1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Doppler Effect and Reflection

  1. Dec 4, 2013 #1


    User Avatar

    Hello everyone,

    Here is something I've discussed with some colleagues recently and that generated a lot of disagreement, so I wanted to bring the topic to Physics Forums to ask what is your understanding of the situations and wheter you agree with the "solution" I provide below or not.

    Imagine that observers [itex]A[/itex] and [itex]B[/itex] are in a motionless media. Observer [itex]A[/itex] moves to the right with velocity [itex]v_{a}[/itex] and observer [itex]B[/itex] moves to the left with velocity [itex]v_{b}[/itex]. At the same time, [itex]A[/itex] emits sound waves of frequency [itex]f_{o}[/itex] in the direction of [itex]B[/itex]. These sound waves are reflected at [itex]B[/itex] and return in the direction of [itex]A[/itex]. The question is: what is the expression that gives the frequency of the reflected wave as measured by [itex]A[/itex].

    I argue the following, and I would like to really appreciate to hear more opinions on this matter to reach a final answer:

    Since both observers are moving with respect to each other, the frequency [itex]f^{'}[/itex] measured by [itex]B[/itex] is given by the Doppler relation (considering that the speed of sound in that media is [itex]u[/itex]):

    [itex]f^{'} = (\frac{u + v_{b}}{u - v_{a}})f_{o}[/itex]

    Now, the wave reflected at [itex]B[/itex] has just the same frequency as that measured by [itex]B[/itex], and therefore has exactly the value [itex]f^{'}[/itex]. However, [itex]A[/itex] observes this reflected wave not as a wave emitted by a moving source, but as if it was being emitted by a source that stationary with respect to the media.

    This is the point where there is a lot of disagreement, so I would like to make it clear beforehand that the reasoning above does not state that [itex]B[/itex] is stationary (this would be a contradition, since we start with a different assumption), it only states that the reflected wave as measured by [itex]A[/itex] appears to be emitted by a source which is stationaty with respect to the media.

    Thus, since [itex]A[/itex] is moving into the reflected wave with velocity [itex]v_{a}[/itex], another Doppler shift piles up on top of the previous one, giving the following frequency [itex]f^{''}[/itex] measured by [itex]A[/itex]:

    [itex]f^{''} = (\frac{u + v_{a}}{u})f^{'}= (\frac{u + v_{a}}{u})(\frac{u + v_{b}}{u - v_{a}})f_{o}[/itex]

    So, this is the story. I hope to hear some comments soon!

    Best regards,
  2. jcsd
  3. Dec 7, 2013 #2

    You may work out the simple problem of having the A observer emiting two sequential particles (with a given time interval of separation), both travelling with the same velocity v. By constructing their equations of motion and (1) considering de time interval between them and (2) the different positions of departure of each one (for A is moving) you will deduce, after the reflections in B and finally after their arrival back at A what the time gap between them has become. With this, you will have a clear statement concerning the frequency change in this process, without having to appeal to the Doppler effect formula.
  4. Dec 7, 2013 #3
    No, that's not right. [itex]A[/itex] will observe the reflected wave as a wave emitted by a moving source.
    The correct expression will be
    [itex]f^{''} = (\frac{u + v_{a}}{u-v_b})f^{'}= (\frac{u + v_{a}}{u-v_b})(\frac{u + v_{b}}{u - v_{a}})f_{o}[/itex]
  5. Dec 7, 2013 #4


    User Avatar

    DaTario, I agree. I will try to work that out and post it here.

    Dauto, I am aware of your result - that is one of the pssible answers my colleagues and I discussed - but I was hoping to hear some argumentation on why you consider the reflected wave to be emitted from a moving source. I'm still not convinced.
  6. Dec 7, 2013 #5


    User Avatar
    Gold Member

    Zag, will you please specify the original locations of A and B? The description you start with, 'Imagine that observers A and B are in a motionless media. Observer A moves to the right with velocity va and observer B moves to the left with velocity vb." is ambiguous. To understand the problem we need to know if A and B are approaching or receding from each other. Thank you.
  7. Dec 10, 2013 #6

    Philip Wood

    User Avatar
    Gold Member

    I've just done it independently and arrived at dauto’s result.

    Let the time between wave crests emitted from A be [itex]T_{em}[/itex], hitting the mirror at B be [itex]T_{ref}[/itex], and received back at A be [itex]T_{rec}[/itex]. Then considering the shorter distances the second crest has to go compared to the first…
    [tex]T_{ref} = T_{em} - \frac{v_A T_{em}}{u} - \frac{v_B T_{ref}}{u}[/tex]
    So re-arranging,
    [tex]T_{ref} = \frac {1 - v_{A}/u}{1 + v_{B}/u} T_{em}[/tex]
    By a similar argument for the return journey
    [tex]T_{rec} = \frac {1 - v_{B}/u}{1 + v_{A}/u} T_{ref}[/tex]
    Hence [itex]\frac{T_{rec}}{T_{em}} = \frac {1 - v_{B}/u}{1 + v_{A}/u} \frac {1 - v_{A}/u}{1 + v_{B}/u}[/itex]

    So [itex]\frac{f_{rec}}{f_{em}} = \frac {1 + v_{A}/u}{1 - v_{B}/u} \frac {1 + v_{B}/u}{1 - v_{A}/u} = \frac{(u +v_{A})(u + v_{B})} {(u - v_{B}) (u - v_{A})}[/itex]
  8. Dec 11, 2013 #7


    User Avatar
    Science Advisor
    Gold Member

    Why do you not just consider the source at A seeing an image source of itself in the reflector at B? This image is receding [edit: moving relative - not necessarily receding] at 2(vA - vB) so the doppler shift will correspond to a velocity of 2(vA - vB). Doesn't any effect of the velocity of the medium cancel?
  9. Dec 11, 2013 #8
    You mean speed vA + 2vB.

    You can't do that because in the mirror you will also see the air itself moving at speed 2vB. Air motion affects Doppler effect.
  10. Dec 11, 2013 #9
    There isn't much to say. From the point of view of B the crests are sent back at the same rate as they are received, but B is moving which causes a Doppler effect.
  11. Dec 11, 2013 #10


    User Avatar
    Science Advisor
    Gold Member

    I'm not sure. If the image is the same distance behind the reflector as the source is in front (that's what you get with mirrors in school), the separation of source and image will be increasing at the twice the separation speed i.e. 2(Va -Vb). The moving source A lays down a modified frequency, due to doppler, in the stationary air and the moving reflector will produce double the doppler effect on reflection. The sound reaching and received by A will also be doppler shifted. This gives the same answer as my original statement, I think. Can you tell me what I've missed in this? It is such a simple answer but I don't see how it's wrong.

    I can't even see how there should be an effect from the air because, in its reference frame, it isn't moving.
  12. Dec 11, 2013 #11
    There are several things going on here.

    1st. the direction of the two objects are being measured in opposite directions (they are both positive when they approach each other). That can lead to confusion but is actually the most common convention when solving problems of Doppler effect. That's why I added the speeds to find the relative velocities.

    2nd. Given the above, the the relative velocities between A and its mirror reflection A' will be 2(vA+vB), but the Doppler formula doesn't take relative velocities between the objects as input (It's not a problem of relativistic Doppler effect of light). The formula takes speeds measured with respect to the air.

    3rd. The speed of the air does play a role because the Doppler formula for a moving source is different than the Doppler formula for a moving observer (again, not a relativistic problem)

    4th. The most serious problem though is that there is a discontinuity at the mirror. On one side the air is at rest while on the other side it appears to move. That discontinuity must be dealt with.

    The sum of the above shows that your mirror idea will cause you more grief than insight.
  13. Dec 11, 2013 #12
    Where do you think the Doppler formula comes from if not the solution to the problem you described?
  14. Dec 12, 2013 #13


    User Avatar
    Science Advisor
    Gold Member

    As you say, there are a few things going on here.

    Firstly, it would be much easier to use the same reference direction for velocities, why not just take motion to the right as positive, in both cases? To avoid "confusion", let's use a consistent convention.

    Then there is the matter of your assertion that it's not just a matter of relative motion of air and source / reflector. If it were not just relative then would you not expect there to be some sort of Doppler effect along the Equator, where A and B and the air are travelling Eastwards at 450m/s (nearly Mach1.5)? As we don't observe it in that case (or in fast aircraft), there must be an explanation that is consistent with our A-B experiment.

    In your response to DaTario, you say that the analysis is, in fact, based on the idea of a stream of particles. Well, in that case, you would agree that Newtonian relativity would apply. How, then, is there any difference between situations of moving air and moving objects? So there is no point in using anything other than the reference frame of a stationary atmosphere. (No mysterious effect involved or neglected)

    What "discontinuity" is there at the mirror, B (assuming that you can apply normal dynamics / kinetic theory to the molecules striking the mirror)? DaTario's particles will hit the reflector (inelastic, infinite mass - to make it easier) and follow simple laws. The resulting frequency of the reflected train of particles will be just as the Doppler formula predicts with Vb relative to the ref frame on receive and another Vb on transmit. As far as the situation at A is concerned, it will emit sound into the atmosphere with a Doppler shift due to Va and there will be an additional shift due to Va on receiving the sound.
    There is no a commonly used 'Doppler Formula' between two moving objects (at least I can't find one).
    What you can do is to consider each boundary at a time and you arrive at a chain of four (u-V)/(u-V) expressions, showing reception or transmission at each. (You can treat the reflection as a microphone and loudspeaker with just the right amount of gain; any constant time delay is irrelevant) You then need to re-arrange them to get a single ratio, in which the V should become the relative velocity Va-Vb.
    Unfortunately, I can't easily write them nicely here and the algebra just got heavier and heavier so I and my pencil dipped out. It would be a simple matter with someone who has access to Mathematica or equivalent.

    f received = f transmitted (u+Va)(u-Vb)(u-Vb)(u+Va) / (u-Va)(u+Vb)(u+Vb)(u-Va)

    I am pretty sure that expression can be reduced to
    f received = f transmitted(u+Va-Vb)/u-Va+Vb)
    Which would prove my point.
  15. Dec 12, 2013 #14

    Philip Wood

    User Avatar
    Gold Member

    It's not that I'm upset at being ignored or anything - perish the thought! - but, in post 6, I derived what I think is the correct formula from first principles. I'm prejudiced, of course, but it seems to me that the logic is inescapable. Was the argument too abbreviated to be followed, I wonder. Or is there some flaw which everyone is too polite to point out?
  16. Dec 12, 2013 #15


    User Avatar
    Science Advisor
    Gold Member

    Sorry about that. It's easy to get neglected here in the heat of debate.
    I see where your derivation comes from but I can't see where my argument is wrong. I can only conclude that the two models are somehow equivalent. After all, the frequency change of the wave that is launched from A and detected by a stationary microphone, is given by (u+V)/(u-V) and this actually occurs four times, effectively. So why isn't that described as the product of four ratios of the same form and why should this not boil down to the velocity difference between the two (ignoring the intermediate measurement of f in the air)?
    By your argument, the 'stretching' of the distance between source and reflector is due to the difference in velocities.

    Just to be sure of this. Are we discussing the sound perceived on a moving source, after reflection by a moving reflector? That's worth checking.

    "Too polite" lol. Just short sighted, in my case.
  17. Dec 12, 2013 #16

    Philip Wood

    User Avatar
    Gold Member

    Thank you, sc, for your kind reply.

    Many years ago, I showed to my own satisfaction that used naively - as by me - virtual image methods gave the wrong answer in Doppler questions. Do we regard the image as both a (moving) receiver which then acts as a moving emitter or what? Keep off images, I say!

    Even more years ago, I was taught at school to analyse Doppler effect in terms of the wavelength changing for a moving source, but the frequency changing for a moving observer. I now think this is unnecessarily complicated, and is anything but helpful when e-m waves are tackled.

    Direct time and distance methods, considering two successive wave crests, win every time for me. And it is the mirror itself and not the virtual image that absorbs and re-emits. The devil is in the detail, of course, and in the case of my derivation (post 6) the detail is which time to use in calculating the shortening (or lengthening) of the distances travelled by the second pulse. See first equation in post 6.
  18. Dec 12, 2013 #17


    User Avatar
    Science Advisor
    Gold Member

    I wouldn't say that the "image" producing the sound. I just used that to tie in with familiar optical ideas - which work correctly. The effect is just the same as that of an image which is moving twice as fast. (Just the same as in optics). If you would agree that you could replace the 'perfect' reflector with a mike and loudspeaker then how can my reasoning not be 'reasonable? Assume a 'stationary' source. The sound that arrives at the reflector has a changed frequency (that is, the observer on B detects a difference in the note that he sees being played on that piano at A. To satisfy the boundary conditions, the frequency of oscillations of the wave leaving the surface must be measured (by B) to be the same. So you have a new emitter, with the changed frequency. This sound is further changed in frequency , in the Earth's frame, and by an additional, identical value. I think your analysis will give the same answer for the frequency heard by B as the frequency of the reflection. How can that be? My reasoning implies that there are four lots of frequency shifting in the whole problem and not just two.
  19. Dec 12, 2013 #18

    Philip Wood

    User Avatar
    Gold Member

    Sorry not to have the time to think about this properly at the moment. But both methods have 4 frequency shifts. Mine has 2 for sound going from moving A to moving B, and 2 for sound going from moving B to moving A.

    At first sight, I don't think it's at all obvious that you can replace a reflector with a mike and loudspeaker. If the reflector is moving, should the mike be moving? Should the loudspeaker be moving? And how fast?
  20. Dec 12, 2013 #19


    User Avatar
    Science Advisor
    Gold Member

    They are on the moving cart instead of the reflector, of course. A straight swap. The loudspeaker just produces a sound in anti phase with the received sound (it produces the same boundary condition as a hard reflector).
  21. Dec 12, 2013 #20
    I agree. I'm just following the convention adopted in the 1st post (which is actually quite common)

    You misunderstood me. What I said was that the objects' speed must be measured relative to the air - not relative to each other.

    There is the discontinuity of the wind speed. Outside of the mirror the air is at rest. Inside of the mirror the image of the air is moving.
    Yes, there is. That's the formula for f' given in the 1st post

    I told you that idea was going to give you nothing but grief.

    There are too many factors there. Look at my 1st post for the correct expression.

    Alas, that's not true. The two expressions are completely different.
    Last edited: Dec 12, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Doppler Effect and Reflection
  1. Doppler Effect (Replies: 1)

  2. The Doppler Effect (Replies: 3)

  3. Doppler Effect (Replies: 1)

  4. Doppler effect (Replies: 10)

  5. Doppler Effect (Replies: 2)