1 files changed, 55 insertions, 27 deletions

diff --git a/content/know/concept/rayleigh-plesset-equation/index.pdc b/content/know/concept/rayleigh-plesset-equation/index.pdc
index ee8622b..9325f3f 100644
--- a/content/know/concept/rayleigh-plesset-equation/index.pdc
+++ b/content/know/concept/rayleigh-plesset-equation/index.pdc
@@ -19,34 +19,29 @@ describes how the radius of a spherical bubble evolves in time
 inside an incompressible liquid.
 Notably, it leads to [cavitation](/know/concept/cavitation/).
 
-
-## Simple form
-
-The simplest version of the Rayleigh-Plesset equation is found
-in the limiting case of a liquid with zero viscosity zero surface tension.
-
-Consider one of the [Euler equations](/know/concept/euler-equations/)
-for the velocity field $\va{v}$,
-where $\rho$ is the (constant) density:
+Consider the main
+[Navier-Stokes equations](/know/concept/navier-stokes-equations/)
+for the velocity field $\va{v}$:
 
 $$\begin{aligned}
     \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
     = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v}
-    = - \frac{\nabla p}{\rho}
+    = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
 \end{aligned}$$
 
 We make the ansatz $\va{v} = v(r, t) \vu{e}_r$,
 where $\vu{e}_r$ is the basis vector;
 in other words, we demand that the only spatial variation of the flow is in $r$.
-The above Euler equation then becomes:
+The above equation then becomes:
 
 $$\begin{aligned}
     \pdv{v}{t} + v \pdv{v}{r}
     = - \frac{1}{\rho} \pdv{p}{r}
+    + \nu \bigg( \frac{1}{r^2} \pdv{r} \Big( r^2 \pdv{v}{r} \Big) - \frac{2}{r^2} v \bigg)
 \end{aligned}$$
 
 Meanwhile, the incompressibility condition
-is as follows in this situation:
+in [spherical coordinates](/know/concept/spherical-coordinates/) yields:
 
 $$\begin{aligned}
     \nabla \cdot \va{v}
@@ -63,42 +58,75 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where $C(t)$ is an unknown function that does not depend on $r$.
-We then insert this result in the earlier Euler equation,
+We then insert this result in the main Navier-Stokes equation,
 and isolate it for $\pdv*{p}{r}$, yielding:
 
 $$\begin{aligned}
     \pdv{p}{r}
-    = - \rho \bigg( \pdv{v}{t} + v \pdv{v}{r} \bigg)
+    = - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2
+    - \nu \Big( \frac{2}{r^4} C - \frac{2}{r^4} C \Big) \bigg)
     = - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2 \bigg)
 \end{aligned}$$
 
 Integrating this with respect to $r$ yields the following expression for $p$,
-where $p_\infty$ is the (possibly time-dependent) pressure at $r = \infty$:
+where $p_\infty(t)$ is the (possibly time-dependent) pressure at $r = \infty$:
+
+$$\begin{aligned}
+    p(r)
+    = p_\infty + \rho \bigg( \frac{1}{r} C' - \frac{1}{2 r^4} C^2 \bigg)
+\end{aligned}$$
+
+From the definition of [viscosity](/know/concept/viscosity/),
+we know that the normal [stress](/know/concept/cauchy-stress-tensor/)
+$\sigma_{rr}$ in the liquid is given by:
+
+$$\begin{aligned}
+    \sigma_{rr}(r)
+    = - p(r) + 2 \rho \nu \pdv{v(r)}{r}
+\end{aligned}$$
+
+We now consider a spherical bubble
+with radius $R(t)$ and interior pressure $P(t)$ along its surface.
+Since we know the liquid pressure $p(r)$,
+we can find $P$ from $\sigma_{rr}(r)$.
+Furthermore, to include the effects of surface tension, we simply add
+the [Young-Laplace law](/know/concept/young-laplace-law/) to $P$:
+
+$$\begin{aligned}
+    P
+    = - \sigma_{rr}(R) + \alpha \frac{2}{R}
+    = p(R) - 2 \rho \nu \Big( \frac{-2}{R^3} C \Big) + \alpha \frac{2}{R}
+\end{aligned}$$
+
+We isolate this for $p(R)$, and equate it to
+our expression for $p(r)$
+at the surface $r\!=\!R$:
 
 $$\begin{aligned}
-    p(r, t)
-    = p_\infty(t) + \rho \bigg( \frac{1}{r} C'(t) - \frac{1}{2 r^4} C^2(t) \bigg)
+    P - \rho \nu \frac{4}{R^3} C - \alpha \frac{2}{R}
+    = p_\infty + \rho \bigg( \frac{1}{R} C' - \frac{1}{2 R^4} C^2 \bigg)
 \end{aligned}$$
 
-We now consider a spherical bubble with radius $R(t)$ and pressure $P(t)$ along the liquid surface.
-To study the liquid boundary's movement, we set $r = R$ and $p = P$,
-and see that $R'(t) = v(t)$, such that $C = r^2 V = R^2 R'$.
-We thus arrive at:
+Isolating for $P$,
+and inserting the fact that $R'(t) = v(t)$,
+such that $C = r^2 v = R^2 R'$,
+yields:
 
 $$\begin{aligned}
     P
-    &= p_\infty + \rho \bigg( \frac{1}{R} \dv{(R^2 R')}{t} - \frac{1}{2 R^4} (R^2 R')^2 \bigg)
+    &= p_\infty + \rho \bigg( \frac{1}{R} \dv{(R^2 R')}{t} - \frac{1}{2 R^4} (R^2 R')^2
+    + \nu \frac{4}{R^3} (R^2 R') \bigg) + \alpha \frac{2}{R}
     \\
-    &= p_\infty + \rho \bigg( 2 (R')^2 + R R'' - \frac{1}{2} (R')^2 \bigg)
+    &= p_\infty + \rho \bigg( 2 (R')^2 + R R'' - \frac{1}{2} (R')^2 + \nu \frac{4}{R} R' \bigg) + \alpha \frac{2}{R}
 \end{aligned}$$
 
-Rearranging this and defining $\Delta p = P - p_\infty$
-leads to the simple Rayleigh-Plesset equation:
+Rearranging this and defining $\Delta p \equiv P - p_\infty$
+leads to the Rayleigh-Plesset equation:
 
 $$\begin{aligned}
     \boxed{
-        R \dv[2]{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2
-        = \frac{\Delta p}{\rho}
+        \frac{\Delta p}{\rho}
+        = R \dv[2]{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2 + \nu \frac{4}{R} \dv{R}{t} + \frac{\alpha}{\rho} \frac{2}{R}
     }
 \end{aligned}$$