# Cosmological constant term in Newtonian gravity [duplicate]

Recently, I came across something I found quite interesting on Wikipedia, which is the addition of the cosmological constant to Newtonian gravity. The Wikipedia page (Alternatives to General Relativity) writes the modified Newton-Poisson equation as follows: $$\nabla^2 \phi + \frac{1}{2}\Lambda c^2 = 4 \pi\rho G .$$

It does not provide relevant sources for this, which has left me scratching my head. So I am wondering why we add the $$\frac{1}{2}\Lambda c^2$$ term instead of, say, just adding $$\Lambda$$. Is this some sort of limit of general relativity, or that this form is perhaps easier to deal with in some way?

P.S. I have at best limited knowledge of general relativity, and perhaps this is something obvious from the GR viewpoint which I have overlooked.

• Have you tried applying dimensional analysis? Commented Dec 15, 2022 at 16:47
• Commented Dec 15, 2022 at 17:14
• I haven't done the research to give a full answer, but I think it's chosen to be congruent with the original definition in Einstein's relativistic equation, en.wikipedia.org/wiki/Cosmological_constant#Equation Other bits: $c^2$ is chosen so that $\Lambda$ has the desired units (m^-2 ?) and because we need a speed to do that. $c$ is one of the most popular speeds, and is clearly relevant to the relativistic justification of $\Lambda$. The $\frac{1}{2}$ will come from an integration of some other equation, but I don't know which one. Commented Dec 15, 2022 at 19:48

Why $${1 \over2}$$ and why $${c^2}$$: the short answer is, so that you don't scratch your head in despair with splintered glass, when it comes to assigning actual numerical values in actual physical units to $$G$$ and $$\Lambda$$ (and yes, to $$\pi$$ as well).
The long answer is, have a look a bit further down, the same § of the WP article: they chute the GR field eq. as $$T^{\mu\nu} = {1 \over {8 \pi G}} (R^{\mu\nu} - {1 \over 2} g^{\mu\nu} R + g^{\mu\nu} \Lambda)$$. Which translates, in covariant coordinates, to $${4 \pi G} T_{\mu\nu} = {1 \over 2} R_{\mu\nu} - {1 \over 4} R + {1 \over 2} \Lambda$$.
Rings any bell? Well, it says to me: $$T_{00}$$ reduces to $$\rho$$, $$g_{00}$$ to $$\phi$$ and $${1 \over 2} R_{00} - {1 \over 4} R$$ to $$\nabla^2 \phi$$ in the Newtonian approximation. That's because, in the GR world, time measures in meters (at a wee bit under $$1.8 \times 10^{10}$$ m per minute); among normal people, $$T_{00}\rightarrow {\rho \over {c^2}}$$ and $${1 \over 2} R_{00} - {1 \over 4} R \rightarrow {1 \over {c^2}} \nabla^2 \phi$$.
That must be why the factor $${1 \over2}$$ and why the factor $${c^2}$$: so that $$G$$ and $$\Lambda$$ denote the two exact same things in two different theories, just like $$4$$ and $$\pi$$ do.