# Michelson–Morley experiment?

Dalespam, harrylin, ghwellsjr, anyone else:

Can you please take one last look at this attached diagram, and rephrased description, and tell me if you agree: (please note that synchronization is always done when the clocks are not apart, and that, in the diagram example, the clocks [A,B] are synchronized and moved apart, only when they are in the stationary frame).
OK then, this is for me the last time that I reply to this in this MMX thread.

Your clocks A and B are synchronized and moved apart when the system is in rest, and next the whole system with clocks is brought to a certain speed - I will assume a speed parallel to the drawing.

Then the one-way speed of light will not appear isotropic in that moving system because the clocks are not synchronized according to the standard SR convention.
What you are saying is, it is absolutely possible to keep them synchronize (A,B) in relation to one another, after they are moved apart, within the stationary frame.
[edit:] If you mean, if they are moved apart, they will remain in sync according to the frame in which they are in rest? Yes. That is however relative, not absolute.
What it is, that makes the device, with the exact same configuration, work the same, in both a moving and stationary frame (in contrast to the diagram) - has to do with what happens in the moving frame, which is: Time dilation of the time delay of the middle clock (C), within the moving frame, coupled with the fact, that both clocks (A,B) are measuring time, at different positions, within the moving frame. The same different positions (A,B) as in the stationary frame, now in the moving frame, makes the difference.
So the difference that compensates for C clock time dilation and makes the device work always the same in both frames, can occur in two ways: 1. When acceleration occurs - if the clocks are synchronized, and moved apart, in the stationary frame (as in the above example and diagram), Or: 2. While the action of moving the clocks apart occurs (in relation to the stationary frame), if synchronization is done in the moving frame.
Huh?! No, the main point of relativity is that you can choose which frame you call "moving" or "stationary"! Correcting your errors, we get:

The difference that compensates for C clock time dilation and makes the device work always the same in both states of motion, can occur in several ways, for example:

1. If the clocks A and B were synchronized, and moved apart before acceleration (as in the above example and diagram), and then re-synchronized by means of light signals.
Or:
2. If those clocks are synchronized and equally moved apart relative tot the system after the acceleration, they will remain synchronized according to the synchronization convention.

And also, you are saying that there is no 'more mechanical' or 'only one ideal spoken words' explanation, from this point on, but at that phase of understanding further, only is a mathematical description available.

Is that correct?

Thanks,

Roi.
No, of course not- just read what we wrote, none of us were saying that!

Success,
Harald

Last edited:
1. If the clocks A and B were synchronized, and moved apart before acceleration (as in the above example and diagram), and then re-synchronized by means of light signals.
Or:
2. If those clocks are synchronized and equally moved apart relative tot the system after the acceleration, they will remain synchronized according to the synchronization convention.

Success,
Harald
1. agree

2. If they move within the rest frame, they are no longer a part of that frame.
The frame has an unknown velocity which will not be zero, therefore each clock will move to its destination at different speeds and be out of synch.

1. agree

2. If they move within the rest frame, they are no longer a part of that frame.
The frame has an unknown velocity which will not be zero, therefore each clock will move to its destination at different speeds and be out of synch.
Clocks that are equally moved apart will remain in sync as measured in the frame in which they were (and are again) in rest. Obviously that means that they must then be out of sync as measured in a frame in which they are moving.

Dale
Mentor
If they move within the rest frame, they are no longer a part of that frame.
I don't like this phrasing. They are material objects and as such they cannot be "a part of that frame" to begin with. Frames are essentially just coordinate systems, so material objects do not go in or out of frames and they are never a part of a frame. I would have said it:

"If they move within the rest frame, they are no longer stationary in that frame."

I don't like this phrasing. They are material objects and as such they cannot be "a part of that frame" to begin with. Frames are essentially just coordinate systems, so material objects do not go in or out of frames and they are never a part of a frame. I would have said it:

"If they move within the rest frame, they are no longer stationary in that frame."
I didn't know we had to submit our posts for review or editing.
On examining your statement, there seems to be redundancy (red or blue).

"If they move within the rest frame, they are no longer stationary in that frame."
Sorry you were not appointed a role model for society, but keep trying!

Coordinate systems have at least one reference object (the origin), right? A frame has a system of clocks, right? Aren't clocks objects?
In the frame with its origin at earth center, is there a clock at Alpha Centauri, no. Do we need one there, no.
A single observer with a single clock and laser is sufficient to make measurements where feasible.

The point being made was physics, not grammar.
If the object is moving relative to the frame or any object that is at rest in that frame, its clock rate will be different, until it arrives at its destination in that frame. It will run at its previous rate but will be out of synch with the master clock which did not move. Synchronization is used here as equal simultaneous readings, which is the purpose of the method used in SR.