Go Back   Physics Help Forum > College/University Physics Help > Special and General Relativity

Special and General Relativity Special and General Relativity Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Mar 27th 2015, 12:28 PM   #1
Junior Member
 
Join Date: Mar 2015
Posts: 4
Expected results from rotating Michelson–Morley like experiment?

I am wondering what results would a fast rotating version of Michelson–Morley experiment produce?
I have been researching this topic for a long time now and I simply do not have any idea what the results might be.
I would very much appreciate your opinion to two proposed thought "gedanken" experiments, or even better, if you could point to already performed experiments that are relevant, I believe this matter has been addressed already in the past...
Those two experiments are:
1st Experiment-
**1- Say I do have a very long steel bar. On one end "A" is a laser pointer fixed so it fires its beam to the other end "B" of the bar. On "B" end I do mark where the beam is. The bar is at rest. What happens when the bar is accelerated to very high speed? The laser pointer is fixed to the bar so it is rotating with it and the centre of rotation is at the pointer "A" end. Will the beam match the mark made when the bar was in rest?
-- I believe there can be only two outcomes
---1a- The beam won't move as Michelson–Morley like experiments suggests. Or
---1b- The beam will move because the speed of light is finite and before it reaches the mark, the mark already changed it's position. I think 1b is the correct answer, even when the pointer and the mark are not in relative motion, the far "B" side with the mark is in fact more accelerated than the "A" pointer side. Not sure if this could be the correct explanation?
2nd Experiment-
***2- Basically the same thing, except that I'd have two synchronized bars far apart, with parallel planes of rotations and centres of rotation at points "A" and "A'", think of it as if those bars were welded at their "A" points to the common shaft. When the bars are at relative rest, I fix the laser pointer to "B" point of one of the bars so it leaves mark at "B'" point of the other. What happens when the bars are accelerated to great speeds (still synchronized)? Will the pointer still match the original mark?
-- Again, I believe only two outcomes are possible
--- 2a- The same as in 1a, the pointer won't move, then I can't reconcile that with finite speed of light
--- 2b- The same as in 1b, except the second part is no more valid - both "B" sides accelerated the same way. In this case I do not have an idea what the explanation would be with respect to Special Relativity.
The most common explanation that I came across is actually more like 2a option, which sounds nonsensical. there is simple graphic explanation on
(http://onlinephys.com/timedilation2.jpg)
which basically says that 2a is the correct answer.
I might be misunderstanding something that Special Relativity says about this and I also believe that this "issue" has already been addressed by some experiment or explanation I did not come across. I also think that Michelson–Morley like experiments could show outcome described in 2b, but results were negative (except D.Miller which is considered as statistical error due to pink noise), so I perhaps am missing something and 2a is a correct answer.
Is there anyone who could explain what it is that I do not understand?
Many thanks
Michael77 is offline   Reply With Quote
Old Mar 27th 2015, 01:22 PM   #2
Forum Admin
 
topsquark's Avatar
 
Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,737
Originally Posted by Michael77 View Post
I am wondering what results would a fast rotating version of Michelson–Morley experiment produce?
I have been researching this topic for a long time now and I simply do not have any idea what the results might be.
I would very much appreciate your opinion to two proposed thought "gedanken" experiments, or even better, if you could point to already performed experiments that are relevant, I believe this matter has been addressed already in the past...
Those two experiments are:
1st Experiment-
**1- Say I do have a very long steel bar. On one end "A" is a laser pointer fixed so it fires its beam to the other end "B" of the bar. On "B" end I do mark where the beam is. The bar is at rest. What happens when the bar is accelerated to very high speed? The laser pointer is fixed to the bar so it is rotating with it and the centre of rotation is at the pointer "A" end. Will the beam match the mark made when the bar was in rest?
-- I believe there can be only two outcomes
---1a- The beam won't move as Michelson–Morley like experiments suggests. Or
---1b- The beam will move because the speed of light is finite and before it reaches the mark, the mark already changed it's position. I think 1b is the correct answer, even when the pointer and the mark are not in relative motion, the far "B" side with the mark is in fact more accelerated than the "A" pointer side. Not sure if this could be the correct explanation?
2nd Experiment-
***2- Basically the same thing, except that I'd have two synchronized bars far apart, with parallel planes of rotations and centres of rotation at points "A" and "A'", think of it as if those bars were welded at their "A" points to the common shaft. When the bars are at relative rest, I fix the laser pointer to "B" point of one of the bars so it leaves mark at "B'" point of the other. What happens when the bars are accelerated to great speeds (still synchronized)? Will the pointer still match the original mark?
-- Again, I believe only two outcomes are possible
--- 2a- The same as in 1a, the pointer won't move, then I can't reconcile that with finite speed of light
--- 2b- The same as in 1b, except the second part is no more valid - both "B" sides accelerated the same way. In this case I do not have an idea what the explanation would be with respect to Special Relativity.
The most common explanation that I came across is actually more like 2a option, which sounds nonsensical. there is simple graphic explanation on
(http://onlinephys.com/timedilation2.jpg)
which basically says that 2a is the correct answer.
I might be misunderstanding something that Special Relativity says about this and I also believe that this "issue" has already been addressed by some experiment or explanation I did not come across. I also think that Michelson–Morley like experiments could show outcome described in 2b, but results were negative (except D.Miller which is considered as statistical error due to pink noise), so I perhaps am missing something and 2a is a correct answer.
Is there anyone who could explain what it is that I do not understand?
Many thanks
Just to be clear, when you are saying "accelerating" do you mean "rotational acceleration?" I'm going to assume that in the following.

Think of the photons leaving their sources as bullets moving at the speed of light. (Many people find it easier to consider it this way for some reason.) Then your answer to 1 is 1b since the "bullets" left with a linear velocity and point B rotates away from that vector.

For 2, the diagram does not seem to me to have much relevance to your question. I don't see any A's or B's and nothing looks like it's rotating. If it's only a linear acceleration you are talking about here then the answer is 2a since the second bar is not moving with the same speed that the "bullet" left the first one at.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.
topsquark is online now   Reply With Quote
Old Mar 27th 2015, 02:17 PM   #3
Junior Member
 
Join Date: Mar 2015
Posts: 4
Thanks for your reply Dan.
Yes, I did mean accelerated in a rotational acceleration sense in both cases, but the actual experiments would be performed in non accelerating reference frames. After acceleration, the bars would be moving/rotating with constant speed. Never the less your answer does make a sense to me.
As for the diagram I've used. I only used it to show how the reference frames are relative, I mean, the observer on the ship would see the beam bounce straight back from the mirror to its origin, meanwhile outside observer would see the light to bounce under an angle to it's source that already moved somewhere else.
In those "experiments" (I think) I am describing similar situations. In both cases the light source is not moving with respect to the mark that was done when the bars were at relative rest. I mean, even after acceleration and then in a non accelerating movement, the relationship between the laser pointer and the mark is similar to the situation with the observer on the ship from the diagram.
I also believe that the beam would not match the original mark, the same as you do.
But I am also aware that this is not in agreement with Special Relativity, -- Obviously unless I am missing something, which is likely--
Then we have those Michelson–Morley like experiments, as I do understand, if the results to my experiments were as I would expect viz. 1b;2b, then those performed experiments (Michelson; Michelson–Morley, Kennedy-Illingworth, etc.) should yield positive results, due to Earth movement and rotation. Which they did not obviously...
I just don't understand why in the motion of the space ship in the diagram, the reference frame is source/mirror or the spaceship, but in those rotating bar/s the reference frame is the outside observer? The Special Relativity is built upon equality of reference frames as far as I do understand.
I am not trying to promote some "crackpot" hypothesis or ether (that would be nonsense explanation to those two experiments as well), I simply do not know at all, what the results to those "experiments" could be, and how it is treated by Special Relativity. I do not know, whether I did indeed misunderstood something in SR and what it is?
Again, thanks for your reply
Mike
Michael77 is offline   Reply With Quote
Old Mar 28th 2015, 07:46 PM   #4
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,347
I agree that the laser mark at B in both cases would be displaced, but this is not at all in disagreement either with Michaelson-Morley or Special Relativity. M-M's classic experiment did not concern itself with acceleration. And the equivalence principle of Special Relativity leads in part to the idea that the displacement of the mark at B is similar to the displacement that would be expected in anequivalent gravitational field. In other words the realization that the mark is displaced due to acceleration leads to the realization that the mark would also be displaced by gravity, because of the equivalence principle. So your hypothesis is correct, but it's not new.
ChipB is offline   Reply With Quote
Old Mar 29th 2015, 06:28 AM   #5
Junior Member
 
Join Date: Mar 2015
Posts: 4
Thanks for your answer Chip.
I do not have any hypothesis to be honest. I really do not know what to think about it, I believe that I am misunderstanding something on SR and cannot find what it is.
I'll try to make simpler explanation of what's bugging me:
If we all agree that the beam would in fact get displaced, then I don't think that the common description as in the case with the observer on the spaceship is correct - viz. picture below. (I am using the same picture again, could upload animation, don't know how to do that ...)

It is not that obvious from the picture what I mean. Let's say that the ship is moving with constant speed 10% of c in an empty space devoid of any other mass. The ship is very large, so the mirror is 0.5 light second from the observer on the ship. Because the ship is not accelerating and there is not any other reference frame outside the ship, the observer is not able to determine how fast is the ship moving, in what direction or if it is moving at all.
So far we all agreed that the beam should not get reflected back to its source, but SR forbids prefered frames of reference, so in the case of the observer on the ship, the observer can just claim that the ship is not moving at all and expect the beam to reflect straight back to its source.
As we all agreed, this is not what would happen, so the observer would in fact detect that the ship is not inertial frame of reference - the beam would not get reflected to its source...
Because of the ability of detecting inertial reference frame and because Michelson–Morley like experiments were designed to do exactly this, why did they failed? I mean, even if the speed of light is constant, the fringes would move nonetheless.
So what actually is the correct answer to the situation with the bars/observer on the ship?
**A - Would the light get reflected to match the mark/source as Michelson-Morley experiments say?
**B - Or as we all expect, the light would not match the mark/source and the depiction on the picture is wrong. Also there would be the possibility to detect primary inertial frame of reference, etc.?
There is Sagnac effect suggesting "B", but also Michelson-Morley suggesting "A". What is it that I do not understand? Can anyone explain it?
Thanks

Last edited by Michael77; Mar 29th 2015 at 06:37 AM.
Michael77 is offline   Reply With Quote
Old Mar 29th 2015, 06:50 AM   #6
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,347
Originally Posted by Michael77 View Post
If we all agree that the beam would in fact get displaced, then I don't think that the common description as in the case with the observer on the spaceship is correct - viz. picture below. (I am using the same picture again, could upload animation, don't know how to do that ...)
Unfortunately the picture is not loaded properly, so I can't see it.

Originally Posted by Michael77 View Post
It is not that obvious from the picture what I mean. Let's say that the ship is moving with constant speed 10% of c in an empty space devoid of any other mass.
Hence it is not accelerating, unlike the situation you described earlier...

Originally Posted by Michael77 View Post
The ship is very large, so the mirror is 0.5 light second from the observer on the ship. Because the ship is not accelerating and there is not any other reference frame outside the ship, the observer is not able to determine how fast is the ship moving, in what direction or if it is moving at all.
So far we all agreed that the beam should not get reflected back to its source,
No, I do not agree with this statement! You have changed the scenario by eliminating the acceleration. With constant velocity and 0 acceleration the beam returns to its starting point, just as M-M shows. Again, your original scenario described a situation with acceleration, not constant velocity. Hope this helps.
ChipB is offline   Reply With Quote
Old Mar 29th 2015, 07:39 AM   #7
Junior Member
 
Join Date: Mar 2015
Posts: 4
Sorry I've misled you Chip.
The bars in the experiments were not to be considered as accelerating. The same as with the ship.
There is the link to the picture (spaceship) : http://onlinephys.com/timedilation2.jpg
"No, I do not agree with this statement! You have changed the scenario by eliminating the acceleration. With constant velocity and 0 acceleration the beam returns to its starting point, just as M-M shows."
As you wrote, the beam would return to its source, as SR claims.
And that is exactly what I do not understand.... Why can't be those rotating (non accelerating) bars considered as equal frame of reference as the spaceship would?
Am I misunderstanding something here?

Last edited by Michael77; Mar 29th 2015 at 08:09 AM.
Michael77 is offline   Reply With Quote
Old Mar 29th 2015, 10:54 AM   #8
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,347
I think I misunderstood your original scenario, assuming that the apparatus was accelerating its spin rate. Under constant spin I would expect the laser spot to return to the origin, point A. In the standard M-M experiment the source and mirrors are stationary with respect to each other. In the variant you propose it would be similar to M-M but with the mirror moving tangentially to the light beam. Under that condition from the point of view of stationary point A the beam reflects straight back to the source (assuming it hits the mirror perpendicularly). From the point of view of the mirror at B the light beam hits at angle of incidence arctan(c/v), and reflects back at 180 minus the angle of incidence, which puts it back torwards the source. Now let's consider the rotating system - the only minor complication introduced by the rotating version is that the light source from A must "lead" point B in order to hit the mirror, by the same amount as above: arctan(c/v). So again, A perceives the light neam hitting the mirror at 90 degrees, while B perceives it hitting at an angle.
ChipB is offline   Reply With Quote
Reply

  Physics Help Forum > College/University Physics Help > Special and General Relativity

Tags
expected, experiment, michelson–morley, results, rotating



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Faulty Michelson-Morley Experiment & Expectations. logandiez Light and Optics 18 Oct 27th 2014 09:26 AM
two technical errors MICHELSON MORLEY tesla2 Theoretical Physics 0 Jun 2nd 2013 04:22 AM
Michelson Morley + Marosz VS MR E tesla2 Quantum Physics 4 Oct 24th 2012 10:17 AM
Michelson-Morley Experiment Aryth Special and General Relativity 1 Sep 1st 2009 03:25 PM


Facebook Twitter Google+ RSS Feed