Understanding the MMX Calculation Problem: Vertical vs. Horizontal Analysis"

[Kanesan]

Thanks Harold. I got it all finally.

Thanks again for all your patience and staying with me until the last point. I greatly appreciate it.

 

[Kanesan]

We have extensively analyzed the experiment from the view of observer in the aether frame. I am able to completely understand it. So, we know what result an observer staying with aether should get.

But, all MMX data are taken from the experiment frame which is not in the aether frame (unless they move at same velocity, which is not the case).

So, how to connect them? I believe that if they are moving, they will not have same result. If both can get same result while having a relative velocity, that adds confusion. It feels similar to seeing both frames having same clock speed...

Can anyone help connect this gap or point out my mistake?

Thanks.

 

[harrylin]

We have extensively analyzed the experiment from the view of observer in the aether frame. I am able to completely understand it. So, we know what result an observer staying with aether should get.

In fact we know much more than that: the Lorentz transformations are symmetrical (not sure if that is the right word), they form a group; the inverse transformations look just the same as the Lorentz transformations. If you work out what a moving observer who assumes to be in rest will observe, those transformations tell you that that observer will get just the same results as if he/she was truly in rest. That is an interesting textbook exercise.
BTW, this is already with classical mechanics and the so-called "Galilean" transformations; the Lorentz transformations are more complex but work just as well between inertial reference systems that are in relative motion..

But, all MMX data are taken from the experiment frame which is not in the aether frame (unless they move at same velocity, which is not the case).

So, how to connect them? I believe that if they are moving, they will not have same result. If both can get same result while having a relative velocity, that adds confusion. It feels similar to seeing both frames having same clock speed...
[..].

The analysis is only more complex. The fact of the matter is that you can transform to a "moving" frame and then pretend that it is a "rest" frame, and all physics works as if it is a true rest frame.

As a matter of fact, that's how the whole development towards SR seriously took off: the MMX null result implied that for such measurements one could pretend that the apparatus is temporarily in rest at a certain time and date, but one could equally well pretend that it is in rest at another time and date - even though its velocity will be different. Just as in classical mechanics, no "ether frame" can be detected.

 

[Kanesan]

Hi Harold,

I understand that we can change frames and whatever frame we are in is the rest frame meaning that all the observations we make within our frame will follow same physics laws. I feel that the use of the word "rest" frame unnecessarily complicates things because everyone/everything is always in its "rest" frame. There is no option! So, there is no problem with that part.

I will try to explain better about what I feel like the problem:

In this MMX case, there are two possible observers because of two frames (but one event/experiment). The observer in the aether frame will see the experiment as moving and the observer in experiment frame will see the aether as moving (if it was there/moving). The observers should see/get two different results, at least expect different results based on real/assumed relative motion. If both observers should get null result, that will be a big surprise because I think no two different frames will get same results. Otherwise it will look as absolute result, common to all frames (at least common to two frames).

We calculated one observer's expected results (that of one in the aether frame) which is not that important I feel. Just a fun exercise!

The experiment's observer (photographic plate) was in the experiment's frame, not moving with aether frame. Since we have the real results in this frame, this is the one for which we should have done all the calculations, but not done so! I don't know how to transform the above calculation result to this frame and even if I could do, the current choice is a puzzle.

So, basically my confusion is that why the aether frame was chosen for calculations rather than the experimenter's frame for which we have results. The understanding of the two frames are clear but the choice selection is not.

I hope this explains my confusion better.

 

[harrylin]

Hi Harold,

I understand that we can change frames and whatever frame we are in is the rest frame meaning that all the observations we make within our frame will follow same physics laws. I feel that the use of the word "rest" frame unnecessarily complicates things because everyone/everything is always in its "rest" frame. There is no option! So, there is no problem with that part.

There is no reason to assume that you are at rest in a "rest frame", you are free to assume that you are co-moving with a "moving frame". As a matter of fact, if you are sitting in your garden chair and you look at your GPS receiver, this receiver calculates with the assumption that you are co-moving with the surface of the Earth. And it provides you with your coordinates in your "moving" frame. GPS uses the "Earth Centered Inertial" frame.

In this MMX case, there are two possible observers because of two frames (but one event/experiment). The observer in the aether frame will see the experiment as moving and the observer in experiment frame will see the aether as moving (if it was there/moving).

Again not so; and this relates to my remark here above. An observer who is co-moving with the apparatus may choose for example the Solar Inertial frame as "rest" frame, which boils down to pretending, for ease of calculation, that the sun is at rest in the ether.
However, in common SR examples, if no precision is given, it is implied that the observer who is co-moving with the apparatus chooses the inertial frame relative to which the apparatus is temporarily in rest, as "rest frame". That is effectively the same as assuming that at that moment the apparatus is in rest in the ether.

The observers should see/get two different results, at least expect different results based on real/assumed relative motion. [..]

That remark is puzzling to me. According to special relativity, we should expect the same results for an interferometer in rest as for a moving interferometer. Perhaps the introduction section of Einstein's 1905 paper will be helpful (just consider the text until the first §): http://www.fourmilab.ch/etexts/einstein/specrel/www/

We calculated one observer's expected results (that of one in the aether frame) which is not that important I feel. Just a fun exercise!

No, that is wrong. We calculated what in two different states of motion the result of the experiment will be, based on certain assumptions. That interference produces for example a certain interference pattern on a photographic plate. The result of such an experiment is an "absolute"; such facts cannot depend on who is looking! Assuming length contraction, you should find how an observer who assumes that the apparatus is in motion will give the same prediction as an observer who assumes the apparatus to be in rest.

The experiment's observer (photographic plate) was in the experiment's frame, not moving with aether frame.

Everything is always in any inertial frame...

[..] So, basically my confusion is that why the aether frame was chosen for calculations rather than the experimenter's frame for which we have results. The understanding of the two frames are clear but the choice selection is not.

I hope this explains my confusion better.

Perhaps the concept of the experiment is not clear. Light was modeled as a wave that propagates at c relative to the ether. Michelson and Morley first repeated an experiment that showed that light propagates independently of matter, and is even little affected by motion of matter through which it propagates. They saw this as a confirmation of the stationary ether model. They expected to be able to detect the velocity relative to the ether because they assumed that moving objects keep the same shape as objects in rest. It was unlikely, but possible, that the Earth was temporarily nearly at rest in the ether, and it was therefore planned to measure at different times of the year. I think that they did not do that, but others did.

Whatever inertial frame you choose as "rest frame", the apparatus will be moving in it at some time. Thus a good theory has to explain the null result for the apparatus in rest and for the apparatus in motion.

 

[Kanesan]

There is no reason to assume that you are at rest in a "rest frame", you are free to assume that you are co-moving with a "moving frame".

I am thinking that all co-movers are in same frame! Not to confuse with too many frames if they are co-moving. Is it wrong?

That is effectively the same as assuming that at that moment the apparatus is in rest in the ether.

I have difficulty grasping this. If we assume rest (temporarily or permanently), doesn't that mean velocity is zero?

We calculated what in two different states of motion the result of the experiment will be

We have calculated only once. Just one experiment (one view) needed two calculations as one experiment can need several individual calculations.

Today I am thinking that may be I am mixing up some aether and SR expectations! May be in aether theory everything is supposed to be absolute and calculation from either view should be same (unlike from SR). Can you see if that was my mistake?

According to special relativity, we should expect the same results for an interferometer in rest as for a moving interferometer.

But this one suggests that my above conclusion was wrong as well!

Thanks Harold.

 

[harrylin]

I am thinking that all co-movers are in same frame! Not to confuse with too many frames if they are co-moving. Is it wrong?

I don't understand that. A "frame" is just jargon for an infinite number of reference systems that are in rest relative to each other (they can have their origin at other positions, etc.). In SR, an inertial reference system tracks the position as function of time of everything in the universe. So, if you consider 10 inertial reference systems that are all in different states of motions, then you are in all those reference systems but you can at most be co-moving with one of them.

I have difficulty grasping this. If we assume rest (temporarily or permanently), doesn't that mean velocity is zero?

Yes. In particular, this implies that all light is supposed to propagate at speed v=c in all directions with respect to that system.

Hopefully you are familiar with classical physics which similarly uses rest systems and moving systems, with the same principle of relativity for mechanics. You can choose any inertial system and call it "rest system"; co-moving objects have zero momentum and zero kinetic energy based on that assumption. It is said that such physical quantities are "relative". In special relativity much more is "relative" in that sense.

We have calculated only once. Just one experiment (one view) needed two calculations as one experiment can need several individual calculations.

Oops yes you are right, in this calculation there's only one state of motion considered. https://en.wikipedia.org/wiki/Michelson–Morley_experiment#Light_path_analysis_and_consequences

However that calculation was meant for an experiment in different states of motion. It's alluded to in that section as follows "Any slight change in the spent time would then be observed as a shift in the positions of the interference fringes." In fact it's explained clearly in the section "Michelson–Morley experiment (1887)":
"The mercury trough allowed the device to turn with close to zero friction, so that once having given the sandstone block a single push it would slowly rotate through the entire range of possible angles to the "aether wind," while measurements were continuously observed by looking through the eyepiece. The hypothesis of aether drift implies that because one of the arms would inevitably turn into the direction of the wind at the same time that another arm was turning perpendicularly to the wind, an effect should be noticeable even over a period of minutes."

Thus, for a change of the spent time they observed:
- the interference pattern of the interferometer fast moving in one direction (at least at one time of the year)
- the same with the interferometer moving in a direction perpendicular to it.
For that means the apparatus was rotated, and the changes in the interference were recorded.

It explains what Michelson expected to find, based on Maxwell's theory of optics, combined with Newton's theory of mechanics. In fact those two theories are incompatible, but neither Maxwell nor Michelson had realized that.

PS. I now notice that also the issue of mirror reflection angle is discussed there, under the header "Mirror reflection".

Today I am thinking that may be I am mixing up some aether and SR expectations! May be in aether theory everything is supposed to be absolute and calculation from either view should be same (unlike from SR). Can you see if that was my mistake? [...]

Probably you mixed them up, but it's just the other way round! In ether theory combined with Newton's mechanics, a detectable effect of motion relative to the ether was predicted, but it was not found. That is explained in the Wikipedia article.

This is also referred to in a general way in Einstein's introduction: "the unsuccessful attempts to discover any motion of the Earth relatively to the “light medium”"
- http://fourmilab.ch/etexts/einstein/specrel/www/

Apart of the introduction, also the first line of §1 is useful, as it uses "stationary system" just as in classical mechanics.