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In considering the doppler effect, why can we not just take the distance between the two as a function of time, divide by the wave velocity, and add this as a shift in our time argument of the sinusoid?

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    $\begingroup$ Maybe include the full derivation so we can follow your steps? $\endgroup$
    – JeffDror
    Commented Feb 3, 2014 at 21:21
  • $\begingroup$ I was just considering having some source, moving with some speed, at some distance from the receiver. Why can't we just add a factor of (d-v_s*t)/v_m where v_m is the speed of the medium, and v_s is the speed of the source? $\endgroup$
    – user24082
    Commented Feb 3, 2014 at 22:14

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You've got the right idea, but are just missing a few details I think.

For concreteness, this is a 1D classical sound wave problem with a stationary receiver and a source having velocity $v_\mathrm{S}$ (positive toward the receiver). The medium may have a velocity, but we'll lump that in with the sound propagation speed, so there is an effective speed $c$ for the wave to propagate from the source to the receiver.1

First, consider a pulse emitted at time $t_0$. If the separation between source and receiver is initially $d_0$, the travel time will simply be $d_0/c$, and so we can map emission time to arrival time: $$ t_0 \to t_0 + \frac{d_0}{c}. $$ Now suppose another pulse is emitted a time $\Delta t$ later. The distance is now $d = d_0 - v_\mathrm{S} \Delta t$, and so we have another emission $\to$ arrival mapping $$ t_0 +\Delta t \to t_0 + \Delta t + \frac{d_0-v_\mathrm{S}\Delta t}{c}. $$

Let an arbitrary time be denoted $t$: $t = t_0 + \Delta t$. Then eliminate $\Delta t$ from the above map: $$ t \to t + \frac{d_0-v_\mathrm{S}(t-t_0)}{c}. $$ Call this mapping $y$: $$ t_\mathrm{R} \equiv y(t_\mathrm{S}) = t_\mathrm{S} + \frac{d_0-v_\mathrm{S}(t_\mathrm{S}-t_0)}{c} $$ So far this matches what you describe.

But how does this get a frequency? Well, you observe a sinusoid of the form $\sin(f_\mathrm{R}t_\mathrm{R}+\phi_\mathrm{R})$, where $f_\mathrm{R}$ is the received frequency and $\phi_\mathrm{R}$ is an arbitrary offset that does not depend on $t_\mathrm{R}$. This is in some sense the definition of $f_\mathrm{R}$. If you want to know the value of the received wave at time $t_\mathrm{R}$, this must match the value of the source at time $$ t_\mathrm{S} = y^{-1}(t_\mathrm{R}) = \frac{c}{c-v_\mathrm{S}} t_\mathrm{R} - \frac{d_0+v_\mathrm{S}t_0}{c-v_\mathrm{S}} $$ Note the inverse, since we are now mapping received times to emitted times. The emitted wave was of the form $\sin(f_\mathrm{S}t_\mathrm{S}+\phi_\mathrm{S})$, and equating the source and received waves with the correct time correction leads to2 \begin{align} \sin(f_\mathrm{S} t_\mathrm{S} + \phi_\mathrm{S}) & = \sin(f_\mathrm{R} t_\mathrm{R} + \phi_\mathrm{R}) \\ f_\mathrm{S} t_\mathrm{S} + \phi_\mathrm{S} & = f_\mathrm{R} t_\mathrm{R} + \phi_\mathrm{R} \\ f_\mathrm{S} \left(\frac{c}{c-v_\mathrm{S}} t_\mathrm{R} - \frac{d_0+v_\mathrm{S}t_0}{c-v_\mathrm{S}}\right) + \phi_\mathrm{S} & = f_\mathrm{R} t_\mathrm{R} + \phi_\mathrm{R} \\ \left(\frac{c}{c-v_\mathrm{S}}\right) f_\mathrm{S} t_\mathrm{R} + \phi_\mathrm{S} - f_\mathrm{S} \left(\frac{d_0+v_\mathrm{S}t_0}{c-v_\mathrm{S}}\right) & = f_\mathrm{R} t_\mathrm{R} + \phi_\mathrm{R}. \end{align} From this we can read off $$ \phi_\mathrm{R} = \phi_\mathrm{S} - f_\mathrm{S} \left(\frac{d_0+v_\mathrm{S}t_0}{c-v_\mathrm{S}}\right), $$ and more importantly $$ f_\mathrm{R} = \left(\frac{c}{c-v_\mathrm{S}}\right) f_\mathrm{S}. $$ This is the correct Doppler shift formula for the case considered.3


1 Note this means $c$ will be a different value for sound propagating away from the receiver if the medium has a bulk velocity.

2 Note in going from the second to the third line, we got rid of the sines. Technically, either they should stay in for the whole calculation, or you should add an arbitrary integer multiple of $2\pi n$. But really the sines have nothing to do with the problem, and are something of a distraction. Any waveform undergoes the same Doppler shift.

3 Though some authors may use a different sign convention. You can check that the sign makes sense in that a source moving subsonically toward you in a subsonic medium ($0 < v_\mathrm{S} < c$) will have a higher frequency.

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You are wondering about the following type of situation. I have a criminal who is in a car being chased by the police. The criminal is moving at a speed $v_c$ and the police are chasing at a speed $v_p$.

Now you have been taught the the frequency shift the criminal will hear in the police's siren is $\frac{c-v_c}{c-v_p}$ where $c$ is the speed of sound. But you are wondering why there isn't a simpler formula that depends only on $v_c-v_p$, say $\frac{c-v_c+v_p}{c}$ or $\frac{c}{c-v_p + v_c}$.

This is a reasonable question and I specifically remember I wanted a similar formula because it would make solving homework problems easier. But there is no such formula. One way to see this is to just do the derivation and see that you get the correct formula (the first one I said), so the other formulas must be wrong.

Another way to see there is no such formula is to consider two different situations where the relative speeds are the same, but the frequency shift is different. The two different situations are as follows.

I am the criminal, and I am sitting still and the cop is driving towards me at one meter per second. In this case there is a very small frequency shift. It is almost imperceptible to me.

In the second case, the relative speed is the same but the cop is moving at the speed of sound (or consider extremely close to the speed of sound if you like). In this case, I would percieve and infinite frequency, and thus the frequency shift must be very large unlike the previous case.

Thus we notice two different frequency shifts even though there was the same relative speed, so the frequency shift must not depend only on the relative speed.

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  • $\begingroup$ Mathematically, I think I understand that it has unexpected results. However, I don't see why the logic is wrong... $\endgroup$
    – user24082
    Commented Feb 3, 2014 at 22:12
  • $\begingroup$ Well if you want help with that you have to explain your logic that you think is right. $\endgroup$ Commented Feb 3, 2014 at 22:21
  • $\begingroup$ I commented up on the question. $\endgroup$
    – user24082
    Commented Feb 3, 2014 at 22:22
  • $\begingroup$ I don't think I am going to be able to help you. Hopefully some one else will answer. $\endgroup$ Commented Feb 3, 2014 at 22:29

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