literature review bdnk theory -- pt 1

tex/grhd.tex

@@ -3280,6 +3280,81 @@ compute $(D, S_i, \tau, A_i)$ from
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+\subsection{Literature review BDNK theory}
+
+\paragraph{\cite{Van:2007pw}}
+
+\begin{itemize}
+  \item Works with Eckart frame. Uses similar notation as \cite{kovtun2012lecture}.
+        Works in SR.
+  \item One derivative (OD) theories are parabolic and are considered acausal.
+    Two derivative (TD) theories are mostly hyperbolic and are considered causal, although,
+    the relation between parabolicity and causality is not straightforward.
+  \item Homogenous equilibrium \question{what does this mean? A: Defined later in the paper.
+    TLDR: $T,\mu\,p,u_\alpha=const$ and no proper time derivatives and zero entropy production.}
+    in OD theories is considered unstable,
+    but it is considered stable in TW theories. This is based on linear stability analysis.
+  \item Argument against parabolic EoMs is that they **some solutions** show infinite propagation speed.
+    More rigorous arguments use a well posedness analysis.
+    However, (IIUC) parabolic PDEs also admit well posed formulations, provided the initial
+    data is given on the characteristic surface of the equations (?).
+  \item The speed of a hyperbolic PDE is in general not infinite, but it can be larger
+    than the speed of light. The actual speed depends on parameters in the equations,
+    cf. wave equation.
+    Hyperbolicity and covariance of PDEs alone does not warrant speeds smaller than the one of light.
+  \item Infinite speeds are not observable, because they do not lie within the regime where the
+    theory is applicable. One can use mean free path length and collision times to give an
+    estimate for the maximum observable speeds.
+  \item They write down the most general form of the energy momentum tensor. They say
+    that it is symmetric, because the internal spin is zero. \question{Why does spin have
+    to be zero here?}
+    They emphasize that due to this symmetry the heat flux and momentum density are equal,
+    however, their physical difference is a key element of their theory, because heat is
+    subjected to dissipation and momentum density not.
+  \item Second law of thermodyamics is given by the divergence of the entropy.
+    Entropy is assumed to be a function of only local rest frame quantities. That is,
+    it only depends on local rest frame energy density, the time-timelike component
+    of the energy-momentum tensor defined according to the velocity field of the material
+    \question{$T^{\mu\nu} u_\mu u_\nu?$}. They say that the thermodynamics \question{do they
+    mean thermodynamics the entropy?} cannot be related to an extrnal observer, hence, there
+    is not dependence on a relative kinetic energy.
+  \item They say that in non-relativistic fluids the internal energy is the difference of
+    conserved total energy and kinetic energy of the material, e.g. thermodynamics depend
+    on an external observer, but nevertheless the correct thermodynamic framework should
+    admit frame independent material equations. This problem is avoided in classical physics
+    \question{classical = non-relativistic?} by introducing concepts like configurational forces
+    and virtual power.
+    Without those \textit{tricks}, one mixes dissipative and non-dissipative effects
+    which in turn can cause generic instabilities.
+  \item They define a relativistic internal energy density as
+    \begin{align}
+      E^\alpha = - u_\beta T^{\alpha\beta} = e u^\alpha + q^\alpha \, ,
+    \end{align}
+    where $q^\alpha$ is transverse to $u^\alpha$. This vector combines rest frame energy density
+    and momentum. A series expansion in the non-relativistic limit for the length of the vector
+    yields
+    \begin{align}
+      \abs{E} = \sqrt{|e^2 - q^\alpha q_\alpha|} \approx e - \frac{\vec{q}^2}{2e} + \dots \, .
+    \end{align}
+  \item They then assume that the entropy density depends on this new internal energy
+    vector as well as the particle number,
+    i.g. $s = s(e, q^\alpha, n) = \hat{s}(\sqrt{|e^2 - q^\alpha q_\alpha|}, n)$.
+    All of this leads to modified thermodynamic (Gibbs) relations and from this they
+    derive the \textit{viscous pressure}.
+  \item They perform a linear stability analysis where one linearizes the PDE system around
+    a equilibrium state. One then uses a plane wave ansatz and analysis the resulting
+    eigen value problem for growing modes. The analysis appears to be straightfoward, but a
+    bit involved.
+    In the end, they showed that the homogeneous equilibrium of the relativistic heat conductin
+    viscous fluid is stable, which is in contrast to the corresponding equations of an
+    Eckart fluid. \question{What is the Eckart fluid?}
+    The interesting result is that besides they only needed thermodynamic inequalitites
+    and stability conditions of
+    the fluid and no other special stability conditions to proof this, which is in
+    contrast to analyses of Isreal-Steward theries.
+\end{itemize}
+
+
 
 \printbibliography