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Michelson Morley Revisited
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Michelson Morley Revisited
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butchmurray
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Michelson Morley Revisited
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27/04/2012 13:35:33 »
Michelson Morley Revisited
Thorntone E. Murray
April 26, 2012
The 1905 Theory of Special Relativity is considered by many as the solution for the null result of the famous 1887 Michelson Morley interferometer experiment.
In the experiment a monochromatic light beam was split into two perpendicular light beams by a half silvered mirror. The two beams were projected onto separate light paths of equal length with reflectors at the end of each. The two reflected beams intersected at and were recombined by the half silvered mirror. The recombined light beam was examined for fringe; there was virtually none (<.02). This indicated that at the point of intersection the two beams were of the same frequency. Also, the light beams are in phase, that is the light waves in the light beams reunited crest to crest and trough to trough. The interferometer (instrument) was rotated for 360 degrees in its plane in increments. At each increment the examination of fringe yielded the same result. The procedure was repeated at various hours of the day and night as well as various times of the year. The result did not vary.
For this reexamination of the experiment the increment of the instrument is with one light path in the direction of motion and one light path perpendicular to the direction of motion only. An imaginary “Fringe >.02” indicator that illuminates if fringe exceeds that level is added to elucidate a fact. The indicator will not illuminate and absolutely does not influence the experiment.
Terms and Equations:
L’x length of the light path that is in the direction of motion in the moving frame
L’y length of the light path that is perpendicular to the direction of motion in the moving frame
L’o proper length of light paths
f frequency of the light in the light paths
c speed of light
l’x wavelength of the light in light path L’x
l’y wavelength of the light in light path L’y
l’o proper wave length of light in light paths L’x and L’y
t’ time in the moving frame
fl=c light frequency multiplied by the wavelength of the light equals the speed of light
L=ct length of the light path equals the speed of light multiplied by time for light to transit the light path
L=Lo(sqrt(1-v2/c2)) the length contraction equation
The Experiment:
Judged from within the moving frame in the laboratory
Observation: The imaginary “fringe >.02” indicator is not illuminated which verifies there is no fringe.
Conclusion: The absence of fringe was verification that the light waves in light paths L’x and L’y were of the same frequency and in phase with each other. It was also confirmation that the time for the light to transit light paths L’x and L’y was equal.
Calculations:
Equations: L=ct and L’y=Lo
L’y/t’=c substituted L with L’y, substituted t with t’ and solved for c
Lo/t’=c substituted L’y with Lo
Equations: fl=c and l’y=l’o
fl’y=c substituted l with l’y
fl’o=c substituted l’y with lo
Equations: L=ct and L’x=Lo
L’x/t’=c substituted L with L’x, substituted t with t’ and solved for c
Lo/t’=c substituted L’x with Lo
Equations: fl=c and l’x=l’o
fl’x=c substituted l with l’x
fl’o=c substituted l’x with lo
The moving frame judged from the rest frame
Observation: The imaginary “fringe >.02” indicator is not illuminated which verifies there is no fringe.
Conclusion: The absence of fringe was verification that the light waves in light paths L’x and L’y were of the same frequency and in phase with each other. It was also confirmation that the time for the light to transit light paths L’x and L’y was equal.
Calculations:
Equations: L=ct and L’y=Lo
L’y/t’=c substituted L with L’y substituted t with t’ and solved for c
Lo/t’=c substituted L’y with Lo
Equations: fl=c and l’y=l’o
fl’y=c substituted l with l’y
flo=c substituted l’y with lo
Equations: L=ct and L’x=Lo and L=Lo(sqrt(1-v2/c2))
L’x/t’=c substituted L with L’x and solved for c
Lo(sqrt(1-v2/c2))/t’<c substituted L’x with Lo(sqrt(1-v2/c2))
L’x/t’<c Violation of the constancy of c
NOTE: If and only if the relative velocity, v=0 then (sqrt(1-v2/c2))=1 and Lo(sqrt(1-v2/c2))=Lo and Lo(sqrt(1-v2/c2))/t’=c. If the value of v is 0<v<c then Lo(sqrt(1-v2/c2))<Lo thus Lo(sqrt(1-v2/c2))/t’<c which is in violation of the constancy of light speed.
Equations: fl=c and l’x=l’o
fl’x=c substituted l with l’x
flo(sqrt(1-v2/c2))<c substituted l’x with lo(sqrt(1-v2/c2))
fl’x<c Violation of the constancy of c
NOTE: If and only if the relative velocity, v=0 then (sqrt(1-v2/c2))=1 and lo(sqrt(1-v2/c2))=lo and flo(sqrt(1-v2/c2))=c. If the value of v is 0<v<c then lo(sqrt(1-v2/c2))<lo thus flo(sqrt(1-v2/c2)<c which is in violation of the constancy of light speed
In the frame in motion judged from relative rest the time for light to transit between two points perpendicular to the direction of motion and the time for light to transit between the two corresponding points in the direction of motion are equal. However, judged from rest the distance between those two corresponding points in the direction of motion is contracted. The result is that light in the direction of motion transits slower than light perpendicular to the direction of motion and the synchrony confirmed by the absence of fringe is maintained. That is a violation of the constancy of the speed of light.
Visualization/Demonstration:
This is intentionally written so that anyone can understand the concept.
You have two glass tubes that are 10 meters long and at right angles to each other. At the intersection of the tubes you send a light pulse into both. At the end of both tubes the light pulses are reflected back to the intersection. Both pulses arrive back at the same time. That is, in essence, what happened in the Michelson Morley experiment. That is what you would see if you were in the moving frame that was moving at any relative velocity. That is also what you would see from the rest frame if the relative velocity of the moving frame was zero.
Now replace the 10 meter glass tube that is in the direction of motion with a one meter long glass tube. That is what it would look like from the rest frame if the moving frame was moving really, really fast. This is because The Theory of Special Relativity says that length in the direction of motion contracts as seen from the rest frame (but not as seen from inside the moving frame) and the faster the moving frame is going the more length in the direction of motion is contracted. In the moving frame both tubes are still 10 meters long. Now in the moving frame another light pulse is sent into both tubes. As seen from inside the moving frame the light pulses arrive back at the intersection at the same time as they did before because both tubes are still 10 meters long as seen from inside the rest frame. So, viewed from the inside the moving frame the light pulses arrive back at the intersection at the same time. But viewed from the rest frame one tube is 10 meters long and the other tube that is in the direction of motion is only one meter long. The light pulse in the tube that is one meter long must travel at one tenth of the speed of the light pulse in the 10 meter long tube for the light pulses to arrive back at the intersection at the same time. After all, they arrived back at the intersection at the same time in the moving frame and you are looking at that same intersection.
But light always moves at the same speed, no matter where you are when you look at it. But, as demonstrated above, length contraction in The Theory of Special Relativity says it doesn’t. So, length contraction in The Theory of Special Relativity is wrong.
Thorntone E. “Butch” Murray
April 27, 2012
“So it was written and a new day dawned”
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Last Edit: 04/05/2012 04:08:34 by butchmurray
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dareo
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Re: Michelson Morley Revisited
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Reply #1 on:
29/04/2012 20:14:42 »
What about the Ce clock experiment, and the Muon experiment?
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butchmurray
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Re: Michelson Morley Revisited
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Reply #2 on:
02/05/2012 14:15:31 »
The Ce clock experiments:
The Ce clock experiments are generally held as proof of time dilation rather than length contraction which is now proven mathematically to violate the constancy of the speed of light. Time dilation and length contraction are, of course, two sides of the same coin. However, at this point I cannot with certainty discount the results of the experiments and can only speculate as to the cause of the apparent disparity.
The muon experiments:
When high energy cosmic rays from space collide with nuclei of the atoms in the air at about 15km above sea level a cascade of secondary particles that include pions are created. The pions decay very quickly and produce muons among other particles. The half-survival distance of an “at rest” muon is about .66km. That means at about 660 meters half of the total muons will decay. After an additional 660 meters half of the remaining muons will decay and so on. It is easy to see that at sea level, 15km from where they were produced, there should be very few if any muons left. However, there are nearly as many muons detected at sea level as detected in the upper atmosphere where they were produced. This is due to the great speed of the muon, very close to the speed of light. The relativistic effect slows the “internal clock” of the muon drastically relative to time in the rest frame of the earth. From the viewpoint of the muon it will still decay in the same amount of time it would if it was at rest. But judged from the rest frame that duration of time is stretched dramatically, stretched to the extent that will decay near sea level rather than 660 meters from where it was produced.
In my research, the size of the nucleus of an atom ranges from 10,000 to 100,000 smaller than the overall size of an atom to its outer shell. So at the altitude of 15km, if the atoms in the air were packed shell to shell in which case the air would be solid, the probability of a cosmic ray impacting a nucleus is between 1 in 10,000 and 1 in 100,000. As only a guess, lets be generous and say that the average distance between atoms at 15km above sea level is about 10 times the size of an atom. The probability of a cosmic ray colliding with a nucleus is now between 1 in 100,000 and 1 in a million. The fact of the matter is that no one can guarantee at what altitude any particular muon was produced. Moreover, no one can say with absolute certainty that the muons counted at sea level were produced just a few meters above the muon detector also at sea level, in the upper atmosphere or somewhere in between.
References:
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The Speed and Decay of Cosmic-Ray Muons: Experiments in Relativistic Kinematics
- The Universal Speed Limit and Time Dilation
MIT Department of Physics
(Dated: February 6, 2012)
Thank you for the question,
Butch Murray
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butchmurray
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Re: Michelson Morley Revisited
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Reply #3 on:
04/05/2012 04:15:23 »
Correction:
I corrected some typos in the calculations. There were some (*) where there should have been (/).
There is some additional material that I haven’t posted that I need dated for the equivalent of “reduction to practice” purposes. Adding them as replies to this post will serve that purpose. I’m sure that will not bother any one.
Thank you,
Butch
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