From: Timo Heister <timo.heister@gmail.com>
Date: Tue, 14 May 2019 21:51:40 +0000 (-0600)
Subject: step-32 text in formula spacing
X-Git-Tag: v9.2.0-rc1~1496^2
X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=27281454a6e981f9fea51594bab0dc32dba9899a;p=dealii.git

step-32 text in formula spacing
---

diff --git a/examples/step-32/doc/intro.dox b/examples/step-32/doc/intro.dox
index d2f56b7882..023636ef5a 100644
--- a/examples/step-32/doc/intro.dox
+++ b/examples/step-32/doc/intro.dox
@@ -229,7 +229,7 @@ quantities involved here have physical units so that the first part of
 the residual has units $\frac{\text{Pa}}{\text{m}}
 \text{m}^{\text{dim}}$ (most easily established by considering the
 term $(\nabla \cdot \mathbf v, p)_{\Omega}$ and considering that the
-pressure has units $\text{Pa}=\frac{\text{kg}}{\text{m\; s}^2}$ and
+pressure has units $\text{Pa}=\frac{\text{kg}}{\text{m}\;\text{s}^2}$ and
 the integration yields a factor of $\text{m}^{\text{dim}}$), whereas
 the second part of the residual has units
 $\frac{\text{m}^{\text{dim}}}{\text{s}}$. Taking the norm
@@ -266,7 +266,7 @@ start at the root and first make sure that everything is mathematically
 consistent first. In our case, this means that if we want to solve the system
 of Stokes equations jointly, we have to scale them so that they all have the
 same physical dimensions. In our case, this means multiplying the second
-equation by something that has units $\frac{\text{Pa\; s}}{\text{m}}$; one
+equation by something that has units $\frac{\text{Pa}\;\text{s}}{\text{m}}$; one
 choice is to multiply with $\frac{\eta}{L}$ where $L$ is a typical lengthscale
 in our domain (which experiments show is best chosen to be the diameter of
 plumes &mdash; around 10 km &mdash; rather than the diameter of the
@@ -1144,9 +1144,9 @@ the following quantities:
   viscosity of the material that flows into the area vacated under the
   rebounding continental plates.
 
-  Using this technique, values around $\eta=10^{21} \text{Pa \; s}
-  = 10^{21} \frac{\text{N\; s}}{\text{m}^2}
-  = 10^{21} \frac{\text{kg}}{\text{m\; s}}$ have been found as the most
+  Using this technique, values around $\eta=10^{21} \text{Pa}\;\text{s}
+  = 10^{21} \frac{\text{N}\;\text{s}}{\text{m}^2}
+  = 10^{21} \frac{\text{kg}}{\text{m}\;\text{s}}$ have been found as the most
   likely, though the error bar on this is at least one order of magnitude.
 
   While we will use this value, we again have to caution that there are many