AMPT + SPheRIO results and discussions

AMPT + SPheRIO results and discussions

Initial Energy
Here we first calculate the total energy in Cartesian coordinates on the surface $$\tau= const.$$ using the energy momentum tensor in hyperbolic coordinates

To calculate the energy, we first transfer the energy momentum tensor into Cartesian coordinates then project it on to the surface and take the "0"-th component. The transformation matrix reads


 * $$\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) \end{align}$$ where the subsrcipt $C$ is used to denote the Cartesian coordinates and $B$ to denote the hyperbolic coordinates. It gives the metric.
 * $$\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}$$

Likewise, one has
 * $$\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}$$ Therefore
 * $$\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}$$

Similarly, we may transform a unitary vector in hyperbolic coordinates into 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_{\mu\lambda}\frac{\partial x_C^{\lambda}}{\partial x_B^{\nu}}e_{B}^{\nu} =\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}{c} 1\\ 0\\ 0\\ 0 \end{array}\right) =\left( \begin{array}{c} \cosh\eta\\ 0\\ 0\\ -\sinh\eta \end{array}\right) \end{align}$$

It is straightforward to verify at this point $$|e_{C\mu}|=|e_{B\mu}|=1$$. The area of the surface element can be calculated as following


 * $$\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. In fact, this also 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, the total energy in Cartesian coordinates reads


 * $$\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_B^{\alpha\beta} =d\sigma_B^{\nu'}{\Lambda^\mu}_{\nu'}\eta_{\mu\nu}{\Lambda^\nu}_{\alpha}T_B^{\alpha\beta}{\Lambda^0}_{\beta} =d\sigma_B^{\nu'}{(\Lambda^T)^{\nu'}}_{\mu}\eta_{\mu\nu}{\Lambda^\nu}_{\alpha}T_B^{\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}$$

In short,
 * $$\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}$$ This is expressed in terms of energy momentum tensor.

The energy momentum tensor
Up to this point, we have not discussed the explicit form of energy momentum tensor. According to ,
 * $$\begin{align}

T^{\mu\nu}_C = \sum_i\frac{p^\mu_i p^\nu_i}{p^0_i}\delta(\vec{x}-\vec{x}_i) \end{align}$$ Integrating on the $$t=t_0$$ surface, one obtains
 * $$\begin{align}

E_C(\text{AMPT})=\int_{\sigma_\mu(t=const.)} T_C^{\mu 0}=\int dxdydz T_C^{00}= \sum_i p^0_i \end{align}$$ which is as expected.

In a coordinate system with a metric, such as hyperbolic coordinates, the above expression can be generalized to
 * $$\begin{align}

T^{\mu\nu}_B = \frac{1}{\sqrt{-g}}\sum_i\frac{p^\mu_{B,i} p^\nu_{B,i}}{p^0_{B,i}}\delta^3(\vec{x}-\vec{x}_i) \end{align}$$ This is because $$\frac{1}{\sqrt{-g}}\delta^4(x-x_i)$$ is the corresponding invariant quantity.

In practical calculations, one uses Gaussian functions to replace the $$\delta$$ function
 * $$\begin{align}

\delta^3(\vec{x}-\vec{x}_i) \rightarrow \frac{1}{\sqrt{2\pi{\sigma_\eta}^2}}\frac{1}{2\pi{\sigma_r}^2}\times \exp \left[-\frac{(x-x_i)^2+(y-y_i)^2}{2{\sigma_r}^2}-\frac{(\eta-\eta_i)^2}{2{\sigma_\eta}^2}\right] \end{align}$$

For completeness, we also write down the momentum components in hyperbolic coordinates in terms of Cartesian components
 * $$\begin{align}

&{p^{\tau}}_B=m_T \cosh(Y-\eta)\\ &{p^x}_B={p^x}_C\\ &{p^y}_B={p^y}_C\\ &{p^{\eta}}_B=\frac{1}{\tau} m_T \sinh(Y-\eta)\\ &Y=\frac{1}{2}\ln\left(\frac{{p^0}_C+{p^z}_C}{{p^0}_C-{p^z}_C}\right) \end{align}$$

The conservation of incident energy
The conservation of incident energy involves several steps.

The first step is the comparison between the sum of incident energy of partons in Cartesian coordinates and that comes from the energy momentum tensor in hyperbolic coordinates constructed from the parton information. The two incident energies, namely $$E_C(\text{AMPT})$$ and $$E_C(\text{IC})$$, should be of the same value, which is a consequence of the conservation of energy momentum tensor.
 * $$\begin{align}

\partial_\mu T_C^{\mu\nu}= 0 \end{align}$$

The conservation of energy momentum tensor for point like particle, in this case, only requires that the particle collisions are localized.

Unfortunately, numerical calculation shows otherwise, I am trying to write an independent script to check whether something was wrong in the numerical implementation.

The second step involves EoS. One diagonalize the energy momentum tensor and obtains the energy density $$\epsilon$$ in the local rest frame, then the pressure $$P$$ is obtained using EoS. By ignoring the viscosity, one can construct the energy momentum tensor again by using these quantities, namely


 * $$\begin{align}

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

One may again calculate the total IC energy and compare it to $$E_C(\text{AMPT})$$. His step justifies the assumptions of local thermalization as well as that of ideal fluid.