Derivation Notes on Freeze-out Surface in SPheRIO

Derivation Notes on Freeze-out Surface in SPheRIO


 * The note is an attempt to reproduce Philipe's note on the same topic as well as do something more

Freeze-out surface in terms of general parameterization
The coordinate transformation from Cartesian to Bjorken is


 * $$\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}$$


 * $$\begin{align}

&g_{\mu\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 & \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 & -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 & -\tau^2 \end{array}\right) \end{align}$$

Or in short notions


 * $$\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 coordinate to Bjorken coordinate


 * $$\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}$$

On freeze-out surface, one needs to calculte


 * $$\begin{align}

p^{\mu}d\sigma_{\mu} \end{align}$$

the surface element is


 * $$\begin{align}

d\sigma_{\mu}=\epsilon_{\mu\nu\lambda\rho}\partial_{y_1}\sigma^{\nu}\partial_{y_2}\sigma^{\lambda}\partial_{y_3}\sigma^{\rho}d^3y \end{align}$$

where $$d^3y=dy_1dy_2dy_3$$  with  $$y_1, y_2, y_3$$  parameterizing the surface and Levi-Civita tensor is defined by  $$\epsilon_{\mu\nu\lambda\rho} \equiv \sqrt{-g}\varepsilon_{\mu\nu\lambda\rho}$$

Now if one assumes the following parameterization for the surface


 * $$\begin{align}

&\tau = \tau(y_1,y_2,y_3) \\ &x=y_1 \\ &y=y_2 \\ &\eta= y_3 \end{align}$$

One may explicitly write down


 * $$\begin{align}

d\sigma_{\mu}= \left( \begin{array}{cccc} 1 \\ -\partial_x\tau \\ -\partial_y\tau \\ -\partial_\eta\tau \end{array}\right) \tau dxdyd\eta \left( \begin{array}{cccc} 1 \\ -\partial_x\tau \\ -\partial_y\tau \\ -\partial_\eta\tau \end{array}\right) \equiv \bar{\partial_{\mu}}\tau \end{align}$$

where  $$\bar{\partial_{\mu}}$$  is not a vector, neither the inner product  $$p^{\mu} \bar{\partial_{\mu}}$$  is a scalar, but the whole expression is indeed a scalar  $$p^{\mu}(\bar{\partial_{\mu}} \tau) \tau dxdyd\eta$$  since we are dealing with a specific coordinate (Bjorken) and specific parameterization of the surface. To compare with Philipe's notes, I do not try to calculate $$p^{\mu}(\bar{\partial_{\mu}} \tau)$$  in Cartesian coordinate, since it might be complicated to do any coordinate transformation. (Though I do believe with some care it can be done, as Philipe did but without showing explicit terms.) Now instead I transform p^{\mu}  into Bjorken coordinate, since we have done the expression for surface element in that coordinate and found a term   $$\bar{\partial_{\mu}}$$  which is not vector, right(?). If I can recover Philipe's expression right above Eq.(12), bingo...


 * $$\begin{align}

&p^{\mu}_B ={\Lambda^{\mu}}_{\nu} p^{\nu}_C  \\ &=\left( \begin{array}{cccc} \cosh\eta & 0 & 0 & -\sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ -1/\tau \sinh\eta & 0 & 0 & 1/\tau\cosh\eta \end{array}\right) \left( \begin{array}{cccc} m_T\cosh Y \\ p_T \cos\phi \\ p_T \sin\phi \\ m_T\sinh Y \end{array}\right)  \\ &=\left( \begin{array}{cccc} m_T\cosh Y\cosh\eta-m_T \sinh Y\sinh\eta \\ p_T \cos\phi \\ p_T \sin\phi \\ -(1/\tau) m_T\cosh Y\sinh\eta+(1/\tau) m_T \sinh Y\cosh\eta \end{array}\right) \end{align}$$


 * $$\begin{align}

&p^{\mu}(\bar{\partial_{\mu}} \tau) \tau dxdyd\eta \\ &=\left( \begin{array}{cccc} m_T\cosh Y\cosh\eta-m_T \sinh Y\sinh\eta \\ p_T \cos\phi \\ p_T \sin\phi \\ -(1/\tau) m_T\cosh Y\sinh\eta+(1/\tau) m_T \sinh Y\cosh\eta \end{array}\right) \left( \begin{array}{cccc} 1 \\ -\partial_x\tau \\ -\partial_y\tau \\ -\partial_\eta\tau \end{array}\right)  \tau dxdyd\eta  \\ &=\left[(m_T\cosh Y\cosh\eta- m_T \sinh Y\sinh\eta)-p_T\cos\phi\partial_x\tau-p_T\sin\phi\partial_y\tau+1/\tau(m_T\cosh Y\sinh\eta- m_T \sinh Y\cosh\eta)\partial_\eta\tau\right]\tau dxdyd\eta \\ &=\left[m_T\cosh (Y-\eta)-p_T\cos\phi\partial_x\tau-p_T\sin\phi\partial_y\tau-(1/\tau) m_T\sinh (Y-\eta)\partial_\eta\tau\right]\tau dxdyd\eta \end{align}$$

If the freeze-out surface is parameterized by $$ T(\tau,x,y,\eta)=T_f $$ such as in Cooper-Fyre, one finds that


 * $$\begin{align}

&\frac{dT}{d\eta}=\frac{d\tau}{d\eta}\frac{\partial T}{\partial\tau}+\frac{\partial T}{\partial\eta} \\ &\partial_{\eta}\tau = -\frac{\partial T}{\partial\eta} /\frac{\partial T}{\partial\tau} \\ &\partial_{x}\tau = -\frac{\partial T}{\partial x} /\frac{\partial T}{\partial\tau} \\ &\partial_{y}\tau = -\frac{\partial T}{\partial y} /\frac{\partial T}{\partial\tau} \end{align}$$

Therefore


 * $$\begin{align}

&p^{\mu}d\sigma_{\mu} \\ &=[m_T\cosh (Y-\eta)+p_T\cos\phi (\frac{\partial T}{\partial x} /\frac{\partial T}{\partial\tau}) +p_T\sin\phi(\frac{\partial T}{\partial y} /\frac{\partial T}{\partial\tau}) \\ &+(1/\tau) m_T\sinh (Y-\eta)(\frac{\partial T}{\partial\eta} /\frac{\partial T}{\partial\tau})]\tau dxdyd\eta \end{align}$$

We see that it does recover Philipe's result.

Freeze-out surface in terms of SPH degree of freedom
In term of SPH degree of freedom, one difficulty is that the surface element is not \tau = const. , while the Kernel functions are. However, the ideal of SPH algorithm is that one introduced a conserved charge in Cartesian coordinate so that


 * $$\begin{align}

(n {u^{\mu}})_{,\mu}=0 \end{align}$$

By Gauss theorem, this implies that


 * $$\begin{align}

\int d\sigma_{\mu}(n {u^{\mu}})= 0 \end{align}$$

or (under certain idealizations)


 * $$\begin{align}

\int_{\tau=\tau_0} d\sigma_{\mu}( n {u^{\mu}})= \int_{freeze-out} d\sigma_{\mu}( n {u^{\mu}}) =const. \end{align}$$

Then one splits the total conserved charge into small quantities and assign them to each SPH particle, which will be conserved during the evolution. Mathematically, this implies


 * $$\begin{align}

d\sigma_{\mu (\tau=\tau_0)}^i(n {u^{\mu}})=d\sigma^i_{\mu (freeze-out)}( n {u^{\mu}}) =\nu^i \end{align}$$

The superscript $$i$$  indicates the index of an SPH particle,  $$\nu^i$$  is the conserved quantity assigned to it from the beginning. The first term in above expression is evaluated on a small piece of $$\tau = \tau_0$$  surface, so that one has  $$nu^0 \tau_0 dxdyd\eta = \nu^i$$  but this is somewhat irrelevant if one assigns a specific value  $$\nu^i$$  by hand. The second piece of above equality actually tells us how to do the surface integral in terms of SPH degree of freedom. We write the surface element as a normal vector $$n_{\mu}$$  multiplying into module, one has


 * $$\begin{align}

n_{\mu}|d\sigma_{\mu}|( n {u^{\mu}}) =\nu^i   |d\sigma_{\mu}|=\frac{\nu^i}{n_{\mu}( n {u^{\mu}}) } \end{align}$$

So that


 * $$\begin{align}

d\sigma_{\mu}=\frac{n_{\mu}\nu^i}{n_{\mu}( n {u^{\mu}}) } =\frac{d\sigma_{\mu}\nu^i}{d\sigma_{\mu}( n {u^{\mu}}) } \end{align}$$

in the right hand side of above expression $$d\sigma_{\mu}$$  is the surface element we calculated above, but not necessarily normalized to the "size" of SPH particle, since its module cancels out in numerator and denominator. So one can calculate the same quantity $$p^{\mu}d\sigma_{\mu}$$  in terms of SPH degree of freedom on any freeze-out surface. If $$\nu^i =1$$  as used by Philipe


 * $$\begin{align}

p^{\mu}d\sigma_{\mu}  =\frac{p^{\mu}d\sigma_{\mu}\nu^i}{d\sigma_{\mu}( n {u^{\mu}}) }   =\frac{p^{\mu}d\sigma_{\mu}}{d\sigma_{\mu}( n {u^{\mu}}) } \end{align}$$

with


 * $$\begin{align}

d\sigma_{\mu}= \left( \begin{array}{cccc} 1 \\ -\partial_x\tau \\ -\partial_y\tau \\ -\partial_\eta\tau \end{array}\right) \end{align}$$


 * $$\begin{align}

p^{\mu} ={\Lambda^{\mu}}_{\nu} p^{\nu}_C  =\left( \begin{array}{cccc} m_T\cosh Y\cosh\eta-m_T \sinh Y\sinh\eta \\ p_T \cos\phi \\ p_T \sin\phi \\ -(1/\tau) m_T\cosh Y\sinh\eta+(1/\tau) m_T \sinh Y\cosh\eta \end{array}\right) \end{align}$$


 * $$\begin{align}

u^{\mu} =\left( \begin{array}{cccc} u^0 \\ u^1 \\ u^2 \\ u^3 \end{array}\right) \end{align}$$

which is different from Philipe's result.

Scaling invariance and integral in rapidity
When considering the scaling invariance, one has $$\partial_\eta \tau =0$$  and the integral in  $$\eta$$  can be carried out analytically...