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”