Subfield: Mechanics


Question: d. After falling for some distance, the water stream usually breaks into smaller droplets. This occurs because small random perturbations to the shape of the stream grow over time, eventually breaking the stream into apart.


For the rest of this problem we ignore the change in the radius of the stream due to changing speed of the water, as considered earlier. Instead, we examine small random variations in the radius of the stream.



Random variations can be broken down into a sum of sinusoidal variations in stream radius, each with a different wavenumber $k$. We can analyze these different sinusoidal variations independently.



Consider a stream of water whose radius obeys



$$

r(y)=r_{0}+A \cos (k y)

$$



where $A \ll r_{0}$ is the perturbation amplitude. To analyze such a stream, it is sufficient to consider only the thickest and thinnest parts of the stream.



Accounting for both sources of curvature, find a condition on $r_{0}$ and $k$ such that the size of perturbations increases with time.



![](https://cdn.mathpix.com/cropped/2023_12_21_514efaf567aad8907c0ag-1.jpg?height=479&width=379&top_left_y=1002&top_left_x=884)

Solution:
If the size of the perturbation increases with time, water must be flowing from the thin parts of the stream to the thick parts. For that to happen, the pressure needs to be higher in the thin parts of the stream than in the thick parts of the stream so that the pressure gradient will force water towards the thick parts, eventually breaking the stream into droplets.



We consider a small patch with side lengths $h$ on the surface of the stream at the thinnest part of the stream. The pressure is



$$

\Delta P_{\text {thin }}=\sigma\left(\frac{1}{r_{\text {thin }}}+\frac{1}{R_{\text {thin }}}\right)

$$



And at the thickest part of the stream,



$$

\Delta P_{\text {thick }}=\sigma\left(\frac{1}{r_{\text {thick }}}+\frac{1}{R_{\text {thick }}}\right)

$$



We are looking for the wavenumbers such that



$$

\Delta P_{\text {thin }}>\Delta P_{\text {thick }}

$$







Using the Young-Laplace equation, this becomes



$$

\sigma\left(\frac{1}{r_{\text {thin }}}+\frac{1}{R_{\text {thin }}}\right)>\sigma\left(\frac{1}{r_{\text {thick }}}+\frac{1}{R_{\text {thick }}}\right)

$$



Dropping the common factor $\sigma$,



$$

\left(\frac{1}{r_{\text {thin }}}+\frac{1}{R_{\text {thin }}}\right)>\left(\frac{1}{r_{\text {thick }}}+\frac{1}{R_{\text {thick }}}\right)

$$



To simplify this further, we will need to find $r$ and $R$ in terms of $A$ and $k$, the variables given in the problem statement.



$r$ is the thickness of the stream, which from the equation given, varies sinusoidally. So



$$

\begin{aligned}

r_{\text {thin }} & =r_{0}-A . \\

r_{\text {thick }} & =r_{0}+A

\end{aligned}

$$



We are going to need $\frac{1}{r}$ to use in the Young Laplace equation, so we make the approximations



$$

\begin{aligned}

& \frac{1}{r_{\text {thin }}} \approx \frac{1}{r_{0}}+\frac{A}{r_{0}^{2}} \\

& \frac{1}{r_{\text {thick }}} \approx \frac{1}{r_{0}}-\frac{A}{r_{0}^{2}}

\end{aligned}

$$



(To find these, recall $\frac{1}{1-\epsilon} \approx 1+\epsilon$ for small $\epsilon$.)



The inequality now becomes



$$

\frac{1}{r_{0}}+\frac{A}{r_{0}^{2}}+\frac{1}{R_{\text {thin }}}>\frac{1}{r_{0}}-\frac{A}{r_{0}^{2}}+\frac{1}{R_{\text {thick }}}

$$



This simplifies to



$$

\frac{2 A}{r_{0}^{2}}>\frac{1}{R_{\text {thick }}}-\frac{1}{R_{\text {thin }}}

$$



Next we need to determine the radius of curvature $R$ of the sinusoidal as a function of $k$ and $A$.







To do this, we compare the sinusoidal function and a circle at small deviations from the thickest part of the stream.



Recall that, for small $\theta$,



$$

\cos \theta \approx 1-\frac{1}{2} \theta^{2}

$$



which means that for small $x$,



$$

y_{\text {sinusoidal }}=A \cos (k x) \approx A\left(1-\frac{1}{2} k^{2} x^{2}\right)

$$



Next we consider a circle of radius $R$. If a particle moves along such a circle at speed $v$, its acceleration is $v^{2} / R$. This means that if the particle moves forward for a short time $t$, it moves forward a distance $v t$ and falls a distance $\frac{1}{2} \frac{v^{2}}{R} t^{2}$. If we set $v t=x$, then the $y$ position of the particle is given by



$$

y_{\text {circle }} \approx y_{0}-\frac{1}{2} \frac{x^{2}}{R} .

$$



Comparing $y_{\text {circle }}$ and $y_{\text {sinusoidal }}$, they give the same motion if $A k^{2}=\frac{1}{R}$.



Then



$$

\begin{aligned}

\frac{1}{R_{\text {thin }}} & =-A k^{2} . \\

\frac{1}{R_{\text {thick }}} & =A k^{2} .

\end{aligned}

$$



Putting these into the inequality,



$$

\frac{2 A}{r_{0}^{2}}>2 A k^{2}

$$



This simplifies to



$$

k<\frac{1}{r_{0}}

$$



So the perturbations will grow as long as they have a wavenumber greater than one over the radius, or equivalently when the wavelength of the perturbation is longer than the circumference of the stream.



This result was discovered experimentally by Plateau and derived theoretically by Rayleigh. The breaking up of a stream into droplets is called the Plateau-Rayleigh instability.

Final answer: $k<\frac{1}{r_{0}}$