Deriving the Cosmological Constant

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I am trying to find a paper which explains the derivation of the Cosmological Constant. I looked several books and sources and it only says " The cosmological constant ##Λ## appears in the Friedmann equation as an extra term" or etc. Sometimes It directly puts it in the Friedmann Equation which that's kind of awkward.

In one book it derived using the Poisson’s equation but is there a kind of simpler way ?
I remember a form of derivation which it was simple and using only integrals and integral constant was appeared as ##Λ##.

Anyone can help me on this ?
Thanks
 

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  • #2
TeethWhitener
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The Einstein field equations were inspired by combining local conservation of the stress-energy tensor (##\nabla_{\nu}T^{\mu\nu} = 0##) with the Bianchi identity (##\nabla_{\nu}G^{\mu\nu} = 0##), where ##G^{\mu\nu} = R^{\mu\nu} - \frac{1}{2}Rg^{\mu\nu}##. This means that if you take the covariant derivative of the Einstein equations:
$$G^{\mu\nu} = \frac{8\pi G}{c^4}T^{\mu\nu}$$
both sides equal zero. The cosmological constant comes from noticing that the covariant derivative of the metric is zero for a Levi-Civita connection ##\nabla_{\nu}g^{\mu\nu} = 0##, so adding a term proportional to the metric does not change the status of local stress-energy conservation:
$$\nabla_{\nu}G^{\mu\nu} = \nabla_{\nu}(G^{\mu\nu}+\Lambda g^{\mu\nu}) = 0$$
So the cosmological constant can be thought of as a constant of integration.
 
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  • #3
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The Einstein field equations were inspired by combining local conservation of the stress-energy tensor (##\nabla_{\nu}T^{\mu\nu} = 0##) with the Bianchi identity (##\nabla_{\nu}G^{\mu\nu} = 0##), where ##G^{\mu\nu} = R^{\mu\nu} - \frac{1}{2}Rg^{\mu\nu}##. This means that if you take the covariant derivative of the Einstein equations:
$$G^{\mu\nu} = \frac{8\pi G}{c^4}T^{\mu\nu}$$
both sides equal zero. The cosmological constant comes from noticing that the covariant derivative of the metric is zero for a Levi-Civita connection ##\nabla_{\nu}g^{\mu\nu} = 0##, so adding a term proportional to the metric does not change the status of local stress-energy conservation:
$$\nabla_{\nu}G^{\mu\nu} = \nabla_{\nu}(G^{\mu\nu}+\Lambda g^{\mu\nu}) = 0$$
So the cosmological constant can be thought of as a constant of integration.
That's really nice but as an freshman student I cant understand that. I remember more simpler explanation but wish I could find it..
 
  • #4
TeethWhitener
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That's really nice but as an freshman student I cant understand that. I remember more simpler explanation but wish I could find it..
Basically, it's a generalization of the fact that taking the derivative of a function ##f(x)## gives the same answer as taking the derivative of ##f(x)+C##, where ##C## is a constant, since the derivative of a constant is zero. But in this case, the covariant derivative of the metric tensor is zero (##\nabla_{\nu}g^{\mu\nu} = 0##), so adding the metric tensor to the Einstein tensor doesn't change its covariant derivative. The cosmological constant is then simply the proportionality constant of the metric tensor.
 
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  • #5
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Basically, it's a generalization of the fact that taking the derivative of a function ##f(x)## gives the same answer as taking the derivative of ##f(x)+C##, where ##C## is a constant, since the derivative of a constant is zero. But in this case, the covariant derivative of the metric tensor is zero (##\nabla_{\nu}g^{\mu\nu} = 0##), so adding the metric tensor to the Einstein tensor doesn't change its covariant derivative. The cosmological constant is then simply the proportionality constant of the metric tensor.
I guess I understand the idea, since, ##\nabla_{\nu}g^{\mu\nu} = 0## we can add any constant in front of ##g^{\mu\nu}## and it will not change anything in the equation. So, we add ##Λ##. Then we take the integral ##Λ## term appears as constant ?

Also in here, The derivation is a bit different (or maybe same, I am not sure). Since Einstein added the ##Λ## to "create a static universe", as in the site said,
We use the fluid equation $$\dot {p}+3H(ρ+P)=0$$

And for ##\dot {p}=0## since density cannot change with the time and ##H## must be constant so we get ##P=-ρ## which simply states that ##w=-1##. So I understand this part I guess but why there's the always 3 in the friedmann equation. ##\frac {Λ} {3}## ? Actually I looked and the density of the ##Λ## shown as,
##ρ_Λ=\frac {Λ} {8πG}## in that case its understandable that where three comes from (just not type ##Λ## in the denisty form) so I should probably ask where, ##ρ_Λ=\frac {Λ} {8πG}## comes from ?
 
  • #6
TeethWhitener
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where, ρΛ=Λ8πGρΛ=Λ8πGρ_Λ=\frac {Λ} {8πG} comes from ?
I'm not really an expert here, but I think it's just a convention to get the units of energy density to agree with the cosmological constant units:
https://en.wikipedia.org/wiki/Einstein's_constant
As far as I can tell, many cosmologists opt to use ##\Omega_{\Lambda}## instead (roughly speaking, the fraction of total energy due to dark energy), since ##\Omega_{\Lambda}## is unitless.
 
  • #7
kimbyd
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That's really nice but as an freshman student I cant understand that. I remember more simpler explanation but wish I could find it..
Have you learned any series expansions?

One of the ways to derive the Einstein equations stems from a series expansion. The idea there is that you can mathematically prove that there is only one thing which a reasonable theory of gravity can depend upon: the curvature. So, the idea is that gravity must stem from some function of the curvature ##R##. You could represent this as a series expansion:

[tex]f(R) = a_0 + a_1 R + a_2 R^2 + a_3 R^3 + ...[/tex]

There's a caveat with the higher powers of ##R## which I'll describe below, but for now the important point is that if you were to try to write down a simple theory of gravity, you could just take the lowest terms. The question is: how many do you need? If you just take the constant value, ##a_0##, you end up with a theory that is overly-simple and can't explain gravity at all. But if you add the linear term, you get General Relativity. The higher-order terms are possible (this is an active area of research, and as I understand it some theories like string theory predict that these will be nonzero).

So, if you look at gravity like a series expansion, the constant term is automatically there just based upon how it's built up. That term becomes the cosmological constant.

Now for the caveat: the curvature itself is defined by rank-4 Ricci curvature tensor, ##R_{\mu\nu\sigma\tau}##. Tensors depend upon the coordinate system you use, and it doesn't make sense to describe the theory in terms of something that depends upon a coordinate system. The terms that appear in the power series expansion above, then, are the Ricci scalar. The Ricci scalar stems from contracting the indices in a particular way to give a single value that is independent of coordinates. This is unique for the ##a_1 R## term, but it turns out there are many ways to get a ##R^2## term, so that those are actually a heck of a lot more complicated than just squaring ##R##: they're actually more complex contractions of the ##R_{\mu\nu\sigma\tau}## tensor.

Long story short: the mathematics of adding terms beyond what General Relativity uses is really, really complicated.
 
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  • #8
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@kimbyd that's really good explanation thanks.

I'm not really an expert here, but I think it's just a convention to get the units of energy density to agree with the cosmological constant units:
Thanks for your help @TeethWhitener

If that's all thats kind of sad..or I should say I expected more then that..maybe it comes from the Poisson’s equation (I dont know exactly what the equation means (it says gravitational potential energy though) but we didnt learn it in our lectures..not yet at least.).

In the book of Ryden Introduction to cosmology (page 73), it writes like
$$∇^2Φ+Λ=4πGρ$$
since ##Λ## term in there appears to make a static universe. It explains in a more detail in the book.
it continues like "...Introducing Λ into Poisson’s equation allows the universe to be static if you set ##Λ = 4πGρ##..." Hows that possible ?

Since
from the ##ρ_Λ=\frac {Λ} {8πG}## we can write

##Λ=ρ_Λ{8πG}## but I couldnt quite make the connection,

If ##∇^2Φ=0## then we should say ##Λ=ρ_Λ{4πG}## then where the 8 come from ?
 
  • #9
TeethWhitener
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@kimbyd that's really good explanation thanks.



Thanks for your help @TeethWhitener

If that's all thats kind of sad..or I should say I expected more then that..maybe it comes from the Poisson’s equation (I dont know exactly what the equation means (it says gravitational potential energy though) but we didnt learn it in our lectures..not yet at least.).

In the book of Ryden Introduction to cosmology (page 73), it writes like
$$∇^2Φ+Λ=4πGρ$$
since ##Λ## term in there appears to make a static universe. It explains in a more detail in the book.
it continues like "...Introducing Λ into Poisson’s equation allows the universe to be static if you set ##Λ = 4πGρ##..." Hows that possible ?

Since
from the ##ρ_Λ=\frac {Λ} {8πG}## we can write

##Λ=ρ_Λ{8πG}## but I couldnt quite make the connection,

If ##∇^2Φ=0## then we should say ##Λ=ρ_Λ{4πG}## then where the 8 come from ?
I don't quite understand what's going on here. This all looks Newtonian. @kimbyd (or another cosmology guru) is much more likely to be helpful.

##\nabla^2\phi=0## is Laplace's equation, which implies empty space (at least within the region of interest). So essentially what this seems to be saying is that if we choose a cosmological constant that is exactly equal to the matter density within a space, we'll get a gravitational potential that looks like empty space in that region. I'm not sure why that's important.

As for the 8 vs. the 4, again, I think it's probably some convention. If you take the Newtonian limit of the Einstein equations, you get ##g_{00} = -1-2\phi## as the relationship between the metric tensor and the Newtonian gravitational potential, so maybe that's where the extra factor of 2 gets picked up. I dunno. Someone more knowledgeable should be able to clear things up.
 
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  • #10
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∇2ϕ=0∇2ϕ=0\nabla^2\phi=0 is Laplace's equation, which implies empty space (at least within the region of interest). So essentially what this seems to be saying is that if we choose a cosmological constant that is exactly equal to the matter density within a space, we'll get a gravitational potential that looks like empty space in that region. I'm not sure why that's important.
In the same book it says,

"The only permissible static universe, in this analysis, is a totally empty universe. If you create a matter-filled universe which is initially static, then gravity will cause it to contract. If you create a matter-filled universe which is initially expanding, then it will either expand forever (if the Newtonian energy U is greater than or equal to zero) or reach a maximum radius and then collapse (if U < 0). Trying to make a matter-filled universe which doesn’t expand or collapse is like throwing a ball into the air and expecting it to hover there.

How did Einstein surmount this problem? How did he reconcile the fact that the universe contains matter with his desire for a static universe? Basically, he added a fudge factor to the equations. In Newtonian terms, what he did was analogous to rewriting Poisson’s equation in the form..."

then comes the part that I asked in the question.
 
  • #11
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As for the 8 vs. the 4, again, I think it's probably some convention. If you take the Newtonian limit of the Einstein equations, you get g00=−1−2ϕg00=−1−2ϕg_{00} = -1-2\phi as the relationship between the metric tensor and the Newtonian gravitational potential, so maybe that's where the extra factor of 2 gets picked up.
Maybe yeah..I dont know too :)
 
  • #12
kimbyd
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@kimbyd that's really good explanation thanks.



Thanks for your help @TeethWhitener

If that's all thats kind of sad..or I should say I expected more then that..maybe it comes from the Poisson’s equation (I dont know exactly what the equation means (it says gravitational potential energy though) but we didnt learn it in our lectures..not yet at least.).

In the book of Ryden Introduction to cosmology (page 73), it writes like
$$∇^2Φ+Λ=4πGρ$$
since ##Λ## term in there appears to make a static universe. It explains in a more detail in the book.
it continues like "...Introducing Λ into Poisson’s equation allows the universe to be static if you set ##Λ = 4πGρ##..." Hows that possible ?

Since
from the ##ρ_Λ=\frac {Λ} {8πG}## we can write

##Λ=ρ_Λ{8πG}## but I couldnt quite make the connection,

If ##∇^2Φ=0## then we should say ##Λ=ρ_Λ{4πG}## then where the 8 come from ?
It's because when you derive the field equations from the above equation in terms of the gravitational potential, the matter term picks up an additional factor of 2 while the ##\Lambda## term does not.

In full General Relativity you don't have this problem because General Relativity also includes pressure terms. You can simply model ##\Lambda## as being a fluid with pressure equal to minus its energy density satisfying the equation ##\Lambda = 8\pi G \rho_\Lambda##. Newtonian mechanics has no concept of pressure having an effect on gravity, so it has weirdness like this when you look at the potential rather than the field equations. In other words, a cosmological constant in the Newtonian limit is a hack that can be complicated to deal with correctly.
 
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  • #13
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It's because when you derive the field equations from the above equation in terms of the gravitational potential, the matter term picks up an additional factor of 2 while the ##\Lambda## term does not.

In full General Relativity you don't have this problem because General Relativity also includes pressure terms. You can simply model ##\Lambda## as being a fluid with pressure equal to minus its energy density satisfying the equation ##\Lambda = 8\pi G \rho_\Lambda##. Newtonian mechanics has no concept of pressure having an effect on gravity, so it has weirdness like this when you look at the potential rather than the field equations. In other words, a cosmological constant in the Newtonian limit is a hack that can be complicated to deal with correctly.
It all makes sense now. Thanks!

I guess thats all for now.
 
  • #14
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Now for the caveat: the curvature itself is defined by rank-4 Ricci curvature tensor, ##R_{\mu\nu\sigma\tau}##.
This is the Riemann tensor, the Ricci is rank 2.

Tensors depend upon the coordinate system you use, and it doesn't make sense to describe the theory in terms of something that depends upon a coordinate system.
The components of a tensor are dependant on the coordinate system the tensor itself isn't.

Cheers
 
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