Approaching Mirrors

by stevmg
Tags: approaching, mirrors
 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 hummingbird-approaching 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
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P: 41,555
 Quote by stevmg My answer is 2.72 years... is that correct?
How did you arrive at this answer? (I presume you are trying to find the time according the 'stationary' observer.)
 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
 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.
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P: 41,555
 Quote by stevmg 1) Closing speed is not 1.2c because that's impossible but is 0.8824c (Velocity Addition formula)
Here's your mistake. As seen by the stationary observer, the closing speed is 1.2c. Why do you think that's impossible?

The velocity addition formula is only relevant if you wanted to know the speed of one mirror in the frame of the other.
P: 1,568
 Quote by stevmg 1) Closing speed is not 1.2c because that's impossible but is 0.8824c (Velocity Addition formula)
Arrrgh, you are loosing all your previous gains. The distance between the two mirrors is covered at the closing speed 1.c in 2.5 years. End.

 2) By length contraction distance = 3 (1 - .8824^2)^(1/2) = 1.411 ltyr
There is no length contraction from the perspective of an observer that notices the mirrors "closing" the 3ly distance at a closing speed of 1.2c

 3) time = 1.411/0.88241.60 = 1.60 year
No.

 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
Closing speed applies exactly the same way in SR as in galilean kinematics. I am quite sure I told you this early on when we got started on the subject of closing speeds.
 P: 651 S-o-o-o-r-r-y OK, looking from mirror 1 would I be right? Steve
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P: 41,555
 Quote by stevmg OK, looking from mirror 1 would I be right?
No. For one thing, the length contraction would be based on 0.6c, not the relative speed of the mirrors.
 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 MMX-3 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.
P: 1,568
 Quote by stevmg 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

You mean FitzGerald.
 At the time of the MMX-3 experiment, I thought that the apparent lack of difference in roundtrip time for these light beams was unexplained using just closing velocities.
No, the FitzGerald contraction has been simply replaced by the "length contraction" of SR.
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P: 41,555
 Quote by stevmg Doc Al. You would agree that the relative speed from M2 onto a "stationary" M1 would be 0.8824. Is that correct?
Yes. The relative speed of the mirrors is 0.8824c.
P: 651
 Quote by starthaus You mean FitzGerald.
Never knew the man.
 Quote by starthaus No, the FitzGerald contraction has been simply replaced by the "length contraction" of SR.
So what? What I was saying that it took the Fitzgerald contraction in the direction of the Earth's motion to explain the LACK of difference in roundtrip time for the light bouncing between the mirrors. Is that right?
 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.
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P: 41,555
 Quote by stevmg Doc Al - The relative speed of the mirrors is 0.8824c. Why would we use 0.6c as you stated?
Because the 3 LY distance was given in the stationary frame. (Until you changed the problem!)
 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.
This is a different problem. Since the distance is now 3 LY in the M1 frame, and we are interested in the time according to the M1 frame, no length contraction is required. 3.4 years would be correct.
P: 651
 Quote by Doc Al Because the 3 LY distance was given in the stationary frame. (Until you changed the problem!) This is a different problem. Since the distance is now 3 LY in the M1 frame, and we are interested in the time according to the M1 frame, no length contraction is required. 3.4 years would be correct.
Doc Al, in the stationary frame, wouldn't one use the 0.6 + 0.6 = 1.2 for the time then (in other words, one would add up the times that the two mirrors would need to travel 1.5 ltyr each to = 3 ltyr.) In other words, it would take 1.5/0.6 = 2.5 yr for each mirror to move the necessary 1.5 ltyr to "meet." Thus, the 2.5 years would be the answer for that scenario. True? Kind of a round about way of looking at it but conceptually is clearer to me.

Now, with regards to the Michelson-Morley 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?
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P: 41,555
 Quote by stevmg Doc Al, in the stationary frame, wouldn't one use the 0.6 + 0.6 = 1.2 for the time then (in other words, one would add up the times that the two mirrors would need to travel 1.5 ltyr each to = 3 ltyr.) In other words, it would take 1.5/0.6 = 2.5 yr for each mirror to move the necessary 1.5 ltyr to "meet." Thus, the 2.5 years would be the answer for that scenario. True? Kind of a round about way of looking at it but conceptually is clearer to me.
Nothing wrong with that.

 Now, with regards to the Michelson-Morley 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?
If there was an ether in which light traveled, then there would be an expected time difference. One way to explain the null result is with Fitzgerald's length contraction proposal.

 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?
Yes.
P: 651
 Quote by Doc Al If there was an ether in which light traveled, then there would be an expected time difference. One way to explain the null result is with Fitzgerald's length contraction proposal.
Fitzgerald's (as I told starthaus after calling him Fitzpatrick - "I never knew the man") length contraction is NOT part of Galilean or Newtonian mechanics or physics, thus, a new concept must be introduced to explain the null difference in order to justify the use of closing velocity (c + u) or (c - u). You have to shorten the distance to make the elapsed roundtrip times come out the same by, coincidently, the same gamma factor of later relativity fame.

Now, where does "closing velocity" (c + u) or (c-u) 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 so-called "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, never-the-less.

A reporter asked, "What did you do?"

The old codger replied, "W-e-l-l - I thought 'What a hell of a way to run a railroad!'"
P: 3,967
 Quote by stevmg 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 hummingbird-approaching 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
This is my initial stab at the problem, but it seems to easy and maybe I am missing something.

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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