# I'm learning about relativity and, by extension, the classic

1. Nov 1, 2009

### fhisicsstudnt

I'm learning about relativity and, by extension, the classic Michelson-Morley setup. I cannot see why a "fringe effect" was expected and could use some discussion. Here is my reasoning.

Firstly we imagine the light signal to propagate from the SM (silvered mirror, in the "center") to M1 and M2 and back at c (distance for each of 2*x), arriving at the detector simultaneously. This is taking the whole thing to be "stationary". The data is recorded in the interferometer's own reference frame, the scientist is in that frame, and so the whole thing considers itself to be at rest consistent with both "new" and "old" mechanics. There is no "fringe effect".

Next we imagine the whole thing to be in uniform translation at velocity v. The only way to accomplish this is to set up a 2nd reference frame which we consider stationary, and which we consider the interferometer to be in uniform translation with respect to. For simplicity we imagine the interferometer to be translating along the axis of one of the arms, call it the a axis.

With respect to the stationary frame the light signal propagates at v1a = c+v one way and v2a = c-v the other. The distance for the signal to travel relative to the stationary frame is x-v*t1a one way and x+v*t2a the other. t1a = (x+v*t1a)/(c+v) = x/c and t2a = (x-v*t2a)/(c-v) = x/c. Total transit time is Ta = 2*x/c

The other signal propagates at vb= (c2+v2)1/2 both ways relative to the stationary frame. The distance for the signal to travel, relative to the stationary frame, is xb = (x2+v2*tb2)1/2 both ways. The transit times are both tb = xb/vb. Tota transit time is Tb = 2*tb = 2*x/c.

So I don't get it. Even without doing the math out, all we are doing is taking the same situation and analyzing it with everything in uniform translation. Whether we are thinking "classically" or in the context of SR, uniformly moving frames and stationary frames are indiscernible from each other. If we consider the interferometer as stationary and the two light beams return to the silvered mirror simultaneously, then any uniformly translating frame will consider the two light beams to return simultaneously also, this is true either classically or relativistically. The only difference between classical and relativistic is that the two beams will classically arrive at M1 and M2 simultaneously in any moving frame whereas relativistically they will not necessarily appear to arrive at M1 and M2 simultaneously, depending on the frame.

2. Nov 1, 2009

### DrGreg

Re: Mmi

Welcome to Physics Forums, fhisicsstudnt!

You've got this the wrong way round. The assumption (which the experiment disproved) was that the speed of light was c relative to the "stationary" aether frame and not c relative to the apparatus.

You can do the calculation two ways: in the supposed aether frame with a speed of c and two modified distances, or in the apparatus frame with a modified speed and a distance of x. You've modified both the distance and the speed!

3. Nov 1, 2009

### fhisicsstudnt

Re: Mmi

Hey, thanks DrGreg, that was really fast!

If we were to take a hypothetical stationary frame in which something moves at c within it and translate it vt relative to another frame, then whatever is moving within the translated frame gains vt. This is just basic classical kinematics isn't it? This is what I learned.

We can grant the velocity v either to the apparatus frame or to the other frame. Since the aether was supposed stationary, it makes sense to grant v to the apparatus frame. Everything in the apparatus frame is then moving at +v relative to the stationary frame.

The problem seems even simpler than this, though. In classical kinematics, two events that are simultaneous in one frame are simultaneous in another uniformly translating frame. So if the two pulses of light strike the silvered mirror simultaneously in the apparatus frame then they must also do so if the apparatus is in uniform translation. Classical and relativistic kinematics agree on this conceptually. What they disagree on is the simultaneity of non colocal events, namely the events wherein the pulses of light strike M1 and M2. If in the apparatus frame the pulses strike M1 and M2 simultaneously then, in relativistic kinematics, the pulses will not strike M1 and M2 simultaneously in a uniformly translating frame. However, since the event wherein the 2 pulses strike the silvered mirror is both colocal and cotemporal, no kinematics that I know of can describe them as non simultaneous in any unaccelerated frame.

Moving frames, viewed from another frame, appear to have different distances and speeds. I modified both because that's how I was taught kinematics. The result, that uniform translation does not alter the outcome, is also consistent with everything I've learned about classical kinematics. Uniformly translating frames are not different, none are special, all should get the "same result" regarding simultaneity, ordering of events, etc.

4. Nov 1, 2009

### JustinLevy

Re: Mmi

Let us make this even simpler.

Consider the same setup with sound waves and the medium as air. The velocity of the sound is $v_{sound}$ with respect to the air. Go into the 'moving frame', and the velocity isn't increased proportionally so that it still appears to be $v_{sound}$. Heck, there is actually a speed in which the sound waves would appear to be stationary in the 'moving frame'. There is a real and physical difference for sound propagation in these two frames.

You keep wanting to remove the medium, which is the very thing the MM experiment was trying to measure.

5. Nov 1, 2009

### JustinLevy

Re: Mmi

My comments were referring to you latest post. Rereading your first post, it seems like this is just an issue with coordinate changes.

Here are the two situations:
the setup as you described, then either
speed of light = c (a constant independent of direction)
OR
speed of light = c(1 - v cos theta), where theta is the angle from the direction of travel in the 'medium'

So for the first case, the time to travel one leg to the mirror and back is: t = x/c + x/c = 2x/c
And it is the same for the other leg.

Now for the second case, the time to travel the 'theta=0' leg is t = x/(v-c) + x/(v+c)
the time to travel the 'theta=90 degrees' leg is t = x/c + x/c = 2x/c

Compare that to what you wrote:
The problem is that you were doing t_(frame b) = x_(frame a) / v_(frame b)

That is why it was not working.
I'm sorry if my previous post confused things because I didn't read through all your math first.

6. Nov 1, 2009

### fhisicsstudnt

Re: Mmi

Comparing something that isn't sound to sound wouldn't, I think, make this simpler. The simplest place to start is, in my opinion, the basic result of classical kinematics:

Events simultaneous in a stationary frame are also simultaneous in any uniformly translating frame.

I don't think this is true. The "medium" is just a hypothetical stationary reference frame. I have invoked such a frame.

Just to be clear, my labels "a" and "b" are not to indicate different frames but rather different pulses. The "a" pulse is the one traveling in the direction of v and the "b" pulse is traveling perpendicular to v. All my quantities are wrt the stationary frame. The distance-traveled for b is just the pythagorean distance (x2+v2*tb2)1/2 and the velocity is just the pythagorean velocity (c2+v2)1/2.

The fundamental issue is whether events judged simultaneous in one frame are also judged simultaneous in a comoving frame, classically. I thought yes, which would rule out any chance of a positive result.

7. Nov 1, 2009

### Staff: Mentor

Re: Mmi

You have to put yourself in the same frame of mind as Michelson, Morley and just about every other physicist of their era. To them, the "medium" (the lumimiferous aether) was a real physical entity, albeit with rather strange properties.

8. Nov 1, 2009

### Cleonis

Re: Mmi

Let me concentrate on the question whether according to pre-relativistic theory of light-propagation frames are discernable.
In pre-relativistic theories there was a supposition of a luminiferous ether (a light bearing medium) and the light was supposed to propagate with a particular velocity with respect to that immovable ether, with no supposition of time dilation or length contraction effects.

I attach two diagrams, each is 256x256 pixels in size (I don't know how speedily they will be approved).
Both are illustrations http://www.cleonis.nl/physics/phys256/special.php" [Broken]

The red lines represent worldlines of clocks that are co-moving. In the diagram newtonian_stationary the red worldlines represent zero velocity reletive to the ether. The yellow lines represent the worldlines of pulses of light. The yellow worldlines are at an angle of 45 degrees, to represent the speed of light relative to the ether. The total procedure takes 10 units of time.

In the diagram newtonian_moving the red worldlines are at an angle, representing that the clocks have a velocity with respect to the immovable ether.
The diagram is a representation of the moving trains thought experiment as considered from a pre-relativistic point of view. The yellow worldlines (the pulses of light) are at an angle of 45 degrees in the diagram. From halfway along the train one lightpulse is sent to the front and the other lightpulse is sent to the rear. Both pulses are returned and they arrive back at halfway the train simultaneously.
The important thing: if the train has a velocity relative to the luminiferous ether then according to a pre-relativistic theory the total time to complete the procedure will be longer than 10 units of time. And the larger the train's velocity relative to the ether, the more the duration will exceed 10 units of time.

This illustrates why according to a pre-relativistic concept of light-propagation frames ought to be discernable.

According to pre-relativistic theory: If there would be multiple trains, all with exactly the same length, moving with a velocity relative to each other, then only onboard the train that is stationary with respect to the ether will the procedure take 10 units of time; on the other trains the procedure will take longer to complete.
It is only in terms of relativistic physics that the prediction is that on all trains the duration will be 10 units of time.

Cleonis

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9. Nov 1, 2009

### JustinLevy

Re: Mmi

You are misunderstanding something very fundamental. The problem is that you are asking the wrong question. We're trying to show you that, but since we are failing to explain it to you, we end up arguing about two different things. Please keep this in mind, when I try to explain again.

(A quick aside to prevent other accidental disagreements:
First, some terminology. Usually in modern physics, classical just means non-quantum. So GR is a classical theory. It is obvious what you mean by it here, and it is convenient, so I will use your terminology here: in this thread, by "classical" I mean "Newtonian"/"pre-relativity".)

YES, you are correct that two events that are simultaneous in one inertial frame will be simultaneous in all inertial frames classically.

But that is NOT the question the MM experiment was looking at. They rotated the experiment to check things, but to make the discussion simpler, let's look at it like this:

1] do the experiment with the interferometer at rest with respect to the 'aether'
then
2] do the experiment with the interferometer moving with respect to the 'aether'

Are the results the same?
The answer, classically with a luminous aether, is NO.
Therefore, classically, one could detect the aether.

Instead, you are doing only one experiment (#1), then just re-analyzing that same experiment in another coordinate system. Of course the results of a single experiment are independent of the coordinate system we use to analyze it.

It took me awhile to realize that is what you were doing. The thread, and ensuing confusion, make much more sense now.

Last edited: Nov 1, 2009
10. Nov 1, 2009

### fhisicsstudnt

Re: Mmi

Hey Cleonis,

Nice site. I saw the figures you referenced.

The scientist on board each train will consider himself in an unmoving frame and measure 10 units of time on any of the trains. Since the scientist considers himself unmoving on the train any results he obtains are valid in any other situation in which he considers himself to be in an unmoving frame. If the pulse moves at c wrt the aether and the train moves at v wrt the aether then each pulse goes c+/-v on the way out and c-/+v on the way in. If the train moves at 2*v then the pulses go c+/-2*v out and c-/+2*v in. In my original post you can replace all the v's with 2*v or 3*v and the transit time remains 2*x/c.The data collected in the MM setup was all collected in the apparatus frame (i.e. within the train itself), so again there seems to be no possibility of a positive result.

In the first diagram we are in the apparatus frame. The pulses all travel identical distances and identical velocities, arriving back simultaneously.

In the 2nd diagram we are in the aether frame, in which the apparatus/train is translating at v. In going to the 2nd diagram you change the angle of the red lines to signify the motion of the apparatus (or train) of v wrt the aether but do not change the angle of the yellow lines. You have to translate the entire frame don't you? You went from the train frame to the aether frame, everything in the train frame has to be translated, not just some things, right? Again that's how I was taught kinematics. If we had cars or baseballs flying around in a cage/garage, and suddenly the whole thing started flying away from me (uniformly), I would have to consider the baseballs and cars (and photons) to be translating also (they'd have some angle in the 2nd diagram). Transit times from a point back to that point would be the same whether it was going at v, 2*v, 3*v, etc.

I still don't understand how. Moving, unmoving, the interferometer itself always considers itself to be "motionless". A thing doesn't move wrt itself. In one experiment the interferometer is motionless wrt the aether and the pulses are simultaneous at the silvered mirror. In another experment the interferometer is translating wrt the aether and, since events simultaneous in one frame are simultaneous in a uniformly translating frame, the pulses are again simultaneous at the silvered mirror.

But what is the aether, except another coordinate system? The data is always collected in the apparatus frame, whether the apparatus is moving wrt the aether or not. The data is "moving" with the apparatus, so it can't "know" if the apparatus is moving wrt the aether or not.

Note that I'm not considering the additional layer of complication involved in the stationary frame actually collecting data from the moving apparatus frame. This would involve accounting for doppler shifts etc. which isn't really pertinent.

I just can't understand how, considering pre-relativistic kinematics, a positive result could possibly have occurred.

11. Nov 1, 2009

### Staff: Mentor

Re: Mmi

At the time when M&M did their experiment, physicists considered the luminiferous aether to be a real physical medium for light waves, just as air is a real physical medium for sound waves, or water for water waves.

In that picture, if the laboratory and the aether are moving with respect to each other, the effect on light waves should be exactly like the effect on sound waves if the laboratory and the air are moving with respect to each other. In the laboratory frame, there is a "wind" of aether or air "blowing" through the laboratory. Light and sound should travel faster when going "downwind" and slower when going "upwind."

12. Nov 1, 2009

### Cleonis

Re: Mmi

I have formed a new hypothesis as to how you arrive at your opinions. I now think you believe that emission of light is in its physics characteristics similar to emission of a bullet from a gun (with velocity composition according to the rules of pre-relativistic mechanics).

Your postings have demonstrated the following two things: you are unaware of how light-propagation is conceptualized (both the pre-relativistic point of view and the relativistic point of view), and most unfortunately: you are immune to explanation. I'm sorry to say that; it's a harsh assessment, but the time has come to make that call.

The irony of it all is that relativistic physics does have a profound bearing on that fundamental question: how does emission and propagation of light relate to mechanics?

Cleonis

13. Nov 1, 2009

### fhisicsstudnt

Re: Mmi

Agreed, if the lab is moving at v wrt the aether then light will move slower one way (c-v upwind) and faster the other way (c+v downwind). I took this into account in my original post. When you do this there is no net effect.

Are you telling me, then, that pre-SR something (*anything*) moving at c in one frame is also moving at c in a relatively moving frame with some velocity v? How can you justify this? It contradicts everything I know of. You are telling me that the water waves on the beach, moving toward the shore at c relative to the water/earth, also move at c relative to the speed boat. You're telling me that when I'm on a moving train and I shout to someone, the sound has velocity c relative to the ground/air and also the train? That the velocity is not c+/-v and the distance-traveled c+/-v*t? How can you justify slanting the red lines but not the yellow lines? When you translate a frame you translate everything moving within it.

Please, help me conceptualize how light propagates (in the pre-relativistic view for now). I doubt I am "immune to explanation" since I have made significant progress already through a physics program and have been a successful student in general.

14. Nov 2, 2009

### JustinLevy

Re: Mmi

The aether is not a coordinate system. It is the MEDIUM in which light propagates.

Is the "air" a coordinate system? No. But there may be an inertial frame in which the air is at rest.

So while the aether is not a coordinate system, there is an inertial frame in which it is at rest. Pre-relativity, this is a special frame in which the speed of light is isotropic and constant. In any inertial coordinate system moving relative to this rest frame, the speed of light would be anisotropic.

Do you understand this?
You need to understand the concept of the aether as a medium before we can continue. You seem to understand (for example you use c+v, and c-v, and your upstream vs downstream comments), but as soon as it comes to calculating anything you incorrectly mix the coordinate systems.

Alright, if you can understand that the aether is a medium, please reread my previous post and think about it for a bit. Then I'll work the math out for you as an example.

...

In the experiment where the aether is at rest with respect to the interferometer, we have:
t(0 degrees) = x/c + x/c = 2x/c
t(90 degrees) = x/c + x/c = 2x/c

Now in the DIFFERENT experiment (not just a different coordinate system, but a DIFFERENT EXPERIMENT) where the aether is moving at speed v with respect to the interferometer, we have:
t(0 degrees) = x/(c-v) + x/(c+v)
t(90 degrees) = x/c + x/c = 2x/c

As you can see, they are different. That is why, if the pre-relativity aether idea was correct, it would be possible to measure the velocity of the aether.

Again, the problem with your math is that you are mixing in coordinates from different frames. Instead of
t_moving = x_moving / velocity of light_moving,
you are using
t_moving = x_aether rest frame / velocity of light_moving

Since in pre-relativity, all inertial frames will agree on the time, you can calculate the time in either coordinate system. But you cannot mix and match the distance and velocity like that. Choose one coordinate system and stick with it through-out the calculation.

Last edited: Nov 2, 2009
15. Nov 2, 2009

### A.T.

Re: Mmi

This is how light behaves in reality and is modeled in SR, not pre-SR.

Physics doesn't make or justify the laws of nature, it just describes them.

Not everything, just the Galilean coordinate transformation, which is replaced by the Lorentz coordinate transformation.

Water waves are not light

In the pre-relativistic view it behaves like water waves. But that was ruled out by the MM experiment.

Last edited: Nov 2, 2009
16. Nov 2, 2009

### Al68

Re: Mmi

For the same reason a positive result will occur for sound waves with the same setup apparatus in motion with respect to the air.

They assumed the aether was the medium for light in the same way air is a medium for sound, not just a coordinate system. And they assumed the speed of light was constant relative to the aether in the same way the speed of sound is constant relative to its medium (air).

17. Nov 2, 2009

### fhisicsstudnt

Re: Mmi

Thank you for the patience Justin.

Yes, mathematically the air can be considered a coordinate system. In fact, to treat it properly we must consider it as a coordinate system/frame. If we consider it at rest then other things move wrt to it (or not).

Okay, the physical aether itself is not a coordinate system. However mathematically we treat it as a stationary frame, a stationary coordinate system.

Yes, I understand the aether as a medium. In math it becomes the stationary frame.

...
Agreed.

I think it is bad setup/form to consider the aether to be moving, since it is physically defined as at rest. To be consistent with the physical hyp (stationary aether) the interferometer should be considered to be moving at v, however of course we can do it this way.

For 0 degrees, the light does indeed now approach the mirror at c-v and return at c+v in the lab frame if, before the lab frame was moving, light was approaching/returning at c. Also, if before the lab was moving light pulses had to traverse the distance x, they now have to traverse the distance x+ct1 and x-ct2. The distance an object must travel, as well as its velocity, are altered by moving the frame. It wouldn't make sense to alter one without the other. If a train moves away from me (and the ground) across a km of track at v, then the track sprouts rockets and the tracks fly away at w, I now consider the train to be movng v+w relative to me and to be traversing more than 1 km (relative to me/the ground). The train now moves much further than 1km across the ground (lab) when it moves 1 km across the track (aether). The time for the train to traverse the km of track is of course identical to before.

Consider:
A conveyor at the airport d long. Jack eschews the conveyor and walks beside it at w relative to the ground (stationary aether) and the lab. He is a light pulse in experiment one. I get on the conveyor (moving aether), which is "conveying" (aether blowing) at v relative to the ground (laboratory), and I walk at w relative to the conveyor/aether, v+w relative to the ground/laboratory. I am a light pulse in experiment 2. When I step onto the conveyor I make a footprint indicating my exact starting location. Jack makes a similar footprint to indicate his exact starting location. This marks the silvered mirror location in the actual experiment. Both of us walk to the end of the conveyor, turn around, and walk back. We will both land on our original footprints simultaneously, we will arrive back at the silvered mirror at the same time. No surprise there, this is just the parallel arm.

Consider a square conveyor moving perpendicular to the other one. I step on the conveyor, making my initial footprint to indicate the exact silvered mirror location, and approach the opposite side at w relative to the ground/lab, while being dragged perpendicularly at v relative to the ground/lab. I will reach the end of the conveyor (M1) after d/w units of time. That's a velocity of (w2+v2)1/2 relative to the moving aether and a distance of (d2+v2*tb2)1/2 relative to the moving aether.

I'll try to anticipate a potential misguided objection that I am granting myself w relative to the lab here, whereas I gave myself w relative to the aether before. The light pulse is moving at w relative to the aether and is then reflected through 90 degrees. All the velocity it originally had in the +y direction is now in the +z direction. It is as if I were walking along at w relative to the conveyor(v+w relative to the ground) and then pivoted 90 degrees to the side and continued my previous stride. I now have velocity w along the +z axis relative to the ground. I certainly cannot possibly retain my previous velocity relative to the aether after pivoting a quarter turn. I must perform an elementary vector analysis to get my new resultant velocity. The distance I travel is also greater relative to the aether as I am both moving the distance d across the conveyor and the distance vtb to the right.

Jack eschews the conveyor and approaches the opposite side at w relative to the stationary aether (and lab). We both reach the opposite side simultaneously (M2), turn around, and come back to the original side Again we each land our foot back onto our original footprint simultaneously. We both come right back to the silvered mirror at the same time.

No matter what we do, we will land right back on our respective starting footprints. We both go and then retrace our own exact steps.

This thought experiment is representative of the MMI.

I believe you are mixing coordinate systems. If light pulses move at c wrt the stationary aether when the lab is stationary, then the lab is moving through the aether at v the pulse in the perpendicular arm traverses a longer distance through the aether at a faster speed, there and back. This is as I illustrated with the conveyors.

Whether sound or light, this setup does not appear capable of returning a positive result. Has anyone done such an experiment with sound?

JustinLevy and Al68:

You are telling me that a frame in uniform translation with respect to the air will lead to different results with sound than a frame that is stationary with respect to the air in a MM setup?

Last edited: Nov 2, 2009
18. Nov 2, 2009

### JustinLevy

Re: Mmi

I just realized I was using the wrong form for the "anisotropic" speed of light for 'pre-relativity'. I think I was essentially still keeping around time dilation. Regardless of the specifics, I made a mistake in the math, which most likely made it difficult for you to understand the qualitative explanations as well.

So let me try again.
x' = x - vt
y' = y
t' = t

For something moving at speed c at angle 0 (with respect to the x axis) in the unprimed frame, it is moving at angle 0 in the primed frame as well, but with speed c-v.

For something moving at speed c at angle theta in the unprimed frame, such that it was moving at angle 90 degrees in the primed frame, then it's speed is:

c'_x = cos(theta) c - v = 0
c'_y = sin(theta) c

so cos(theta) = v/c, which gives sin(theta) = sqrt(1 - (v/c)^2)
so we have:
c_x = v
c_y = c sqrt(1-(v/c)^2)
c'_x = 0
c'_y = c sqrt(1-(v/c)^2)

=======================
Okay, now let's work out the prediction for the experiments again.
You can use whatever coordinate system you want. But choose one and stick with it through out the calculation.

In the experiment where the aether is at rest with respect to the interferometer, and using the rest frame of the interferometer (which is also the rest frame of the aether), we have:
t(0 degrees) = x/c + x/c = 2x/c
t(90 degrees) = x/c + x/c = 2x/c

Now in the DIFFERENT experiment (not just a different coordinate system, but a DIFFERENT EXPERIMENT) where the aether is moving at speed v with respect to the interferometer, we have:

In the rest frame of the aether (so the interferometer is moving with velocity v)
and since we are using the rest frame of the aether the speed of light is isotropic in this frame
t(90 degrees) = x/ (c sqrt(1-(v/c)^2)) + x/(c sqrt(1-(v/c)^2)) = (2x/c) * (1/sqrt(1-(v/c)^2))

t(0 degrees) = t_away + t_back
t_away = (x + v t_away)/c
so, t_away = x/(c - v)
t_back = (x - v t_away)/c
so, t_back = x/(c + v)
so, t(0 degrees) = x/(c - v) + x/(c + v)

In the rest frame of the interferometer (so the aether is moving with velocity v according to this frame) and since the aether is moving in this frame the speed of light is anisotropic in this frame
t(90 degrees) = x/(c sqrt(1-(v/c)^2)) + x/(c sqrt(1-(v/c)^2)) = (2x/c) * (1/sqrt(1-(v/c)^2))
t(0 degrees) = x/(c-v) + x/(c+v)
Which of course matches the results obtained when calculating this same experiment in the aether rest frame.

Notice however, that the result is DIFFERENT from the experiment where the interferometer was at rest with respect to the aether.
Hopefully you can see more clearly now how you were mixing the different coordinate systems incorrectly. (If you are using the distances according to the aether rest frame, then you must use the velocities according to that frame. Instead you used the velocities according to the interferometer rest frame.)

19. Nov 2, 2009

### yuiop

Re: Mmi

That would be right.

Here is a setup that might be easier to analyse or visualise. Imagine a large L shaped device being towed across a calm lake. The arms of the "L" shape are the same length as each other, as in the MM experiment. There are 2 swimmers. One has to swim from the fulcrum of the L alongside the arm that is parallel to the direction the L is being towed in, and then back to the fulcrum. The other swimmer has to swim a diagonal path to get from the fulcrum of the L to the end of the arm that is orthogonal to the direction the L is being towed in and then back to the fulcrum again. Even though the two swimmers swim at exactly the same speed as each other relative to the water and both start at the same time, the swimmmer that swims the diagonal path returns to the fulcrum before the other swimmer. In pre relativistic times, photons were thought to move in a manner somewhat similar to the way the swimmers move in the water and that is why they expected the light signals in the MM experiment to return at different times and cause an interferance pattern.

There is a nice flash animation of what I have just described here: http://www.upscale.utoronto.ca/PVB/Harrison/SpecRel/Flash/MichelsonMorley/MichelsonMorley.html

20. Nov 2, 2009

### fhisicsstudnt

Re: Mmi

Hey guys, I see it now. Thank you so much, this has been quite a learning experience for me. The math and the animation all helped.

I even did it out on graph paper. I redrew the cross displaced half a box over repeatedly with the dots always displaced by one box length. Quite a surprising result!

You guys rock.