Derivation Notes on Initial Energy Conservation in SPheRIO

Derivation Notes on AMPT initial condition and energy in Cartesian coordinates

Definitions of coordinates and their transformations
Hyperbolic coordinates $$x^\mu_B\equiv (\tau,x,y,\eta)$$ are defined as following
 * $$\begin{align}

&\tau=\sqrt{t^2-z^2}\\ &\eta=\frac{1}{2}\ln\frac{t+z}{t-z} \end{align}$$ Note that $$\tau$$ is a time-like coordinate, which is not the proper time.

In the follows, 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}$$

Surface element and initial energy
Similarly, we may transform (the orientation of) a surface element in hyperbolic coordinates $$\tau=const.$$, which is 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.

As a side note, in the above, the surface element was calculated by (1) assuming that it s a vector, therefore its components in Cartesian coordinates was obtained by coordinate transformation (2) the module of the surface element was obtained by using the fact the surface $$\tau = const.$$ is parameterized in terms of orthogonal coordinates, therefore all displacements corresponding to each variables ($$\tau,\eta,x,y$$) are orthogonal, the total area of the 3-surface element is simply a product of the module of three displacements.

On the other hand, the surface element can be calculated by using some more general formulae, which reads
 * $$\begin{align}

d\sigma_{\mu} = \epsilon_{\mu\alpha\beta\gamma}e_1^\alpha e_2^\beta e_3^\gamma d^3y \end{align}$$ where $$\epsilon_{\mu\alpha\beta\gamma}=\sqrt{-g}[\mu,\alpha,\beta,\gamma]$$ is the Levi-Civita tensor, $$y_1,y_2,y_3$$ parameterize the 3-surface, and $$e_1^\alpha=\partial x^\alpha/\partial y_1,e_2,e_3$$ describe the connection between the coordinates and its parameters. For instance, see the discussion near (3.2.2) of. In our case of the Cartesian coordinates, $$\sqrt{-g}=1$$, therefore $$d\sigma_{C\mu}=[\mu,\alpha,\beta,\gamma]e_1^\alpha e_2^\beta e_3^\gamma d^3y$$. It is straightforward to verify that one obtains the same result, namely,
 * $$\begin{align}

d\sigma_{C\mu}= dxdyd\eta \left( \begin{array}{c} \tau\cosh\eta\\ 0\\ 0\\ -\tau\sinh\eta \end{array}\right) \end{align}$$

Transform back into Cartesian coordinates
One may readily verify that, if one will, the above result can be transform back into Cartesian coordinates by writing $$T^{\mu\nu}_B$$ in terms of $$T^{\mu\nu}_C$$. For instance
 * $$\begin{align}

&T^{00}_B = \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} 1\\ 0\\ 0\\ 0 \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\\ -\frac{1}{\tau}\sinh\eta & 0 & 0 & \frac{1}{\tau}\cosh\eta \end{array}\right) \left( \begin{array}{cccc} T_C^{00} & T_C^{01} & T_C^{02} & T_C^{03}\\ T_C^{10} & \ddots & & \\ T_C^{20} &  &  & \\ T_C^{30} &  &  & \end{array}\right) \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & -\frac{1}{\tau}\sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ -\sinh\eta & 0 & 0 & \frac{1}{\tau}\cosh\eta \end{array}\right) \left( \begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right) \\ &=\cosh^2\eta T_C^{00}-\cosh\eta\sinh\eta T_C^{30}-\cosh\eta\sinh\eta T_C^{03}+\sinh^2\eta T_C^{33} \end{align}$$ where
 * $$\begin{align}

{\left(\Lambda^{-1}\right)^{\mu}}_{\nu}\equiv \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & -\sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ -\frac{1}{\tau}\sinh\eta & 0 & 0 & \frac{1}{\tau}\cosh\eta \end{array}\right) \end{align}$$. Similarly, one gets
 * $$\begin{align}

T^{03}_B=\frac{1}{\tau}[-\cosh\eta\sinh\eta T_C^{00}+\sinh^2\eta T_C^{30}+\cosh^2\eta T_C^{03}-\cosh\eta\sinh\eta T_C^{33}] \end{align}$$ Therefore
 * $$\begin{align}

\cosh\eta T_B^{00}+\tau\sinh\eta T^{03}_B=\cosh\eta T_C^{00}-\sinh\eta T_C^{30} \end{align}$$ or
 * $$\begin{align}

\int_\sigma\tau dxdyd\eta(\cosh\eta T_B^{00}+\tau\sinh\eta T^{03}_B)=\int_\sigma\tau dxdyd\eta(\cosh\eta T_C^{00}-\sinh\eta T_C^{30}) \end{align}$$ Note that it is consistent with the above expressions. For the hypersurface $$\tau=const.$$, the energy density also involves $$T_C^{30}$$.

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^3(\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.)} d\sigma_{\mu} 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 of energy momentum tensor 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.

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

&{p^{0}}_B\equiv {p^{\tau}}_B=m_T \cosh(Y-\eta)\\ &{p^{1}}_B\equiv {p^x}_B={p^x}_C\\ &{p^{2}}_B\equiv {p^y}_B={p^y}_C\\ &{p^{3}}_B\equiv {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}$$

It is straightforward to verify that $$p_B^0=p_C^0(\cosh\eta-v_z\sinh\eta)$$ where the extra factor is the same one that comes from the $$\delta$$ function transformation in the above discussions.

A simple example
Now let us discuss a simple example. Considering a single free particle with 4-momentum $$p^\mu$$ in Cartesian coordinates. In the following, we will try to verify explicitly for a closed surface, one has
 * $$\begin{align}

\mathcal{I}=\int d\sigma_{C\mu} T_C^{\mu 0}= 0 \end{align}$$ where the closed 3-surface $$\sigma_\mu$$ is defined by $$\tau=\tau_1,\tau_2$$, $$\eta=\eta_1,\eta_2$$, $$x=x_1,x_2$$ and $$y=y_1,y_2$$. Since the energy momentum tensor is evaluated in the Cartesian coordinates, the expression is not covariant, the vanishing contraction corresponds to the energy conservation in Cartesian coordinates. We use the following notations
 * $$\begin{align}

\mathcal{I}=\mathcal{I}_\tau+\mathcal{I}_\eta+\mathcal{I}_x+\mathcal{I}_y \end{align}$$

As shown above, the 3-surface element at $$\tau=const$$ reads
 * $$\begin{align}

d\sigma_{C\mu}= dxdyd\eta \left( \begin{array}{c} \tau\cosh\eta\\ 0\\ 0\\ -\tau\sinh\eta \end{array}\right) \end{align}$$ and the integral on the 3-surface $$\tau=\tau_1, \tau_2$$ yields
 * $$\begin{align}

\mathcal{I}_\tau=\int_{\tau_1}\tau dxdyd\eta(\cosh\eta T_C^{00}-\sinh\eta T_C^{30})-\int_{\tau_2}\tau dxdyd\eta(\cosh\eta T_C^{00}-\sinh\eta T_C^{30}) \end{align}$$

Similarly, the 3-surface element at $$\eta=const$$ reads
 * $$\begin{align}

d\sigma_{C\mu}= dxdyd\eta \left( \begin{array}{c} \sinh\eta\\ 0\\ 0\\ -\cosh\eta \end{array}\right) \end{align}$$ and the integral on the 3-surface $$\eta=\eta_1, \eta_2$$ yields
 * $$\begin{align}

\mathcal{I}_\eta=\int_{\eta_1}dxdyd\tau(\sinh\eta T_C^{00}-\cosh\eta T_C^{30})-\int_{\eta_2}dxdyd\tau(\sinh\eta T_C^{00}-\cosh\eta T_C^{30}) \end{align}$$

Expressions for the other two 3-surfaces are more intuitive, one has
 * $$\begin{align}

\mathcal{I}_x=\int_{x_1}dtdydz T_C^{10}-\int_{x_2}dtdydz T_C^{10} \end{align}$$ and
 * $$\begin{align}

\mathcal{I}_y=\int_{y_1}dtdxdz T_C^{20}-\int_{y_2}dtdxdz T_C^{20} \end{align}$$

The energy momentum tensor of a point-like particle $$q$$ in Cartesian coordinates possesses the following form
 * $$\begin{align}

T^{\mu\nu}_C = \frac{p^\mu p^\nu}{p^0}\delta^3(\vec{x}-\vec{x}_q)= \frac{p^\mu p^\nu}{p^0}\delta(x-x_q)\delta(y-y_q)\delta(z-z_q) \end{align}$$ where the $$\delta$$ functions deserve some special attention. For instance, $$\delta(z-z_q)$$ is a function of $$z$$ and $$\eta$$ when $$\tau$$ remains constant, where $$z_q \equiv z_q(t) \equiv z_q(t(\eta,\tau))$$, following this line of thought, one has
 * $$\begin{align}

&\delta(z-z_q)=\frac{\delta(\eta-\eta_q(\tau=const.))}{|(z-z_q)'|}=\frac{\delta(\eta-\eta_q(\tau=const.))}{|\frac{\partial z}{\partial \eta}-\frac{dz_q}{dt}\frac{\partial t}{\partial \eta}|}=\frac{\delta(\eta-\eta_q(\tau=const.))}{|\tau\cosh\eta-v_z\tau\sinh\eta|}\\ &=sgn(1-\tanh(\eta)\tanh(Y))\frac{\delta(\eta-\eta_q(\tau=const.))}{\tau\cosh\eta-v_z\tau\sinh\eta}=\frac{\delta(\eta-\eta_q(\tau=const.))}{\tau\cosh\eta-v_z\tau\sinh\eta} \end{align}$$ when the 3-surface is parameterized in $$\eta$$ with $$\tau = const.$$, and $$\eta_q(\tau)$$ is determined by the particle's worldline $$z_q(t)$$. On the other hand,
 * $$\begin{align}

\delta(z-z_q)=sgn(\eta-Y)\frac{\delta(\tau-\tau_q(\eta=const.))}{\sinh\eta-v_z\cosh\eta} \end{align}$$ when the 3-surface is parameterized in $$\tau$$ with $$\eta = const.$$. Also the other two $$\delta$$ functions, for instance, $$\delta(x-x_q)=\delta(x-x_q(t(\eta,\tau=const.)))$$ are also function of $$\eta,\tau$$, but once the integral is carried out first in terms of these variables, they do not affect the results in the cases of $$\mathcal{I}_\tau$$ and $$\mathcal{I}_\eta$$ (and they do in the case of $$\mathcal{I}_x$$ and $$\mathcal{I}_y$$, one should use for instance $$\delta(x_1-x_q(t))=sgn(v_x)\delta(t-t_q(x_1))/v_x$$ and $$\delta(y_1-y_q(t))=sgn(v_y)\delta(t-t_q(y_1))/v_y$$ respectively). Substituting into the expressions of $$\mathcal{I}$$, one gets
 * $$\begin{align}

\mathcal{I}_\tau= \int_{\tau_1} d\eta \delta(\eta-\eta_a)(p^0\cosh\eta -p^3\sinh\eta)-\int_{\tau_2}d\eta \delta(\eta-\eta_b)(p^0\cosh\eta -p^3\sinh\eta)=\left\{\begin{matrix} p^0-p^0=0&\textrm{the world-line goes across both surfaces}\\ 0-0=0&\textrm{the world-line goes across neither surface}\\ +p^0&\textrm{if the particle only crosses } \tau_1\\ -p^0&\textrm{if the particle only crosses } \tau_2 \end{matrix}\right. \end{align}$$
 * $$\begin{align}

\mathcal{I}_\eta=\int_{\eta_1}d\tau\delta(\tau-\tau_a)(p^0\sinh\eta - p^3\cosh\eta)-\int_{\eta_2}d\tau\delta(\tau-\tau_b)(p^0\sinh\eta - p^3\cosh\eta)=\left\{\begin{matrix} p^0-p^0=0&\textrm{the world-line goes across both surfaces}\\ 0-0=0&\textrm{the world-line goes across neither surface}\\ +sgn(\eta-Y)p^0&\textrm{if the particle only crosses } \eta_1\\ -sgn(\eta-Y)p^0&\textrm{if the particle only crosses } \eta_2 \end{matrix}\right. \end{align}$$
 * $$\begin{align}

\mathcal{I}_x=\int_{x_1}dt p^1 \delta(x-x_q(t))-\int_{x_2}dt p^1 \delta(x-x_q(t))=\left\{\begin{matrix} p^0-p^0=0&\textrm{the world-line goes across both surfaces}\\ 0-0=0&\textrm{the world-line goes across neither surface}\\ +sgn(v_x)p^0&\textrm{if the particle only crosses } x_1\\ -sgn(v_x)p^0&\textrm{if the particle only crosses } x_2 \end{matrix}\right. \end{align}$$
 * $$\begin{align}

\mathcal{I}_y=\int_{y_1}dt p^2 \delta(y-y_q(t))-\int_{y_2}dt p^2 \delta(y-y_q(t))=\left\{\begin{matrix} p^0-p^0=0&\textrm{the world-line goes across both surfaces}\\ 0-0=0&\textrm{the world-line goes across neither surface}\\ +sgn(v_y)p^0&\textrm{if the particle only crosses } y_1\\ -sgn(v_y)p^0&\textrm{if the particle only crosses } y_2 \end{matrix}\right. \end{align}$$ where $$\eta_a \equiv \eta_q(\tau_1)$$, $$\eta_b\equiv \eta_q(\tau_2)$$, $$\tau_a \equiv \tau_q(\eta_1)$$, $$\tau_b \equiv \tau_q(\eta_2)$$ are the values of $$\eta,\tau$$ coordinates when the particle does go across the surfaces $$\tau_1, \tau_2, \eta_1, \eta_2$$. Among the eight possible surfaces discussed above, one notes that the particle always goes across two of them. In practice, one may always choose a big enough area of $$\tau_1, \tau_2$$ surface, so that the particle goes across $$\tau_1, \tau_2$$ surfaces and, one ignores the integral on the $$\eta_1, \eta_2, x_1, x_2, y_1, y_2$$ surfaces. In a more general case, if the system consists of many point-like particles, one can usually find big enough area of $$\tau_1, \tau_2$$ so that all the particles go through the $$\tau_1, \tau_2$$ area. From the above results, one sees that the energy is conserved, and the value of the contraction of the 0-th component of the energy momentum tensor with the 3-surface gives exactly the energy in Cartesian coordinates.

The conservation of incident energy in AMPT model
Now we are in a good position to discuss the initial energy in AMPT model.

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}$$

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}$$ In the Cartesian coordinates, the covariant derivative is simply derivative, therefore one has
 * $$\begin{align}

\int d\sigma_{C\mu} T_C^{\mu\nu}= 0 \end{align}$$ where $$\sigma_\mu$$ is any closed surface. Energy conservation corresponds to the case when $$\nu=0$$.

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

In a typical Pb+Pb collision, AMPT model generates 40,000 partons. The energy momentum tensor was calculated using the 4-momentum of these partons in Cartesian coordinates and, using the $$\delta$$ function approximation above with $$\sigma=0.6$$fm. The energy momentum tensor is evaluated on the grid, a typical grid setting is 25*25*25, for $$\tau_0=0.2$$ one obtains the initial energy up to 8%-10% of discrepancy. If one increase the grids number to 50*50*50, precision improves and one gets 3-5% discrepancy.

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 $$E_C(\text{AMPT})$$. This step justifies the assumptions of local thermalization as well as that of ideal fluid. However, during the diagonalizing process, one gets numerical problem that at certain grid point, the resulting energy density is negative and, has to be ignored. The resulting initial energy, for case of $$\tau_0=0.2$$ and grid setting of 25*25*25, is 100% more than the incident energy.