Derivation Notes on Energy Conservation Problem in SPheRIO

Derivation Notes on ﻿Energy Conservation Problem in SPheRIO

When one talks about energy conservation in SPheRIO, it is inevitably connected to the following list consisting of four points, any one of them goes wrong, energy can not be considered as conserved.


 * NEXUS/EPOS IC energy check
 * Energy conservation during hydro evolution in hyperbolic $$\{x,y,\eta,\tau\}$$  coordinates
 * Energy conservation during hydro evolution in Cartesian $$\{x,y,z,t\}$$  coordinates
 * Energy conservation during freeze-out procedure

Energy conservation during hydro evolution in hyperbolic coordinates
Energy conservation lies in the expression $${T^{\mu\nu}}_{;\mu}=0$$ (for any coordinate system). Since the expression is covariant, theoretically, if it is valid in one specific coordinate system, it should be also hold true in any other coordinate system. On the other side, energy is not a scalar, though conserved, it should have different value in different coordinates. Furthermore, numerically, one always has to check whether energy conserves in different coordinate systems. In this context, only the second point in list has been indeed checked. To be specific, under hyperbolic coordinate, for ideal hydro, using entropy flow conservation as granted, one may explicitly check if the energy conservation is numerically valid. In order to do this, one firstly writes the time component of the energy momentum conservation$$^*$$,


 * $$\begin{align}

{T^{\mu0}}_{;\mu}=0 \end{align}$$

and, instead of making use of equation of state, one rewrites directly the expression in terms of SPH degree of freedom as follows (from Gabriel's notes)


 * $$\begin{align}

&\frac{1}{\tau}\partial_{\mu}(\tau T^{\mu0})+\Gamma^0_{\lambda\mu}T^{\lambda\mu}=0 \\ &\frac{1}{\tau}\partial_{\mu}(\tau T^{\mu0})+\tau T^{33}=0\\ &\frac{1}{\tau}\partial_{\tau}(\tau T^{00})+\frac{1}{\tau}\partial_i(\tau T^{i0})+\tau T^{33}=0 \end{align}$$ where we made use of the fact that the only three non-zero connections are


 * $$\begin{align}

\Gamma^0_{33}=\tau, \Gamma^3_{30}=\Gamma^3_{03}=\frac{1}{\tau} \end{align}$$

Here comes the not-so-trivial part, one wants to integrate. The integral will be carried out for a 4-dimensional volume between two (infinite) surfaces on which one thinks of energy and wants to compare their values. During hydro evolution, both the two surfaces are at constant "time", therefore space like. The freeze-out surface is involved when one wants to compare the energy of system during evolution with final state hadrons/SPH particles. Since freeze-out surface is not necessarily space time, the following expression inspired by the usual 3D Gauss Law does not trivially vanish.


 * $$\begin{align}

\int d^4x\frac{1}{\sqrt{-g}}\partial_i(\sqrt{-g} T^{i0})=\int d\tau d\eta dxdy \sqrt{-g}\frac{1}{\sqrt{-g}}\partial_i(\sqrt{-g} T^{i0})=\int d\tau d\eta dxdy \partial_i(\tau T^{i0})\ne 0 \end{align}$$

where $$d^4x=d\tau d\eta dxdy \sqrt{-g}$$  and  $$\sqrt{-g}=\tau$$. The correct thing to do is to integral on the expression resulting from energy momentum conservation, i.e.


 * $$\begin{align}

\int d^4x \frac{1}{\tau} \left\lbrace\partial_{\tau}(\tau T^{00})+\partial_i(\tau T^{i0})+\tau^2 T^{33}\right\rbrace =\int d^4x \frac{1}{\tau} \left\lbrace\partial_{\tau}(\tau T^{00})+\tau^2 T^{33}\right\rbrace=\int d\tau d\eta dxdy\left\lbrace\partial_{\tau}(\tau T^{00})+\tau^2 T^{33}\right\rbrace \end{align}$$

In last step, the spacial part of the integral is replaced by the summation over SPH particles at equal time. The time integral in the last term cannot be further simplified, so one has to keep it


 * $$\begin{align}

\int{d\tau d\eta dx dy} \left\lbrace\partial_{\tau}(\tau T^{00})+\tau^2 T^{33}\right\rbrace =\left[\tau \int{ d\eta dx dy} T^{00}\right]_{\tau_i}^{\tau_f}+\int{d\tau}\tau^2 \left(\int d\eta dx dy T^{33} \right)=0 \end{align}$$

Therefore one arrives in an expression (B for hyperbolic)


 * $$\begin{align}

E_{B}(\text{Evolution})=\int d\eta dxdy\left\lbrace\tau T^{00}+\int_{\tau_0}^{\tau} d\tau' \tau'^2T^{33}\right\rbrace \end{align}$$ (in terms of energy momentum tensor)

One may rewrite energy momentum tensor in terms of SPH particles, one has


 * $$\begin{align}

&(T^{00})_i=\sum_j\left(\frac{T^{00}}{s^*}\right)_j\nu_jW_{ij}=\sum_j\left(\frac{T^{00}}{s\gamma\tau}\right)_j\nu_jW_{ij} \\ &(T^{33})_i=\sum_j\left(\frac{T^{33}}{s^*}\right)_j\nu_jW_{ij}=\sum_j\left(\frac{T^{33}}{s\gamma\tau}\right)_j\nu_jW_{ij} \end{align}$$

together with the normalization condition of the kernel


 * $$\begin{align}

\int{d\eta dx dy} W_{ij} = 1 \end{align}$$ one finally gets


 * $$\begin{align}

E_{B}(\text{Evolution})=\sum_j\left(\frac{\nu}{s\gamma}\right)_jT^{00}_j +\int^{\tau} \tau d\tau \sum_j\left(\frac{ \nu}{s\gamma}\right)_jT^{33}_j \end{align}$$ (in terms of SPH degree of freedom)

defined as the "energy" in hyperbolic coordinates

However, if one looks closely the process of derivation, it does not even involves EOM, the energy conservation during the hydro evolution should be quite an easy task to be accomplished as long as the EOS is ok, which was pointed out by Kodama ages ago. To clarify the word "easy", let us imagine that someone modifies the EOM (spacial component of energy momentum conservation) by adding or removing a term by mistake (which might lead to violation of causality or even worse), however, the "energy conservation" check shall not show any problem, since the two equations are independent. Besides, numerically, the integral in \tau  in the second term of  $$E_B$$  actually might introduce some visible deviation from energy conservation. As a matter of fact, if one keeps the same size of $$h_{xy}   h_{\eta}$$  and in the meantime decreases  d\tau, purely due to better precision of the integral in the second term, energy conservation gets "better".

In this context, it gives a second reason to verify the conservation of energy in Cartesian coordinates, namely, $$E_C$$. Since the time component of energy momentum tensor conservation in Cartesian coordinates can not be derived solely from that in hyperbolic coordinates. In other words, the energy conservation in Cartesian coordinates is not independent of the EOM in hyperbolic coordinates. And the first reason for doing that is the final observed energy is usually measured in terms of emitted hadrons in the laboratory system, which is Cartesian.

$$^*$$ Theoretically, it is equivalent to write down the parallel part of $${T^{\mu\nu}}_{;\mu}=0$$, namely, $$u_{\nu}{T^{\mu\nu}}_{;\mu}=0$$ which shall not make any difference.

Similar argument but for the $$\eta$$  component of energy momentum tensor gives


 * $$\begin{align}

P^3_B(\text{Evolution})\rightarrow \int{d\tau d\eta dx dy} \left\lbrace\partial_{\tau}(\tau T^{03})+2 T^{03}\right\rbrace =\left[\tau \int{ d\eta dx dy} T^{03}\right]_{\tau_i}^{\tau_f}+2\int{d\tau} \left(\int d\eta dx dy T^{03} \right)=0 \end{align}$$

for the $$x-y$$  component, the result is quite trivial


 * $$\begin{align}

P^{1,2}_B(\text{Evolution})\rightarrow \int{d\tau d\eta dx dy} \left\lbrace\partial_{\tau}(\tau T^{0(1,2)})\right\rbrace =\left[\tau \int{ d\eta dx dy} T^{0(1,2)}\right]_{\tau_i}^{\tau_f}=0 \end{align}$$

The total energy of IC in Cartesian coordinates
The total energy of IC can be verified using the expression of $$E_B$$, there is no ambiguity since the IC are defined on  $$\tau=\tau_0=const$$


 * $$\begin{align}

E_{B}(\text{IC})=\int d\eta dxdy\left\lbrace\tau T^{00}\right\rbrace \end{align}$$ (in terms of energy momentum tensor)


 * $$\begin{align}

E_{B}(\text{IC})=\tau \sum_j\left(\frac{T^{00}}{s\gamma\tau}\right)_j\nu_j \end{align}$$ (in terms of SPH degree of freedom)

What is the expression of energy of IC in Cartesian coordinates? We may similarly write down the equation corresponding to energy conservation


 * $$\begin{align}

&\partial_{\mu} T^{\mu0}=0 \\ &\int d^4x \partial_{\mu} T^{\mu0}=0 \end{align}$$

By using Gauss Law in 4d space, one can instead doing a surface integral


 * $$\begin{align}

\int_{\sigma_0+\sigma_f} d^3\sigma_{\mu} T^{\mu0}=0 \end{align}$$

where $$\sigma_0$$  is the initial surface where one defines IC, in our model it is  $$\tau=\tau_0$$, and  $$\sigma_f $$ is the freeze-out surface Therefore by introducing properly the normal direction of the surface, one defines energy in Cartesian coordinates (C for Cartesian)


 * $$\begin{align}

E_C(\text{IC})= \int_{\sigma_0}d\sigma_{\mu} T^{\mu0}=\int_{\sigma_f}d\sigma_{\mu} T^{\mu0} \end{align}$$

To calculate the energy, two thing has to be taken care of, firstly, to correctly calculate surface and its normal vector, secondly the energy momentum tensor should be transformed into Cartesian coordinates. Following expressions are just for code implementation, one might want to skip them when reading

Using the convention $$v^\mu \to [v]$$. Therefore, $$v_\mu \to [g][v]$$, with  $$A^{\mu\nu} \to [A]$$  and  $$A_{\mu\nu} \to [g] [A] [g]$$


 * $$\begin{align}

{\Lambda^\mu}_{\nu'}=  \frac{\partial x_C^{\mu}}{\partial x_B^{\nu'}}=\left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \tau\sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right)  =[\Lambda] \end{align}$$


 * $$\begin{align}

g_{\mu\nu} =\eta_{\mu'\nu'}{\Lambda^{\mu'}}_{\mu}{\Lambda^{\nu'}}_{\nu} =\eta_{\mu'\nu'}\frac{\partial x_C^{\mu'}}{\partial x_B^{\mu}}\frac{\partial x_C^{\nu'}}{\partial x_B^{\nu}} =\left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \tau\sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right) \left( \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right) \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \tau\sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right) =\left( \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -\tau^2 \end{array}\right) \end{align}$$


 * $$\begin{align}

&[g] = [\Lambda][\eta][\Lambda] \end{align}$$


 * $$\begin{align}

T_C^{\mu\nu} ={\Lambda^\mu}_{\mu'}{\Lambda^\nu}_{\nu'}T_B^{\mu'\nu'} =\frac{\partial x_C^{\mu}}{\partial x_B^{\mu'}}\frac{\partial x_C^{\nu}}{\partial x_B^{\nu'}}T_B^{\mu'\nu'} =\left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \tau\sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right) \left( \begin{array}{cccc} T_B^{00} & T_B^{01} & T_B^{02} & T_B^{03}\\ T_B^{10} & \ddots &  & \\ T_B^{20} &  &  & \\ T_B^{30} &  &  & \end{array}\right)  \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \tau\sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right) \end{align}$$


 * $$\begin{align}

&[T_C] = [\Lambda][T_B][\Lambda] \end{align}$$


 * $$\begin{align}

T_C^{\mu0}=\frac{\partial x_C^{\mu}}{\partial x_B^{\mu'}}\frac{\partial x_C^{0}}{\partial x_B^{\nu'}}T_B^{\mu'\nu'} =\left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \tau\sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right)  \left( \begin{array}{cccc} T_B^{00} & T_B^{01} & T_B^{02} & T_B^{03}\\ T_B^{10} & \ddots &  & \\ T_B^{20} &  &  & \\ T_B^{30} &  &  & \end{array}\right)  \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \tau\sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right)  \left( \begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right) \end{align}$$

Then we calculate what is the orientation of a unitary vector in hyperbolic coordinates when we transform into in Cartesian coordinates


 * $$\begin{align}

e_{B}^{\mu}= \left( \begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right) \end{align}$$


 * $$\begin{align}

&e_{C\mu}=\eta_{\lambda\mu}e_{C}^{\lambda} =\eta_{\lambda\mu}\frac{\partial x_C^{\lambda}}{\partial x_B^{\nu}}e_{B}^{\nu} \\ &=\left( \begin{array}{cccc} 1, & 0, & 0, & 0 \end{array}\right) \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \tau\sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right)  \left( \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right)  \\ &=\left( \begin{array}{cccc} 1, & 0, & 0, & 0 \end{array}\right) \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & -\sinh\eta\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ \tau \sinh\eta & 0 & 0 & -\tau\cosh\eta \end{array}\right)  =\left( \begin{array}{cccc} \cosh \eta, & 0, & 0, & -\sinh\eta \end{array}\right) \end{align}$$

It is straightforward to verify at this point $$|e_{C\mu}|=|e_{B\mu}|=1$$, they are indeed unitary vectors and the module does not change. On the other hand, what happens to an infinitesimal piece of surface? Does the size of surface change, surface elements can not be understood as covariant 4-vectors. Intuitively, since we are describing the same physical quantity in different coordinate systems, though the Jacobian could be different for different surface elements, the size of the surface element should not change when one adopt a different coordinate system. Mathematically, it is more complicated. One can define the surface element in a way that it is a 4-vector, but then one should prove Gauss theorem explicitly, or, one can define the surface element in a way that Gauss theorem is satisfied, but again, in this case one has to show explicitly that the surface element is indeed a vector. A more detailed discussion can be found in "Lecture Notes of An Advanced Course in GR by E. Poisson". Here we are only going to illustrate this by a simple exampe. We will show explicitly below that the area of a surface element for constant \tau  does not depends on coordinates. a 3D volume in 4D space for constant \tau, is determined by three vectors


 * $$\begin{align}

d\vec{r}_1\equiv \frac{\partial x_C^{\mu}}{\partial \eta}d\eta =\left( \begin{array}{c} \tau\sinh\eta d\eta \\ 0 \\ 0 \\ \tau\cosh\eta d\eta \end{array}\right) \end{align}$$


 * $$\begin{align}

d\vec{r}_2\equiv \frac{\partial x_C^{\mu}}{\partial x}dx=\left( \begin{array}{c} 0 \\ dx \\ 0 \\ 0 \end{array}\right) \end{align}$$ and


 * $$\begin{align}

d\vec r_3\equiv \frac{\partial x_C^{\mu}}{\partial y}dy=\left( \begin{array}{c} 0 \\ 0 \\ dy \\ 0 \end{array}\right) \end{align}$$ for completeness, we also write down here


 * $$\begin{align}

d\vec{r}_0\equiv \frac{\partial x_C^{\mu}}{\partial \tau}d\tau =\left( \begin{array}{c} \cosh\eta d\tau \\ 0 \\ 0 \\ \sinh\eta d\tau \end{array}\right) \end{align}$$


 * $$\begin{align}

\end{align}$$
 * d\sigma_{C\mu}|_{(\tau=const.)}\equiv |d\sigma_{C\mu}|=|d\sigma^{\mu}_{C}|=|d\vec{r}_1||d\vec{r}_2||d\vec{r}_3|=\tau  dxdyd\eta

which had used the fact that the three vectors are mutually orthogonal, so this implies


 * $$\begin{align}

\end{align}$$
 * d\sigma_{B\mu}|= \tau dxdyd\eta \rightarrow |d\sigma_{C\mu}|=\tau dxdyd\eta

where $$d\vec r$$  are the 4-vectors in Cartesian coordinates. One sees that the size of the surface element defined by $$dx, dy, d\eta$$  does not change. Putting all pieces together and bearing in mind the convention that $$ {\Lambda^\mu}_{\nu}$$ is the  $$\mu$$-th row and  $$\nu$$-th column of a matrix, or  $$\mu$$-th column and  $$\nu$$-th row of its transpose.


 * $$\begin{align}

d\sigma_{C\mu}T_C^{\mu0} = ([\eta] [d\sigma_{C}])^T [T_C] [ \hat{t} ] = \left([\eta] [\Lambda] [d\sigma_{B}] \right)^T  [\Lambda] [T_B] [\Lambda] [ \hat{t} ] = [d\sigma_{B}]^T [\Lambda] [\eta] [\Lambda] [T_B] [\Lambda] [ \hat{t} ] = [d\sigma_{B}]^T [g] [T_B] [\Lambda] [ \hat{t} ]   = [d\sigma_{B}]^T [T_B] [\Lambda_0] \end{align}$$


 * $$\begin{align}

&d\sigma_{C\mu (\tau=const.)}T_C^{\mu0} \equiv d\sigma_{C\mu}T_C^{\mu 0}  =\eta_{\mu\nu}d\sigma_C^{\mu}T^{\nu0}_C =\eta_{\mu\nu}{\Lambda^\mu}_{\nu'}d\sigma_B^{\nu'}{\Lambda^\nu}_{\alpha}{\Lambda^0}_{\beta}T_C^{\alpha\beta} =d\sigma_B^{\nu'}{\Lambda^\mu}_{\nu'}\eta_{\mu\nu}{\Lambda^\nu}_{\alpha}T_C^{\alpha\beta}{\Lambda^0}_{\beta}  =d\sigma_B^{\nu'}{(\Lambda^T)^{\nu'}}_{\mu}\eta_{\mu\nu}{\Lambda^\nu}_{\alpha}T_C^{\alpha\beta}{(\Lambda^T)^\beta}_{0} \\ &=\tau dxdy \tau d\eta \left( \begin{array}{cccc} 1, & 0, & 0, & 0 \end{array}\right) \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \tau\sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right)  \left( \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right)  \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \tau\sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right)  \left( \begin{array}{cccc} T_B^{00} & T_B^{01} & T_B^{02} & T_B^{03}\\ T_B^{10} & \ddots &  & \\ T_B^{20} &  &  & \\ T_B^{30} &  &  & \end{array}\right)  \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \tau\sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right)  \left( \begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right) \\ &=\tau dxdy d\eta \left( \begin{array}{cccc} 1, & 0, & 0, & 0 \end{array}\right) \left( \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -\tau^2 \end{array}\right)  \left( \begin{array}{cccc} T_B^{00} & T_B^{01} & T_B^{02} & T_B^{03}\\ T_B^{10} & \ddots &  & \\ T_B^{20} &  &  & \\ T_B^{30} &  &  & \end{array}\right)  \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \tau\sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right)  \left( \begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right)  \\ &=\tau dxdy d\eta \left( \begin{array}{cccc} 1, & 0, & 0, & 0 \end{array}\right) \left( \begin{array}{cccc} T_B^{00} & T_B^{01} & T_B^{02} & T_B^{03}\\ T_B^{10} & \ddots &  & \\ T_B^{20} &  &  & \\ T_B^{30} &  &  & \end{array}\right)  \left( \begin{array}{c} \cosh\eta \\ 0 \\ 0 \\ \tau\sinh\eta \end{array}\right)  \\ &=\tau dxdy d\eta \left( \begin{array}{cccc} 1, & 0, & 0, & 0 \end{array}\right) \left( \begin{array}{c} \cosh\eta T_B^{00}+\tau\sinh\eta T_B^{03} \\ \cosh\eta T_B^{10}+\tau\sinh\eta T_B^{13} \\ \cosh\eta T_B^{20}+\tau\sinh\eta T_B^{23} \\ \cosh\eta T_B^{30}+\tau\sinh\eta T_B^{33} \end{array}\right)  \\ &=\tau dxdy d\eta (\cosh\eta T_B^{00}+\tau\sinh\eta T_B^{03}) \end{align}$$


 * $$\begin{align}

E_C(\text{IC})=\int_{\sigma_i(\tau=\tau_0)} \tau dxdy d\eta (\cosh\eta T_B^{00}+\tau\sinh\eta T_B^{03}) \end{align}$$ (In terms of energy momentum tensor)


 * $$\begin{align}

E_C(\text{IC})=\sum_j \left(\frac{\nu}{\gamma s} \right)_j (\cosh\eta T_B^{00}+\tau\sinh\eta T_B^{03})_j \end{align}$$ (In terms of SPH particles)

This expression also serves to calculate the energy in Cartesian coordinates during the hydro evolution,


 * $$\begin{align}

E_C(\text{Evolution})=\int_{\sigma_i(\tau=const)} \tau dxdy d\eta (\cosh\eta T_B^{00}+\tau\sinh\eta T_B^{03}) \end{align}$$ (In term of energy momentum tensor)

which can be expressed in terms SPH degree of freedom


 * $$\begin{align}

&d\sigma_{C\mu}T_C^{\mu0} = \tau dxdy d\eta (\cosh\eta T_B^{00}+\tau\sinh\eta T_B^{03}) \\ &E_C(\text{Evolution}) =\int \tau dxdy d\eta (\cosh\eta T_B^{00}+\tau\sinh\eta T_B^{03}) =\sum_j\frac{\nu_j}{\gamma_j s_j}(\cosh\eta T_B^{00}+\tau\sinh\eta T_B^{03})_j \end{align}$$ (In term of SPH particles)

where we have used following SPH relations in Boost Invariant frame:


 * $$\begin{align}

\int d\eta dx dy\, W = 1 f(x) = \sum_j \frac{ \nu_j}{\tau \gamma_j s_j} f_j W_j(x) \end{align}$$

Integrating both sides on \tau = const  surface, one obtains


 * $$\begin{align}

\int \tau d\eta dxdy f = \sum_{j} \frac{\nu_j}{ \gamma_j s_j } f_j \end{align}$$ therefore,


 * $$\begin{align}

\int \tau d\eta dx dy \to \sum_j \frac{\nu_j}{\gamma_j s_j} \end{align}$$ (Cartesian coordinates)

Similar argument but for the \eta  component of energy momentum tensor gives


 * $$\begin{align}

P^{3}_C(\text{Evolution})=\int_{\sigma_i(\tau=const)} \tau dxdy d\eta (\sinh\eta T_B^{00}+\tau\cosh\eta T_B^{03}) \end{align}$$

for the $$x-y$$  component, the result is quite trivial, it possesses the same form as in hyperbolic coordinates


 * $$\begin{align}

P^{1,2}_C(\text{Evolution})=\int_{\sigma_i(\tau=const)} \tau dxdy d\eta T_B^{0(1,2)} \end{align}$$

In the preceding derivations, we did not take into account surface element of constant $$\eta$$ ,


 * $$\begin{align}

\end{align}$$
 * d\sigma_{C\mu}|_{(\eta=const.)} = d\tau dxdy

with its normalized vector for constant $$ \eta$$ reads,


 * $$\begin{align}

n^{\mu}= \frac{1}{\tau}\left( \begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array}\right) \end{align}$$

the conservation of the $$t$$  component of energy momentum tensor reads


 * $$\begin{align}

&d\sigma_{C\mu (\eta=const.)}T_C^{\mu 0} \\ &=d\tau dxdy \frac{1}{\tau}\left( \begin{array}{cccc} 0, & 0, & 0, & 1 \end{array}\right) \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \tau\sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right)  \left( \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right)  \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \tau\sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right)  \left( \begin{array}{cccc} T_B^{00} & T_B^{01} & T_B^{02} & T_B^{03}\\ T_B^{10} & \ddots &  & \\ T_B^{20} &  &  & \\ T_B^{30} &  &  & \end{array}\right)  \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \tau\sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right)  \left( \begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right) \\ &=d\tau dxdy \frac{1}{\tau}\left( \begin{array}{cccc} 0, & 0, & 0, & 1 \end{array}\right) \left( \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -\tau^2 \end{array}\right)  \left( \begin{array}{cccc} T_B^{00} & T_B^{01} & T_B^{02} & T_B^{03}\\ T_B^{10} & \ddots &  & \\ T_B^{20} &  &  & \\ T_B^{30} &  &  & \end{array}\right)  \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \tau\sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right)  \left( \begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right) \\ &= d\tau dxdy \frac{1}{\tau}\left( \begin{array}{cccc} 0, & 0, & 0, & \tau^2 \end{array}\right) \left( \begin{array}{cccc} T_B^{00} & T_B^{01} & T_B^{02} & T_B^{03}\\ T_B^{10} & \ddots &  & \\ T_B^{20} &  &  & \\ T_B^{30} &  &  & \end{array}\right)  \left( \begin{array}{c} \cosh\eta \\ 0 \\ 0 \\ \tau\sinh\eta \end{array}\right)  \\ &=\tau d\tau dxdy \left( \begin{array}{cccc} 0, & 0, & 0, & 1 \end{array}\right)  \left( \begin{array}{c} \cosh\eta T_B^{00}+\tau\sinh\eta T_B^{03} \\ \cosh\eta T_B^{10}+\tau\sinh\eta T_B^{13} \\ \cosh\eta T_B^{20}+\tau\sinh\eta T_B^{23} \\ \cosh\eta T_B^{30}+\tau\sinh\eta T_B^{33} \end{array}\right)  \\ &= \tau d\tau dxdy (\cosh\eta T_B^{30}+\tau\sinh\eta T_B^{33}) \end{align}$$

It is ready to verify that the resulting expression has the same dimension as $$d\sigma_{C\mu (\tau=const.)}T_C^{\mu0}$$. Now we see that this term vanishes when one consider the like of IC provided by NEXUS. In these cases the energy momentum tensor T_B^{\mu\nu}  vanishes for large  $$\eta$$, therefore the integral will not contribute. However, this term does not vanish when one considers, for instance, Bjorken scaling case, which implies $$T_B^{\mu\nu}$$  is not a function of  $$\eta$$  while this expression is a function of  $$\eta$$  due to hyperbolic terms. It is noting here, the time component of energy momentum tensor conservation on $$\eta = const.$$  surface in Cartesian coordinates does not vanish identically even with such strong assumptions, very different from that in hyperbolic coordinates. Even though they are different quantities, it is a very very strange result. Similarly, if the surface element is for constant $$ \eta$$, the conservation of the $$\eta$$  component of energy momentum tensor reads


 * $$\begin{align}

d\sigma_{C\mu (\eta=const.)}T_C^{\mu3} = dxdy d\tau \frac{1}{\tau} \left( \begin{array}{cccc} 0, & 0, & 0, & \tau^2 \end{array}\right)  \left( \begin{array}{cccc} T_B^{00} & T_B^{01} & T_B^{02} & T_B^{03}\\ T_B^{10} & \ddots &  & \\ T_B^{20} &  &  & \\ T_B^{30} &  &  & \end{array}\right)  \left( \begin{array}{c} \sinh\eta \\ 0 \\ 0 \\ \tau\cosh\eta \end{array}\right)  = \tau d\tau dxdy  (\sinh\eta T_B^{30}+\tau\cosh\eta T_B^{33}) \end{align}$$

As was pointed out at the very beginning, one should be able to somehow show the conserved energy in hyperbolic coordinates is a linear combination of conserved quantities in Cartesian coordinates. Now we first try to show this for scaling invariant solution. We will take the interval of $$\eta$$  being  $$(-\eta_0,\eta_0)$$  and notice that the normal vectors have opposite sign on these two surfaces. In this case


 * $$\begin{align}

&E_C(\text{Evolution}) =d\sigma_{C\mu (\tau=const.)}T_C^{\mu0}+d\sigma_{C\mu (\eta=const.)}T_C^{\mu0}  \\ &=\int_{\eta=-\eta_0}^{\eta=\eta_0} \tau dxdy d\eta (\cosh\eta T_B^{00}+\tau\sinh\eta T_B^{03}) +\int_0^{\tau} \tau' d\tau' dxdy (\cosh\eta T_B^{30}+\tau'\sinh\eta T_B^{33})|_{\eta=\eta_0, \eta=-\eta_0} \\ &=\int_{\eta=-\eta_0}^{\eta=\eta_0} \tau dxdy d\eta \cosh\eta T_B^{00} +\int_0^{\tau} \tau' d\tau' dxdy \cosh\eta T_B^{30}|_{\eta=\eta_0, \eta=-\eta_0}  \\ &=2 \tau dxdy \sinh\eta_0 T_B^{00}+\int_0^{\tau} 2\tau' d\tau' dxdy  \cosh\eta_0 T_B^{30} \end{align}$$


 * $$\begin{align}

&P^{3}_C(\text{Evolution})=d\sigma_{C\mu (\tau=const.)}T_C^{\mu3}+d\sigma_{C\mu (\eta=const.)}T_C^{\mu3}\\ &=\int_{\eta=-\eta_0}^{\eta=\eta_0} \tau dxdy d\eta (\sinh\eta T_B^{00}+\tau\cosh\eta T_B^{03}) +\int_0^{\tau} \tau' d\tau' dxdy (\sinh\eta T_B^{30}+\tau'\cosh\eta T_B^{33})|_{\eta=\eta_0, \eta=-\eta_0} \\ &=\int_{\eta=-\eta_0}^{\eta=\eta_0} \tau^2 dxdy d\eta \cosh\eta T_B^{03} +\int_0^{\tau} \tau'^2 d\tau' dxdy \cosh\eta T_B^{33}|_{\eta=\eta_0, \eta=-\eta_0}  \\ &=2 \tau^2 dxdy \sinh\eta_0 T_B^{03}+\int_0^{\tau} 2\tau'^2 d\tau' dxdy  \cosh\eta_0 T_B^{33} \end{align}$$

Well it seems that no linear combination of the two terms is equivalent to the foregoing
 * $$\begin{align}

E_{B}(\text{Evolution})=\int d\eta dxdy\left\lbrace\tau T^{00}_B+\int_{\tau_0}^{\tau} d\tau' \tau'^2T^{33}_B\right\rbrace \end{align}$$

Why? Here I try to give an explanation of what is happening, but I am not sure at all about the validity of the following statement. There is no Gauss Theorem in curved space, in the sense that the time component of the covariant derivative of energy momentum tensor leads to an integral form which is not covariant, that is to say, it depends on specific coordinate. This is in line with the well known argument that there is conceptual problem of the energy conservation in general relativity. As a result, if one has two sets of different coordinates, the attempt to explore the equivalence between different integral forms of Gauss theorem in respective coordinates, or in other words, of the conservation law, must fail. Even though our problem here is in fact in flat space-time, the same conclusion holds. (This does not imply that one can therefore seek for new invariant integral by introducing new coordinate system, then leads to contradict to the total number of invariant integrals the system possesses, since these quantities are not indeed invariant in time.)

Energy on freeze-out surface in Cartesian coordinates
In the first place, one should once more derive the expression of energy from the time component of $${T^{\mu\nu}}_{;\mu}=0$$, namely,


 * $$\begin{align}

&\frac{1}{\tau}\partial_{\mu}(\tau T^{\mu0})+\Gamma^0_{\lambda\mu}T^{\lambda\mu}=0 \\ &\frac{1}{\tau}\partial_{\mu}(\tau T^{\mu0})+\tau T^{33}=0\\ &\partial_{\mu}(\tau T^{\mu0})+\tau^2 T^{33}=0\\ &\int d\tau d\eta dxdy \sqrt{-g} \frac{1}{\tau}\partial_{\mu}(\tau T^{\mu0})+\int d\tau d\eta dxdy \sqrt{-g} \tau T^{33}=0 \end{align}$$

where $$d^4x=d\tau d\eta dxdy \sqrt{-g} $$. Using again Gauss Law, we define energy on freeze-out surface in hyperbolic coordinates as in the way we had defined it during hydro evolution, except the second term cannot be further simplified


 * $$\begin{align}

E_B(\text{freeze-out})=\int_{\sigma_f}d\sigma_{B\mu}(\tau T^{\mu0})+\int d^4x \tau^2 T^{33} \end{align}$$ The corresponding formula in Cartesian coordinate takes a simpler form


 * $$\begin{align}

E_C(\text{freeze-out})=\int_{\sigma_f}d\sigma_{C\mu} T^{\mu0} \end{align}$$ (in term of energy momentum tensor)

To bring some usefulness to the above expression, one has to rewrite it in terms of SPH degree of freedom. The key task here is to calculate $$d\sigma_C^{\mu}$$, which is complicated and error-prone. This is because:
 * Here the integral is no longer on a surface $$\tau = const$$, of which we know an analytic expression, so that $$d\sigma_C^{\mu}= dxdyd\eta$$  and  $$d\sigma_B^{\mu}=\tau dxdyd\eta$$ can no longer be explicitly made use of.
 * More seriously, the SPH relations $$\int  d\eta dx dy \to \sum_j \frac{\nu_j}{\tau s_j}$$  (hyperbolic coordinates) has to be reconsidered since it came from
 * $$\begin{align}

(su^{\mu})_{;\mu}=0 \rightarrow (\tau su^{\mu})_{,\mu}=0 \rightarrow\int dxdyd\eta \tau\gamma s = const \end{align}$$ which was actually derived only on $$\tau = const$$ surface in hyperbolic coordinates.

Now the first point seems purely mathematical, we will leave it for the moment. Let us consider the second point, namely, how to rewrite the entropy conservation in hyperbolic coordinates for any infinitesimal surface $$d\sigma_B^{\mu}$$  and apply it to entropy representation of SPH method. The entropy conservation can be written down using a surface integral


 * $$\begin{align}

\int d\sigma_{B\mu} (su^{\mu}) = const \end{align}$$

where we adopt the form of Gauss theorem introduced in the book "An Advanced Course in GR by E. Poisson" (3.3.1), and the surface element is defined by (3.2.2) in the same book. It is very important to note that (though it can be different due to how things are defined mathematically) there is no $$\tau$$  factor in the conserved flow(!). For a general expression of surface element on freeze-out surface parameterized as $$T(\tau,x,y,\eta) = T_f$$, please see Derivation Notes on Freeze-out surface in SPheRIO which was derived by Philipe in the first place


 * $$\begin{align}

\end{align}$$
 * d\sigma_{B\mu}| = \tau dxdyd\eta

In $$\tau=const$$ case we had before


 * $$\begin{align}

\int d\sigma_{B\mu} (su^{\mu}) = \int dxdyd\eta (\tau \gamma s) =const = S_{total} \end{align}$$

As a matter of fact, the spirit of entropy representation of SPH method lies in dividing the total amount of conserved entropy into each SPH particles


 * $$\begin{align}

S_{total}=\sum_j\nu_j \end{align}$$

where each $$ \nu_j$$ is defined at initial  $$\tau =\tau_0$$  surface,


 * $$\begin{align}

\nu_j \equiv (dx_0dy_0d\eta_0 \tau_0 \gamma_0 s_0)_j =const \end{align}$$

Entropy of each individual SPH particle is then held constant during the evolution. So it is reasonable to suppose that if one splits the total entropy into pieces and assgin them to individual SPH particles, the entropy $$ \nu_j$$ of each individual SPH particle is then held constant during the evolution even on non-timelike surface  $$d\sigma_B^{\mu}$$, namely,


 * $$\begin{align}

\int_{\partial V} d\sigma_{B\mu} ( su^{\mu}) = S_{total} \Rightarrow    \left( d\sigma_{B\mu} ( su^{\mu}) \right)_j = \nu_j \end{align}$$ (on any surface including freeze-out surface)

which is consistent with the conservation of total entropy of the system, therefore


 * $$\begin{align}

\left(d\sigma_{B\mu}\right)_j u^{\mu}\equiv d\sigma_{Bj} \cdot u_j=\frac{\nu_j}{s_j}\equiv V_j \end{align}$$

By introducing $$n_{j\mu}$$  as an unitary 4-vector in the direction of  $$\sigma_{Bj\mu}$$,  $$\frac{\nu_j}{s_j}\frac{1}{|n\cdot u|_j}$$  is the module of the normal vector in hyperbolic coordinates, therefore  $$\sigma_{Bj\mu}=\frac{\nu_j}{s_j}\frac{n_{j\mu}}{|n\cdot u|_j}$$. The expression has made use of four velocity but mathematically it has nothing to do with it. The module of $$n_{j\mu}$$, if not unitary, cancels out on the numerator and denominator.

Up to this point, we had only used the notions of vector and tensor, they can be in any coordinate, namely, Cartesian or hyperbolic. However, it is actually not a trivial fact that $$d\sigma_{\mu}$$ is indeed a 4-vector: its orientation transform as a vector but its not obvious whether its module is invariant. The point is, if $$d\sigma_{\mu}$$  is not a covariant 4-vector, we should always indicate in which coordinate we are discussing when refering to it. We had seen in the above a simple case, and actually it can be generally proved that surface element $$d\sigma_{\mu}$$  is indeed a 4-vector. See "Lecture Notes of An Advanced Course in GR by E. Poisson" for a more detailed discussion.

Again putting every piece together and noting the $$\tau$$  factor cancels, one finally arrives to the desired expression


 * $$\begin{align}

&E_C(\text{freeze-out})=\sum_j \frac{\nu_j}{s_j}\frac{n_{j\mu}}{|n\cdot u|_j}[(\varepsilon+P)u^{\mu}u^0-Pg^{\mu0}] \\ &=\sum_j \frac{\nu_j}{s_j}\frac{1}{|n\cdot u|_j}\left[((\varepsilon+P)u^0u^0-P)n^0-(\varepsilon+P)u^0\vec{u}\cdot\vec{n}\right]_j \\ &=\sum_j \left(\frac{\nu_j}{s_j}(\varepsilon+P)\gamma-\frac{\nu_j}{s_j}\frac{Pn^0}{(n\cdot u)_j}\right) \end{align}$$

where $$\alpha$$  stands for different species of particles and  $$j$$  enumerates all SPH particles, and all vectors and tensors should be expressed in Cartesian coordinates. Also one made use of the expression of energy momentum tensor in terms of thermodynamic quantities


 * $$\begin{align}

T^{\mu\nu}=(\varepsilon+P)u^{\mu}u^{\nu}-Pg^{\mu\nu} \end{align}$$

This is a good place to make some remarks.
 * Since the surface element $$d\sigma_{\mu}$$  is a 4-vector, entropy conservation (of non-voscious case) is not like energy conservation. The conserved quantity possess the same value in any coordinate system. Therefore we can calculate the entropy in hyperbolic coordinates and use it to compare the results obtained in Cartesian coordinates.
 * On the other hand, the expression of $$E_C(\text{freeze-out})$$ can also be derived intuitively (from discussion notes with Philipe and Kodama). The strategy in the follows is more of making generalization by constructing Lorentz invariant terms rather than deriving it straightforwardly from a certain conservation law


 * $$\begin{align}

E_{\alpha}\left(\frac{d^3N}{dp^3}\right)_{\alpha} =\int_{\sigma_f}d\sigma_{\mu} p_{\alpha}^{\mu}f(u_j\cdot p_{\alpha}) \end{align}$$ (with everything in Cartesian coordinates)


 * $$\begin{align}

\tilde{E}_C(\text{freeze-out})=\sum_{\alpha}\int d^3p_{\alpha}E_{\alpha}\left(\frac{d^3N}{dp^3}\right)_{\alpha} =\sum_{\alpha}\sum_j\frac{\nu_j}{s_j}\frac{1}{|n\cdot u|_j}\int\frac{d^3p_{\alpha}}{p_{\alpha 0}}(n_{j\mu} p_{\alpha}^{\mu})p_{\alpha}^0f(u_j\cdot p_{\alpha})  =\int_{\sigma_f}d\sigma_{\mu} T^{\mu0} \end{align}$$

where one made use of another way of expression energy momentum tensor in terms of individual particles


 * $$\begin{align}

T^{\mu\nu}=\sum_{\alpha}\int\frac{d^3p_{\alpha}}{P^0_{\alpha}}p^{\mu}_{\alpha}p^{\nu}_{\alpha}f_{\alpha}(u\cdot p_{\alpha}) \end{align}$$

One comment on why we called it intuitive, this is because one cannot write down similarly an expression of freeze-out energy in hyperbolic coordinates, the reason is again, the energy, though conserved, is not an invariant neither covariant quantity, in this context it is actually not well defined in hyperbolic coordinates, therefore there is no easy way one can intuitively write it down solely by its physical content.
 * It can be shown straightforwardly, that when the surface element is time like in hyperbolic coordinates, the above expression goes to the previous simple form that we have derived.

Transformation Matrix Formulae

 * $$\begin{align}

\left( \begin{array}{cccc} d\tau \\ d\eta \end{array}\right)  = \left( \begin{array}{cccc} \partial_t \tau & \partial_z \tau \\ \partial_t \eta & \partial_z \eta \end{array}\right) \left( \begin{array}{cccc} dt \\ dz \end{array}\right)   = [\Lambda] \left( \begin{array}{cccc} dt \\ dz \end{array}\right) \end{align}$$


 * $$\begin{align}

dx^{\mu'} = \frac{ \partial x^{\mu'} }{ \partial x^\nu } d x^\nu = {\Lambda^{\mu'}}_\nu d x^\nu \end{align}$$


 * $$\begin{align}

\left( \begin{array}{cccc} \partial_\tau \\ \partial_\eta \end{array}\right)  = \left( \begin{array}{cccc} \partial_\tau t& \partial_\tau z \\ \partial_\eta t& \partial_\eta z \end{array}\right) \left( \begin{array}{cccc} \partial_t \\ \partial_z \end{array}\right)   =[\Lambda']  \left( \begin{array}{cccc} \partial_t \\ \partial_z \end{array}\right) \end{align}$$


 * $$\begin{align}

\partial_{\mu'}=\frac{ \partial }{ \partial x^{\mu'} }= \frac{ \partial x^{\nu} }{ \partial x^{\mu'} } \frac{ \partial }{ \partial x^\nu } = {\Lambda^\nu}_{\mu'} {\partial_\nu} \end{align}$$


 * $$\begin{align}

\frac{ \partial x^{\nu} }{ \partial x^{\mu'} } \frac{ \partial x^{\mu'} }{ \partial x^{\lambda} } = \delta^\nu_\lambda \end{align}$$


 * $$\begin{align}

{\left(\Lambda^{-1}\right)^{\nu}}_{\mu'}  {\Lambda^{\mu'}}_{\lambda} = \delta^\nu_\lambda \end{align}$$


 * $$\begin{align}

[ \Lambda^{-1} ] [\Lambda'] = 1 \end{align}$$


 * $$\begin{align}

{\Lambda^{\mu}}_{\nu'} \equiv \frac{\partial x_C^{\mu}}{\partial x_B^{\nu'}}=\left( \begin{array}{cccc} \partial_\tau t & \partial_\eta t\\ \partial_\tau z & \partial_\eta z \end{array}\right)  =\left( \begin{array}{cccc} \cosh \eta & \tau\sinh \eta \\ \sinh \eta & \tau \cosh \eta \end{array}\right) \end{align}$$

Therefore, the inverse transformation


 * $$\begin{align}

{\left(\Lambda^{-1}\right)^{\mu}}_{\nu'} \equiv \frac{\partial x_B^{\mu}}{\partial x_C^{\nu'}}  =\left( \begin{array}{cccc} \partial_\tau t & \partial_\eta t\\ \partial_\tau z & \partial_\eta z \end{array}\right)^{-1}  = \left( \begin{array}{cccc} \partial_t \tau & \partial_t \eta\\ \partial_z \tau & \partial_z \eta \end{array}\right) =\left( \begin{array}{cccc} \cosh \eta & -\sinh \eta \\ -1/\tau \sinh \eta & 1/\tau \cosh \eta \end{array}\right) \end{align}$$

The above matrix, as defined, is connected with inverse transformation from Cartesian coordinates to hyperbolic coordinates


 * $$\begin{align}

\left( \begin{array}{cccc} \partial_t \\ \partial_z \end{array}\right)  =\left( \begin{array}{cccc} \frac{\partial\tau}{\partial t}\partial_\tau+\frac{\partial\eta}{\partial t}\partial_\eta \\ \frac{\partial\tau}{\partial z}\partial_\tau+\frac{\partial\eta}{\partial z}\partial_\eta \end{array}\right)   = \left( \begin{array}{cccc} \partial_t \tau & \partial_t \eta\\ \partial_z \tau & \partial_z \eta \end{array}\right) \left( \begin{array}{cccc} \partial_\tau \\ \partial_\eta \end{array}\right) \end{align}$$

When transform the energy momentum tensor from Bjoken coordinate to Cartesian coordinates, one involves the transpose of the matrix


 * $$\begin{align}

T_B^{\mu\nu}\rightarrow T_C^{\mu\nu} = {\Lambda^{\mu}}_{\mu'} {\Lambda^{\nu}}_{\nu'} T_B^{\mu'\nu'} ={\Lambda^{\mu}}_{\mu'} ({\Lambda^{T})^{\nu'}}_{\nu} T_B^{\mu'\nu'}  ={\Lambda^{\mu}}_{\mu'}  T_B^{\mu'\nu'}({\Lambda^{T})^{\nu'}}_{\nu}  = \left( \begin{array}{cccc} \partial_t \tau & \partial_t \eta\\ \partial_z \tau & \partial_z \eta \end{array}\right) \left( \begin{array}{cccc} T_B^{00} & T_B^{01} & T_B^{02} & T_B^{03}\\ T_B^{10} & \ddots &  & \\ T_B^{20} &  &  & \\ T_B^{30} &  &  & \end{array}\right)  \left( \begin{array}{cccc} \partial_\tau t & \partial_\tau z\\ \partial_\eta t & \partial_\eta z \end{array}\right)^T \end{align}$$

where


 * $$\begin{align}

\left( \begin{array}{cccc} \partial_\tau t & \partial_\tau z\\ \partial_\eta t & \partial_\eta z \end{array}\right)^T =\left( \begin{array}{cccc} \partial_\tau t & \partial_\eta t\\ \partial_\tau z & \partial_\eta z \end{array}\right)  \ne \left( \begin{array}{cccc} \partial_t \tau & \partial_t \eta\\ \partial_z \tau & \partial_z \eta \end{array}\right)  =\left( \begin{array}{cccc} \partial_\tau t & \partial_\tau z\\ \partial_\eta t & \partial_\eta z \end{array}\right)^{-1} \end{align}$$

the transpose is different from inverse simple because the transformation is not orthogonal like a spacial rotation.