# I Asymmetry in Doppler effect

1. Jul 11, 2016

### DaTario

Hi All,

What is the explanation for the fact that doppler effect has an asymmetry with relation to the relative movement fo the source and the observer? For instance,
$\frac{v + 5}{v} \neq \frac{v}{v -5}$
The version I have seems to be somewhat truncated to me; it simply relates this fact (asymmetry) to the existence of a special system of reference, which is attached to the medium where the wave propagates.

Best Regards,

DaTario

2. Jul 11, 2016

### Staff: Mentor

Why should it be symmetrical?

3. Jul 11, 2016

### mathman

What is the context, sound or light?

4. Jul 12, 2016

### DaTario

Hi mathman,

I am trying to get a more complete argument to work with my students the idea that, within a material medium, in the case of waves, one are not expected to claim that under a change of reference frame, the physics should be the same.

The source getting near, at a velocity of 5 m/s, with the observer at rest does not produce the same result as in the situation where the source is standing still and the observer are getting near to the source at 5 m/s.

Dale: the answer to your question is in the above comment to mathman.

Thank you, both.
Best wishes,

DaTario

Last edited: Jul 12, 2016
5. Jul 12, 2016

### DaTario

Dale: perhaps your question is my most fundamental doubt.

I would like to explain in a clear way why there is a difference between changing the reference frame in a collision setup and changing the reference frame in this doppler effect experiment.

Best Regards,

DaTario

6. Jul 12, 2016

### pixel

Let’s consider initially that the source, the medium and the observer are at rest with respect to each other.

In the first case, let the observer start moving toward the source. As he moves, he encounters wavefronts more frequently than when he was at rest so the frequency increases. When his speed is equal to the speed of sound, the observed frequency is twice what it was when he was at rest. In fact, he can move arbitrarily fast (subject to being < speed of light) and the frequency will increase correspondingly.

In the second case, let the source move toward the observer. It is now following the sound that it emits. As it moves faster, the emitted wavefronts get bunched closer and closer. All it has to do is move at the speed of sound for all the wavefronts to be bunched up in the same place (we could say the wavelength is now 0 and the frequency is ∞).

Because there is a medium, it depends which of the observer or source is moving with respect to the medium.

7. Jul 12, 2016

### mathman

In the case of sound waves, the medium is the reference frame of interest, so the situation is not symmetrical.

8. Jul 12, 2016

### DaTario

I think it is appealing to put in equivalence these two situations:

1) the observer moving toward the source, which is at rest with relation to the ground (with no wind)

2) the observer at rest with relation to the ground and a wind going from the source to the observer.

9. Jul 12, 2016

### DaTario

But do you agree that being " the reference of interest" does not seem to be something strong enough to produce a solid explanation?

I think we need a stronger statement. Something like: "in the special reference frame in which the air is still (whenever it is a possibility) there is a formation of a specific space and time pattern of wave amplitudes if the source is at rest (pattern I). If not, there is another pattern (pattern II). When the observer moves toward the source at rest, he sees the pattern I modified by a drift."

I am not content with this, however.

Last edited: Jul 12, 2016
10. Jul 12, 2016

### A.T.

11. Jul 12, 2016

### Staff: Mentor

The physics is the same under a change of reference frame, that is one of the fundamental symmetries of nature.

The problem that students have is that they don't know how to properly transform the reference frames. You have to transform the velocity of the source, receiver, and the medium. Usually they forget that step. So in most frames you need to know the acoustical wave equation in the presence of wind. So a moving source and an at rest observer in still air transforms to an at rest source and a moving observer being carried along by the wind. It is a much more complicated scenario to analyze, but it is physically equivalent.

12. Jul 12, 2016

### Staff: Mentor

There is wind in one and not the other. That makes all the asymmetry.

13. Jul 12, 2016

### DaTario

14. Jul 12, 2016

### DaTario

Ok, thank you. Your text helps clearifying.

15. Jul 12, 2016

### DaTario

Just to register, I found recently a nice way to express the doppler formula, which seems easier to memorize:
$\frac{f_r}{v \pm u_r} = \frac{f_s}{v \pm u_s}$