Prove that the value of the cosmological constant equals the energy density of the vacuum

Question

I know that Einstein introduced his cosmological constant assuming it as an independent parameter, something characteristic of the Universe, in itself, but the term of it in the field equations can be moved "to the other side" of equality, written as a component of the energy-tension tensor $T$ for vacuum:

$$T_{\mu\nu}^{(vac)}=-\cfrac{\Lambda c^4}{8\pi G}g_{\mu\nu}.$$

Since this result would correspond directly to the energy density $ρ$ in the energy-tension tensor $T$, this has the inevitable consequence that we are already talking about the vacuum energy given by the following relationship according to General Relativity:

$$\rho_{vac}=\cfrac{\Lambda c^2}{8\pi G}.$$

Thus, the existence of a cosmological constant Λ different from zero is equivalent to the existence of a vacuum energy different from zero; there is no way in which we can escape this conclusion. This is why the terms cosmological constant and vacuum energy are used interchangeably in General Relativity.

But I do not understand if it is correct to say that the cosmological constant and the energy density of the vacuum have the same value or how to prove that they actually have the same value, could someone help me?

Answer 1

The standard einstein field equation is given through: \begin{align*} R_{{\mu}{\nu}}-\frac{1}{2}{R}{g_{{\mu}{\nu}}}+{\Lambda}{g_{{\mu}{\nu}}}=\frac{{8}{\pi}{G}}{{c^{4}}}{T_{{\mu}{\nu}}} \end{align*} This can be interpreted as an equation of motion for the metric field. If we now consider a vacuum in which spacetime is flat, it concludes that the metric is therefore provided by the static minkowsky line element. However, a constant metric directly implies that all terms containing it's partial derivative must vanish, hence the Christoffel symbol and correspondingly to that, also the Riemann curvature tensor must be zero. Now we can obtain the vacuum field equation: \begin{align*} {T^{\Lambda}_{{\mu}{\nu}}}=\frac{{\Lambda}{c}^{4}}{{8}{\pi}{G}}{{\eta}_{{\mu}{\nu}}} \end{align*} The time-time component of the energy momentum tensor is the energy density, the space components are given by the pressure of the stress tensor. Now the resulting equations are: \begin{align*} \Lambda=\frac{{8}{\pi}{G}{\rho_{\Lambda}}}{{c}^{2}}{\qquad}{{\rho}_{\Lambda}}{{c}^{2}}+p_{\Lambda}=0 \end{align*} Q.E.D.