Suppose there is a device which is producing and listening to sound (sonar), a reflector is moving with respect to device and the velocity of reflector and device with respect to the air is not known, can we determine the relative velocity with doppler effect?
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$\begingroup$ Can you assume that neither the device nor the reflector are moving faster than $c$ with respect to the air? $\endgroup$– BowlOfRedCommented Jul 12 at 22:55
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$\begingroup$ yes definitely, $\endgroup$– user324939Commented Jul 13 at 11:46
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$\begingroup$ @Pradyuman I have updated my answer with derivation of an exact analytical solution. $\endgroup$– KDPCommented Jul 15 at 13:54
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2 Answers
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can we determine the relative velocity with doppler effect?
Yes we can. This is similar to the principle that that police speed radar works on, except that the police radar uses electromagnetic waves. The formula for the relative velocity is:
$$\Delta v = \frac{(f_r -f_0)}{f_0 } \frac{c}{2} = \frac{\Delta f}{f_0 } \frac{c}{2}$$
where $f_r$ is the frequency returned and $f_0$ is the original emitted frequency and c is the speed of the wave in the medium. For the sonar radar specified in the question, c is the speed of sound in air. See Wikipedia
EDIT: It turns out that he Wikipedia formula for the relative velocity calculated from the reflected signal is an approximation that only works for velocities much lower than the speed of sound. Here is the derivation of the correct formula that works right up to the speed of sound. (If the object is going away from us at greater than the speed of sound the sonar signal cannot catch up with it and if it is coming towards us at greater than the peed of sound the object will arrive before the reflected sound wave.)
Lets assume we are observer $S_1$ with velocity $v_1$ going to the right and on the right of us is observer S2 also going to the right with velocity v2. Using the Doppler formula, the frequency $f_2$ received by $S_2$ is given by|: $$ f_2 = f_1 \frac{(c-v_2)}{(c-v_1)} .$$ Upon reflection, $S_2$ becomes the source and $S_1$ become the receiver so the velocities are swapped and the signs are reversed. The frequency $f_3$ returned to $S_1$ is given by :$$ f_3 = f_2 \frac{(c+v_1)}{(c+v_2)} = f_1 \frac{(c-v_2)}{(c-v_1)}\frac{(c+v_1)}{(c+v_2)} .$$
We can solve the above formula for $v_2$ and we get: $$v_2 = \frac{c (f_1 - f_3) + v_1(f_1 + f_3)}{ (f_1 + f_3) + v_1(f_1 - f_3)/c }$$
The relative velocity is then:
$$\Delta v = (v_2-v_1) = \frac{c (f_1 - f_3) + v_1(f_1 + f_3)}{ (f_1 + f_3) + v_1(f_1 - f_3)/c } -v_1 $$
This result for the relative velocity requires us to know our own velocity $(v_1)$ relative to the medium. This is not a big problem because a source vehicle such as a car or aircraft can measure its own air speed using a fan driven device or a Venturi tube. Similarly it is not difficult for a submarine to measure its own velocity relative to the water it is moving through and use this information to calculate the velocity of the object it is tracking. Unfortunately there is no way to write the Doppler equation in terms of relative velocity using sound, independently of the velocity of the source through the medium. However the relativistic Doppler equation can (has to) be written purely in terms of relative velocity. When the source is stationary relative to a sound carrying medium so that $v_1=0$, the equation becomes much simpler:
$$\Delta v = c \frac{ (f_1 - f_3) }{ (f_1 + f_3) } $$.
A positive solution means the object is going away from us and a negative solution means it is coming towards us.
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$\begingroup$ Well I am asking about sound, $\endgroup$– user324939Commented Jul 14 at 7:36
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$\begingroup$ See the last sentence: "c is the speed of sound in air". Just replace the c in the equation for the speed of sound in air. $f_0$ is the frequency of the emitted sound. and $\Delta f$ is the difference between the sound frequency received back and the emitted sound frequency. $\endgroup$– KDPCommented Jul 14 at 8:58
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$\begingroup$ @Pradyuman Also see my answer to a related question here: physics.stackexchange.com/a/821574/388464 $\endgroup$– KDPCommented Jul 16 at 5:51
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$\begingroup$ let me tell you a few more things about doppler effect the quantity $\frac{c-v}{f}$ is constant for all sources and observers, and in a radar measurement if you take the geometric mean of $\frac{c-v}{f'} =\frac{c}{f}$ and $\frac{c+v}{f''}=\frac{c}{f}$ which is outgoing and incoming you get $\frac{c^2-v^2}{f'f''}=\frac{c^2}{f^2}$ is also invariant and if you assume the Einstein synchronisation convention $f' = f''$ you'll get $(c^2 -v^2)t'^2 = c^2t^2$ $\endgroup$– user324939Commented Jul 16 at 6:09
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$\begingroup$ @Pradyuman Just curious. What is the "$f$" in $\frac{c-v}{f}$? Who measures it? $\endgroup$– KDPCommented Jul 16 at 6:37
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In a manner it is possible as Doppler effect dictates that the received frequency of a wave changes based on the relative motion between the source and observer. In this case, the sonar acts as both the source and receiver of sound waves. By analyzing the shift in frequency of the reflected wave compared to the emitted wave, we can determine the relative velocity.
The specific nature of the shift (higher or lower frequency) depends on whether the reflector is moving towards or away from the sonar. However, the magnitude of the shift is directly proportional to the relative velocity. By measuring the frequency shift and using the known properties of sound in air, we can calculate the relative velocity between the sonar and the reflector.
Derivation of Relative Velocity Using the Doppler Effect
Given
$$ \begin{align*} v & = \text{speed of sound in air}, \\ v_s & = \text{velocity of the sonar device}, \\ v_r & = \text{velocity of the reflector}, \\ f & = \text{emitted frequency}, \\ f' & = \text{frequency received by the reflector}, \\ f'' & = \text{frequency received back by the sonar after reflection}. \end{align*} $$Derivation
1. Frequency shift due to the moving reflector (reflector as receiver, sonar as source)
When the reflector is moving relative to the sonar source, the frequency $ f' $ received by the reflector is given by:
$$ f' = f \left( \frac{v + v_r}{v - v_s} \right) $$2. Frequency shift due to the moving reflector (reflector as source, sonar as receiver)
The reflector now acts as a source reflecting the frequency $ f' $. The sonar, moving relative to this source, receives a frequency $ f'' $:
$$ f'' = f' \left( \frac{v + v_s}{v - v_r} \right) $$3. Substitute $ f' $ into the expression for $ f'' $
$$ f'' = \left( f \left( \frac{v + v_r}{v - v_s} \right) \right) \left( \frac{v + v_s}{v - v_r} \right) $$Simplifying the expression:
$$ f'' = f \left( \frac{(v + v_r)(v + v_s)}{(v - v_s)(v - v_r)} \right) $$4. Relative velocity determination
The relative velocity $v_{\text{rel}} $ between the sonar and the reflector affects the frequency shift. The relationship between the emitted frequency $ f $ and the observed frequency $ f'' $ after reflection can be simplified to:
$$ f'' = f \left( \frac{v + v_{\text{rel}}}{v - v_{\text{rel}}} \right) $$where $ v_{\text{rel}} $ is the effective relative velocity between the sonar and the reflector.
5. Solving for the relative velocity $ v_{\text{rel}} $
The observed frequency shift $ \Delta f = f'' - f $. Expressing the frequency shift in terms of $ v_{\text{rel}} $:
$$ \Delta f = f \left( \frac{v + v_{\text{rel}}}{v - v_{\text{rel}}} \right) - f $$ $$ \Delta f = f \left( \frac{(v + v_{\text{rel}}) - (v - v_{\text{rel}})}{v - v_{\text{rel}}} \right) $$ $$ \Delta f = f \left( \frac{2v_{\text{rel}}}{v - v_{\text{rel}}} \right) $$Solving for $ v_{\text{rel}} $:
$$ \Delta f = \frac{2f v_{\text{rel}}}{v - v_{\text{rel}}} $$ $$ v_{\text{rel}} = \frac{\Delta f \cdot v}{2f + \Delta f} $$Thus, the relative velocity $ v_{\text{rel}} $ between the sonar device and the moving reflector can be determined from the observed frequency shift $ \Delta f $.
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1$\begingroup$ Any derivation would be helpful, $\endgroup$– user324939Commented Jul 14 at 7:35
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$\begingroup$ i am assuming $v_\text{rel} = v_r -v_s$ and if $v_r =v_s$ then answer by 3rd and 4th does not match. Moreover, in the 4th section the formula comes from nowhere, this is precisely what am asking to derive $\endgroup$– user324939Commented Jul 15 at 10:20