in a paper (hep-th/9905012)says that de (0,0) einstein tensor is(adsbygoogle = window.adsbygoogle || []).push({});

[tex]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}])[/tex] (eq1)

and [tex]T_{00}=\frac{\rho\delta(y) n(t,y)^2 }{b(t,y)}[/tex]

where

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

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

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

other relation are:

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

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

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

[tex] \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})[/tex]

with [tex]a_0=a(t,y=0)[/tex]

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

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# (0,0) einstein tensor

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**