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The Doppler effect equation is: $$f_{\text{observer}} = f_{\text{source}} \left( \frac{v \pm v_{\text{observer}}}{v \pm v_{\text{source}}} \right)$$ where we take the velocity positive when it is from the observer to the listener, otherwise we use the minus sign. Click here for the derivation.

Now my question is what would happen when both the source and the observer would be in a velocity but not in a straight line but in a 2D plane? For example, for moving source and a stationary observer in a plane, the formula is: $$f_{\text{observer}} = f_{\text{source}} \left( \frac{v}{v - v_{\text{source}}\cos{\theta_{s}}} \right)$$ where $\theta_{s}$ is the angle between the sound velocity and the source velocity. Click here for the derivation.

I want to know how to derive the formula when even the observer has its own velocity in the planar case. I found here on the internet that the formula is: $$f_{\text{observer}} = f_{\text{source}} \left( \frac{v - v_{\text{observer}}\cos{\theta_{o}}}{v - v_{\text{source}}\cos{\theta_{s}}} \right)$$ where $\theta_{o}$ is the angle between the sound velocity and observer velocity. But how to derive this formula?

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    $\begingroup$ If you have a derivation for the second formula, could you perhaps modify that derivation to arrive at the third formula? $\endgroup$ Nov 13, 2021 at 18:53
  • $\begingroup$ No, I couldn't. I have obtained the formula when either the observer is moving or when the source is moving, but not when both the observer and the source are moving. $\endgroup$ Nov 13, 2021 at 19:00
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    $\begingroup$ Both formulas are using the components of velocity along the line connecting the source and observer. $\endgroup$
    – R.W. Bird
    Nov 13, 2021 at 19:49
  • $\begingroup$ Related : Deriving relativistic Doppler shift in terms of wavelength $\endgroup$
    – Frobenius
    Nov 14, 2021 at 5:00

1 Answer 1

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If you know the Doppler factor for stationary emitter and moving receiver and vice versa, you can just imagine a stationary repeater in the middle that intercepts the signal and emits another signal at the same frequency that is detected by the real receiver. That shows that the Doppler shift factor between a general emitter and receiver is the product of the factors for the moving emitter, stationary receiver and the stationary emitter, moving receiver.

Here's a way to derive it directly. Suppose the source emits a pulse at time $0$ and position $\mathbf 0$, and another signal at time $δt_s$ and position $\mathbf v_s δt_s$, and the observer detects the pulses at times $t$ and $t+δt_o$ and positions $\mathbf x$ and $\mathbf x + \mathbf v_o δt_o$. Then you have

$$\begin{eqnarray} \lVert \mathbf x \rVert &=& vt \\ (\mathbf x + \mathbf v_o δt_o - \mathbf v_s δt_s)^2 &=& v^2 (t+δt_o-δt_s)^2 \end{eqnarray}$$

Substitute $t \to \lVert \mathbf x \rVert / v$ in the second equation, expand it, and discard terms that are second order in $δt_s$ and $δt_o$, to get

$$\mathbf x \cdot \mathbf v_o\,δt_o - \mathbf x \cdot \mathbf v_s\,δt_s = \lVert \mathbf x \rVert v (δt_o-δt_s)$$

Solve for $δt_s/δt_o = f_o/f_s$ to get

$$\frac{f_o}{f_s} = \frac{v - \mathbf v_o\cdot \mathbf{\hat x}}{v - \mathbf v_s\cdot \mathbf{\hat x}}$$

where $\mathbf{\hat x} = \mathbf x / \lVert \mathbf x \rVert$ is a unit vector pointing from the location of the source at the time of emission toward the location of the observer at the time of reception.

Incidentally, you can derive the special-relativistic Doppler shift for light in a similar way, getting $\displaystyle \frac{f_o}{f_s} = \frac{\mathbf v_o\cdot \mathbf x}{\mathbf v_s\cdot \mathbf x}$ where the $\mathbf v$s are four-velocities and $\mathbf x$ is a null four-vector.

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