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Time slicing and apparent horizons

To treat the collision of particles as time evolutional process, we consider the following slice of spacetime:

  region I$\displaystyle : t=z, t\le T,$    
  region II$\displaystyle : z=-t, t\le T,$ (9)
  region III$\displaystyle : t=T, -T\le z\le T,$    

where $ T\le 0$ and particles collide at $ T=0$. In order to find apparent horizon on the above slice, we first prepare surfaces with zero expansion in region I and III, then connect them smoothly by requiring that the null normal coincides at the junction of region I and III, $ t=z=T$. In region I, the surface which have zero expansion is given by

$\displaystyle v=-\Phi+$const.$\displaystyle ,$ (10)

and its null normal $ k_1^a$ is

  $\displaystyle k_1^u=\left({\rho_0}/{\rho}\right)^{-(D-3)},$    
  $\displaystyle k_1^v=\left({\rho_0}/{\rho}\right)^{D-3},$ (11)
  $\displaystyle k_1^\rho=1.$    

In region III, the surface which have zero expansion is given by

$\displaystyle az=\pm f(a\rho),$ (12)

where $ a$ is a constant of integration determined by the matching condition at the junction. For $ D=4$, the function $ f(x)$ is given by

$\displaystyle f(x)=\cosh^{-1} x,$ (13)

and for $ D>4$,

$\displaystyle f(x)$ $\displaystyle =-\frac{x^{-D+4}}{D-4} {}_2F_1\left(\frac{1}{2},\frac{D-4}{2(D-3)}, \frac{3D-10}{2(D-3)},x^{2(3-D)}\right)$    
  $\displaystyle \qquad\qquad\qquad\qquad -\sqrt{\pi}\frac{\varGamma\left(\frac{D-4}{2(D-3)}\right)}{\varGamma\left(\frac{1}{2(3-D)}\right)},$ (14)

where $ {}_2F_1$ is the Gauss' hyper-geometric function. The null normal $ k_3^a$ of the surface is given by

  $\displaystyle k_3^u=(a\rho)^{D-3}-\sqrt{(a\rho)^{2(D-3)}-1},$    
  $\displaystyle k_3^v=(a\rho)^{D-3}+\sqrt{(a\rho)^{2(D-3)}-1},$ (15)
  $\displaystyle k_3^\rho=1.$    

Matching these surfaces and null normals at the junction $ t=z=T$, we have

$\displaystyle f(a\rho_b)$ $\displaystyle =-aT,$ (16)
$\displaystyle \left({\rho_0}/{\rho_b}\right)^{D-3}$ $\displaystyle =(a\rho_b)^{D-3}+\sqrt{(a\rho_b)^{2(D-3)}-1}.$ (17)

where $ \rho _b$ is the radius of the surface at the junction. From this, the relation between $ T$ and $ \rho _b$ can be given parametrically as

  $\displaystyle \frac{T}{\rho_0}=-\xi f\left(\frac{1}{\xi}(2\xi^{3-D}-1)^{1/2(3-D)}\right) ,$ (18)
  $\displaystyle \frac{\rho_b}{\rho_0}=\left(2\xi^{3-D}-1\right)^{1/2(3-D)}.$ (19)

where $ 0\le\xi\le 1$.
Figure 1: The relation between $ T$ and the horizon radius $ \rho _b$ for $ D=4,\cdots , 11$. The intersection of the dotted line and $ T/\rho _0$-axis is the time $ T_{c}$ when the apparent horizon appears. The value of $ \vert T_c/\rho _0\vert$ decreases as $ D$ increases.
\includegraphics[width=0.5\linewidth]{fig1.eps}
FIG. 1 shows the relation between $ T$ and $ \rho _b$ for each $ D$. We denote the time when the apparent horizon appears as $ T=T_{c}$. The value of $ \vert T_c/\rho _0\vert$ becomes small as $ D$ increases. For large $ D$, we have

$\displaystyle \rho_b/\rho_0\approx D^{-1/2D},\quad T_c/\rho_0\approx -1/D$ (20)

at $ T=T_c$. The intersection of the $ z=$const. plane and the surface in region III is a $ D-3$-dimensional sphere, of which expansion is positive and proportional to $ D-3$. Thus the surface has negative expansion on $ (\rho/\rho_0,z/\rho_0)$-plane and its curvature on this plane increases with the increase of space-time dimension $ D$. This leads to the decrease in the distance of two particles at the horizon formation. The shape of apparent horizons for $ D=4$ and $ D=5$ are shown in FIG. 2 and FIG. 3.

Figure 2: The apparent horizon for $ D=4$ at $ T/\rho _0=-0.278,-0.225,0$. The dark line is the horizon at $ T=T_c=-0.278\rho _0$, and light line is the horizon at $ T=0$. The unit of the axis is $ \rho _0$.
\includegraphics[width=0.5\linewidth]{fig2.eps}


next up previous
Next: Hoop conjecture Up: High-energy head-on collisions of Previous: High-energy particle collisions at
Yasusada Nambu
2002-08-23