Nice way to show that plane wave is a limit of spherical wave

Question

I am looking for some nice way to show the (university) students that a plane wave is a limit of a spherical wave $\frac{A}{r}e^{i\omega t - ikr}$ for large $r$ (or, of course, that it is a reasonable approximation for a large $r$ domain). It is for purposes of acoustics, but it should be shown for general case.

There are multiple approaches certainly, such as:

• showing that acoustic pressure and velocity get in phase for large $r$ which is typical plane wave feature.
• showing that amplitude rate of change converges faster than magnitude of amplitude

but I find these "not elegant". Isn't there something better?

Answer 1

I would start with moving into a coordinate system where the centre of the wave is far away, let's say at $-x_0$. So $$r = \sqrt{(x+x_0)^2+y^2+z^2}$$ If we now assume that $x_0 \gg x,y,z$, then we can make a Taylor expansion and end up with $$r \approx x_0 + x$$ where we have neglected higher orders of $x$, $y$, and $z$. This approximation also makes sense intuitively. Moving along the $x$ direction changes the radius by the amount we move. Moving perpendicular to the $x$ direction will, to first order, not change the distance to the centre of the wave.

With this the wave now looks like $$\frac{A}{x_0+x}e^{i\omega t-ik(x_0+x)}$$ The $-ikx_0$ term in the exponent can be factored out and gives a constant phase shift. The $x_0 + x$ in the denominator can be approximated by $x_0$ because $x_0\gg x$. So we end up with $$\frac{A}{x_0}e^{-ikx_0}e^{i\omega t-ikx}$$ This is a plane wave with constant amplitude $A/x_0$ and an additional phase shift of $e^{-ikx_0}$.

Answer 2

I do not know if you will consider this not rigorous enough or indeed totally wrong?

There are two part to your equation $\displaystyle \frac{A}{r}e^{i\omega t - ikr}$ which perhaps is better written as $\displaystyle \frac{A}{r}e^{i\omega t - i\vec k \cdot \vec r}$.

For a spherical wave $\vec k \cdot \vec r = kr$ if $\vec r$ is the position vector from the source.

For a plane wave moving in the positive $x$ direction $\vec k \cdot \vec r = kr \cos \theta = kx$ where $\theta$ is the angle between the $k$ vector and the $r$ vector.

For small angles $\theta, \cos \theta \approx 1$ which is equivalent to $x \approx r$.

The other part to consider is $\dfrac 1 r$ and for this look at

$\dfrac{d(r^{-1})}{dr} = -r^{-2} \Rightarrow \Delta (r^{-1}) \approx - r^{-2} \Delta r \Rightarrow \Delta (x^{-1}) \approx - x^{-2} \Delta x$

which is your "showing that amplitude rate of change converges faster than magnitude of amplitude (changes)" and is equivalent to doing a Taylor expansion?

So you end up with $\displaystyle B\; e^{i\omega t - ikx}$ where $B$ changes less and less over a given range of $x$ values as $x$ gets larger and larger.

Answer 3

For plane wave and spherical wave, $\vec{k}$ and $\vec{r}$ are parallel, i.e., wave k propagates in the radial direction r. Using direction cosines, the spherical wave equation can be expanded to: $$\vec{k} \cdot \vec{r}=k cos\alpha\cdot r cos\alpha + k cos\beta\cdot r cos\beta +k cos\gamma\cdot r cos\gamma=kr(cos^2\alpha+\cos^2\beta+cos^2\gamma)=kr$$

For a vector $\vec{r}$=(x,y,z) close to the x-axis, i.e., x >> y and x >> z, we have $\alpha \approx 0$, $\beta\approx 90^o$, $\gamma\approx 90^o$. Since $x=rcos\alpha$, thus $r\approx x$ and $\vec{k} \cdot \vec{r}\approx kx$. Another way to look at it is that from the above equation, we have $\vec{k} \cdot \vec{r}\approx kr cos^{2}\alpha = kxcos\alpha\approx kx$. For the amplitude $\frac{A}{r}$, the derivative or the slope of $\frac{1}{r}=-r^{-2}$. Therefore, a larger r will have a smaller amplitude, but a more constant amplitude. $$\psi(\mathbf r,t)= (\frac{A}{r}) e^{i(\mathbf k \cdot \mathbf r - \omega t)}= (\frac{A}{x})e^{i( k \cdot x - \omega t)} $$ The graph below shows angle $\alpha$ gets smaller as x increases or the wave arcNear moves farther away from the origin O to become arcFar and less curved. The length difference between arcN and line AB can be a measure of the curvature = $r\alpha - rsin\alpha \approx r(\frac{\alpha^3}{6}-\frac{\alpha^5}{120})$, which approaches to 0 as $\alpha$ -> 0 as in the arcF. We can also calculate the arcN curved bottom area of the cone OAB $AR_\alpha$ and the flat circular bottom area AB of the cone OAB $AF_\alpha$. For the solid angle $\alpha$, $AR_\alpha = 4 r^2 \alpha \cdot sin\alpha$ and $AF_\alpha = \pi (r \cdot sin\alpha)^2$. The ratio of $$\frac{AR_\alpha}{AF_\alpha} = \frac{4 r^2 \alpha \cdot sin\alpha}{\pi r^2 \cdot sin^2\alpha}= (\frac{\pi}{4}\frac{sin\alpha}{\alpha})^{-1}$$ Therefore, as $\alpha$ -> 0, this ratio decreases from > $\frac{4}{\pi}$ to $= \frac{4}{\pi}$ like what the arcF shows it is flattened, more like a plane wave.

A good example is that the Sun, which is $d=1.5\cdot10^{8}$ km away from the Earth whose radius is $r=6.4\cdot10^{3}$ km, shines light to the Earth and makes an angle of $tan\alpha = \frac{r}{d}=4.27\cdot10^{-5} \approx \alpha$. For the closest planet Mercury, $r=2.44 \cdot10^{3}$ km and $d=5.8\cdot10^{7}$ km, $tan\alpha = \frac{r}{d}=4.21\cdot10^{-5} \approx \alpha$. From a point on either the Earth or Mercury, the sunshine is like a plane wave.