# (0,0) einstein tensor

1. Mar 19, 2010

### alejandrito29

in a paper (hep-th/9905012)says that de (0,0) einstein tensor is

$$G_{00}=3(\frac{\dot{a^2}}{a^2}-\frac{n^2}{b^2}[\frac{a''}{a}+\frac{a'^2}{a^2}-\frac{a'b'}{ab}])$$ (eq1)

and $$T_{00}=\frac{\rho\delta(y) n(t,y)^2 }{b(t,y)}$$

where

$$a=a_0+(\frac{|y|}{2}-\frac{y^2}{2})[a']_0-\frac{y^2[a']_{1/2}}{2}$$ (eq2)

and $$a''=[a']_0 (\delta(y)-\delta(y-1/2) ) + ([a']_0 +[a']_{1/2} ) (\delta(y-1/2) - 1 ) )$$

with $$[a']_0$$ is the jump of a detivate of a in y=0....

other relation are:

$$\frac{[a']_0}{a_0b_0}=-\frac{k^2 \rho}{3}$$ (Eq 2.5)

$$b=b_0+2|y|(b_{1/2}-b_0)$$ (Eq3)

i don't understan why the (0,0) component of Einstein equation at y=0 is:

$$\frac{\dot{a_0^2}}{a_0^2}=\frac{n_0^2}{b_0^2}(-\frac{[a']_{1/2}}{a_0}-\frac{b_{1/2}[a']_0}{b_0a_0}+\frac{[a']_0^2}{4a_0^2})$$

with $$a_0=a(t,y=0)$$

i calculate $$k^2T_{00}=-\frac{3n_0^2[a']_0\delta(y)}{b_0^2a_0}$$

but my mind question is that, i don't understand how i evaluate $$a'_0$$ end $$b'_0$$...if i derivate Eq 2, the $$\frac{d|y|}{dy}$$ are not defined on y=0¡¡¡¡¡¡