The cosmological constant $\Lambda$ can be introduced into the gravitational action like this : \begin{equation} S = \frac{1}{2 \kappa} \int_{\Omega} (R - 2 \Lambda) \sqrt{-g} \; d^4 x + \text{matter terms}. \end{equation} The spacetime region $\Omega$ is arbitrary here. Now, what amaze me is that we can also write this : \begin{equation}\tag{1} -\: \frac{\Lambda}{8 \pi G} \int_{\Omega} \sqrt{-g} \; d^4 x = -\: \frac{\Lambda \, \mathcal{V}_4}{8 \pi G}, \end{equation} where $\mathcal{V}_4$ is the 4-volume of the spacetime region $\Omega$. So, the cosmological constant $\Lambda$ may be interpreted as the conjugate "variable" to $\mathcal{V}_4$, and as a Lagrange multiplier associated to a 4-volume constraint. We could suppose that since $\mathcal{V}_4$ should be very large and the action $S$ "reasonable", then $\Lambda$ should be small. My intuition tells me that there should be an equation like this one : \begin{equation}\tag{2} \Lambda \, \mathcal{V}_4 \sim \Lambda_{\text{max}} \mathcal{V}_{\text{min}}, \end{equation} where $\mathcal{V}_{\text{min}}$ is the smallest 4-volume that is physically meaningfull ; $\mathcal{V}_{\text{min}} \approx \ell_{\text{P}}^4$ ($\ell_{\text{P}}$ is the Planck length), and $\Lambda_{\text{max}} \approx \ell_{\text{P}}^{-2}$ is the "natural" value associated to the quantum vacuum. We then get \begin{equation}\tag{3} \Lambda \sim \frac{\ell_{\text{P}}^2}{\mathcal{V}_4}, \end{equation} which is thus very small. The relation $\Lambda \, \mathcal{V}_4 \approx \ell_{\text{P}}^2 \propto \hbar$ is also similar to an Heisenberg uncertainty relation ; $\Delta t \, \Delta E \ge \hbar$, which isn't surprising since the cosmological constant is introduced at the level of the action !

Can we make the previous idea more "rigorous" ? Does it make sense to interpret $\Lambda$ as a Lagrange multiplier associated to a constrained 4-volume introduced into the action ?

If the universe is spatially closed ($k = 1$) and also closed in time (especially if $\Lambda$ was negative), then the 4-volume of the whole universe would be finite.

Any opinion on this ?


1 Answer 1


Just a partial and naive "answer", using the uncertainty principle.

An observer makes an energy mesurement in some empty volume $V_3$ during a time intervall $\Delta t$. According to Heisenberg uncertainty principle, he wil get an uncertainty $\Delta E$ on the energy mesurement : \begin{align} \Delta t \; \Delta E &\approx \Delta t \; \rho_{\text{vac}} \, \Delta V_3 \\[12pt] &= \Delta t \; \frac{\Lambda \, c^4}{8 \pi G} \; \Delta V_3 \\[12pt] &= \frac{\Lambda \; \Delta\mathcal{V}_4 \; c^3}{8 \pi G} \ge \frac{\hbar}{2}, \end{align} where I have inserted the uncertainty on 4-volume of the spacetime region the observer is studying ; $\Delta\mathcal{V}_4 = c \, \Delta t \; \Delta V_3$. Thus \begin{equation}\tag{1} \Lambda \; \Delta \mathcal{V}_4 \ge 4 \pi \, \ell_{\text{P}}^2. \end{equation} Apparently, the cosmological constant $\Lambda$ depends on the size of the spacetime region that is sampled. This is weird !

We could also invert the result, saying that given some experimental value of $\Lambda$, then the accesible portion of spacetime to any observer would have an uncertainty constrained by \begin{equation}\tag{2} \Delta\mathcal{V}_4 \ge \frac{4 \pi \, \ell_{\text{P}}^2}{\Lambda_{\text{exp}}}. \end{equation} The Planck length is $\ell_{\text{P}} \approx 1.6 \times 10^{-35} \text{m}$. The current value of the cosmological constant is $\Lambda_{\text{exp}} \sim 10^{-52} \text{m}^{-2}$, which gives the smallest uncertainty on the 4-volume : \begin{equation}\tag{3} \Delta \mathcal{V}_{4 \, \text{min}} \sim 3 \times 10^{-17} \text{m}^4 \sim (0.0757 \text{mm})^4. \end{equation} (Note : the 4-volume of our observable universe is $\mathcal{V}_4 = c \, \Delta t \; \mathcal{V}_3 \sim (5 \times 10^{26} \text{m})^4$)

The problem with this "answer" is that it doesn't say why $\Lambda$ could be interpreted as a Lagrange multiplier to be added to the total action of the universe.

  • $\begingroup$ There's a mistake in the reasoning above : the volume $V_3$ and 4-volume $\mathcal{V}_4$ should be replaced with their uncertainties ; $\Delta V_3$ and $\Delta \mathcal{V}_4$, respectively. Then the $\Delta\mathcal{V}_4 \sim 10^{-17} \text{m}^4$ is making sense, actually ! I'll make the changes later. $\endgroup$
    – Cham
    Commented Nov 28, 2015 at 17:23

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