How to calculate cosmological constant?

In summary: OPs question is no.In summary, the cosmological constant can be determined from the curvature of space through the Einstein tensor, using the Einstein Field Equations and experimental data. It is not related to the radii of curvature and cannot be calculated solely from the metric.
  • #1
Ans
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I reading book, part about gravity. It is Landau-Lifschiz, "Theory of field". And I trying to understand how to calculate cosmological constant. Re-read part with introduction of cosmological constant several times, still not understood.
Let's imagine function of space curvature, without presence of mass, is known. In such case it is easy to calculate Cristoffel symbols, Richi tensor etc. But how to find value of cosmological constant from them?
I have impression what cosmological constant = 1/(R1*R2), where R1 and R2 are main radiuses of curvature. Is it correct?
 
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  • #2
Ans said:
Let's imagine function of space curvature, without presence of mass, is known

I'm not sure I understand what this function is supposed to be. Is it the metric? Is it the stress-energy tensor? Is it something else?
 
  • #3
PeterDonis said:
I'm not sure I understand what this function is supposed to be. Is it the metric? Is it the stress-energy tensor? Is it something else?
I saying about geometry, so metric.
 
  • #4
Ans said:
I saying about geometry, so metric.

Then you compute the Einstein tensor of that metric, and divide it by ##8 \pi G / c^4## to convert it to a stress-energy tensor, in accordance with the Einstein Field Equation. If that stress-energy tensor has a part that is a constant multiple of the metric, then that is the cosmological constant. (More precisely, it's minus the cosmological constant, since it's on the RHS of the EFE.)
 
  • #5
PeterDonis said:
Then you compute the Einstein tensor of that metric, and divide it by ##8 \pi G / c^4## to convert it to a stress-energy tensor, in accordance with the Einstein Field Equation. If that stress-energy tensor has a part that is a constant multiple of the metric, then that is the cosmological constant. (More precisely, it's minus the cosmological constant, since it's on the RHS of the EFE.)

If tensor divide by something, it still would be tensor.
If calculate cosmological constant from metric, it is equal to
[itex]\Omega=\frac{1}{2} g^{lm} g^{ik} R_{iklm}[/itex]
However, I think the equation is too complex. May it be found from curvature of space, radiuses of curvature?
 
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  • #6
Ans said:
If calculate cosmological constant from metric, it is equal to
[itex]\Omega=\frac{1}{2} g^{lm} g^{ik} R_{iklm}[/itex]
However, I think the equation is too complex. May it be found from curvature of space, radiuses of curvature?

I don't understand how you could come up with that expression from the metric (which metric?)... But if you take advantage of the symmetries ##g_{ij}=g_{ji}## and ##R_{ijkl}=-R_{jikl}## that expression simplifies down to zero.
 
  • #7
Ans said:
If tensor divide by something, it still would be tensor.

Yes, just with different units--stress-energy units in this case, instead of curvature units.

Ans said:
If calculate cosmological constant from metric

That's not what I said. I said, if the stress-energy tensor is a constant multiple of the metric, i.e., ##\Lambda g_{ab}## with ##\Lambda## a constant. You don't have to "calculate" a constant from the metric or anything else.
 
  • #8
Is it possible to identify a CC from the metric alone ? I mean, if the stress energy tensor is not otherwise constrained, you can always add a term ## \Lambda g_{ab} ## to both sides, is there a condition that identifies a preferred value of ## \Lambda ##?
 
  • #9
wabbit said:
Is it possible to identify a CC from the metric alone ?

Yes. You do what I described in post #4. If you know the metric, you know its Einstein tensor (just compute it). If the Einstein tensor is a multiple of the metric, then that multiple is the cosmological constant. (The bit in post #4 about dividing by ##8 \pi G / c^4## is just to change the units, as I noted in post #7. You can tell whether you have a multiple of the metric without doing that.)

For example, take the metric for de Sitter spacetime (there are actually several different ones, depending on which chart you choose). Compute the Einstein tensor of that metric. You will find that it is a multiple of the metric--i.e., as matrices, the Einstein tensor is just some ##\Lambda## times the metric tensor. That multiple ##\Lambda## is the cosmological constant (with a minus sign).

wabbit said:
if the stress energy tensor is not otherwise constrained, you can always add a term ##\Lambda g_{ab}## to both sides

Yes, but that changes the Einstein tensor, which changes the metric--i.e., the metric that gives rise to this modified Einstein tensor is different from the metric that gave rise to the original Einstein tensor (before you added ##\Lambda g_{ab}## ).
 
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  • #10
What I meant is, the metric ## g ## gives you the Einstein tensor ## G ##, and the EFE says ## G+\Lambda g = 8\pi T ## . If ## (\Lambda, T) ## is a solution, so is ## (\Lambda+x,T+\frac{x}{8\pi} g) ##. My question was, is there a condition that fixes x ?

(I do not argue with your statement that if the Einstein tensor is precisely a multiple of the metric, then there is a natural choice with T=0, but this is about the general case - is there a natural choice there too ? )
 
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  • #11
wabbit said:
If ##(\Lambda, T)## is a solution, so is ##(\Lambda+x,T+\frac{x}{8\pi} g)## . My question was, is there a condition that fixes x ?

Experimental data does, because experimental data tells you the stress-energy tensor. Your transformation changes the stress-energy tensor from ##T## to ##T + g x / 8 \pi##.

To put it another way, ##\Lambda## has a physical meaning: it tells you the degree to which the Einstein tensor fails to vanish if ##T = 0##, i.e., if there is no matter or energy present (other than ##\Lambda## itself). That can be measured experimentally, though of course it isn't easy (the only measurement of it we have currently is the data that tells us the expansion of the universe is accelerating). And once you have that value for ##T = 0##, you use the same value in regions of spacetime where ##T \neq 0##.
 
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  • #12
OK so going back to OP original question, I suppose we can say

- the CC is unrelated to radii of curvature,

- you can determine it from the curvature, through Einstein's tensor, using the EFE, provided you have a little bit of information about the stress-energy tensor and its sources. If for instance the sources are known/assumed to be matter and radiation, it is enough to measure G anywhere far away from any source and use the fact that T=0 there : the Einstein tensor as a whole is proportional to the metric at any such point, and ##\Lambda=-\frac{G_{00}}{g_{00}} ##.
 
  • #13
wabbit said:
the CC is unrelated to radii of curvature

I'm not sure I would say that, since, as you note, you can determine the CC from the curvature. You just need to measure the curvature in a region where there are no sources present.
 
  • #14
Ah yes you're right, thanks for the correction.

This would be a much better, coordinate-free way to express the CC than using ## \Lambda=-\frac{G_{00}}{g_{00}} ## at such a point.

I am not sure what the exact result is, is it as simple as CC=scalar curvature ?

And even if it's not exactly practical, I like this recipe for measuring the CC : go to outer space, find a really quiet place away from any matter, electric or magnetic field and such, and measure the sum of the angles, and surface, of a triangle. A little easy formula and here's your CC : )
 
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  • #15
wabbit said:
I am not sure what the exact result is, is it as simple as CC=scalar curvature ?

Almost. The Ricci scalar ##R## is equal to ##4 \Lambda## for de Sitter spacetime.

wabbit said:
go to outer space, find a really quiet place away from any matter, electric or magnetic field and such, and measure the sum of the angles, and surface, of a triangle.

One triangle won't be enough, and they won't all be ordinary spatial triangles; remember that you're measuring spacetime curvature, not space curvature.
 
  • #16
PeterDonis said:
Almost. The Ricci scalar ##R## is equal to ##4 \Lambda## for de Sitter spacetime.
OK that's simple enough for me thanks : )
One triangle won't be enough, and they won't all be ordinary spatial triangles; remember that you're measuring spacetime curvature, not space curvature.
Spatial curvature should be OK though since we're at a point where T=0, no ? In suitable coordinates I suppose we must have for Ricci curvature, ## R=diag(-\Lambda,\Lambda,\Lambda,\Lambda) ## or something close to that so I thought one triangle would be enough. If not, one tetrahedron then ?
 
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  • #17
wabbit said:
Spatial curvature should be OK though since we're at a point where T=0, no ?

No. Space curvature is coordinate-dependent. Spacetime curvature is not.

For example, consider de Sitter spacetime in the flat slicing, i.e., each spatial slice has zero curvature. The spacetime curvature is not zero: as I said before, the Ricci scalar is ##R = 4 \Lambda##. But in this slicing, since the space curvature is zero, the only way to measure ##R## is to look at things that happen over time--for example, to see that initially "comoving" test particles diverge: they accelerate away from each other.
 
  • #18
Thanks, I see. No way around that de Sitter case, we do need both... OK I am now aiming a laser beam from one fixed point towards the 4 vertices of my tetrahedron and measuring the aiming angles required to hit the targets - that ought do do it this time - unless those zero-length paths break it somehow, then I'll throw billiard balls : )
 

1. How is the cosmological constant calculated?

The cosmological constant, denoted by the Greek letter lambda (λ), is calculated using the Einstein field equations. This equation relates the curvature of space-time to the matter and energy present in the universe. The value of lambda is determined by measuring the expansion rate of the universe and the density of matter and energy within it.

2. What units are used to measure the cosmological constant?

The units used to measure the cosmological constant are typically expressed in terms of energy density, such as joules per cubic meter (J/m^3) or electron volts per cubic meter (eV/m^3). This is because lambda represents the energy density of the vacuum of space-time.

3. Can the cosmological constant change over time?

According to the current understanding of the universe, the value of the cosmological constant is believed to be a constant and does not change over time. However, some theories suggest that it may vary in different regions of the universe or in different eras of the universe's history.

4. How does the cosmological constant affect the expansion of the universe?

The cosmological constant is a key factor in determining the expansion rate of the universe. A larger value of lambda would result in a faster expansion, while a smaller value would result in a slower expansion. The current accepted value of lambda is extremely small, meaning that the universe is expanding at an accelerating rate.

5. What implications does the cosmological constant have for the fate of the universe?

The value of the cosmological constant has significant implications for the ultimate fate of the universe. If the value of lambda remains constant or continues to increase, the universe will continue to expand at an accelerating rate and eventually reach a state of maximum entropy known as the "heat death" of the universe. However, if the value of lambda decreases over time, the expansion of the universe could slow down and eventually reverse, leading to a possible "big crunch" scenario.

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