0

In $2D$ it's easy to deal with doppler effect given the velocity of the source and the velocity of the observer like in fig (1) where we have the observer at A static and the source in direction of $\vec{u}$ emitting waves with velocity $v$ and frequency $f$ and moving with velocity $v_s$ we will say that

$$f'=\frac{v}{v-v_s\cos(\theta)}*f$$ Where theta is the angle between the $x$-axis and the $\vec{u}$.

Now if we have have a vector in space with the same frequency and velocity and a static point in space how do we represent the actual frequency $f'$ that the observer receives?

fig (1) enter image description here

fig (2) fig1

Qmechanic
  • 201,751
Youssef Mohamed
  • 129
  • Related : Doppler effect equation derivation. – Frobenius Nov 21 '22 at 16:17
  • Two vectors define a plane. You have two vectors: The vector from the moving particle to the fixed particle at t=0, and the vector of the moving particle's velocity. You can rotate the 3-D case into the 2-D case. First place the separation vector arbitrarily somewhere in a 2-D surface. Then rotate around this until the velocity is in the same 2-D surface. – Boba Fit Nov 21 '22 at 22:45

2 Answers2

1

enter image description here

2D case
$$\vec R=\begin{bmatrix} a-u\,t \\ b \\ \end{bmatrix}$$ and $~\vec R~$ with polar coordinate

$$\vec R_p=r\,\begin{bmatrix} \cos(\phi) \\ \sin(\phi) \\ \end{bmatrix}$$

with

$$\vec R=\vec R_p\quad \Rightarrow\\ r=\sqrt{b^2+(a-\,t)^2}\quad, \tan(\phi)=\frac{b}{a-u\,t}\\ v=\dot r=-\frac{u\,(a-u\,t)}{\sqrt{b^2+(a-u\,t)^2}}$$

substitute $~(a-u\,t)=\frac b{\tan(\phi)}~$ in $~v~$ you obtain $~v=-u\,\cos(\phi)$

$$ f' = \frac{c}{c -u\,\cos(\phi) } f $$

3D case

$$\vec R=\begin{bmatrix} a-u\,t \\ b \\ c\\ \end{bmatrix}\quad, \vec R_p= r\left[ \begin {array}{c} \cos \left( \phi \right) \sin \left( \theta \right) \\ \sin \left( \phi \right) \sin \left( \theta \right) \\ \cos \left( \theta \right) \end {array} \right] $$

$\vec R=\vec R_p\quad\Rightarrow$

$$r=\sqrt{b^2+(a-u\,t)^2+c^2}\\ \tan(\phi)=\frac{b}{a-u\,t}\\ \tan(\theta)=\frac{\sqrt{b^2+(a-u\,t)^2}}{c}\\ v=\dot r=-\frac{u\,(a-u\,t)}{\sqrt{b^2+(a-u\,t)^2+c^2}}=-u\,\cos(\phi)\,\sin(\theta) $$

$$ f' = \frac{c}{c -u\,\cos(\phi)\,\sin(\theta) } f $$

Eli
  • 11,878
1

If $\hat{r}$ is a unit vector pointing from the source to the receiver, then $v_s \cos \theta = \hat{r} \cdot \vec{v}_s$, where $\vec{v}_s$ is the velocity vector of the source. So the natural generalization is $$ f' = \frac{c}{c - \hat{r} \cdot \vec{v}_s} f $$ which is perfectly well-defined in 3D.

Michael Seifert
  • 48,772