question

[alejandrito20]

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

 

 

[math]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}])[/math] (eq1)

 

 

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

 

where

 

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

 

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

 

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

 

other relation are:

 

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

 

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

 

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

 

[math] \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})[/math]

 

 

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

 

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

 

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

[Amr Morsi]

The R.H.S. of Eq. 2 is only a function in y, the rest are constants. So, differentiate it with respect to y, and then substitute a time with y=0 and another time with y=1/2. You will get 2 relations for the constants a'o and a'1/2 and thus get both in terms of ao.

 

The same can be done with eq. 3.

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I think you will have to integrate Eq.1 and T00 with respect to y in order to get rid of the delta function present in T00. Integrate it from -epsilon to +epsilon and thus some term in G00 will vanish and then substitute with the constants-relations you obtained previously.

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If Eq. 2 is NOT differentiable at y=0, try to integrate the equation following it(containing a''), it is the origin of Eq. 2...... It will be easy to evaluate.