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Cosmical Constant

  • Thread starter Sissy
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  • #1
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Hello,

I don't understand how to solve this problem. I do not want a solution because I will calculate it by my own but I need some hints how to start.

I don't know what to do :frown: Here is the exercise:


Show that, if the cosmical constant term is retained in Einstein's equation, it reduces in empty space to [tex]R_{ij}+\Lambda g_{ij} = 0[/tex]. Deduce that the spherically symmetric Schwarzschild solution (cf. equation (52.9)) is given by

[tex]b = 1- \dfrac{2m}{r} - \dfrac{1}{3} \Lambda r^2[/tex]

Using the approximate equation (48.17), show that this implies the existance of an additional force of repulsion from the centre proportional to the radius r.

With

Equation (52.9) [tex] b = e^\beta = 1- \dfrac{2m}{r} [/tex]

Equation (48.17) [tex] g_{44} = 1 + \dfrac{2U}{c^2} [/tex]

[tex]U[/tex] is probably the Newtonian potential function.


I have no idea how to start.

Thank you for giving me some hints.

greetings
 

Answers and Replies

  • #2
Pengwuino
Gold Member
4,989
15
Have you derived the Schwarzchild solution? Contract the field equations with the metric with upper indices to give you conditions on the Ricci scalar. Then the derivation should follow in the same manner.
 
  • #3
6
0
In lecture we derived the Schwarzschild metric

[tex] \mathrm{d} s^2 = \dfrac{\mathrm{d} r^2}{1 - (2m/r)} + r^2 ( \mathrm{d} \Theta ^2 + sin^2 \Theta \mathrm{d} \Phi^2 ) -c^2 \left( 1 - \dfrac{2m}{r} \right) \mathrm{d} t^2 [/tex]

and we derived

[tex] b = 1 - \dfrac{2GM}{c^2 r} [/tex]

and called [tex]r_s = \dfrac{2GM}{c^2}[/tex] the Schwarzschild radius.

For the Einstein field equation we had

[tex] R_{ij} - \dfrac{1}{2} g_{ij} R = - \kappa T_{ij} [/tex]

Because the space is empty in this exercise, the energy-momentum tensor vanish and I got the equation

[tex] R_{ij} - \dfrac{1}{2} g_{ij} R = 0 [/tex]

From the Schwarschild metric I got the metric tensor

[tex]g_{ij} = \begin{pmatrix}
\dfrac{1}{1- (2m/r)} & 0 & 0 & 0 \\
0 & r^2 & 0 & 0 \\
0 & 0 & r^2 ~ sin^2 \Theta & 0 \\
0 & 0 & 0 & -c^2 + \frac{c^2 2m}{r}
\end{pmatrix}
[/tex]

And what shoul I contract now?

Thank you for helping me

Greetings
 
  • #4
6
0
Sorry, I made a mistake:

The field equation for empty space with cosmical constant ist

[tex] R_{ij} - \dfrac{1}{2} R g_{ij} + \Lambda g_{ij} = 0 [/tex]

And the exercise said, that [tex]R_{ij} + \Lambda g_{ij} = 0 [/tex]

So this means, that [tex]\dfrac{1}{2} R g_{ij}[/tex] is [tex]0[/tex]

How this becomes zo zero?

Is my metric tensor correct?

Thanks for help
greetings
 

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