# Non-relativistic limit of the cosmological constant

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": $$$$\Lambda = \begin{cases} \text{case \mathcal{A} :} \quad \Lambda_0 \sim \mathrm{L}^{-2}. \\[12pt] \text{case \mathcal{B} :} \quad \displaystyle{\frac{\Lambda_0}{c^2}}, \qquad \text{where \Lambda_0 \sim \mathrm{T}^{-2}.} \\[12pt] \text{case \mathcal{C} :} \quad \displaystyle{\frac{\Lambda_0}{c^4}}, \qquad \text{where \Lambda_0 \sim \mathrm{A}^2 (squared acceleration).} \end{cases}$$$$ The vacuum mass density is defined by the following expression: $$$$\rho_{\text{vac}} = \frac{\Lambda c^2}{8 \pi G} = \begin{cases} \displaystyle{\frac{\Lambda_0 c^2}{8 \pi G}} \sim \frac{\mathrm{M}}{\mathrm{V}}. \qquad \text{There's no non-relativisitc limit in this case, if \Lambda_0 \ne 0.} \\[12pt] \displaystyle{\frac{\Lambda_0}{8 \pi G}} \sim \frac{\mathrm{M}}{\mathrm{V}}, \qquad \text{\Lambda_0 survives the non-relativistic limit.} \\[12pt] \displaystyle{\frac{\Lambda_0}{8 \pi G c^2}} \sim \frac{\mathrm{M}}{\mathrm{V}}, \qquad \text{\Lambda_0 is removed under the non-relativistic limit.} \end{cases}$$$$ I guess case $$\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.

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:
• 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. – Cham Dec 30 '18 at 18:28
• The Schwarzschild-deSitter metric in the weak field limit gives $$g_{tt} = 1 - \frac{2 G M}{r c^2} + \frac{\Lambda}{3} \, r^2 \approx 1 + 2 \frac{\phi}{c^2}.$$ Using $\Lambda = \Lambda_0 / c^2$ gives the newtonian potential $$\phi = -\, \frac{G M}{r} + \frac{\Lambda_0}{6} \, r^2,$$ where $\Lambda_0 \sim \mathrm{T}^{-2}$. – Cham Dec 30 '18 at 19:15