Who's moving towards whom in Doppler effect situations?

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    Doppler Doppler effect
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SUMMARY

The discussion centers on the interpretation of the Doppler effect in a scenario involving a skydiver emitting sound waves while descending at terminal velocity. Participants clarify that the classical Doppler shift formulas are dependent on the relative motion of the source and observer, specifically emphasizing that sound waves require a medium (air) for propagation. The correct formulas are established: for a source moving towards an observer, the formula is fo = (c + vo)/c * fs, and for an observer moving towards a source, it is fo = (c/c - vs) * fs. Misunderstandings regarding the application of these formulas are addressed, leading to a consensus on their correct usage.

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  • Familiarity with sound wave propagation and the role of the medium.
  • Knowledge of classical and relativistic Doppler shift formulas.
  • Basic grasp of frequency, wavelength, and their relationship in wave mechanics.
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azzarooni88
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Hello all, (1st post)

I would like some help with deciding if the source is moving towards the observer or vice versa in Doppler effect situations.

The example that I am confused about is a skydiver holding a constant frequency emitting source during the descent. The skydiver is traveling at terminal velocity. His friend is on the ground and hears a frequency that is higher.
I am wondering why we can't treat the example as the skydiver is stationary since he is not accelerating and that his friend (and the Earth) is moving towards him i.e. Observer is moving towards source?

Source is moving towards observer

Formula: fo = (c + vo)/c) fs

where c=speed of sound

Observer is moving towards source

Formula: fo = (c/c-vs) fs

Cheers
 
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If it's the speed of sound, then you need to consider who is moving relative to the air.
 
PeroK said:
If it's the speed of sound, then you need to consider who is moving relative to the air.
Can you elaborate? My lecturer said it had something to do with the medium (air)
 
azzarooni88 said:
Can you elaborate? My lecturer said it had something to do with the medium (air)

The speed of sound (through air) is independent of the speed of the source.
 
azzarooni88 said:
I am wondering why we can't treat the example as the skydiver is stationary since he is not accelerating and that his friend (and the Earth) is moving towards him i.e. Observer is moving towards source?
The classical Doppler shift formula is given in the rest frame of the medium. The relativistic Doppler shift formula is symmetrical so it doesn’t matter.

http://mathpages.com/rr/s2-04/2-04.htm
 
Thanks. But how can we add the velocity of the observer when its the skydiver that's moving?
 
azzarooni88 said:
The example that I am confused about is a skydiver holding a constant frequency emitting source during the descent.

Is it a source of sound waves or a source of electromagnetic waves? Sound waves require a medium, and so the speed of the medium relative to both the emitter and the receiver has an effect. Electromagnetic waves do not require a medium.
 
Mister T said:
Is it a source of sound waves or a source of electromagnetic waves? Sound waves require a medium, and so the speed of the medium relative to both the emitter and the receiver has an effect. Electromagnetic waves do not require a medium.
Right that makes sense and yes it is a source of sound waves. But when we calculate the velocity of the skydiver, why do we use that value as the velocity of the observer?
 
azzarooni88 said:
Source is moving towards observer

Formula: fo = (c + vo)/c) fs

where c=speed of sound

Observer is moving towards source

Formula: fo = (c/c-vs) fs

Cheers

You have got these the wrong way round, by the way.
 
  • #10
PeroK said:
You have got these the wrong way round, by the way.
I have not actually.
 
  • #11
azzarooni88 said:
I have not actually.

Then I am unable to help you any further.
 
  • #12
PeroK said:
Then I am unable to help you any further.
look up online doppler equation if you don't believe me
 
  • #13
azzarooni88 said:
look up online doppler equation if you don't believe me

You have a formula where the source is moving with a ##v_o## (velocity of observer) in it. And a formula where the observer is moving with a ##v_s## (velocity of source in it). That is the wrong way round!
 
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  • #14
azzarooni88 said:
Source is moving towards observer

Formula: fo = (c + vo)/c) fs
There is a simple sanity check to see whether this is the right formula. Suppose that the source is moving at the speed of sound. The source will exactly keep up with the transmitted signal. All of the peaks and valleys will pile up on top of one another. Wavelength: 0. Frequency: infinite.

The above formula does not have that feature. It merely doubles the received frequency when the velocity of the source is at the speed of the signal.
Observer is moving towards source

Formula: fo = (c/c-vs) fs
This formula (if properly parenthesized) does have the required feature. If you set v=c then the frequency becomes infinite.

This suggests that @PeroK objected correctly that the formulas are reversed.
 
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  • #15
jbriggs444 said:
There is a simple sanity check to see whether this is the right formula. Suppose that the source is moving at the speed of sound. The source will exactly keep up with the transmitted signal. All of the peaks and valleys will pile up on top of one another. Wavelength: 0. Frequency: infinite.

The above formula does not have that feature. It merely doubles the received frequency when the velocity of the source is at the speed of the signal.

This formula (if properly parenthesized) does have the required feature. If you set v=c then the frequency becomes infinite.

This suggests that @PeroK objected correctly that the formulas are reversed.

Okay I agree. My bad. Thank you for fixing it. I've been wrong about a lot recently
 
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