cwes99_03 Posted October 18, 2005 Report Share Posted October 18, 2005 I've been going over the Doppler effect, attempting to understand if there are any missed concepts by either Doppler himself when he wrote the equations for acoustic waves, or modern scientists who applied the theories of Doppler to light. I find one thing to stump me in my math, so I'm hoping some experienced physicists will give me some advice. Ok, I started working on the doppler effect from scratch, much as Doppler himself must have years ago, and have come to the following equations. Let t_0 be the start of time and t_1 be the end of time. Let x_1 = v_s * (t_1 - t_0), where v_s is the velocity of the source w/ respect to the medium Let x_2 = v_o * (t_1 - t_0), where v_o is the velocity of the observer w/ respect to the medium Let x_3 = v_w * (t_1 - t_0), where v_w is the velocity of the wave through the medium (v_w is what I chose to keep constant across my calculations because sound waves travel at a constant velocity through air of a continuous nature) Let (t_1 - t_0) = {1}/{f_a}, the time for one wave to travel the distance of one emitted wavelength from a non-moving source as seen by a non-moving observer. then f_e = f_a * {v_w}/{(v_w - v_s + v_o)}, the effective frequency heard by an observer. Serway-Beichner Physics for Scientists and Engineers fifth ed. p.531-533 has two separate equations, one for an observer in motion with respect to a stationary source and one visa versa. They chose to hold the wavelength as a constant for the equation with a stationary source and moving observer. Why would they be inconsistent? By doing this they allow the apparent velocity of the wave to be v'>v_w. This of course can't be so because we know there to be a limit to what velocity sound can travel through air. I have recently also looked into an observer traveling in front of the source at the same speed of the source. According to my equation, this observer should hear the sound being emitted at the same frequency as the emitter is emitting. {f_e} = {f_a}*{ (v_w - v_o) }/{ (v_w + v_s) } comes from Serway, and would provide this result, where my equation above provides the same result. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted October 19, 2005 Report Share Posted October 19, 2005 I have recently also looked into an observer traveling in front of the source at the same speed of the source. According to my equation, this observer should hear the sound being emitted at the same frequency as the emitter is emitting.Quite right. :) The waves are "compressed" at the source and expanded at the receiver. Quote Link to comment Share on other sites More sharing options...
cwes99_03 Posted October 19, 2005 Author Report Share Posted October 19, 2005 Ok, but which equation shows the correct understanding of the propagation of the wave? Namely, can sound travel over great distances above v_w, the velocity of the wave in the medium. I have heard that waves can initially travel above the natural velocity of sound in the medium if the medium is pumped/driven hard enough at the start, but that depending on the elastic restoration force it will reduce down to the natural speed of sound in that medium. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted October 19, 2005 Report Share Posted October 19, 2005 That's what a shock wave is. Quote Link to comment Share on other sites More sharing options...
cwes99_03 Posted October 19, 2005 Author Report Share Posted October 19, 2005 Quite good. Now will someone answer the posted question? Quote Link to comment Share on other sites More sharing options...
CraigD Posted October 19, 2005 Report Share Posted October 19, 2005 … f_e = f_a * {v_w}/{(v_w - v_s + v_o)}… {f_e} = {f_a}*{ (v_w - v_o) }/{ (v_w + v_s) } comes from Serway…but which equation shows the correct understanding of the propagation of the wave?Per my derivation, for an observer is in front of the source:Fwf= Fs*Vw/(Vw-Vs)Fo= Fwf*(Vw-Vo)/Vw = Fs*(Vw-Vo)/(Vw-Vs)Where: Vw is the speed of sound in the sound medium; Fw is the “absolute” frequency at rest with respect to the medium; Fs is the frequency emitted by the source moving at Vs (0<=Vs<Vw); and Fo is the frequency observed by an observer moving at Vo (Vs<Vw), where positive values mean in the same direction, negative the opposite direction, of the source; For an observer behind the source, the sign of Vs is reversed, giving:Fwb= Fs*Vw/(Vw+Vs)Fo= Fwb*(Vw-Vo)/Vw = Fs*(Vw-Vo)/(Vw+Vs) This agrees with {f_e} = {f_a}*{ (v_w - v_o) }/{ (v_w + v_s) }, the one from Serway. When f_a=Fw=Fs and v_s=Vs=0, the f_e = f_a * {v_w}/{(v_w - v_s + v_o)} to be a special case of the Serway equation. It gives incorrect results when v_s<>0 unless v_s=v_o, and can be indeterminate for reasonable values of v_s and v_o. … can sound travel over great distances above v_w, the velocity of the wave in the medium.I don’t think so.I have heard that waves can initially travel above the natural velocity of sound in the medium if the medium is pumped/driven hard enough at the start, but that depending on the elastic restoration force it will reduce down to the natural speed of sound in that medium.This sounds reasonable, if you hypothesize that the medium undergoes some sort of increase in temperature or a phase transition from non-dispersive (gas) to dispersive (liquid) in the medium near whatever’s doing the pumping. I’d guess this to be a short-range effect only. Quote Link to comment Share on other sites More sharing options...
cwes99_03 Posted October 20, 2005 Author Report Share Posted October 20, 2005 It is like hitting a gong. Most gongs when struck hard, will have an initial frequency higher than the natural frequency. A professor at NIU in Illinois has studied all kinds of percussive instruments, and demonstrated this just last week. He also demonstrated the opposite effect, where the initial frequency is lower than the natural frequency when struck hard. But, that is a bit off topic. What you have done here is rehash Serway. I understand Serway's derivation, which I guess is Doppler's derivation. I've been looking at this problem for about a month now with several days work under my belt, searching the internet for a discussion of this, and find none. My question therefore still stands, perhaps I should clarify what exactly I'm looking for. They chose to hold the wavelength as a constant for the equation with a stationary source and moving observer. Why would they be inconsistent? By doing this they allow the apparent velocity of the wave to be v'>v_w. This of course can't be so because we know there to be a limit to what velocity sound can travel through air. So other than the initial shock wave, which wouldn't happen in this case as we'll let v_s < v_w and v_s=k a constant, why would current theory allow the wavelength to remain unchanged by the observer and the velocity as seen by the observer to be above the v_w limit? This is exactly opposite of what is allowed by Einstein's SR, which does not allow the apparent velocity of light to be any different than c Quote Link to comment Share on other sites More sharing options...
CraigD Posted October 20, 2005 Report Share Posted October 20, 2005 … My question therefore still stands…I don’t completely understand the questions, but will attempt to answer to the best of my understanding. I haven’t seen the texts cwes99_03 references, so can’t speak to anything that hasn’t been posted to this thread, but the Physics doesn’t depend on the particular text describing it, so that shouldn’t be a problem. I believe cwes99_03 is asking “how is it possible for a moving observer to measure a wave speed greater (or less) than Vw, the constant wave speed for the medium?” and “does a moving observer measuring the wave speed to be different than Vw contradict Special Relativity?” How is it possible for a moving observer to measure a wave speed greater than Vw?An observer moving at Vo directly measures the wave speed to be Vw+Vo.Eg: On a windless day, 2 experimenters, Alice and Bob, synchronize their accurate watches, and board a long, fast (moving at V) train. Alice goes to the frontmost car, Bob, to the rearmost (separated by distance D). At precisely time 0, Alice fires a starter’s pistol, and, short time later, Bob hears it and records the time (T). Adjusting for Bob’s previously measured reaction time (how long it takes him to hear a sound and press the “freeze” button on his watch, recording the time), they find that D/T = Vw +V, where Vw is the known speed of sound for the current air temperature. Note that the wave did not travel through the still air faster than Vw. Due to the motion of the rearmost car of train through the air toward the first wave peak, arrived in less time, giving a measured velocity greater than Vw. Does a moving observer measuring the wave speed to be different than Vw contradict Special Relativity?No. SR explicitly applies to light (electromagnetic radiation, photons, or, more generally, bosons), not sound, ocean waves, or other waves that propagate through a material medium. As experimentally demonstrated by experiments such as the Michelson-Morley experiment, light does not behave the same way as sound. Were the previous “train” experiment carried out using light instead of sound, D/T would = Vw, regardless of the speed of the train V. In short, sound travels through a medium (air) at a constant speed, independently from an observer moving with respect to the medium. Contrary to the expectations of most pre-MM experiment Physicists (and a few non-mainstream scientists to this day), light does not require a medium (“luminescent ether”), so will be observed to have the same speed regardless of the movement of an observer. Quote Link to comment Share on other sites More sharing options...
cwes99_03 Posted October 20, 2005 Author Report Share Posted October 20, 2005 Your reply, my friend is what I expected, but could not quite come to say to myself. I suppose that the MM experiment, since I myself have not actually done it, should be enough to satisfy my indulgence here. I guess that leads me to my next bit of work, determining any errors in the MM-experiment (I know sounds cocky, but it is such fun.) Does the aether have to exist for light to travel at c vs. c+v=c'? That's rhetorical, and shall be my next study. Thanks again. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted October 21, 2005 Report Share Posted October 21, 2005 Does the aether have to exist for light to travel at c vs. c+v=c'? That's rhetorical, and shall be my next study.Actually, not necessarily but it's a veeeeeeery subtle point. It isn't so much the existence of æther that would have been against the principle of relativity but, rather, an absolute coordinate system. Quote Link to comment Share on other sites More sharing options...
cwes99_03 Posted October 21, 2005 Author Report Share Posted October 21, 2005 Yes, the other question that was asked and apparently answered is, Is the aether dragged with the rotation of the earth? Quote Link to comment Share on other sites More sharing options...
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