Theory of relativity/Friedmann–Lemaître–Robertson–Walker metric

In the 1920s-1930s Friedmann, Lemaître, Robertson, and Walker independently modeled a universe with the premise that the matter in it would be observed to be uniform from anywhere in it. Since they did so independently and expressed their results with different choices of coordinates there are different coordinate expressions for the resulting line element and different sources will refer to the line element with different subsets of these names depending on who the source was relying upon as its reference matterial. RW and FRW are very common. The line element expressed in a very common choice of coordinates can be written

$$ds^2 = dct^2 - a^2 \left(ct\right)\frac{\left(d\rho ^2 + \rho ^2 d\theta ^2 + \rho ^2 sin^2 \theta d\phi ^2 \right)}{\left(1+\frac{1}{4}k\frac{\rho ^2 }{R_{0}^2}\right)}$$

where k = -1 is the case of a negatively curved universe, k=0 is the case of a flat universe, and k = 1 is the case of a positively curved universe.

The flat universe
Observations of the angular size of fluctuations in the cosmic microwave background radiation are consistent with flat universe models. As such the FLRW model best supported by the evidence is $$ds^2 = dct^2 - a^2 \left(ct\right)\left(d\rho ^2 + \rho ^2 d\theta ^2 + \rho ^2 sin^2 \theta d\phi ^2 \right)$$

Dark Energy
General relativity/Einstein equations with the inclusion of a nonzero cosmological constant are $$G^{\mu \nu}-\lambda g^{\mu \nu}=kT^{\mu \nu}$$

Where $$k=\frac{8\pi G}{c^4}$$ and this latter G is the gravitational constant of the universe.

A positive cosmological constant yields accelerating universe solutions and is sometimes refered to as "dark energy". Such an acceleration term has been observed in the expansion of the universe and so there is evidence that there exists a positive $$\lambda $$, dark energy. So lets look at the flat universe FLRW line element for the case of a positive cosmological constant. First let us define a function $$f$$ of time as $$f\left(ct\right) = a^2 \left(ct\right)$$

Then the Einstein tensor for this line element can be written as $$ \frac{3}{4}\frac{\left(\frac{df}{dct}\right)^2 }{f^2}    & 0 & 0 & 0 \\ 0 & -\frac{1}{4}\frac{\left(\frac{df}{dct}\right)^2 -4f\frac{d^2 f}{dct^2 }}{f^2}     & 0   & 0    \\ 0 & 0   & -\frac{1}{4}\frac{\left(\frac{df}{dct}\right)^2 -4f\frac{d^2 f}{dct^2 }}{f^2}      & 0   \\ 0 & 0  & 0    & -\frac{1}{4}\frac{\left(\frac{df}{dct}\right)^2 -4f\frac{d^2 f}{dct^2 }}{f^2} \end{bmatrix} $$
 * G ^{\mu } _{\nu}|| = \begin{bmatrix}

And yields a Ricci-scalar of $$R =-3\frac{\left(\frac{d^2 f}{dct^2 }\right)}{f} =-4\lambda $$

The solution from this equation for the Ricci-scalar for a positive cosmological constant is $$f = Asinh\left(2\sqrt{\frac{\lambda}{3}}ct\right)+Bcosh\left(2\sqrt{\frac{\lambda}{3}}ct\right)$$

Putting the big bang at $$t=0$$ yields $$B=0$$, and choosing A appropriately results in the flat FLRW line element for a positive cosmological constant as $$ds^2 = dct^2 - \left(\frac{sinh\left(2\sqrt{\frac{\lambda}{3}}ct\right)}{sinh\left(2\sqrt{\frac{\lambda}{3}}cT\right)}\right)\left(d\rho ^2 + \rho ^2 d\theta ^2 + \rho ^2 sin^2 \theta d\phi ^2 \right)$$

where $$T$$ is the age of the universe in this choice of time coordinate. The resulting Einstein tensor is $$ \lambda+\frac{\lambda }{sinh^2 \left(2\sqrt{\frac{\lambda}{3}}ct\right)}    & 0 & 0 & 0 \\ 0 & \lambda-\frac{1}{3}\frac{\lambda }{sinh^2 \left(2\sqrt{\frac{\lambda}{3}}ct\right)}     & 0   & 0    \\ 0 & 0   & \lambda-\frac{1}{3}\frac{\lambda }{sinh^2 \left(2\sqrt{\frac{\lambda}{3}}ct\right)}      & 0   \\ 0 & 0  & 0    & \lambda-\frac{1}{3}\frac{\lambda }{sinh^2 \left(2\sqrt{\frac{\lambda}{3}}ct\right)} \end{bmatrix} $$
 * G ^{\mu } _{\nu}|| = \begin{bmatrix}

The dark energy tensor part of this then is $$ \lambda    & 0 & 0 & 0 \\ 0 & \lambda    & 0   & 0    \\ 0 & 0   & \lambda      & 0   \\ 0 & 0  & 0    & \lambda \end{bmatrix} $$
 * g ^{\mu } _{\nu}\lambda || = \begin{bmatrix}

And the stress-energy tensor of the electromagnetic radiation in the universe is then given by $$ \frac{c^4 }{8\pi G}\frac{\lambda }{sinh^2 \left(2\sqrt{\frac{\lambda}{3}}ct\right)}    & 0 & 0 & 0 \\ 0 & -\frac{c^4}{24\pi G}\frac{\lambda }{sinh^2 \left(2\sqrt{\frac{\lambda}{3}}ct\right)}     & 0   & 0    \\ 0 & 0   & -\frac{c^4}{24\pi G}\frac{\lambda }{sinh^2 \left(2\sqrt{\frac{\lambda}{3}}ct\right)}      & 0   \\ 0 & 0  & 0    & -\frac{c^4}{24\pi G}\frac{\lambda }{sinh^2 \left(2\sqrt{\frac{\lambda}{3}}ct\right)}      \end{bmatrix} $$
 * T ^{\mu } _{\nu}|| = \begin{bmatrix}

which for very small times near the big bang becomes $$ \frac{3c^2}{32\pi G t^2}    & 0 & 0 & 0 \\ 0 & -\frac{c^2}{32\pi G t^2}    & 0   & 0    \\ 0 & 0   & -\frac{c^2}{32\pi G t^2}          & 0   \\ 0 & 0  & 0    & -\frac{c^2}{32\pi G t^2} \end{bmatrix} $$
 * T ^{\mu } _{\nu}|| = \begin{bmatrix}

The behavior of $$a\left(ct\right)$$ was derived for a Ricci-scalar equation above modeling the universe to contain only dark energy and electromagnetic radiation. In the early history of the real universe, much of the electromagnetic radiation made a phase transition to ordinary matter and dark matter. As such the amount of electromagnetic radiation actually observed left over in the cosmic microwave background radiation is only about 10-5 of the electromagnetic radiation energy density represented in this model.

Hubble's "Constant"
A ruler distance between comoving galaxies may be defined by $$ L=a\int _{0} ^{\rho }\frac{d\rho }{\left(1+\frac{1}{4}k\frac{\rho ^2 }{R_{0} ^{2}}\right)}$$

Such that galaxies at constant coordinate position travel away from each other with a ruler velocity of $$ v=\frac{dL}{dt}$$

One can write this $$ v=HL$$

where H is a function of time referred to as Hubble's "constant" and is given by $$ H=\frac{\left(\frac{da}{dt}\right)}{a}$$

For the above flat space model we have $$ H=c\sqrt{\frac{\lambda }{3}}\frac{cosh\left(2\sqrt{\frac{\lambda }{3}}ct\right)}{\sqrt{sinh\left(2\sqrt{\frac{\lambda }{3}}ct\right)sinh\left(2\sqrt{\frac{\lambda }{3}}cT\right)}}$$

The current value for Hubble's constant is then $$ H_{0}=c\sqrt{\frac{\lambda }{3}}coth\left(2\sqrt{\frac{\lambda }{3}}cT\right)$$

where T is the age of the universe. Our measured value of Hubble's constant in 1/years is $$ H_{0}=\left(6.934\pm0.079\right)\times 10^{-11}y^{-1}$$

Our estimate for the age of the universe is $$ T=\left(13.798\pm 0.037\right)\times 10^{9}y$$

Using these in the model and a function solver we can make a prediction for the value of the cosmological constant. In units of one over the square of light years we then find that it is $$ \lambda \sim 10^{-20}ly^{-2}$$

Which in SI units is $$ \lambda \sim 10^{-52}m^{-2}$$