Who's moving towards whom in Doppler effect situations?

[azzarooni88]

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: f_o = (c + v_o)/c) f_s

where c=speed of sound

Observer is moving towards source

Formula: f_o = (c/c-v_s) f_s

Cheers

 

[PeroK]

If it's the speed of sound, then you need to consider who is moving relative to the air.

 

[azzarooni88]

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)

 

[PeroK]

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.

 

[Dale]

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

 

[azzarooni88]

Thanks. But how can we add the velocity of the observer when its the skydiver that's moving?

 

[Mister T]

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.

 

[azzarooni88]

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?

 

[PeroK]

Source is moving towards observer

Formula: f_o = (c + v_o)/c) f_s

where c=speed of sound

Observer is moving towards source

Formula: f_o = (c/c-v_s) f_s

Cheers

You have got these the wrong way round, by the way.

 

[azzarooni88]

You have got these the wrong way round, by the way.

I have not actually.

 

[PeroK]

I have not actually.

Then I am unable to help you any further.

 

[azzarooni88]

Then I am unable to help you any further.

look up online doppler equation if you don't believe me

 

[PeroK]

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!

 

[jbriggs444]

Source is moving towards observer

Formula: f_o = (c + v_o)/c) f_s

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: f_o = (c/c-v_s) f_s

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.

 

[azzarooni88]

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