Understanding the Doppler Effect with Moving Sources and Observers

AI Thread Summary
The discussion centers on understanding the Doppler Effect when both the source and observer are in motion. Participants express confusion about how to derive the observed frequency from the individual cases of moving sources and observers, particularly regarding the squaring of frequencies. The equations for observed frequency when either the source or observer is stationary are provided, but participants struggle with the implications of combining these scenarios. Key points include the importance of relative speeds to the medium and the suggestion to analyze the situation from a stationary observer's perspective. The conversation highlights the complexity of the Doppler Effect in dynamic systems and the need for clarity in mathematical transformations.
windowofhope
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Homework Statement


Can't figure how to prove when both source and observer are moving that the observed frequency is the product of the two independent cases below..

Homework Equations


We know individual cases when source is moving and observer is still, the observed frequency is equal to fs/(1-(vs/c)*cos(theta_s)) where fs is the source frequency and theta_s is the angle between the source velocity and observer

Conversely when observer is moving, the observed frequency is fs*(1+(vo/c)*cos(theta_o))

The Attempt at a Solution


When you multiply both out, you get fo2=fs2*(c+vocos(theta_o))/(c-vscos(theta_s)) but they shouldn't be squared??
 
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Transform into a frame where one of them is still. What do you get?
 
well of course either times
(1+(vo/c)*cos(theta_o))

or divided by
(1-(vs/c)*cos(theta_s))

I just can't comprehend how the product of the two shifts leads to the real equation..
 
Khashishi said:
Transform into a frame where one of them is still. What do you get?
That doesn't work, since then the medium will be moving. The speeds relative to the medium are important.
@windowofhope , consider a stationary observer between them. What frequency will the observer hear? If the observer were to emit a tone at that frequency, what frequency would the other receiver hear?
 
I assumed it was in vacuum, since the equations listed above have no "n".
 
haruspex said:
That doesn't work, since then the medium will be moving. The speeds relative to the medium are important.
@windowofhope , consider a stationary observer between them. What frequency will the observer hear? If the observer were to emit a tone at that frequency, what frequency would the other receiver hear?
what youre alluding to isn't really what I'm getting at since swapping fo and fs wouldn't solve it either. The math is clearly right - I'm just confused as to how you can simplify fo^2=fs^2 *((1+(vo/c)*cos(theta_o)))/((1-(vs/c)*cos(theta_s)))
 
windowofhope said:
I'm just confused as to how you can simplify fo^2=fs^2 *((1+(vo/c)*cos(theta_o)))/((1-(vs/c)*cos(theta_s)))
I'm confused as to how you get
##f_o^2=f_s^2 \frac{(1+\frac{v_o}{c}\cos(\theta_o))}{(1-\frac{v_s}{c}\cos(\theta_s))}##.
Following my own hint in post #4 I get a rather simpler result.
 
haruspex said:
I'm confused as to how you get
##f_o^2=f_s^2 \frac{(1+\frac{v_o}{c}\cos(\theta_o))}{(1-\frac{v_s}{c}\cos(\theta_s))}##.
Following my own hint in post #4 I get a rather simpler result.
Honestly I'm not following what you're suggesting... :-/

I got the squares simply by taking each individual formula and multiplying them together.. It then becomes clear that you get the proper term in parentheses (1+vocoso/c)/(1-vscoss/c) but the fo and fs are squared as a result of that as well...
 
windowofhope said:
I got the squares simply by taking each individual formula and multiplying them together...
ok, but unfortunately that's not a good guess.
Try following my reasoning.
Put a stationary observer O on the straight line between source S and receiver R. What frequency does O hear? Call that frequency f'.
Now forget S and consider O generating a frequency f'. What, according to the Doppler equations, is the frequency R will hear?
 
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