# Approaching Mirrors

1. Aug 2, 2010

### 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

Last edited: Aug 2, 2010
2. Aug 2, 2010

### Staff: Mentor

How did you arrive at this answer? (I presume you are trying to find the time according the 'stationary' observer.)

3. Aug 2, 2010

### stevmg

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. Aug 2, 2010

### yossell

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. Aug 2, 2010

### Staff: Mentor

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.

6. Aug 2, 2010

### starthaus

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.

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

No.

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.

7. Aug 2, 2010

### stevmg

S-o-o-o-r-r-y

OK, looking from mirror 1 would I be right?

Steve

8. Aug 2, 2010

### Staff: Mentor

No. For one thing, the length contraction would be based on 0.6c, not the relative speed of the mirrors.

9. Aug 2, 2010

### 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 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.

10. Aug 2, 2010

### starthaus

You mean FitzGerald.
No, the FitzGerald contraction has been simply replaced by the "length contraction" of SR.

11. Aug 2, 2010

### Staff: Mentor

Yes. The relative speed of the mirrors is 0.8824c.

12. Aug 2, 2010

### stevmg

Never knew the man.
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?

Last edited: Aug 2, 2010
13. Aug 2, 2010

### stevmg

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. Aug 2, 2010

### Staff: Mentor

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.

15. Aug 2, 2010

### 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.

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?

16. Aug 3, 2010

### Staff: Mentor

Nothing wrong with that.

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.

Yes.

17. Aug 3, 2010

### stevmg

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!'"

18. Aug 3, 2010

### yuiop

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.

19. Aug 3, 2010

### stevmg

Thanks, kev. Doc Al and starthaus have hammered my poor brain with, essentially, what you have said, but please look at what I wrote about closing velocities in my post above
https://www.physicsforums.com/showpost.php?p=2824889&postcount=17
to Doc Al. Any comments would be appreciated!

BTW - what DO they call "relativistic mass increase" these days?

Thanks,

stevmg

20. Aug 3, 2010

### yuiop

Fitzgerald contraction and the length contraction of SR are mathmatically the same thing. They only differ philosophically in how "real" the length contraction is.

We are not talking about the closing velocity of light. Closing velocity applies to two objects seen by one observer. In your OP example, the closing speed of the two mirrors is simply 0.6+0.6 = 1.2c (yes, even in relativity). At a closing velocity of 1.2c the mirrors collide in time 3/1.2 = 2.5 years. Relativistic velocity addition is not required for closing velocities. The velocity addition formula is used when we wish to find the velocity of a single object according to one observer, when we know the velocity of the object according to a different observer (and we know the velocity of the second observer relative to the first). The important thing is that no single object exceeds the speed of light relative to any given inertial observer. If we want to know the speed of m2 from the point of view of m1 then that is a different thing to closing velocity and that is when we would use the relativistic velocity addition formula. This of course means that the total travel time and distance of the light signal is different according to an observer comoving with one of the mirrors, but of course we expect that in relativity.

21. Aug 3, 2010

### yuiop

They don't. Its use has been banned! Sometimes it is called the quantity that is obtained when you divide the the momentum of an object by its velocity (p/v) or the enrgy increase divided by c^2, when its use is unavoidable in polite circles. :tongue:

Last edited: Aug 3, 2010
22. Aug 3, 2010

### Staff: Mentor

Why do you mean by 'your' reasoning? The closing speed is a simple matter of kinematics--nothing to do with "relativistic mass increase" or even relativity at all.

23. Aug 3, 2010

### stevmg

"Your reasoning" doesn't mean anything pejorative or "out of thin air." "Your reasoning" refers to accepted state-of-art current modern physics concepts or kinematics, as is stated.

I APOLOGIZE for the use of "relativistic mass increase." I hope this doesn't mean that I get banished to some dark corner of the universe. It is mentioned by kev that sometimes it is used in "polite circles." I also run around in circles, but unfortunately, my "circle" doesn't have any people in it.

So, again, as far as Michelson-Morley III is concerned there is no closing velocity and c is the only velocity (in a vacuum) that light either approaches or eminates. There is no need to adjust with the 1/$$\gamma$$ correction factor to make-up for lost time by shortening the distance because light going parallel to the Earth's tangental orbit and light going perpendicular to it are going at the same speed relative to the MM apparatus.

That apparatus bangs the light back and forth for 22 miles (35 km), doesn't it?

24. Aug 3, 2010

### starthaus

In the frame of the lab, there is no "closing speed" but in any frame at rest wrt the lab, you need both closing speed and length contraction in order to explain the null result.

No, it doesn't. This would be a huge distance. The longest light trip in MMX is about 32m, three orders of magnitude smaller than what you are thinking. See here

25. Aug 3, 2010

### yuiop

By now you have probably figured this out, but I will show you another way to do it without explicitly using closing velocity that might help.

Imagine we have a rod AB of length L going to the right with velocity u.

|--ut--><-------ct------|
A----------L------------B

The mirror at A travels to the right with velocity u and the light signal from B travels to the left at velocity c. We can obtain the time t for the "collision" of the light particle with the mirror by dividing the distance L by the "closing velocity" (c+u) so that t = L/(c+u). Note that the closing velocity is not the speed of the light, but the speed with which the light and the mirror are approaching each other. However if you are dubious about the notion of closing speeds, then you can reason like this. The distance ut+ct must equal L as can be seen in the diagram, so we can say ut+ct=L => t(u+c)=L => t = L/(u+c).

After the light reflects off the mirror

|-------------------ct-------------------->|
A----------L------------B--------ut------->|

the light chases after the B mirror which has a head start of L and we can use similar reasoning to conclude that ct=L+ut => t(c-u)=L => t=L/(c-u). (or we could just say the distance is L and the closing velocity is (c-u) and t=L/(c-u).)

It turns out that when you work out the diagonal path of the signal traveling along the vertical arm, using good old Pythagorous, that the signal traveling the vertical path returns before the signal travelling the horizontal path. This contradicts what was actually measured in the MM experiment. Now if the speed of light is constant and independent of the velocity of the source and since the time is measured by a common clock at the fulcrum, the only variable left that can explain the null result is if L is shorter by a factor of gamma when it moving. The only other explanation is a ballistic theory of light in which the velocity of light depends on the velocity of the source (and this case the velocity of light really would be c+/-u) and then length contraction would not be required. Since length contraction was a pretty radical concept at the time it shows the high degree of confidence that the likes of Lorentz and Fitzgerald had in Maxwell's equations (and the constant speed of light) and ruling out the ballistic theory.

Hope that helps some.

Note that I have been sloppy with velocity signs and just use magnitudes, but it works out the same if you do it properly.

Last edited: Aug 3, 2010