While studying the classic Doppler effect equation, I was curious to how the traditional equation of the doppler effect for sound, where the observer and sources are facing each other, gets effected once the observer is stationary and the source moves at an angle from the horizontal?
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2$\begingroup$ I ran into this recently myself. Apparently the main change is that one substitutes the observer and source speeds with their components in the direction of the wave. In the case of a stationary source, see kirkmcd.princeton.edu/examples/wave_velocity.pdf. For the case of a stationary observer, see section 2 of scholar.harvard.edu/files/schwartz/files/lecture21-doppler.pdf. (For my own purposes I'd like to see a unified treatment of such, but haven't seen it yet...) $\endgroup$– SemiclassicalCommented Nov 25, 2020 at 4:55
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2 Answers
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Here's a derivation of a general formula, which turns out to be just as simple as the commonly quoted formulas that work only in special cases.
Choose the rest frame of the medium as your reference frame. Let $c$ be the speed of sound. Say that an emitter with velocity $\def\v{\mathbf v} \v_e$ emits a wavefront at $\def\x{\mathbf x} t=0,\x=0$, and a second wavefront at $t=δt, \x = δt\,\v_e$, and a receiver with velocity $\v_r$ detects the first wavefront at $t=Δt, \x=Δ\x$ and the second at $t=Δt+k\,δt, \x=Δ\x+k\,δt\,\v_r$. The ratio of time intervals, $\frac{k\,δt}{δt} = k$, in the $δt\to 0$ limit, is the Doppler (red)shift factor. (I picked $k$ at random because, weirdly, there is no standard letter for this ratio.)
Both wavefronts travel at $c$, so
$$\begin{align} Δ\x^2 &= c^2 Δt^2 \\ (Δ\x + k\,δt\,\v_r - δt\,\v_e)^2 &= c^2 (Δt + k\,δt - δt)^2 \end{align}$$
Subtracting the first equation from the second gives you
$$2 Δ\x \cdot (k\,δt\,\v_r - δt\,\v_e) + (k\,δt\,\v_r - δt\,\v_e)^2 = 2 c^2 Δt \, (k\,δt - δt) + c^2 (k\,δt - δt)^2$$
In the $δt\to 0$ limit, the second term on each side is negligible. If you drop those terms, it's easy to solve for $k$:
$$\boxed { k = \frac{c^2 - \mathbf c \cdot \v_e}{c^2 - \mathbf c \cdot \v_r} }$$
where $\mathbf c = Δ\x/Δt$ is the "velocity of sound": a vector of length $c$ pointing from the emitter toward the receiver.
You can derive the special-relativistic Doppler shift for light in a similar way. The result is even simpler:
$$k = \frac{\x \cdot \v_e}{\x \cdot \v_r}$$
where the $\v$s are four-velocities, and $\x$ is a null four-vector pointing from emitter to receiver.
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The equation will change as the moving source/observer changes position. You will only get a constant frequency if the source and observer are both moving with the same velocity (unless $\theta=0$ or $\theta=\pi$).
