Usually, when we apply the non-relativistic limit ($c \rightarrow \infty$) to relativistic equations, the cosmological constant $\Lambda \sim \mathrm{L}^{-2}$ is simply offhandedly neglected by putting $\Lambda = 0$. I don't remember having saw other limits defined to $\Lambda$. However, it is theoretical possible to consider that this constant could depend on $c$ in such a way that a part of it could survive the non-relativistic limit. As basic examples, consider the following three "reparametrisations": \begin{equation} \Lambda = \begin{cases} \text{case $\mathcal{A}$ :} \quad \Lambda_0 \sim \mathrm{L}^{-2}. \\[12pt] \text{case $\mathcal{B}$ :} \quad \displaystyle{\frac{\Lambda_0}{c^2}}, \qquad \text{where $\Lambda_0 \sim \mathrm{T}^{-2}$.} \\[12pt] \text{case $\mathcal{C}$ :} \quad \displaystyle{\frac{\Lambda_0}{c^4}}, \qquad \text{where $\Lambda_0 \sim \mathrm{A}^2$ (squared acceleration).} \end{cases} \end{equation} The vacuum mass density is defined by the following expression: \begin{equation} \rho_{\text{vac}} = \frac{\Lambda c^2}{8 \pi G} = \begin{cases} \displaystyle{\frac{\Lambda_0 c^2}{8 \pi G}} \sim \frac{\mathrm{M}}{\mathrm{V}}. \qquad \text{There's no non-relativisitc limit in this case, if $\Lambda_0 \ne 0$.} \\[12pt] \displaystyle{\frac{\Lambda_0}{8 \pi G}} \sim \frac{\mathrm{M}}{\mathrm{V}}, \qquad \text{$\Lambda_0$ survives the non-relativistic limit.} \\[12pt] \displaystyle{\frac{\Lambda_0}{8 \pi G c^2}} \sim \frac{\mathrm{M}}{\mathrm{V}}, \qquad \text{$\Lambda_0$ is removed under the non-relativistic limit.} \end{cases} \end{equation} I guess case $\mathcal{C}$ should be the most natural formulation to properly apply the non-relativistic limit to Einstein's equation and recover Newton's gravitation theory. In this case, the parameter $\Lambda_0 \sim \mathrm{A}^2$ is associated to a purely relativistic phenomenon.

I now wonder if this kind of analysis has been discussed before and would like to know some references on it.

Can we define other limits as well (non-relativistic $c \rightarrow \infty$, non-gravitational $G \rightarrow 0$ or non-quantum $\hbar \rightarrow 0$ limits), by extracting more powers of $c$, and also extracting appropriate factors of $G$ and $\hbar$ from $\Lambda$? The resulting "classical" $\Lambda_0$ could then have some interesting interpretation.


1 Answer 1


Dark energy acts like a term $(\Lambda/8\pi)g^{\mu\nu}$ in the stress-energy tensor. Nonrelativistic matter ("dust"), by definition, is a stress-energy tensor that, in Minkowski coordinates in its rest frame, has the stress-energy tensor $\operatorname{diag}(\rho,0,0,0)$. There is no frame in which the metric looks like $\operatorname{diag}(\text{const.},0,0,0)$, and therefore there is no nonrelativistic description of dark energy.

This is similar to the fact that there is no nonrelativistic description of electromagnetic waves. Note that you don't get a nonrelativistic theory of electromagnetism by letting $c\rightarrow0$. For more on this kind of thing, see:

Le Bellac M and Levy-Leblond J M 1973, "Galilean electromagnetism," Nuov. Cim. B 14 217-233

Marc De Montigny, Germain Rousseaux, "On the electrodynamics of moving bodies at low velocities," http://arxiv.org/abs/physics/0512200

  • $\begingroup$ We still can allow the cosmological tensor to have a non-trivial limit in Einstein's equation, by using the case $\mathcal{B}$ that I defined. Poisson's equation gets a $\Lambda_0$ term in this case. "Dark energy" (disguised as $\Lambda_0$) can enter a non-relativistic equation. $\endgroup$
    – Cham
    Dec 30, 2018 at 18:28
  • $\begingroup$ The Schwarzschild-deSitter metric in the weak field limit gives \begin{equation}g_{tt} = 1 - \frac{2 G M}{r c^2} + \frac{\Lambda}{3} \, r^2 \approx 1 + 2 \frac{\phi}{c^2}.\end{equation} Using $\Lambda = \Lambda_0 / c^2$ gives the newtonian potential \begin{equation}\phi = -\, \frac{G M}{r} + \frac{\Lambda_0}{6} \, r^2,\end{equation} where $\Lambda_0 \sim \mathrm{T}^{-2}$. $\endgroup$
    – Cham
    Dec 30, 2018 at 19:15

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