update grhd notes

tex/grhd.tex

@@ -2348,7 +2348,7 @@ We find
 \end{align}
 Two commonly used frames are
 \begin{enumerate}
-  \item Eckhart frame: $\delta u_\mu$ chosen such that $j_\mu = 0$ which corresponds to no charge flow in the local rest frame,
+  \item Eckart frame: $\delta u_\mu$ chosen such that $j_\mu = 0$ which corresponds to no charge flow in the local rest frame,
   \item Landau frame: $\delta u_\mu$ chosen such that $q_\mu = 0$ which corresponds to no energy flow in the local rest frame.
 \end{enumerate}
 
@@ -2683,11 +2683,11 @@ These invariants hold in a generic unspecified frame.
 They depend on the coefficients $\eta,\zeta,\sigma,\chi_T$ which
 in turn depend on the zero-derivative hydro variables.
 \cite{kovtun2012lecture} says that $\chi_T$ is supposed to be zero so that the
-equations derived in the Eckhart frame provide the right non-relativistic limit
+equations derived in the Eckart frame provide the right non-relativistic limit
 (I suppose he means the Navier-Stokes equations).
 However, I think this term needs to be kept around when making frame transformations.
 
-From \eqref{eq:frame-invariants} we can derive the Eckhart frame, which is characterized
+From \eqref{eq:frame-invariants} we can derive the Eckart frame, which is characterized
 by $j^\mu = 0, \EE = \epsilon, \NN = n$.
 Using \eqref{eq:invariant-l-mu} and $j^\mu = 0$ we find
 \begin{align}
@@ -2713,7 +2713,7 @@ and because of $\EE = \epsilon, \NN = n$ we find
 
 
 
-\subsubsection{Eckhart frame}
+\subsubsection{Eckart frame}
 
 This is probably not the frame we want to use with the AV method, because
 there $j^\mu = 0$ which in turn implies that the continuity equation of hydro
@@ -2743,7 +2743,7 @@ for the equilibrium quantities
   \epsilon = -p + T s + \mu n,
 \end{align}
 where $s$ is the specific entropy.
-Because for the Eckhart frame discussed here we have $\EE = \epsilon, \NN = n$ as well as
+Because for the Eckart frame discussed here we have $\EE = \epsilon, \NN = n$ as well as
 \begin{align}
   s &= \pdv{\PP}{T} + \OO(\partial^2),
   &
@@ -2755,7 +2755,7 @@ Because for the Eckhart frame discussed here we have $\EE = \epsilon, \NN = n$ a
   \right) + \OO(\partial^2).
 \end{align}
 \question{Is the choice $\EE = \epsilon, \NN = n$ always implied
-when talking about the Eckhart frame?}
+when talking about the Eckart frame?}
 With this we can rewrite \eqref{eq:div-t-munu-transverse} as
 \begin{align}
   \Delta_{\mu\nu} u^\lambda \partial_\lambda u^\mu &=
@@ -3050,34 +3050,39 @@ First, we simplify \eqref{eq:transverse-traceless-symmetric-tensor}:
   - \frac{1}{2} \tensor{\Delta}{_\mu^\nu} \partial_\lambda u^\lambda \, .
 \end{align}
 
-Next we derive identities with which we can eliminate temporal derivatives of $u^\mu$
-when we are given $v^i, a^\mu$.
-From \eqref{eq:four-acceleration} we derive
-\begin{align}
-  \partial_t u^t &= \frac{1}{u^t} ( v_j a^j - u^j \partial_j u^t )
-  &
-  \partial_t u_i &= \frac{1}{u^t} ( a_i - u^j \partial_j u_i )
-  \nonumber\\
-  &= \frac{1}{W} ( v_i a^i - W v^i \partial_i W ) \, ,
-  &
-  &= \frac{1}{W} (a_i - W v^j \partial_j (W v_i) ) \, .
-\end{align}
-With this we can rewrite the divergence as
-\begin{align}
-  \partial_\lambda u^\lambda &=
-  \frac{1}{u^t} ( v_i a^i - u^i \partial_i u^t ) + \partial_i u^i \, .
-\end{align}
+% Next we derive identities with which we can eliminate temporal derivatives of $u^\mu$
+% when we are given $v^i, a^\mu$.
+% From \eqref{eq:four-acceleration} we derive
+% \begin{align}
+%   \partial_t u^t &= \frac{1}{u^t} ( v_j a^j - u^j \partial_j u^t )
+%   &
+%   \partial_t u_i &= \frac{1}{u^t} ( a_i - u^j \partial_j u_i )
+%   \nonumber\\
+%   &= \frac{1}{W} ( v_i a^i - W v^i \partial_i W ) \, ,
+%   &
+%   &= \frac{1}{W} (a_i - W v^j \partial_j (W v_i) ) \, .
+% \end{align}
+% With this we can rewrite the divergence as
+% \begin{align}
+%   \partial_\lambda u^\lambda &=
+%   \frac{1}{u^t} ( v_i a^i - u^i \partial_i u^t ) + \partial_i u^i \, .
+% \end{align}
 
 
 The Euler equation in SR and with diffusive terms included now reads
 \begin{align}
   0 &= \partial_t(S_i - s_i) + \partial_j(f_{S,i}^j - f_{s,i}^j)\, ,
   \\
-  S_i &= (\epsilon + p) v_i W^2 \, ,
+  S_i &= W^2 (\epsilon + p) v_i \, ,
   \\
   s_i &= \zeta v_i W^2 \partial_\lambda u^\lambda
   + \eta (
     \partial_i W + \eta^{t\mu} \partial_\mu u_i + W a_i + u_i v_j a^j - \frac{1}{2} v_i W^2 \partial_\lambda u^\lambda
+  )\,
+  \nonumber\\
+  &= \zeta v_i W^2 \partial_\lambda u^\lambda
+  + \eta (
+    \partial_i W - \partial_t (W v_i) + W a_i + W v_i v_j a^j - \frac{1}{2} v_i W^2 \partial_\lambda u^\lambda
   )\, ,
 \end{align}
 
@@ -3247,6 +3252,13 @@ However, its not yet clear if this part is really susceptible to errors, or whet
 such a variable transformation should not already be incorporated in
 the zero-derivative hydro equations and, therefore, in the \constoprim{} routine?}
 
+The above procedure was first proposed in \cite{pandya2022conservative} where they
+observed that the primitive recovery in the BDNK formulation can be done analytically,
+however, it suffers from the problem that the recovery  becomes indeterminant in the
+case of vanishing physical viscosities. To this end, they recovered the time derivatives
+of the primitives of the zero-derivative hydro equations ($\pdvt \rho_0, \pdvt \epsilon, \pdvt v_i$)
+in the above way, because it ensures that in the limit of vanishing viscosity the equations
+are consistent.
 
 
 \printbibliography