EFE with cosmological constant

In summary, "EFE with cosmological constant" refers to the Einstein Field Equations (EFE) of general relativity, which incorporate a cosmological constant (Λ) to account for the accelerated expansion of the universe. The cosmological constant represents a uniform energy density filling space homogeneously, influencing the dynamics of cosmic structures. Its inclusion modifies the equations, allowing for solutions that describe various cosmological models, such as de Sitter and anti-de Sitter spaces, which help explain observations like the cosmic microwave background and galaxy formation.
  • #1
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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  • #2
Just grind out the Einstein tensor, subtract ##\Lambda g_{ab}##, then divide by ##8\pi G/c^4##.
 
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  • #3
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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  • #4
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.
 
  • #5
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.
 
  • #6
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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  • #7
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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FAQ: EFE with cosmological constant

What is the Einstein Field Equation with a cosmological constant?

The Einstein Field Equation (EFE) with a cosmological constant is a modification of Einstein's original field equations of General Relativity. It is given by \( G_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu} \), where \( G_{\mu\nu} \) is the Einstein tensor, \( \Lambda \) is the cosmological constant, \( g_{\mu\nu} \) is the metric tensor, \( G \) is the gravitational constant, \( c \) is the speed of light, and \( T_{\mu\nu} \) is the stress-energy tensor.

What is the significance of the cosmological constant in the EFE?

The cosmological constant \( \Lambda \) represents an energy density filling space homogeneously. It was originally introduced by Einstein to allow for a static universe, but it is now understood to be related to dark energy, which is responsible for the accelerated expansion of the universe.

How does the cosmological constant affect the solutions to the EFE?

The inclusion of the cosmological constant changes the dynamics of spacetime. For example, in cosmological models, it can lead to solutions that describe an accelerating universe. In the context of black holes, it can modify the Schwarzschild solution to the Schwarzschild-de Sitter or anti-de Sitter solutions, depending on the sign of \( \Lambda \).

Why was the cosmological constant originally introduced by Einstein?

Einstein introduced the cosmological constant to his field equations to achieve a static universe, which was the prevailing belief at the time. He added \( \Lambda \) to counteract the attractive force of gravity and achieve a stable, unchanging universe. However, after the discovery of the universe's expansion by Edwin Hubble, Einstein reportedly referred to the introduction of the cosmological constant as his "biggest blunder."

What is the current understanding of the cosmological constant in modern cosmology?

In modern cosmology, the cosmological constant is associated with dark energy, which constitutes about 68% of the total energy density of the universe. It is responsible for the observed acceleration in the expansion of the universe. The exact nature of dark energy remains one of the biggest mysteries in physics.

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