From 7f65c526132ee98d59d1a2b53d08c4b49330af03 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 28 Jan 2023 11:04:09 +0100
Subject: Improve knowledge base

---
 source/know/concept/euler-equations/index.md | 141 +++++++++++++++++----------
 1 file changed, 90 insertions(+), 51 deletions(-)

(limited to 'source/know/concept/euler-equations')

diff --git a/source/know/concept/euler-equations/index.md b/source/know/concept/euler-equations/index.md
index 3730ea3..2654d2b 100644
--- a/source/know/concept/euler-equations/index.md
+++ b/source/know/concept/euler-equations/index.md
@@ -11,14 +11,13 @@ layout: "concept"
 
 The **Euler equations** are a system of partial differential equations
 that govern the movement of **ideal fluids**,
-i.e. fluids without viscosity.
-There exist several forms, depending on
-the surrounding assumptions about the fluid.
+i.e. fluids without [viscosity](/know/concept/viscosity/).
 
 
-## Incompressible fluid
 
-In a fluid moving according to the velocity vield $$\va{v}(\va{r}, t)$$,
+## Incompressible fluids
+
+In a fluid moving according to the velocity field $$\va{v}(\va{r}, t)$$,
 the acceleration felt by a particle is given by
 the **material acceleration field** $$\va{w}(\va{r}, t)$$,
 which is the [material derivative](/know/concept/material-derivative/) of $$\va{v}$$:
@@ -33,14 +32,15 @@ This infinitesimal particle obeys Newton's second law,
 which can be written as follows:
 
 $$\begin{aligned}
-    \va{w} \dd{m}
+    \va{w} m
     = \va{w} \rho \dd{V}
     = \va{f^*} \dd{V}
 \end{aligned}$$
 
-Where $$\dd{m}$$ and $$\dd{V}$$ are the particle's mass volume,
-and $$\rho$$ is the fluid density, which we assume, in this case, to be constant in space and time.
-Then the **effective force density** $$\va{f^*}$$ represents the net force-per-particle.
+Where $$m$$ and $$\dd{V}$$ are the particle's mass and volume,
+and $$\rho$$ is the fluid density, which we assume
+to be constant in space and time in this case.
+Now, the **effective force density** $$\va{f^*}$$ represents the net force-per-particle.
 By dividing the law by $$\dd{V}$$, we find:
 
 $$\begin{aligned}
@@ -51,13 +51,13 @@ $$\begin{aligned}
 Next, we want to find another expression for $$\va{f^*}$$.
 We know that the overall force $$\va{F}$$ on an arbitrary volume $$V$$ of the fluid
 is the sum of the gravity body force $$\va{F}_g$$,
-and the pressure contact force $$\va{F}_p$$ on the enclosing surface $$S$$.
+and the pressure contact force $$\va{F}_p$$ on the enclosing surface $$\partial V$$.
 Using the divergence theorem, we then find:
 
 $$\begin{aligned}
     \va{F}
     = \va{F}_g + \va{F}_p
-    = \int_V \rho \va{g} \dd{V} - \oint_S p \dd{\va{S}}
+    = \int_V \rho \va{g} \dd{V} - \oint_{\partial V} p \dd{\va{S}}
     = \int_V (\rho \va{g} - \nabla p) \dd{V}
     = \int_V \va{f^*} \dd{V}
 \end{aligned}$$
@@ -76,31 +76,28 @@ Dividing this by $$\rho$$,
 we get the first of the system of Euler equations:
 
 $$\begin{aligned}
-    \boxed{
-        \va{w}
-        = \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
-        = \va{g} - \frac{\nabla p}{\rho}
-    }
+    \va{w}
+    = \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+    = \va{g} - \frac{\nabla p}{\rho}
 \end{aligned}$$
 
-The last ingredient is **incompressibility**:
+The last ingredient is incompressibility:
 the same volume must simultaneously
-be flowing in and out of an arbitrary enclosure $$S$$.
+be flowing in and out of an arbitrary enclosure $$\partial V$$.
 Then, by the divergence theorem:
 
 $$\begin{aligned}
     0
-    = \oint_S \va{v} \cdot \dd{\va{S}}
+    = \oint_{\partial V} \va{v} \cdot \dd{\va{S}}
     = \int_V \nabla \cdot \va{v} \dd{V}
 \end{aligned}$$
 
-Since $$S$$ and $$V$$ are arbitrary,
-the integrand must vanish by itself everywhere:
+Since $$V$$ is arbitrary,
+the integrand must vanish by itself,
+leading to the **continuity relation**:
 
 $$\begin{aligned}
-    \boxed{
-        \nabla \cdot \va{v} = 0
-    }
+    \nabla \cdot \va{v} = 0
 \end{aligned}$$
 
 Combining this with the equation for $$\va{w}$$,
@@ -118,62 +115,104 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The above form is straightforward to generalize to incompressible fluids
-with non-uniform spatial densities $$\rho(\va{r}, t)$$.
-In other words, these fluids are "lumpy" (variable density),
-but the size of their lumps does not change (incompressibility).
 
-To update the equations, we demand conservation of mass:
+
+## Compressible fluids
+
+If the fluid is compressible,
+the condition $$\nabla \cdot \va{v} = 0$$ no longer holds,
+so to update the equations we demand that mass is conserved:
 the mass evolution of a volume $$V$$
-is equal to the mass flow through its boundary $$S$$.
+is equal to the mass flow through its boundary $$\partial V$$.
 Applying the divergence theorem again:
 
 $$\begin{aligned}
     0
-    = \dv{}{t}\int_V \rho \dd{V} + \oint_S \rho \va{v} \cdot \dd{\va{S}}
+    = \dv{}{t}\int_V \rho \dd{V} + \oint_{\partial V} \rho \va{v} \cdot \dd{\va{S}}
     = \int_V \dv{\rho}{t} + \nabla \cdot (\rho \va{v}) \dd{V}
 \end{aligned}$$
 
 Since $$V$$ is arbitrary, the integrand must be zero.
-This leads to the following **continuity equation**,
-to which we apply a vector identity:
+The new **continuity equation** is therefore:
 
 $$\begin{aligned}
     0
     = \dv{\rho}{t} + \nabla \cdot (\rho \va{v})
-    = \dv{\rho}{t} + (\va{v} \cdot \nabla) \rho + \rho (\nabla \cdot \va{v})
+    = \dv{\rho}{t} + \va{v} \cdot \nabla \rho + \rho \nabla \cdot \va{v}
+    = \frac{\mathrm{D} \rho}{\mathrm{D} t} + \rho \nabla \cdot \va{v}
 \end{aligned}$$
 
-Thanks to incompressibility, the last term disappears,
-leaving us with a material derivative:
+When the fluid gets compressed in a certain location, thermodynamics
+states that the pressure, temperature and/or entropy must increase there.
+For simplicity, let us assume an *isothermal* and *isentropic* fluid,
+such that only $$p$$ is affected by compression, and the
+[fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/)
+reduces to $$\dd{E} = - p \dd{V}$$.
+
+Then the pressure is given by a thermodynamic equation of state $$p(\rho, T)$$,
+which depends on the system being studied
+(e.g. the ideal gas law $$p = \rho R T$$).
+However, the quantity in control of the dynamics
+is not $$p$$, but the internal energy $$E$$.
+Dividing the fundamental thermodynamic relation by $$m \: \mathrm{D}t$$,
+where $$m$$ is the mass of $$\dd{V}$$:
 
 $$\begin{aligned}
-    \boxed{
-        0
-        = \frac{\mathrm{D} \rho}{\mathrm{D} t}
-        = \dv{\rho}{t} + (\va{v} \cdot \nabla) \rho
-    }
+    \frac{\mathrm{D} e}{\mathrm{D} t}
+    = - p \frac{\mathrm{D} v}{\mathrm{D} t}
 \end{aligned}$$
 
-Putting everything together, Euler's system of equations
-now takes the following form:
+With $$e$$ and $$v$$ the specific (i.e. per unit mass)
+internal energy and volume.
+Using that $$\rho = 1 / v$$,
+and substituting the above continuity relation:
+
+$$\begin{aligned}
+    \frac{\mathrm{D} e}{\mathrm{D} t}
+    = - p \frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{1}{\rho} \Big)
+    = \frac{p}{\rho^2} \frac{\mathrm{D} \rho}{\mathrm{D} t}
+    = - \frac{p}{\rho} \nabla \cdot \va{v}
+\end{aligned}$$
+
+It makes sense to see a factor $$-\nabla \cdot \va{v}$$ here:
+an incoming flow increases $$e$$.
+This gives us the time-evolution of $$e$$ due to compression,
+but its initial value is another equation of state $$e(\rho, T)$$.
+
+Putting it all together,
+Euler's system of equations now takes the following form:
 
 $$\begin{aligned}
     \boxed{
         \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
         = \va{g} - \frac{\nabla p}{\rho}
-        \qquad
-        \nabla \cdot \va{v}
-        = 0
-        \qquad
+        \qquad \quad
         \frac{\mathrm{D} \rho}{\mathrm{D} t}
-        = 0
+        = - \rho \nabla \cdot \va{v}
+        \qquad \quad
+        \frac{\mathrm{D} e}{\mathrm{D} t}
+        = - \frac{p}{\rho} \nabla \cdot \va{v}
     }
 \end{aligned}$$
 
-Usually, however, when discussing incompressible fluids,
-$$\rho$$ is assumed to be spatially uniform,
-in which case the latter equation is trivially satisfied.
+What happens if the fluid is actually incompressible,
+so $$\nabla \cdot \va{v} = 0$$ holds again? Clearly:
+
+$$\begin{aligned}
+    \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+    = \va{g} - \frac{\nabla p}{\rho}
+    \qquad \quad
+    \frac{\mathrm{D} \rho}{\mathrm{D} t}
+    = 0
+    \qquad \quad
+    \frac{\mathrm{D} e}{\mathrm{D} t}
+    = 0
+\end{aligned}$$
+
+So $$e$$ is constant, which is in fact equivalent to saying that $$\nabla \cdot \va{v} = 0$$.
+The equation for $$\rho$$ enforces conservation of mass
+for inhomogeneous fluids, i.e. fluids that are "lumpy",
+but where the size of the lumps is conserved by incompressibility.
 
 
 
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