# I 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 travelling 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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#### PeroK

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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

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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

Mentor
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 thats moving?

#### Mister T

Gold Member
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

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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.

#### azzarooni88

You have got these the wrong way round, by the way.
I have not actually.

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#### azzarooni88

Then I am unable to help you any further.
look up online doppler equation if you dont believe me

#### PeroK

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look up online doppler equation if you dont 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 and berkeman

#### jbriggs444

Homework Helper
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.

• berkeman and PeroK

#### 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

• jbriggs444 and berkeman

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