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MrYes
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- TL;DR Summary
- Does the proportional relationship between frequency, wavelength, and light speed offer a simple explanation of the Michelson-Morley experiment's null result?
The Michelson-Morley experiment (MMX) was set up to detect an effect associated with the luminiferous ether, a substance that doesn't exist. On the other hand unmoving space does exist in the sense that a spectrometer can be placed in a position where it doesn't detect Doppler redshifts or blueshifts from the cosmic microwave background (CMB). Anchored at the position of this spectrometer a coordinate system can be drawn that provides a standard reference frame against which the speeds of light and their sources can be measured relative to this position. Then this question, without an ether to slow the light from the moving source, wouldn't there still be a fringe shift in the direction of earth's motion? The unrestricted light in that direction would travel at c+v, while the light in the perpendicular direction would travel at c + a transverse velocity component, but slower, exactly the inverse of the MMX expectation. Even so, the two rays still united with perfect precision at the instrument focus. What in the relationship of frequency, wavelength and speed between the two rays in their journey through their moving light paths traveling through the Doppler-neutral coordinate system of the CMB, could explain how one ray moving faster could match the other going slower?
My guess is the relationship between lightspeed (c), wavelength (lambda), and frequency (f) is found in the equation, c divided by lambda equals f. With units of distance and time of centimeters (cm) and nanoseconds (ns), for MMX, c is 30.0 cm/ns, and v is 0.003 cm/ns. To solve for frequency, f(lab), as it was in the MMX lab, I set f(lab) equal to 30.0 cm/ns divided by 0.00006 cm, the wavelength of the MMX ray, and found the ray's frequency, 500,000 beats per ns. Since the lab was moving, its light source was sending wavelengths at the speed c+v relative to a coordinate system anchored at rest with the CMB. To find how long wavelengths would be in the unmoving coordinate system compared to the moving coordinate system I set lambda(CMB) equal to (c+v) divided by f(lab), 30.003 cm/ns divided by 500,000 beats per ns. Out popped the length of lambda(CMB), 0.000060006 cm.
The result could be an interference pattern identical to the pattern created when the source isn't moving. In other words, even though one ray is moving faster relative to a position Doppler-neutral to the CMB, lambda(CMB) traveling at c+v would match the frequency of lambda(lab) traveling at c, 500,000 beats per ns. Light traveling at 30.003 cm/ns divided by 0.000060006 cm is 500,000 beats per ns, the same frequency as in the lab. It also means each ray's wavelength takes the same time to transit its wavelength as the other ray transits its wavelength. 0.000060006 cm divided by 30.003 cm/ns, (lambda(CMB)/(c+v)), is 0.000002 ns. And 0.00006 cm divided by 30.0 cm/ns is also 0.000002 ns. The longer wave moving faster hurries into the shorter wave moving slower producing an interference pattern the same as if each wave was the same length traveling at the same speed. Thus in both coordinate systems, the one anchored in the Doppler-neutral to the CMB position and the one traveling with the earth through that system MMX produced a null result. In the light of this simple elaboration of the role of frequency in the MMX null result, what is the necessity of the Einstein-Lorentz transformation?
My guess is the relationship between lightspeed (c), wavelength (lambda), and frequency (f) is found in the equation, c divided by lambda equals f. With units of distance and time of centimeters (cm) and nanoseconds (ns), for MMX, c is 30.0 cm/ns, and v is 0.003 cm/ns. To solve for frequency, f(lab), as it was in the MMX lab, I set f(lab) equal to 30.0 cm/ns divided by 0.00006 cm, the wavelength of the MMX ray, and found the ray's frequency, 500,000 beats per ns. Since the lab was moving, its light source was sending wavelengths at the speed c+v relative to a coordinate system anchored at rest with the CMB. To find how long wavelengths would be in the unmoving coordinate system compared to the moving coordinate system I set lambda(CMB) equal to (c+v) divided by f(lab), 30.003 cm/ns divided by 500,000 beats per ns. Out popped the length of lambda(CMB), 0.000060006 cm.
The result could be an interference pattern identical to the pattern created when the source isn't moving. In other words, even though one ray is moving faster relative to a position Doppler-neutral to the CMB, lambda(CMB) traveling at c+v would match the frequency of lambda(lab) traveling at c, 500,000 beats per ns. Light traveling at 30.003 cm/ns divided by 0.000060006 cm is 500,000 beats per ns, the same frequency as in the lab. It also means each ray's wavelength takes the same time to transit its wavelength as the other ray transits its wavelength. 0.000060006 cm divided by 30.003 cm/ns, (lambda(CMB)/(c+v)), is 0.000002 ns. And 0.00006 cm divided by 30.0 cm/ns is also 0.000002 ns. The longer wave moving faster hurries into the shorter wave moving slower producing an interference pattern the same as if each wave was the same length traveling at the same speed. Thus in both coordinate systems, the one anchored in the Doppler-neutral to the CMB position and the one traveling with the earth through that system MMX produced a null result. In the light of this simple elaboration of the role of frequency in the MMX null result, what is the necessity of the Einstein-Lorentz transformation?
