Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

(0,0) einstein tensor

  1. Mar 19, 2010 #1
    in a paper (hep-th/9905012)says that de (0,0) einstein tensor is

    [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]


    [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¡¡¡¡¡¡
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted