From 5ed7553b723a9724f55e75261efe2666e75df725 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Tue, 8 Nov 2022 18:14:21 +0100
Subject: The tweaks and fixes never stop

---
 source/know/concept/stokes-law/index.md | 33 +++++++++++++++++----------------
 1 file changed, 17 insertions(+), 16 deletions(-)

(limited to 'source/know/concept/stokes-law/index.md')

diff --git a/source/know/concept/stokes-law/index.md b/source/know/concept/stokes-law/index.md
index 3a02a83..40212ef 100644
--- a/source/know/concept/stokes-law/index.md
+++ b/source/know/concept/stokes-law/index.md
@@ -50,10 +50,10 @@ $$\begin{gathered}
     \\
     v_r
     = U f(r) \cos\theta
-    \qquad
+    \qquad \quad
     v_\theta
     = - U g(r) \sin\theta
-    \qquad
+    \qquad \quad
     v_\phi
     = 0
 \end{gathered}$$
@@ -76,7 +76,7 @@ $$\begin{aligned}
     \\
     &= U \dv{f}{r} \cos\theta - \frac{U g}{r} \cos\theta + \frac{2 U f}{r} \cos\theta - \frac{U g}{r} \cos\theta
     \\
-    &= U \cos\theta \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big)
+    &= U \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big) \cos\theta
 \end{aligned}$$
 
 The parenthesized expression must be zero for all $$r$$,
@@ -103,16 +103,16 @@ which is as follows for our ansatz $$p(r, \theta)$$:
 $$\begin{aligned}
     0
     = \nabla^2 p
-    &= \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{p}{r} \Big) + \frac{1}{r^2 \sin\theta} \pdv{}{\theta}\Big( \sin\theta \pdv{p}{\theta} \Big)
+    &= \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{p}{r} \Big) + \frac{1}{r^2 \sin\theta} \pdv{}{\theta}\Big( \pdv{p}{\theta} \sin\theta \Big)
     \\
     0
     &= \frac{\eta U \cos\theta}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big)
     - \frac{\eta U q}{r^2 \sin\theta} \pdv{}{\theta}\Big( \sin^2\theta \Big)
     \\
-    &= \frac{\eta U \cos\theta}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big)
+    &= \frac{\eta U}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big) \cos\theta
     - \frac{2 \eta U q}{r^2 \sin\theta} \sin\theta \cos\theta
     \\
-    &= \eta U \cos\theta \Big( \dvn{2}{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big)
+    &= \eta U \Big( \dvn{2}{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big) \cos\theta
 \end{aligned}$$
 
 Again, the parenthesized expression must be zero for all $$r$$,
@@ -129,7 +129,7 @@ The pressure is therefore:
 
 $$\begin{aligned}
     p
-    = \eta U \cos\theta \Big( \frac{C_3}{r^2} + C_4 r \Big)
+    = \eta U \Big( \frac{C_3}{r^2} + C_4 r \Big) \cos\theta
 \end{aligned}$$
 
 Consequently, its gradient $$\nabla p$$ in spherical coordinates is as follows:
@@ -137,7 +137,7 @@ Consequently, its gradient $$\nabla p$$ in spherical coordinates is as follows:
 $$\begin{aligned}
     \nabla p
     = \vu{e}_r \pdv{p}{r} + \vu{e}_\theta \frac{1}{r} \pdv{p}{\theta}
-    = \vu{e}_r \Big( \eta U \cos\theta \dv{q}{r} \Big) - \vu{e}_\theta \Big( \eta U \sin\theta \frac{q}{r} \Big)
+    = \vu{e}_r \Big( \eta U \dv{q}{r} \cos\theta \Big) - \vu{e}_\theta \Big( \eta U \frac{q}{r} \sin\theta \Big)
 \end{aligned}$$
 
 According to the Stokes equation, this equals $$\eta \nabla^2 \va{v}$$.
@@ -148,24 +148,24 @@ $$\begin{aligned}
     &= \pdvn{2}{v_r}{r} + \frac{1}{r^2} \pdvn{2}{v_r}{\theta} + \frac{2}{r} \pdv{v_r}{r}
     + \frac{\cot\theta}{r^2} \pdv{v_r}{\theta} - \frac{2}{r^2} \pdv{v_\theta}{\theta} - \frac{2}{r^2} v_r - \frac{2 \cot\theta}{r^2} v_\theta
     \\
-    &= U \cos\theta \Big( \dvn{2}{f}{r} - \frac{1}{r^2} f + \frac{2}{r} \dv{f}{r} - \frac{1}{r^2} f
-    + \frac{2}{r^2} g - \frac{2}{r^2} f + \frac{2}{r^2} g \Big)
+    &= U \Big( \dvn{2}{f}{r} - \frac{1}{r^2} f + \frac{2}{r} \dv{f}{r} - \frac{1}{r^2} f
+    + \frac{2}{r^2} g - \frac{2}{r^2} f + \frac{2}{r^2} g \Big) \cos\theta
     \\
-    &= U \cos\theta \Big( \dvn{2}{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big)
+    &= U \Big( \dvn{2}{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big) \cos\theta
 \end{aligned}$$
 
 Substituting $$g$$ for the expression we found from incompressibility lets us simplify this:
 
 $$\begin{aligned}
     \eta (\nabla^2 \va{v})_r
-    &= \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big)
+    &= \eta U \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big) \cos\theta
 \end{aligned}$$
 
 The Stokes equation says that this must be equal to the $$r$$-component of $$\nabla p$$:
 
 $$\begin{aligned}
-    \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big)
-    = \eta U \cos\theta \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big)
+    \eta U \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big) \cos\theta
+    = \eta U \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big) \cos\theta
 \end{aligned}$$
 
 Where we have inserted $$\idv{q}{r}$$.
@@ -213,14 +213,15 @@ $$\begin{gathered}
     \\
     \boxed{
         v_r
-        = U \cos\theta \Big( 1 + \frac{a^3}{2 r^3} - \frac{3 a}{2 r} \Big)
+        = U \Big( 1 + \frac{a^3}{2 r^3} - \frac{3 a}{2 r} \Big) \cos\theta
         \qquad
         v_\theta
-        = - U \sin\theta \Big( 1 - \frac{a^3}{4 r^3} - \frac{3 a}{4 r} \Big)
+        = - U \Big( 1 - \frac{a^3}{4 r^3} - \frac{3 a}{4 r} \Big) \sin\theta
     }
 \end{gathered}$$
 
 
+
 ## Drag force
 
 From the definition of [viscosity](/know/concept/viscosity/),
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