1
Aug210, 03:48 PM

P: 651

Say that two mirrors are 3 ltyr apart and perfectly facing each other. Say they approach at 0.6c (relative to a "stationary" observer)
M1>0.6 0.6<M2 If a lightbeam is flashed at M1 towards M2 and bounces back and forth until the mirrors cross each other, how far will the light beams travel and how long will it take for the light beams to travel until the mirrors do cross. This is a variant of the hummingbirdapproaching trains problem but now we have relativity mixed in. Maybe this belongs in the homework section, so, if so, please move it there. My answer is 2.72 years... is that correct? If we were Newtonian or Galilean it would be 2.5 years. stevmg 



#2
Aug210, 04:44 PM

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#3
Aug210, 04:56 PM

P: 651

1) Closing speed is not 1.2c because that's impossible but is 0.8824c (Velocity Addition formula)
2) By length contraction distance = 3 (1  .8824^2)^(1/2) = 1.411 ltyr 3) time = 1.411/0.88241.60 = 1.60 year 4) If we were Galilean, it would be 3.0/(0.6 + 0.6) = 2.5 years I think I screwed up before. Hope I am right now. stevmg 



#4
Aug210, 04:56 PM

P: 341

Approaching Mirrors
stevmg  it's late so I'm confused  but why is relativity even relevant to this question? It looks as though all the information and the calculation to be done wholly from the stationary observer's point of view and I can't see any funny stuff going on that calls for anything particularly relativistic.




#5
Aug210, 05:01 PM

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P: 40,877

The velocity addition formula is only relevant if you wanted to know the speed of one mirror in the frame of the other. 



#6
Aug210, 05:22 PM

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#7
Aug210, 05:31 PM

P: 651

Sooorry
OK, looking from mirror 1 would I be right? Steve 



#8
Aug210, 05:35 PM

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#9
Aug210, 06:00 PM

P: 651

Now, here's we I need a class on this subject as I am getting lost.
Doc Al. You would agree that the relative speed from M2 onto a "stationary" M1 would be 0.8824. Is that correct? starthaus  I am really confused about closing speed. In the MMX .pdf file you referred me to, you explained that the lack of difference in roundtrip times for the reflecting beams of light (head on versus perpendicular) was due to a contraction factor in the direction of motion of the Earth. But that contraction factor was explained by Lorentz and Fitzpatrick by a later 1898 experiment. At the time of the MMX3 experiment, I thought that the apparent lack of difference in roundtrip time for these light beams was unexplained using just closing velocities. By closing velocity assumption, there should have been a difference in roundtrip times. At least, that's what I got out of your paper. the time dilation part at the end was more icing on the cake or another way of proving it. You would have to theorize a length contraction in the direction of the Earth orbital velocity to account for the similarity of the roundtriptimes. 



#10
Aug210, 06:28 PM

P: 1,568

You mean FitzGerald. 



#12
Aug210, 06:44 PM

P: 651





#13
Aug210, 06:51 PM

P: 651

Doc Al 
The relative speed of the mirrors is 0.8824c. Why would we use 0.6c as you stated? If we assume that the two mirrors are 3 ltyr apart in the M1 FR (that wasn't the original problem, but let's use that), would it take 3/0.8824 = 3.40 year in the M1 FR or do we length contract the 3 ltyr. I know this is getting basic but I have to get out of my "black hole" of misunderstanding somehow. If the answer is yes, then I have a little more. 



#14
Aug210, 07:04 PM

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#15
Aug210, 07:27 PM

P: 651

Now, with regards to the MichelsonMorley experiment, if we did not use length contraction which is not part of Galilean mechanics, there WOULD be a difference in roundtrip time for the perpendicular and parallel light beams, right? See the figure below: Because of the mathematics of inverses, the total travel time back and forth parallel to the relative motion of an "ether" would be greater than the back and forth travel time for light perpendicular to the "ether." That would be true in a Galilean or Newtonian Universe, right? 



#16
Aug310, 04:48 AM

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#17
Aug310, 05:46 AM

P: 651

Now, where does "closing velocity" (c + u) or (cu) fit in? In other words, either the "u" doesn't exist (i.e., no ether) and closing velocity just remains a mathematic theoretical entity which has no place here. Light always approaches at c and there is no closing velocity  at least not with light unless one calls c the universal closing velocity for light, no pluses or minuses. By "your" reasoning, one can look at two objects approaching each other, say two trains steaming at each other one from the west and one from the east. The total closing velocity would be <2c but each individual's speed would be <c (because of socalled "relativistic mass increase"  an outmoded term these days.) One farmer once saw such an event in his field  two trains on the same track blasting towards each other at way subrelativistic speeds, but fast, nevertheless. A reporter asked, "What did you do?" The old codger replied, "Well  I thought 'What a hell of a way to run a railroad!'" 



#18
Aug310, 05:46 AM

P: 3,966

In the frame of the stationary observer one mirror travels at 0.6c and travels 1.5 light years to arrive at the centre. The time is then 1.5/0.6= 2.5 years and of course the other mirror arrives at the centre at the same time. In 2.5 years the light signal travels 2.5 lightyears. That answers both your questions. 


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