International Journal of Pure and Applied Physics. ISSN 0973-1776 Volume 14, Number 2 (2018), pp. 73-85 © Research India Publications http://www.ripublication.com The Doppler Effect on Light Waves Waldemar Friesen Technical University Vienna, D-60528 Frankfurt, Im Mainfeld 7 / 5 OG 1, Germany. Abstract After the Doppler effect on sound, where the sound of a passing vehicle is getting lower and lower, I calculate that in light the change in frequency due to the Doppler effect would only be about 5/3 10-6 of the frequency. Because of this, one has to go to the wavelength and look at a distance of about 106 wavelengths. Because of the high velocities, the rectilinearly moved receiver is replaced by one rotating on one wheel, in fact two, if you include the opposite one rotating with the opposite direction of speed. Specifically, light from a light source falls on two parallel mirrors mounted opposite each other on a wheel. At a certain angle of rotation, the reflected light comes together in a telescope lens fixed outside in the room and the interference of the two rays is considered enlarged. If the propagation of the light depends on the receiver, the mirrors, and if the distance from the light source to the mirrors is much greater than that from the telescope, the first is predominant and there is interference in contrast to static mirrors. If the propagation of the light depends on the transmitter, the interference occurs with a very large distance from the mirrors to the telescope. Possible dimensions are 500 revolutions/min, wheel radius = 0.5m, distance mirror-wheel = 2.5m. The experiment is only conceived, not executed. I expect the light to propagate depends on the receiver, the mirrors. Subsequently, several, even more difficult, tests are given of how to decide that the propagation is rotated with the receiver as it rotates around the transmitter or when one of them is rotated on its own. Keywords. Doppler Effect, Light Propagation. 1. The Doppler Effect in General The aim of this essay is to devise an experiment that proves that there is no Doppler effect on electromagnetic waves, especially light waves. If a source generates a signal 74 Waldemar Friesen F(t) = (f1(t), f2(t), f3(t)) at the point x = (x1, x2, x3) of the three-dimensional space, it is conventionally assumed that it radially propagates in the vacuum in the three directions at the velocity of light c. Here, it does not matter that the amplitude decreases, for example with 1/r2, where r is the distance from the source, r = ||y - x||. Because the amplitude depends only on the distance, it is a one-dimensional problem and I denote with y further any arbitrary but fixed spatial direction and with f the field F in that direction. The signal f(t) is given at point x. At y I get a constant times f(t) with infinitely fast light propagation. Since I do not care about the amplitude and the signal is proportional to that at x, at y I also say f(t). Because of the finite propagation velocity c, I get the signal with the time difference T, which it takes to go through the distance y - x. y - x = cT, I get f(t - T) = f(t - (y - x)/c). Now I always fix the place of the source and choose it as the origin of the coordinate system, x = 0. I only look at different receivers y. The signal simplifies to f(t - y/c). As mentioned at the beginning, I consider the signal only in one direction y. In the Doppler effect, the frequency is changed by a velocity v in that direction. This requires a periodic function f. Therefore, I now go to f(t) = sin ωt as the simplest representative of such a function with the signal sin ω(t - y/c). The Doppler effect is based on a signal dependent only on the source, which propagates in space. It does not matter if it happens in a vacuum or in a medium. Now, as a receiver moves away from the source at velocity v, he sees a slowed propagation velocity of c - v as he moves away from the signal peaks and they follow him, moving towards the source he sees the velocity of c + v. With T = 1/μ I denote the oscillation period, ω = 2π/T, and with μ = ω/2π the frequency. The signal has traversed the path d = cT after T. For the same distance d, if one moves with v from the source d = (c - v)T-, corresponding to the angular frequency ω- = 2π/T- = 2π(c - v)/d = 2π(c - v)/(cT) = ω(c - v)/c with the frequency μ- = ω-/2π = (ω/2π)(c -v)/c = μ(c - v)/c = μ - μv/c. Similarly, if one moves to the source with -v the increased propagation velocity c + v yields, the same path d = cT and d = (c + v)T+ with the angular frequency ω+ = 2π/T+ = ω(c + v)/c with frequency μ+ = ω+/2π = (ω/2π)(c + v)/c = μ(c + v)/c = μ + μv/c. The frequency μ is essentially reduced by -μv/c or set high by μv/c. 2. The Doppler Effect on Sound and Electromagnetic Waves In the following, I give a numerical example of the Doppler effect. If you drive by bus at night on the highway, the traffic has calmed down and a lorry meets you only every 100 meters, you can hear the car sound from a distance with a high note, which is getting closer, louder and louder and gets lower, loudest, when the truck drives past one, then quietens again and gets still lower, until it disappears with a dark tone. This can be explained by the Doppler effect. At the beginning you move towards the source, the frequency is increased by μv/c, later decreased by -μv/c. The total difference is 2μv/c. Only the velocity in the direction to the source, that is in the radial direction, enters in v. That the sound is getting lower, follows from the fact that this v at the beginning corresponds to the speed of the two vehicles on the highway, as they approach, the proportion in the direction of the distance gets always smaller until it is The Doppler Effect on Light Waves 75 after all almost 0 when they lie side by side, since the distance stands perpendicular to the velocity, and then it gets bigger again, but now you move away from the source. Overall, the speed v to be considered becomes smaller and smaller, so that the sound frequency decreases more and more. In numbers: the trucks drive on the highway about 80km/h, so that it can be calculated easier afterwards, I say 90km/h, because they come from opposite directions the double speed counts, v = 180km/h = 180/3.6 m/sec = 50m/sec. The propagation speed of the sound is 330m/sec. v/c = 50/330 = 0.1̅5…. The difference between the highest frequency μ+ and the lowest frequency μ- is 2μv/c = 2μ 0.1̅5… = 0.3̅0...μ. The value of v/c = 50/330 = 5/33 delivers μ+ = 38/33 μ and μ- = 28/33 μ, the ratio of the two frequencies amounts to 38/28, ie about 4/3. In the case of electromagnetic waves, for a ratio of v/c = 5/33 at v, it would be necessary to generate a speed of 300,000 5/33 km/sec, ie almost 50,000km/sec, which is in reality bad to manage. I now calculate which frequency shift you can really expect, if there is a Doppler effect. I'm assuming a wheel with a radius of 1 meter, which rotates with 5000 revolutions per minute. The circumference moves at a speed of 2π 1m 5000/min. π = 3.1415..., 2π = 6.283... ≈ 6. I'll cancel that for the 60 seconds and get the speed 500m/sec. It is equivalent to 3.6 times the kilometers per hour, ie 1800 kilometers per hour. For v/c I get 500/(3 108) = 5/3 10-6. The Doppler effect assumes that the signal propagates in space at fixed velocity c. If I move to the transmitter with speed v, the wavelength λ remains unchanged. Only because of my velocity v do I see a propagation velocity of c - v and thus a frequency μ- = (c - v)/λ. Regarding the frequency shift, I notice nothing at the small value 5/3 10-6. You can also let the transmitter, as above the truck, move with v. Then the propagation velocity c remains, the frequency changes to μ+ and thus due to c = λ + μ+ the wavelength. Now, one can prove a change in frequency over the change in wavelength by tracking the signal over so many wavelengths λ that the total change at the end of the path is approximately in the range of one wavelength. We have c = λ+μ+ = λ+μ(1 + v/c). It follows that λ+ = c / (μ(1+v/c)) ≈ (c/μ)(1 - v/c) = λ(1 - v/c). The wavelength change -λv/c is therefore calculated roughly the same as the frequency change μv/c only with the sign reversed. To determine the sum of all wavelength changes nλv/c, it must be in the range λ, that is n = c/v = 3/5 10-6 = 0.6 10-6. I have to track the signal roughly over a distance of 106 wavelengths.