Solving Einstein's Equation with Cosmological Constant

[Sissy]

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 Here is the exercise:


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

$$b = 1- \dfrac{2m}{r} - \dfrac{1}{3} \Lambda r^2$$

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

With

Equation (52.9) $$b = e^\beta = 1- \dfrac{2m}{r}$$

Equation (48.17) $$g_{44} = 1 + \dfrac{2U}{c^2}$$

$$U$$ is probably the Newtonian potential function.


I have no idea how to start.

Thank you for giving me some hints.

greetings

 

[Pengwuino]

Have you derived the Schwarzschild 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.

 

[Sissy]

In lecture we derived the Schwarzschild metric

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

and we derived

$$b = 1 - \dfrac{2GM}{c^2 r}$$

and called $$r_s = \dfrac{2GM}{c^2}$$ the Schwarzschild radius.

For the Einstein field equation we had

$$R_{ij} - \dfrac{1}{2} g_{ij} R = - \kappa T_{ij}$$

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

$$R_{ij} - \dfrac{1}{2} g_{ij} R = 0$$

From the Schwarzschild metric I got the metric tensor

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

And what shoul I contract now?

Thank you for helping me

Greetings

 

[Sissy]

Sorry, I made a mistake:

The field equation for empty space with cosmical constant ist

$$R_{ij} - \dfrac{1}{2} R g_{ij} + \Lambda g_{ij} = 0$$

And the exercise said, that $$R_{ij} + \Lambda g_{ij} = 0$$

So this means, that $$\dfrac{1}{2} R g_{ij}$$ is $$0$$

How this becomes zo zero?

Is my metric tensor correct?

Thanks for help
greetings

 

[paranormal]

Hello,

The first step in solving this problem would be to understand the concept of the cosmological constant and its role in Einstein's equation. The cosmological constant, denoted as Λ, is a term that Einstein added to his equation to account for the observed expansion of the universe. It is a constant value that represents the energy density of the vacuum of space. In the absence of any matter or energy, the cosmological constant acts as a repulsive force, causing the universe to expand at an accelerating rate.

To solve this problem, you will need to use the spherically symmetric Schwarzschild solution, which describes the gravitational field around a spherical, non-rotating mass. This solution is given by b = 1- \dfrac{2m}{r}, where m is the mass of the object and r is the distance from the center of the object. This solution does not take into account the cosmological constant.

To incorporate the cosmological constant, we need to modify the Schwarzschild solution to include the term Λr^2. This results in the equation b = 1- \dfrac{2m}{r} - \dfrac{1}{3} \Lambda r^2. This new equation is the solution to Einstein's equation in empty space, when the cosmological constant is taken into account.

Using this new solution, you can then show that there is an additional force of repulsion from the center, proportional to the radius r. This can be done by using the approximate equation g_{44} = 1 + \dfrac{2U}{c^2}, where U is the Newtonian potential function. This additional force is due to the presence of the cosmological constant and is responsible for the observed accelerating expansion of the universe.

I hope this helps to give you some hints on how to approach this problem. Good luck with your calculations!