EFE with cosmological constant

  • Context: Undergrad 
  • Thread starter Thread starter grav-universe
  • Start date Start date
  • Tags Tags
    Energy density
Click For Summary
SUMMARY

The discussion focuses on the formulation of the Einstein Field Equations (EFE) with a cosmological constant (Λ) using a specific metric. The energy density (ρ) and isotropic pressure (p) are modified to ρ_{(Λ)} = ρ + Λ/(8πG/c^4) and p_{(Λ)} = p - Λ/(8πG/c^4), respectively. Participants emphasize the importance of using General Relativity (GR) textbooks over Wikipedia for accurate computations and suggest utilizing computational tools like Maxima for complex calculations. The conversation highlights the necessity of deriving the Einstein tensor from the given metric to establish relationships between pressure, density, and the functions A, B, and C.

PREREQUISITES
  • Understanding of General Relativity (GR) principles
  • Familiarity with Einstein Field Equations (EFE)
  • Knowledge of cosmological constants and their implications
  • Experience with computational tools like Maxima for tensor calculations
NEXT STEPS
  • Study the derivation of the Einstein tensor from specific metrics
  • Learn about the implications of the cosmological constant in GR
  • Explore the use of Maxima for solving complex differential equations in GR
  • Review GR textbooks for detailed methodologies on EFE computations
USEFUL FOR

Physicists, mathematicians, and students of General Relativity seeking to deepen their understanding of the Einstein Field Equations, particularly in the context of cosmological constants and their effects on energy density and pressure.

grav-universe
Messages
460
Reaction score
1
Given the metric
$$c^2 d\tau^2 = B(r) c^2 dt^2 - A(r) dr^2 - C(r) r^2 d\Omega^2$$

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 ##\Lambda##?
 
Last edited:
Physics news on Phys.org
Just grind out the Einstein tensor, subtract ##\Lambda g_{ab}##, then divide by ##8\pi G/c^4##.
 
  • Like
Likes   Reactions: vanhees71
Ibix said:
Just grind out the Einstein tensor, subtract ##\Lambda g_{ab}##, then divide by ##8\pi G/c^4##.
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 ##\rho## and ##p##, then with the cosmological constant, these become just ##\rho_{(\Lambda)} = \rho + \Lambda / (\frac{8 \pi G}{c^4})## and ##p_{(\Lambda)} = p - \Lambda / (\frac{8 \pi G}{c^4})## 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?
 
Last edited:
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 ##\rho## and ##p##, then with the cosmological constant, these become just ##\rho_{(\Lambda)} = \rho + \Lambda / (\frac{8 \pi G}{c^4})## and ##p_{(\Lambda)} = p - \Lambda / (\frac{8 \pi G}{c^4})## respectively?
That is one way to treat the cosmological constant, yes: as part of the stress-energy tensor. Basically this means moving the ##\Lambda## 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 ##\rho## and ##p##, then with the cosmological constant, these become just ##\rho_{(\Lambda)} = \rho + \Lambda / (\frac{8 \pi G}{c^4})## and ##p_{(\Lambda)} = p - \Lambda / (\frac{8 \pi G}{c^4})## 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 ##\Lambda g_{ab}## 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, ##\rho## and ##p## are the density and pressure and the subscripted versions are something you can compute but isn't really physically meaningful. In the second case, ##\rho_{(\Lambda)}## and ##p_{(\Lambda)}## 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 ##p##, ##\rho## and ##\Lambda## to your ##A##, ##B## and ##C##. That's where it starts getting difficult.
 
Last edited:
Ibix said:
so it's a screwdriver that switches between flathead and Phillips with no warning.
That would be so cool! I want one!
 
  • Haha
Likes   Reactions: vanhees71

Similar threads

  • · Replies 26 ·
Replies
26
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
732
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 0 ·
Replies
0
Views
2K