Hydrodynamics Entropy Function
from Trangenstein, 4.4.2 through 4.4.4
Thermodynamics
$\rho$ = density
$P$ = pressure
$e_{int}$ = internal specific energy
$T$ = temperature
$S$ = entropy
$i,j,k, ...$ spans $x,y,z$
$I,J,K, ...$ spans all elements
Second law of thermodynamics:
$de_{int} = T dS - P d (1/\rho)$
therefore
$P = \partial_{1/\rho} e$
$T = \partial_S e_{int}$
Ideal gas relation:
$P = \rho R T$
$e_{int} = c_v T$
where
$R$ = gas constant
$c_v$ = heat capacity
polytropic gas heat capacity:
$c_v = \frac{R}{\gamma-1}$
where
$\gamma$ = heat capacity ratio
polytropic ideal gas:
$e_{int} = c_v T = \frac{RT}{\gamma-1}$
therefore
$(\gamma - 1) e_{int} = R T$
therefore
$P / \rho = R T = (\gamma - 1) e_{int}$
$P = (\gamma - 1) \rho e_{int}$
specific enthalpy:
$h = e_{int} + P/\rho = e_{int} + (\gamma - 1) e_{int} = \gamma e$
where
$h$ = enthalpy
second law, in terms of $dS$:
$dS = \frac{1}{T}(de_{int} + P d(1/\rho))$
$d(1/\rho) = -1/\rho^2 d\rho$
therefore
$dS = \frac{1}{T} de_{int} - \frac{1}{T \rho^2} P d\rho $
second law for polytropic ideal gas:
using $e_{int} = \frac{P}{\rho (\gamma - 1)}$
therefore $de_{int} = \frac{dP}{\rho (\gamma - 1)} - \frac{P d\rho}{\rho^2 (\gamma-1)}$
substitute into second law:
$dS = \frac{1}{T}de_{int} - \frac{1}{T \rho^2} P d\rho $
$dS = \frac{1}{T}(\frac{dP}{\rho (\gamma - 1)} - \frac{P d\rho}{\rho^2 (\gamma-1)}) - \frac{1}{T \rho^2} P d\rho $
$dS = \frac{1}{T \rho (\gamma - 1)} dP - \frac{P}{T \rho^2 (\gamma-1)} d\rho - \frac{1}{T \rho^2} P d\rho $
$dS = \frac{1}{T \rho (\gamma - 1)} dP - \frac{\gamma}{\gamma-1} \frac{P}{T \rho^2} d\rho $
second law partials for polytropic ideal gas:
$\partial_P S = \frac{1}{T \rho (\gamma - 1)}$
$\partial_\rho S = -\frac{\gamma}{\gamma - 1} \frac{P}{T \rho^2}$
substitute $T = \frac{e_{int}}{c_v} = \frac{P}{(\gamma - 1)\rho c_v}$
our partials of $S$ become:
$\frac{\partial S}{\partial P} = \frac{c_v}{P}$
$\frac{\partial S}{\partial \rho} = -\gamma \frac{c_v}{\rho}$
solve to find the entropy function:
$dS = \frac{c_v}{P} dP - \gamma \frac{c_v}{\rho} d\rho$
$S = S_0 + c_v (ln (P) - ln (P_0)) - \gamma c_v (ln(\rho) - ln(\rho_0))$
$S = S_0 + c_v ln ( \frac{P}{P_0} (\frac{\rho_0}{\rho})^\gamma )$
The entropy function should be constant (i.e. $\frac{dS}{dt} = 0$ except in the presence of shock waves.
Notice that velocity does not appear in the entropy function.
Nor does it appear later in the entropy flux function $\rho S$.
This might be why the entropy flux function gradient with respect to primitives matches the diagonalized quasilinear solution,
which is equal to the acoustic matrix plus identity times $v_x$.
This $v_i$ variables might be distinct of the entropy flux, but they are also distinct of what its gradient computes.
Characteristics
primitives:
$W_I = \downarrow I \left[\matrix{
\rho \\
v_i \\
P
}\right]$
conservatives:
$U_I = \downarrow I \left[\matrix{
\rho \\
\rho v_i \\
E_{total}
}\right]$
for
$E_{total} = \rho e_{total}$ = total energy density
$e_{total} = e_{int} + e_{kin} = e_{int} + \frac{1}{2}v^2$ = total specific energy
Derivative of conservative wrt primitive:
$\frac{\partial U_I}{\partial W_J} = \downarrow I \overset{\rightarrow J}{ \left[\matrix{
1 & 0 & 0 \\
v_i & \rho \delta_{ij} & 0 \\
\frac{1}{2} v^2 & \rho v_j & \frac{1}{\gamma - 1}
}\right]}$
Derivative of primitives wrt conservatives:
$\frac{\partial W_J}{\partial U_I} = \downarrow J \overset{\rightarrow I}{ \left[\matrix{
1 & 0 & 0 \\
-\frac{v_j}{\rho} & \delta_{ij} \frac{1}{\rho} & 0 \\
\frac{1}{2} (\gamma-1) v^2 & -(\gamma-1) v_i & \gamma-1
}\right] }$
Flux vector:
$F_{Ij} = \downarrow I \left[\matrix{
\rho v_j \\
\rho v_i v_j + \delta_{ij} P \\
H_{total} v_j
}\right]$
for
$H_{total} = \rho h_{total}$ = total entropy density
$h_{total} = e_{total} + P$ = total specific entropy
Derivative of flux vector with respect to primitive variables:
$\frac{\partial F_{Ij}}{\partial W_K} = \downarrow I \overset{\rightarrow K}{ \left[\matrix{
v_j &
\rho \delta_{jk} &
0 \\
v_i v_j &
\rho (\delta_{ik} v_j + v_i \delta_{jk}) &
\delta_{ij} \\
\frac{1}{2} v^2 v_j &
\rho v_j v_k + H_{total} \delta_{jk}&
\frac{\gamma}{\gamma-1} v_j
}\right] }$
Derivative of flux vector with respect to conservative variables:
$\frac{\partial F_{Ij}}{\partial U_K}
= \frac{\partial F_{Ij}}{\partial W_L} \frac{\partial W_L}{\partial U_K}$
$= \downarrow I \overset{\rightarrow L}{ \left[\matrix{
v_j &
\rho \delta_{jl} &
0 \\
v_i v_j &
\rho (\delta_{il} v_j + v_i \delta_{jl}) &
\delta_{ij} \\
\frac{1}{2} v^2 v_j &
\rho v_j v_l + (E_{hydro} + P) \delta_{jl} &
\frac{\gamma}{\gamma-1} v_j
}\right] } \cdot \downarrow L \overset{\rightarrow K}{ \left[\matrix{
1 & 0 & 0 \\
-\frac{1}{\rho} v_l & \frac{1}{\rho} \delta_{kl} & 0 \\
\frac{1}{2} (\gamma-1) v^2 & -(\gamma-1) v_k & \gamma-1
}\right] }$
$= \downarrow I \overset{\rightarrow K}{ \left[\matrix{
0 &
\delta_{jk} &
0 \\
-v_i v_j + \frac{1}{2} \delta_{ij} (\gamma-1) v^2 &
\delta_{ik} v_j + \delta_{jk} v_i - \delta_{ij} (\gamma-1) v_k &
\delta_{ij} (\gamma-1) \\
v_j (\frac{1}{2} (\gamma-1) v^2 - h_{total}) &
-(\gamma-1) v_j v_k + \delta_{jk} h_{total} &
\gamma v_j
}\right] }$
Hyperbolic conservation law:
$\frac{\partial U_I}{\partial t} + \frac{\partial F_{Ij}}{\partial x_J} = 0$
with respect to conservative variables:
$\frac{\partial U_I}{\partial t} + \frac{\partial F_{Ij}}{\partial U_K} \frac{\partial U_K}{\partial x_j} = 0$
Rewritten in terms of primitives:
$\frac{\partial U_I}{\partial W_K} \frac{\partial W_K}{\partial t} + \frac{\partial F_{Ij}}{\partial U_K} \frac{\partial U_K}{\partial x_j} = 0$
$\frac{\partial W_L}{\partial U_I} \frac{\partial U_I}{\partial W_K} \frac{\partial W_K}{\partial t} + \frac{\partial W_L}{\partial U_I} \frac{\partial F_{Ij}}{\partial U_K} \frac{\partial U_K}{\partial x_j} = 0$
$\delta_{LK} \frac{\partial W_K}{\partial t} + \frac{\partial W_L}{\partial U_I} \frac{\partial F_{Ij}}{\partial U_K} \frac{\partial U_K}{\partial x_j} = 0$
$\frac{\partial W_L}{\partial t} + \frac{\partial W_L}{\partial U_I} \frac{\partial F_{Ij}}{\partial U_M} \frac{\partial U_M}{\partial W_K} \frac{\partial W_K}{\partial x_j} = 0$
reindex
$\frac{\partial W_I}{\partial t} + \frac{\partial W_I}{\partial U_L} \frac{\partial F_{Lj}}{\partial U_M} \frac{\partial U_M}{\partial W_K} \frac{\partial W_K}{\partial x_j} = 0$
Therefore the equivalent flux matrix for the primitive variables - the quasilinear flux matrix - is given by:
$\frac{\partial W_I}{\partial U_L} \frac{\partial F_{Lj}}{\partial U_M} \frac{\partial U_M}{\partial W_K}
= \frac{\partial W_I}{\partial U_L} \frac{\partial F_{Lj}}{\partial W_K}$
$= \downarrow I \overset{\rightarrow L}{ \left[\matrix{
1 & 0 & 0 \\
-\frac{v_i}{\rho} & \delta_{il} \frac{1}{\rho} & 0 \\
\frac{1}{2} (\gamma-1) v^2 & -(\gamma-1) v_l & \gamma-1
}\right] }
\cdot
\frac{\partial F_{Lj}}{\partial W_K}$
$= \downarrow I \overset{\rightarrow L}{ \left[\matrix{
1 & 0 & 0 \\
-\frac{v_i}{\rho} & \delta_{il} \frac{1}{\rho} & 0 \\
\frac{1}{2} (\gamma-1) v^2 & -(\gamma-1) v_l & \gamma-1
}\right] }
\cdot
\downarrow L \overset{\rightarrow K}{ \left[\matrix{
v_j &
\rho \delta_{jk} &
0 \\
v_l v_j &
\rho (\delta_{lk} v_j + v_l \delta_{jk}) &
\delta_{lj} \\
\frac{1}{2} v^2 v_j &
\rho v_j v_k + H_{total} \delta_{jk}&
\frac{\gamma}{\gamma-1} v_j
}\right] }$
$= \downarrow I \overset{\rightarrow K}{\left[\matrix{
v_j & \rho \delta_{jk} & 0 \\
0 & \delta_{ik} v_j & \delta_{ij} \frac{1}{\rho} \\
0 & \gamma P \delta_{jk} & v_j
}\right]}$
$= \downarrow I \overset{\rightarrow K}{\left[\matrix{
0 & \rho \delta_{jk} & 0 \\
0 & 0 & \delta_{ij} \frac{1}{\rho} \\
0 & \gamma P \delta_{jk} & 0
}\right]} + \delta_{IK} v_j$
$= A_{IjK} + \delta_{IK} v_j$
for
$A_{IjK} = \downarrow I \overset{\rightarrow K}{\left[\matrix{
0 & \rho \delta_{jk} & 0 \\
0 & 0 & \delta_{ij} \frac{1}{\rho} \\
0 & \gamma P \delta_{jk} & 0
}\right]} = $ the acoustic matrix in direction $x_j$
Let $c_{snd}^2 = \frac{\gamma P}{\rho}$ be the speed of sound.
Right eigenvectors of $A_{IjK}$:
$A_{IjK} y_K = \lambda y_K$
$\downarrow I \overset{\rightarrow K}{\left[\matrix{
0 & \rho \delta_{jk} & 0 \\
0 & 0 & \delta_{ij} \frac{1}{\rho} \\
0 & \gamma P \delta_{jk} & 0
}\right]} \cdot y_K$
has the following solutions:
$\left\{\matrix{
\rho \delta_{jk} y_k = \lambda y_\rho \\
\frac{1}{\rho} \delta_{ij} y_P = \lambda y_i \\
\gamma P \delta_{jk} y_k = \lambda y_P
}\right\}$
$\left\{\matrix{
y_j = \lambda y_\rho / \rho \\
y_i = \delta_{ij} y_P / (\rho \lambda) \\
y_j = \lambda y_P / (\gamma P)
}\right\}$
combining $y_j$ of the 1st and 3rd constraints:
$\lambda y_\rho / \rho = \lambda y_P / (\gamma P)$
divide by $\lambda$:
$(\gamma P) / \rho = y_P / y_\rho$
Let $y_\rho = 1$
Therefore $y_P = \frac{\gamma P}{\rho} = c_{snd}^2$
substitute $y_P$ into the 2nd constraint:
$y_i = \delta_{ij} \frac{\gamma P}{\rho^2 \lambda}$
equate this with the 2nd constraint:
$y_j = \frac{\gamma P}{\rho^2 \lambda} = \lambda / \rho$
solve for $\lambda$:
$\lambda^2 = \frac{\gamma P}{\rho} = c_{snd}^2$
$\lambda = \pm c_{snd}$
use $\lambda$ and $y_i = \delta_{ij} \lambda / \rho$ to solve for $y_i$:
$y_i = \pm \delta_{ij} \frac{c_{snd}}{\rho}$
Then there is $\lambda = 0$ (TODO show where this comes from)
$\left\{\matrix{
\rho \delta_{jk} y_k = 0 \\
\frac{1}{\rho} \delta_{ij} y_P = 0 \\
\gamma P \delta_{jk} y_k = 0
}\right\}$
This is true for $y_j = 0$ by the 1st and 3rd constraint, and $y_P = 0$ by the 2nd constraint
Therefore we are left with three degrees of freedom: one of $y_\rho = 1$, and two of $y_{j'} = 1$ for $j' \ne j$
Put these together and we get:
$A_{IjK} y_{KL} = \Lambda_{IK} y_{KL}$
For $\Lambda_{IK} = \delta_{IK} \lambda_K$ is the diagonal matrix of eigenvalues
and $y_{KL}$ are the collection of eigenvectors $y_K$, for the index $L$ denoting the distinct eigenvector.
Written out:
$\downarrow I \overset{\rightarrow K}{\left[\matrix{
0 & \rho \delta_{jk} & 0 \\
0 & 0 & \frac{1}{\rho} \delta_{ij} \\
0 & \gamma P \delta_{jk} & 0
}\right]}
\cdot
\downarrow K \overset{\rightarrow L}{\left[\matrix{
1 &
1 &
0 &
1 \\
-\delta_{ij} \frac{c_{snd}}{\rho} &
0 &
\delta_{j'j} &
\delta_{ij} \frac{c_{snd}}{\rho} \\
c_{snd}^2 &
0 &
0 &
c_{snd}^2
}\right]}
=
\downarrow I \overset{\rightarrow K}{
\left[\matrix{
-c_{snd} & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & c_{snd}
}\right]}
\cdot
\downarrow K \overset{\rightarrow L}{\left[\matrix{
1 &
1 &
0 &
1 \\
-\delta_{ij} \frac{c_{snd}}{\rho} &
0 &
\delta_{j'j} &
\delta_{ij} \frac{c_{snd}}{\rho} \\
c_{snd}^2 &
0 &
0 &
c_{snd}^2
}\right]}$
Now we can find the eigenvalues of the change in flux with respect to change in conservative as:
$= \frac{\partial F_{Ij}}{\partial W_L} \frac{\partial W_L}{\partial U_K}$ by chain rule
$= \delta_{IN} \frac{\partial F_{Nj}}{\partial W_L} \frac{\partial W_L}{\partial U_K}$ by inserting a delta
$= \frac{\partial U_I}{\partial W_M} \frac{\partial W_M}{\partial U_N} \frac{\partial F_{Nj}}{\partial W_L} \frac{\partial W_L}{\partial U_K}$ by replacing the delta with a change in coordinates and its inverse
Now we can replace the $\frac{\partial W_M}{\partial U_N} \frac{\partial F_{Nj}}{\partial W_L}$ with the acoustic matrix $A_{Njl} + \delta_{NL} v_j$.
$= \frac{\partial U_I}{\partial W_M} (A_{NjL} + \delta_{NL} v_j) \frac{\partial W_L}{\partial U_K}$
So the eigenvectors of $\frac{\partial F_{Ij}}{\partial U_K}$ are equal to those of $A_{IjK}$ plus $v_j$ (why? because $A_{IjK}$ is traceless?).
And (only if the eigenvectors of $A_{IjK}$ are equal to those of $\delta_{IK} v_j$. why again?) the right eigenvectors of $\frac{\partial F_{Ij}}{\partial U_K}$
are equal to $\frac{\partial U_I}{\partial W_K} y_{KL}$.
Entropy Function
Entropy recap:
$S = S_0 + c_v ln ( \frac{P}{P_0} (\frac{\rho_0}{\rho})^\gamma $
Derivative of $S$ wrt primitives:
$\partial_\rho S = -\gamma c_v / \rho$
$\partial_{v_i} S = 0$
$\partial_P S = c_v / P$
$\partial_I S = \overset{\rightarrow I}{\left[\matrix{
-\frac{\gamma c_v}{\rho} &
0 &
\frac{c_v}{P}
}\right]}$
Derivative of densitised entropy wrt primitives:
$\partial_\rho (\rho S) = S - \gamma c_v$
$\partial_{v_i} (\rho S) = 0$
$\partial_P (\rho S) = c_v \rho / P$
$\partial_I S = \overset{\rightarrow I}{\left[\matrix{
S - \gamma c_v & 0 & \frac{c_v \rho}{P}
}\right]}$
...transformed by the transpose quasilinear flux matrix...
$\partial_I (\rho S) \frac{\partial W_I}{\partial U_L} \frac{\partial F_{Lj}}{\partial W_K}$
$= \overset{\rightarrow I}{\left[\matrix{
S - \gamma c_v & 0 & \frac{c_v \rho}{P}
}\right]}
\cdot
\downarrow I \overset{\rightarrow K}{\left[\matrix{
v_j & \rho \delta_{jk} & 0 \\
0 & \delta_{ik} v_j & \delta_{ij} \frac{1}{\rho} \\
0 & \gamma P \delta_{jk} & v_j
}\right]}$
$=\overset{\rightarrow K}{\left[\matrix{
v_j (S - \gamma c_v) &
\delta_{jk} \rho S &
\frac{v_j c_v \rho}{P}
}\right]}$
$= \rho v_j \partial_I S + S \left[\matrix{ v_j & \delta_{jk} \rho & 0 }\right]$
$= \rho v_j \partial_I S + S \partial_I (\rho v_j)$
$= \partial_I (S \rho v_j)$
So the gradient of the densitized entropy function, times the quasilinear flux matrix, is equal to the gradient of the combined densitized entropy function times the velocity in the flux direction.
Is that the conditions for the definition of the entropy flux?
Second derivative of densitized entropy with respect to primitives:
$(\rho S)_{,\rho\rho} = S_{,\rho} = -\frac{\gamma c_v}{\rho}$
$(\rho S)_{,\rho P} = S_{,P} = \frac{c_v}{P}$
$(\rho S)_{,v_i K} = 0$
$(\rho S)_{,P P} = (\frac{c_v \rho}{P})_{,P} = -\frac{c_v \rho}{P^2}$
$(\rho S)_{IK} = c_v \cdot \downarrow I \overset{\rightarrow K}{\left[\matrix{
-\frac{\gamma}{\rho} & 0 & \frac{1}{P} \\
0 & 0 & 0 \\
\frac{1}{P} & 0 & -\frac{\rho}{P^2}
}\right]}$
eigenvalues of second derivative of entropy with respect to primitives:
$\lambda = 0$ times 3
and
$(-c_v \frac{\gamma}{\rho} - \lambda)(-c_v \frac{\rho}{P^2} - \lambda) - \frac{{c_v}^2}{P^2} = 0$
$\lambda^2 + c_v (\frac{\gamma}{\rho} + \frac{\rho}{P^2}) \lambda + {c_v}^2 \frac{\gamma - 1}{P^2} = 0$
$\lambda = \frac{1}{2} {c_v}^2 \left(
-(\frac{\gamma}{\rho} + \frac{\rho}{P^2}) \pm \sqrt{(\frac{\gamma}{\rho} + \frac{\rho}{P^2})^2 - 4 \frac{\gamma - 1}{P^2}}
\right)$
$\lambda = -\frac{1}{2} {c_v}^2 \left(
\frac{\gamma}{\rho} + \frac{\rho}{P^2} \mp \sqrt{\frac{\gamma^2}{\rho^2} + 2 \frac{\gamma}{P^2} + \frac{\rho^2}{P^4} - 4 \frac{\gamma - 1}{P^2}}
\right)$
$\lambda = -\frac{1}{2} {c_v}^2 \left(
\frac{\gamma}{\rho} + \frac{\rho}{P^2} \mp \sqrt{\frac{\gamma^2}{\rho^2} - 2 \frac{\gamma}{P^2} + \frac{\rho^2}{P^4} + \frac{4}{P^2}}
\right)$