# Einstein's Mirrors Experiment

1. Dec 17, 2011

### Mike Hobbs

*** Had a chance to rethink this and still believe it makes sense, so I need somebody to put me straight :-) ***

Ok. Having been beaten senseless untill I can see how the maths behind Einstein's Mirrors Experiment works, another question to help with my journey.

http://en.wikipedia.org/wiki/File:Time-dilation-002.svg

The above diagram is a representation of the setup for the experiment.

As pointed out in another thread, a speed for the moving observer of (1/sqrt(2))c, or 0.707c, generates a 45 degree angle for the path of the light signal in the stationary observer's frame.

As the stationary observer, we know where this beam is heading. If we place an opaque tube in line with the expected path, the signal will travel straight through the tube and hit the final mirror.

In the moving observer's frame, a 45 degree tube approaching at 0.707c, wouldn't allow the signal to pass straight through it to the bottom mirror. However, as length dilation is equivalent to the time dilation, the tube, rather than being defined by an x * x box, giving a 45 degree angle, is defined by a 0.707x * x box, giving a 54.74 degree angle. The signal now meets the tube at exactly the right time for it to travel along the angled tube and still hit the mirror directly below.

Apologies. This is going somewhere, hopefully.

Is this right so far ?

Now, in the stationery observer's frame, we can add a mirror perpendicular to the tube and a distance from the end of it, to reflect the signal straight back along the tube.

Space dilation should squash the dimensions of the mirror in the same way as the tube, changing it from a diagonal across a box at y * y, giving 45 degrees, to a diagonal acroos a box of dimension 0.707y * y, giving an angle of 54.74 degrees, but in the opposite direction to the tube.

In the 'moving' observer's frame, the light still travels down the tube as described above, but then hits a mirror pointing away from the tube. The signal bounces off nowhere near the tube.

This does rather assume that light hitting a moving mirror rebounds at the same angle as light hitting a stationery mirror. Is this correct ?

What's wrong with the above ? Both can't be true, so there must be an error in the 'logic' somewhere ?

Thanks.

2. Dec 17, 2011

### Staff: Mentor

The tube is straight in both frames. Remember, the tube is an extended object moving at v<c, and the light pulse is a "point" object moving at c.

3. Dec 24, 2011

### Mike Hobbs

Sorry, didn't make myself clear. The tubes are straight in both FoRs, but appear at different angles. 45 degrees in the 'stationery' observer's FoR and 55 degrees in the 'moving' observer's FoR. I'm fairly sure this works okay. The question is, what happens in the 'moving' obverver's FoR when the signal hits the mirror at the end of the tube ?

In the 'stationery' observer's FoR, the difference in angle between the tube and mirror is 45 + 45 degrees, ie. 90 degrees and the signal bounces straight back along the tube. In the 'moving' observer's FoR, the difference in angle between the tube and mirror is 55 + 55 degrees, ie. 110 degrees (approx), so the signal doesn't bounce back along the tube.

Both can't be correct, so what's wrong ?

4. Dec 24, 2011

### Staff: Mentor

Remember, the mirror is subject to length contraction also, so it is not at a 135º angle in the frame where it is moving. Find the angle of the mirror, find the angle of the pulse of light (not the same as the angle of the tube), use those to find the angle of incidence, that will give you the angle of reflection, and the light will go back down the 55º tube. It might be worthwhile to actually work through the math on this one because it depends on details of angles and speeds.

Last edited: Dec 24, 2011
5. Jan 3, 2012

### Mike Hobbs

If you imagine that in the 'stationery' observer's FoR, the light signal ascends from the bottom mirror at 45 degrees, bounces off the top mirror at 90 degrees and then hits the bottom mirror, again at 45 degrees, this gives you a triangle representing the path of the signal.

Surrounding both the upward and downward signals with a tube at 45 degrees still allows the signal to complete it's path.

In the 'moving' observer's FoR, the signal travels vertically between the two mirrors and back again. The tubes appear to approach at 0.707c, but with length contraction, the 'horizontal' distance traversed by the tubes is reduced by a factor of 0.707. This allows for the signal to travel along the moving tubes and still reach it's destination, so still works in both FoRs.

Adding a mirror at the end of the 2nd tube, perpendicular to the tube, reflects the signal back along the tube in the 'stationery' FoR.

The 'horizontal' distance for the mirror is contracted by a factor of 0.707 in the same as for the tubes. However, as the tube & mirror are pointing in different directions, in the 'stationery' FoR, within the 'moving' FoR, they're 'twisted' in different directions by the contraction. They no longer meet at 90 degrees to allow the signal to bounce back along the tube.

Even at an angle of 45 degress. A vertical signal hitting the mirror would rebound parallel to the mirrors and never reach the moving tube. At the contracted 55 degrees, it should head down toward the baseline rather than head back up to the top mirror ?

6. Jan 3, 2012

### Staff: Mentor

I am sorry, I am not following your description. Could you draw a diagram of the light path and the angles of all of the mirrors/tubes in the rest frame of the mirrors and tubes?

You are guaranteed by the Lorentz invariance of Maxwell's equations that everything will work out, but I cannot follow the description to be more explicit.

7. Jan 3, 2012

### Mike Hobbs

A diagram would make things a lot easier. You seem to be able to include a link to an existing diagram, but not add a new diagram to any posts.

What's the best way of getting a new diagram into a post ?

Thanks.

8. Jan 3, 2012

### Staff: Mentor

If you can put the image on a hosted image site, like imageshack, then you can post a link to it. Otherwise you can always upload a file as an attachment. Both options are available on the little tool-bar above the area where you edit your post after hitting the "Go Advanced" or "Quote" buttons.

9. Jan 4, 2012

### Mike Hobbs

Cool. Thanks.

Ok. This is the setup as viewed by the stationery observer :

http://img854.imageshack.us/img854/6678/sr00101.png [Broken]

The signal travels between the two mirrors at 45 degrees, hitting mirror D at 90 degrees, allowing it to bounce back through the tubes.

... and this is the 'overview' of the angles for the 'moving' observer :

http://img824.imageshack.us/img824/1425/sr00102.png [Broken]

... this shows the tubes and mirror D contracted. Note that mirror D turns to the same angle as the opposite tube.

In the 'moving' observer's FoR, the tubes & mirror approach at 0.707c, but the signal travels vertically between his two stationery mirrors :

http://img836.imageshack.us/img836/8738/sr00103.png [Broken][PLAIN]http://img850.imageshack.us/img850/2917/sr00104.png [Broken] [Broken][PLAIN]http://img850.imageshack.us/img850/2917/sr00104.png [Broken] [Broken]

http://img542.imageshack.us/img542/5128/sr00106.png [Broken]http://img830.imageshack.us/img830/1589/sr00107.png [Broken]http://img718.imageshack.us/img718/108/sr00108.png [Broken]

The signal travels vertically between the two mirrors, traversing the two moving tubes, but hitting mirror D at an angle that doesn't allow the signal to return along the tubes.

Where is this wrong ?

Last edited by a moderator: May 5, 2017
10. Jan 4, 2012

### Staff: Mentor

Are all of the pieces of the apparatus stationary wrt each other or are the different pieces moving wrt each other?

It looks like A and B form a standard light clock with D and the tubes moving wrt A and B. I am not sure about C, is that just another name for A?

If B is moving wrt the tubes and D then it is simply not possible for the light to leave B, go through the tube, bounce off D, go back through the tube, and hit B again. By the time the light returned B would have moved away.

Can you get at your question with an apparatus that has no moving parts? E.g, what about getting rid of A and C and replacing A with another 90 deg mirror, with all parts stationary wrt each other.

Last edited: Jan 4, 2012
11. Jan 5, 2012

### Mike Hobbs

In the 'stationery' observer's FoR, the tubes & mirror D are stationery and mirrors B and A/C (same mirror, different pos) are moving at 0.707c.

In the 'moving' observer's FoR, the mirror's A & B are stationery and the tubes and mirror D are approaching / passing at 0.707c.

In the 'moving' FoR, the signal travelling vertically between the two mirrors, would be timed exactly to match the movement of the tubes, to allow for it to pass through the tubes on it's route between the two mirrors and back. This is ensured by placing the tubes in the correct position within the 'stationery' FoR and has to be the case for both to see the same 'result'.

The example can't be presented with all parts stationary. The point of the experiment is that the mirrors (A&B) are moving with respect to the 'stationery' observer ?

The tubes are probably irrelevant. They just emphasize the path of the light and the length contraction ? Mind you they also serve to confirm that mirror D is 'twisted' in the correct direction and to the correct angle, as it must match the angle of the opposite tube in both FoRs ?

12. Jan 5, 2012

### Staff: Mentor

Sure, I understand that, but that is not what I was asking. Clearly you can have the device moving relative to some observer.

I was asking if the example really needs to have the different parts moving wrt each other. I.e. because you have different parts moving wrt each other, the device itself does not have a single "rest" frame from which the analysis is clear. Half of the device is stationary wrt the stationary observer and the other half is stationary wrt the moving observer.

Frankly, if you really need that complicated a scenario it will be easier to simply prove the invariance of Maxwell's equations, as I have mentioned already. I don't think that will be satisfactory to you but there is a big danger in making mistakes whenever your scenario is overly complicated. For instance, I can already see that if B isn't stationary wrt D and the tubes then the device doesn't work in any frame.

13. Jan 6, 2012

### Mike Hobbs

Switching between FoRs to interpret what the observers see is the whole basis of the original experiment. It's where we derive the time dilation factor from. I'm not changing that, just adding a few additional objects.

The additional parts within the 'stationary' observers FoR, seem fairly simple. Is the setup in anyway unreasonable ? I'd say 'unrealistic', but we are talking about someone travelling at 0.707c. We could do away with the tubes and have just the additional mirror, but I don't think this would simplify things and mirror D is an integral part of what the original query was, so it's difficult to remove that.

I understand your concern about switching between FoRs. At first, it doesn't seem that the signal can travel along the moving tubes within the 'moving' observer's FoR. If you picture the signal as a single photon (as in the original experiment ?), rather than a beam of light, it's easier. My arrows are a bit misleading, it's really just the tip of the arrow that represents the signal. I'm no artist ... nor physicist for that matter.

I believe that this works okay ? The speed of the signal being c, with the speed of the tubes being 0.707c and the tubes being at approx. 55 degrees, provides for the perfect timing for the signal to travel in a vertical line and always be in the appropriate gap within the tube. Again, this isn't tweaked to prove a point, it's just wrapping the 'stationery' signal with the tubes and allowing the length contraction to do it's bit.

Not sure why you're concerned about mirror B and mirror D moving with respect to each other in the 'stationary' FoR. Are you saying that the original setup wouldn't work in the stationary FoR ?

... and I must stop typing 'stationary' as 'stationery' !!

14. Jan 6, 2012

### Staff: Mentor

They don't seem simple to me.

I have no concern about switching FORs. It is a standard part of relativity. And that is how I understood your arrows. You may not be an artist, but the graphics are well done.

If it won't work in one FOR then it won't work in any FOR. The problem is easiest to see in the stationary observer FOR. At some t=t1 the light pulse from A hits mirror B and is bounced through the tube. It travels some distance d, hits mirror D, and returns. That takes a time 2d/c. But at t=t1+2d/c mirror B is no longer at that location but has moved 2d/c(.707) to the right.

15. Jan 9, 2012

### Mike Hobbs

Ah. Quite right. I'm being a numpty. Went wrong somewhere. Let me go back over it to see where the brain fart happened.

Thanks.

16. Jan 9, 2012

### Mike Hobbs

Hold on ...

Gets a bit confusing switching between examples. For the 'moving' observer's signal to travel at 45 degrees in the 'stationary' observer's FoR, the motion of the light across x, y must be equal. The signal must travel at 1c, along the hypotenuse, so the velocity in each direction must be sqrt(1/2)c = 0.707c.

At 0.707c, the 'moving' observer's signal would travel at a 45 degree angle between the mirrors, at a speed of c, in the 'stationary' observer's FoR.

The original diagram is okay ?

It's confusing because you'd expect a signal bounced vertically between the two central mirrors to travel at c and showing the height as c and the base as 0.707c, doesn't give you a 45 degree angle. Showing the speed as the units is misleading. The hypotenuse is acually c*t0, and the height and base are 0.707c*t1.

So, in the 'stationary' observer's FoR, if the signal takes 1 sec to travel between the two mirrors, at a 45 degree angle, the mirrors are 0.707c*t1 apart. The moving observer takes a full second to travel from the central point to the final mirror position, but a signal from the central point to the final mirror position would only take 0.707 secs and a signal between the two mirrors would only take 0.707 secs.

Does this sound okay ?

Last edited: Jan 9, 2012
17. Jan 9, 2012

### Staff: Mentor

Hi Mike, the objection I made still stands. With the different parts of the system moving B can only be in the proper place once, not twice. However, even if you patched that up, I am just not motivated enough to analyze such a complicated device.

If you decide to investigate a device which has no moving parts then I will gladly help you analyze it from a stationary and a moving reference frame. Otherwise, below is the best introductory-level site that I have found to show that Maxwell's equations are Lorentz invariant in general. The nice thing about that is that you don't need to analyze any specific device, you know that it automatically holds for all possible devices which obey Maxwell's equations.

http://www.relativity.li/en/maxwell2/str-maxwell-equations/ [Broken]

Last edited by a moderator: May 5, 2017
18. Jan 9, 2012

### zonde

Light does not reflects from moving mirror the same way as from stationary mirror.
Consider picture like that
Code (Text):
+_____________
\ \ \ \
__+__________
\ \ \ \
_____+_______
\ \ \ \
________+____
\ \ \ \

horizontal lines are wavefront of light at different times (moving from top to bottom)
skewed lines are mirror at different times (moving from left to right)
Different points of wavefront will be reflecting from mirror at different times (marked with pluses). So in order to find angle of reflection you have to take line that goes trough pluses as surface of mirror.

19. Jan 9, 2012

### alexg

In the 'moving' frame, the light is bouncing straight across from one mirror to the other.

If the tubes are fixed at an angle in the frame of the light and mirrors, and are at rest with respect to that frame, the light cannot traverse the tube.

20. Jan 10, 2012

### Mike Hobbs

Thanks Dalespam. I'll read through the website.

Last edited by a moderator: May 5, 2017