EFE with cosmological constant

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Discussion Overview

The discussion revolves around the formulation of the Einstein field equations (EFE) with a cosmological constant in the context of a specific metric. Participants explore how to express energy density and pressures in terms of the metric functions, while considering the implications of including the cosmological constant.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asks how to algebraically express the EFE for energy density and pressures using the metric functions A, B, and C, including a cosmological constant.
  • Another participant suggests calculating the Einstein tensor, adjusting for the cosmological constant, and dividing by a constant factor.
  • A participant proposes that the energy density and pressure with the cosmological constant can be expressed as modified versions of those without it, questioning if this approach is valid for both vacuum and non-vacuum metrics.
  • Some participants emphasize the importance of using reliable sources, such as GR textbooks, over Wikipedia for solving the EFE.
  • There is a discussion on whether the cosmological constant should be treated as part of the stress-energy tensor or as a separate property of spacetime, leading to different interpretations of the energy density and pressure.
  • One participant notes that generating the Einstein tensor leads to complex differential equations relating pressure, density, and the metric functions, indicating a challenging aspect of the problem.

Areas of Agreement / Disagreement

Participants express differing views on how to treat the cosmological constant and its implications for the energy density and pressure. There is no consensus on the best approach, and the discussion remains unresolved regarding the treatment of the cosmological constant.

Contextual Notes

Participants mention the potential for confusion arising from different sign conventions in literature, as well as the complexity of deriving the Einstein tensor from the given metric.

grav-universe
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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##?
 
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Just grind out the Einstein tensor, subtract ##\Lambda g_{ab}##, then divide by ##8\pi G/c^4##.
 
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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?
 
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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.
 
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Ibix said:
so it's a screwdriver that switches between flathead and Phillips with no warning.
That would be so cool! I want one!
 
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