EFE with cosmological constant

Given the metric
$$

how would the Einstein field equations be spelled out algebraicly for the energy density and radial and tangent pressures in terms of the unknown functions A, B, and C, while also including a cosmological constant $$?

Just grind out the Einstein tensor, subtract $$, then divide by $$.

Ibix said:

Just grind out the Einstein tensor, subtract $$, then divide by $$.

Okay, so let's see. From what you said and from what little bit I can find on Wiki, if the energy density and isotropic pressure without the cosmological constant are $$ and $$, then with the cosmological constant, these become just $$ and $$ respectively? Could it be that simple? To be clear, this applies to both the exterior vacuum metric and the interior non-vacuum metric of a body?

grav-universe said:

From what you said and from what little bit I can find on Wiki

Why are you looking at Wikipedia? The place to learn how to solve the EFE is GR textbooks. If the computations get too cumbersome to do by hand, there are computer tools like Maxima that can help you out.

It seems like you are trying to build a house when your only tool is a small screwdriver.

grav-universe said:

if the energy density and isotropic pressure without the cosmological constant are $$ and $$, then with the cosmological constant, these become just $$ and $$ respectively?

That is one way to treat the cosmological constant, yes: as part of the stress-energy tensor. Basically this means moving the $$ term from the LHS to the RHS of the EFE. But you still need to compute the Einstein tensor from the metric you wrote down.

PeterDonis said:

It seems like you are trying to build a house when your only tool is a small screwdriver.

And I don't think Wiki uses a consistent sign convention, so it's a screwdriver that switches between flathead and Phillips with no warning.

grav-universe said:

Okay, so let's see. From what you said and from what little bit I can find on Wiki, if the energy density and isotropic pressure without the cosmological constant are $$ and $$, then with the cosmological constant, these become just $$ and $$ respectively? Could it be that simple?

I don't really know what you mean by "energy density and isotropic pressure without the cosmological constant...[and]...with the cosmological constant". The point is that if you believe that the cosmological constant is a property of spacetime then it makes sense to have the $$ term separate from the stress-energy tensor. If you believe it is a result of some kind of dark energy then you should include it in the stress energy tensor. In the first case, $$ and $$ are the density and pressure and the subscripted versions are something you can compute but isn't really physically meaningful. In the second case, $$ and $$ are the density and pressure and the unsubscripted versions are the contributions from everything except dark energy.

And yes, that bit is simple. And the only challenge in generating the Einstein tensor is the book-keeping. However, once you've done it you get three (actually four, but one will be clearly degenerate) simultaneous second order differential equations relating your $$, $$ and $$ to your $$, $$ and $$. That's where it starts getting difficult.

Ibix said:

so it's a screwdriver that switches between flathead and Phillips with no warning.

That would be so cool! I want one!

You find the Schwarzschild solution including a non-zero cosmological constant here:

https://itp.uni-frankfurt.de/~hees/art-ws16/lsg06.pdf