Determined c in S v. Result of the Michelson_Morley Exp.

  • Thread starter arbol
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In summary, the conversation discusses the concept of a moving coordinate system S' in relation to a stationary system S, and the path of a ray of light emitted by S' in both systems. The length of the path of the ray of light is determined to be the length of a rigid rod, and the time it takes to complete the path is also calculated. The conversation then mentions the result of the Michelson-Morley experiment, which shows that the length of the rigid rod in the stationary system S is affected by its motion in relation to the moving system S'. However, there is a problem with the result as it does not account for the determined velocity of the ray of light in the stationary system. This leads to a discussion about
  • #1
arbol
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1. Let S' be an x'y'-coordinate system. Let the x'-axis of S' coincide with the x-axis of an xy-coordinate system S. Let the y'-axis of S' be parallel to the y-axis of S. Let S' move in system S along the x-axis of S with constant velocity v in the direction of increasing x, and let the origin of S' coincide with the origin of S at the time t = t' = 0s.

2. Let a ray of light emitted by the moving system S' depart from x' = 0m at the time t' = 0s towards x' = x'1 and reach x'1 at the time t' = t'1, and let it be reflected at x'1 back to x' = 0m, reaching 0m at the time t' = 2*t'1.

3. Let the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 be, in the moving system S', the length L of a rigid rod.

4. Let a ray of light emitted by the moving system S' depart from y' = 0m at the time t' = 0s towards y' = y'1 and reach y'1 at the time t' = t'1, and let it be reflected at y'1 back to y' = 0m, reaching 0m at the time t' = 2*t'1.

5. Let the length of the path of the ray of light emitted by the moving system S' from y' = 0m to y' = y'1 be also (independently), in the moving system S', the length L of the rigid rod.

6. Let a ray of light emitted by the stationary system S depart from x = 0m at the time t = 0s towards x = x1 and reach x1 at the time t = t1, and let it be reflected at x1 back to x = 0m, reaching 0m at the time t = 2*t1.

7. Let the length of the path of the ray of light emitted by the stationary system S from x = 0m to x = x1 be also (independently), in the stationary system S, the length L of the rigid rod.


8. Let a ray of light emitted by the stationary system S depart from y = 0m at the time t = 0s towards y = y1 and reach y1 at the time t = t1, and let it be reflected at y1 back to y = 0m, reaching 0m at the time t = 2*t1.

9. Let the length of the path of the ray of light emitted by the stationary system S from y = 0m to y = y1 be also (independently), in the stationary system S, the length L of the rigid rod.

10. Let the time (t'1 - 0s) = (t1 - 0s) = L/c.

11. Let the Length of the path of the moving system S' from x = 0m to x = v*L/c be, in the stationary system S, the length D of a rigid rod.

12. In the moving system S', the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m is

2*L.

13. In the moving system S', the length of the path of the ray of light emitted by the moving system S' from y' = 0m to y' = y'1 and back to y' = 0m is

2*L.

14. In the stationary system S, the length of the path of the ray of light emitted by the stationary system S from x = 0m to x = x1 and back to x = 0m is

2*L.

15. In the stationary system S, the length of the path of the ray of light emitted by the stationary system S from y = 0m to y = y1 and back to y = 0m is

2*L.

16. In the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m is

D + L + D + (L - 2*D) = 2*L.

17. In the stationary system S, the length of the path of the ray of light emitted by the moving system S' from y' = 0m to y' = y'1 and back to y' = 0m is

sqrt(L^2 + D^2) + sqrt (L^2 + D^2) = 2*sqrt(L^2 + D^2).

18. The ray of light emitted by the moving system S' moves in the stationary system S with the determined velocity c.

19. The time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1 and back to x' = 0m is

2*L/c.

20. The time in the stationary system S the ray of light emitted by the moving system S' takes to move from y' = 0m to y' = y'1 and back to y' = 0m is

2*sqrt(L^2 + D^2)/c.

21. By the result of the Michelson-Morley Experiment,

2*L/c = 2*sqrt(L^2 + D^2)/c, or

L = L*sqrt(1 + (D/L)^2).

22. The problem with the result of the Michelson-Morley experiment is that (1) L in both sides of the equation "L = L*sqrt(1 + (D/L)^2)" is the length of the rigid rod in the stationary system S, and (2) the ray of light emitted by the moving system S' moves in the stationary system S with the determined velocity c.

23. The problem with the ray of light emitted by the moving system S' moving in the stationary system S with the determined velocity c is that (1) L in both sides of the equation "L = L*sqrt(1 + (D/L)^2)" is the length of the rigid rod in the stationary system S, and (2) the equation "L = L*sqrt(1 + (D/L)^2)" is the result of the Michelson-Morley experiment.
 
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  • #2
Hi arbol,

I am not going to go through each step in a 23-point derivation. Let me just say that you are correct that the emitter is moving at the same velocity as the detector in any frame.

However, remember that light (especially in this experiment) is a wave. Most waves that had been studied up until the M&M experiment propagated through a medium. If you were to build a similar apparatus for acoustical waves you could use it to detect the motion of the medium wrt your device. This is the velocity the the M&M experiment surprisingly failed to detect: the velocity of the medium through which light propagates, aka the aether.
 
  • #3
DaleSpam said:
Hi arbol,

I am not going to go through each step in a 23-point derivation. Let me just say that you are correct that the emitter is moving at the same velocity as the detector in any frame.

However, remember that light (especially in this experiment) is a wave. Most waves that had been studied up until the M&M experiment propagated through a medium. If you were to build a similar apparatus for acoustical waves you could use it to detect the motion of the medium wrt your device. This is the velocity the the M&M experiment surprisingly failed to detect: the velocity of the medium through which light propagates, aka the aether.


I was going to suggest the following:

I propose (1) that in the stationary system S, the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 moves with the determined velocity c and (2) that in the stationary system S, the ray of light emitted by the moving system S' from y' = 0m to y' = y'1 moves with velocity sqrt(c^2 + v^2).

The purpose for my proposition (if it has not been proposed already) is to have

(1) The time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1 and back to x' = 0m be

2*L/c, and

(2) The time in the stationary system S the ray of light emitted by the moving system S' takes to move from y' = 0m to y' = y'1 and back to y' = 0m be

2*sqrt(L^2 + D^2)/sqrt(c^2 + c^2) = 2L/c = 2*t'1 (the result of the Michelson-Morley experiment).
 
  • #4
arbol said:
I was going to suggest the following:

I propose (1) that in the stationary system S, the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 moves with the determined velocity c and (2) that in the stationary system S, the ray of light emitted by the moving system S' from y' = 0m to y' = y'1 moves with velocity sqrt(c^2 + v^2).

The purpose for my proposition (if it has not been proposed already) is to have

(1) The time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1 and back to x' = 0m be

2*L/c, and

(2) The time in the stationary system S the ray of light emitted by the moving system S' takes to move from y' = 0m to y' = y'1 and back to y' = 0m be

2*sqrt(L^2 + D^2)/sqrt(c^2 + c^2) = 2L/c = 2*t'1 (the result of the Michelson-Morley experiment).


I think the following is a more appropriate and a more comprehensive proposition:

I propose (1) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 is

L + D, then


the time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1
to be

(L + D)/(c + v), and

the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 to move in the stationary system S with velocity (c + v),

(2) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = x'1 back to x' = 0m is

L - D, then


the time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = x'1 back to x' = 0m
to be

(L - D)/(c - v), and

the ray of light emitted by the moving system S' from x' = x'1 back to x' = 0m to move in the stationary system S with velocity (c - v),


(3) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m is

2*L, then


the time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1
and back to x' = 0m to be

2*L/c, and

the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m to move in the stationary system S with velocity c,

(4) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m is

2*t'1*(c^2 - v^2), then


the time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1
and back to x' = 0m to be

2*t'1*(c^2 - v^2)/(c^2 - v^2), and

the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m to move in the stationary system S with velocity (c^2 - v^2), and

(5) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' (1) from y' = 0m to y' = y'1 or (2) from y' = y'1 back to y' = 0m is

sqrt(L^2 + D^2), and

the length of the path of the ray of light emitted by the moving system S' from y' = 0m to y' = y'1 and back to y' = 0m is

2*sqrt(L^2 + D^2), then


the time in the stationary system S the ray of light emitted by the moving system S' takes to move (1) from y' = 0m to y' = y'1, (2) from y' = y'1 back to y' = 0m, or (3) from y' = 0m to y' = y'1 and back to y' = 0m to be

sqrt(L^2 + D^2)/sqrt(c^2 + v^2), sqrt(L^2 + D^2)/sqrt(c^2 + v^2), or 2*sqrt(L^2 + D^2)/sqrt(c^2 + v^2) respectively, and

the ray of light emitted by the moving system S' (1) from y' = 0m to y' = y'1, (2) from y' = y'1 back to y' = 0m, or (3) from y' = 0m to y' = y'1 and back to y' = 0m to move in the stationary system S with velocity sqrt(c^2 + v^2),
 
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  • #5
arbol said:
(4) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m is

2*t'1*(c^2 - v^2), then


the time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1
and back to x' = 0m to be

2*t'1*(c^2 - v^2)/(c^2 - v^2), and

the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m to move in the stationary system S with velocity (c^2 - v^2),

This part ought to be written as

(4) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m is

t'1*(c^2 - v^2), then


the time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1
and back to x' = 0m to be

t'1*(c^2 - v^2)/(c^2 - v^2), and

the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m to move in the stationary system S with velocity (c^2 - v^2),
 

1. What is the significance of the Michelson-Morley Experiment?

The Michelson-Morley Experiment was a groundbreaking scientific experiment that aimed to detect the presence of an ether medium through which light waves were believed to travel. Its failure to do so ultimately led to the development of Albert Einstein's theory of special relativity.

2. What is the difference between S and c in the context of this experiment?

S refers to the speed of the Earth's motion through the proposed ether medium, while c refers to the speed of light. The experiment aimed to measure the difference between these two speeds, which was expected to be significant if the ether medium existed.

3. How did the Michelson-Morley Experiment challenge the prevailing scientific beliefs at the time?

The experiment's results went against the expected outcome based on the widely accepted theories of light and motion at the time. This challenged the idea of an ether medium and prompted a reevaluation of our understanding of light and motion.

4. What were the implications of the Michelson-Morley Experiment for the development of modern physics?

The experiment's failure to detect the ether medium ultimately led to the development of Einstein's theory of special relativity, which revolutionized our understanding of space, time, and the relationship between matter and energy.

5. Can the Michelson-Morley Experiment be replicated today?

Yes, the experiment has been replicated numerous times with increasingly precise instruments, always yielding the same result. It remains an important experiment in the history of science and continues to be studied and referenced in modern physics research.

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