- #1

Sissy

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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 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