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Q & A: Doppler shift formulas

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Most recent answer: 07/14/2013
Q:
Why does the commonly used doppler effect equation f=f0(vr+c)/(vs+c) give different results if you have the receiver (vr) moving away from the source and the source stationary (vs=0),versus if you have the source moving away from the receiver and the receiver is stationary? I find the less commonly used equation f=f0(1+((vr-vs)/c)), where vr-vs is positive if they are moving towards each other, to be more accurate and more useful. If you have trouble reading the equations in this format, I am referring to the first two equations in the wikipedia article for doppler effect.
- Juan (age 23)
St Louis, MO, USA
A:

I think, since you use "c", you're asking about the Doppler shift of light. There the principle of relativity says that only the relative velocity (v) of the source and receiver (as measured by either one) matters, just as you sensed. For sound, on the other hand, there is a medium, and the velocities relative to the medium also matter.

For light, the correct Doppler shift expression is f=f0((c-v)/(c+v))1/2 where f is the frequency measured by the receiver, f0 is the frequency measured by the sender, and v is the velocity with which the sender is going away from the reciever, as measured by the receiver. The expression gets slightly more complicated if the sender also has velocity at right angles to that. (see ) At low speeds, v << c, this expression is very close to the one you gave. At high relative speeds your expression doesn't have clear meaning, because the relative velocity of the sender and receiver as seen by some other observer depends on which other observer is chosen.

For sound the first expression you gave looks ok, with "c" being the speed of sound, so long as it's understood that the velocities are measured in the frame in which the medium is at rest. Here the actual results do indeed depend on whether the receiver is moving away from the sender or vice-versa, in the medium frame. If the sender is moving away at c, the frequency goes to zero, but if the receiver is moving away, it just drops a factor of two. You can check with some simple pictures.

Mike W. 


(published on 07/13/2013)

Follow-Up #1: sound Doppler shift

Q:
Sorry, maybe my question wasn't clear. I was in fact referring to sound, but my question is why do the two equations give different answers and which one is right? In the second equation I gave it doesn't matter whether the source or receiver or both are moving, you get the same answer. In the first equation you will get different answers depending on who is moving.
- Juan (age 23)
St. Louis, MO, USA
A:

Sorry, I forgot to specify before that it's your first expression which works for sound. Again, think of the receiver leaving the sender at the speed of sound. The receiver will be riding along with one part of the wave, and see no oscillations. It sees zero frequency. That's very different from what happens if the sender is moving away at the speed of sound and the receiver is at rest in the medium. There the distance between wave crests is just stretched a factor of two. Your first expression gets that right.

Mike W.


(published on 07/14/2013)

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