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Einstein initially added the Cosmological Constant because (if I get this right) it seemed to him that the universe should be static. I agree that back then this would have been an obvious assumption. I'm curious now, before Hubble, where there any opinions/debates about whether the universe would be expanding or contracting?

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<a href="'_paradox">Olber's paradox</a> is a pretty decent argument that they universe is not static, uniform and eternal, and was known in the 19th Century. – Jerry Schirmer Mar 14 '11 at 18:27
I think before Einstein space was not dynamic so the question about static or not would have been strange. Space was the scene where things were happening. – MBN Mar 14 '11 at 18:43
In which equation should Newton add a constant? – Georg Mar 14 '11 at 19:47
Was there even a notion of structure and dynamics of the cosmos outside the solar system at the time? Obviously universal gravitation implies the possibility of these things, but I don't recall reading that anyone was thinking about them. If there was, well, the solar system is stabilized against collapse by angular momentum; so put the star far enough away and make their mass low enough, and compute their proper motion... – dmckee Mar 14 '11 at 19:52
@Georg-- In the gravitational force equation, or equivalently in Laplace's equation, of course. Where else? It appears that Newton did believe in an infinite, homogeneous, static Universe (e.g., , ). In that case, it follows that there must be a modification of the law on large scales. I thought Newton didn't realize this but Laplace did, but according to some of the answers below I'm wrong: Newton did realize it. – Ted Bunn Mar 16 '11 at 1:02

It's a very good question but the answer is that Newton's Universe actually doesn't have to expand or contract so no cosmological constant is needed. Well, it's a bit more subtle.

The right non-relativistic gravitational equation where one should add the vacuum energy density is the Poisson equation $$ \nabla^2 \phi_g = 4\pi G \rho + \Lambda_{{\rm Newton}}. $$ I added a Newtonian cosmological constant term. For a Newtonian cosmology with a uniform mass distribution at the cosmological scale (e.g. above hundreds of megaparsecs), you actually have to add this term (with a negative sign), to neutralize the mass density at the very long distance scale. If you omit this term, the $\phi_g$ potential has to have a Laplacian with a constant term, so $\phi_g$ itself will have to contain something like $\vec x^2$ which will inevitably be minimized at some point of the Universe - $\vec x =0$ in my conventions.

Amusingly enough, one may describe Newton's gravitational forces without any $\phi_g$, by manifestly summing the forces from other point masses in the Universe. It's an equivalent approach to calculate the acceleration but it allows us to avoid the problem with the preferred point in the Universe. I may just claim that the forces acting on the Earth that are caused by very distant objects cancel. This is equivalent to saying that the Earth is the $\vec x = 0$ point - and one may say the same thing for any other object in the Universe (a method to regulate the infrared divergences from the forces caused by very distant objects).

Needless to say, the assumption that we choose a "uniform cutoff" around every probe in the Universe is totally equivalent to adding the neutralizing Newtonian cosmological constant above. Also, you won't be able to invent any non-equivalent yet consistent Newtonian cosmologies with a uniform Universe but a nonzero cosmological constant. That's really because the Newtonian spacetime is flat and the cosmological constant is the curvature of the empty spacetime - which vanishes in Newton's theory by definition.

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This is correct, but should read, "uniform, static Universe". If you want a spatially uniform universe, you just have to do the same trick that you do in GR to go from Oppenheimer-Snyder collapse to a flat FRW cosmology--start with the former, and let the initial radius of the dust cloud go to infinity. – Jerry Schirmer Mar 14 '11 at 20:40
The right equation should be of the form $\Delta\phi+\lambda\phi=\rho$, which gives a Yukawa-type potential. This alteration seems to have been considered to resolve Zeeliger's paradox (which is essentially the gravitational analogue of Olbers' paradox.) – timur Apr 19 '11 at 6:00

This is really a good question and indeed questions were raised at that time about how Newton's universe could give rise to a static universe if gravity is always attractive. Newton argued, if the universe were infinite in extent and matters were distributed more or less uniformly throughout this infinite universe then collapse could be avoided since there would be no center for the universe to collapse. However, this argument is not flawless since one can show that this kind of universe will be highly unstable. A slight perturbation can break the balance and inevitably lead to collapse of the universe. Surprisingly no body argued at that time that the universe could be expanding (or contracting). Instead some people really tried to modify Newton's law of gravity so that gravity may be repulsive at large distances. So you see, a kind of repulsive gravity was indeed proposed much earlier than GR.

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Newton used this as a proof that the universe was infinite.

If all the matter in the universe is attracting every other bit of matter then the universe should collapse into a single point. He reasoned that the only way around this was if the universe was infinite, then every bit of matter would have an equal force in every direction - from the attraction of every other object around it.

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In the 1800's, astronomers believed in an island universe model, in which our galaxy was alone in infinite space. In a Newtonian framework, this actually seems like a superior model, since it can be dynamically stable against radial gravitational collapse. The real problem is the second law of thermodynamics, which says that the universe can't be infinitely old -- but that didn't come until ca. 1850. (Once they understood some thermodynamics, they could also reason that stars would gradually evaporate from the galaxy because the velocity distribution has tails.) – Ben Crowell Oct 19 '13 at 19:30
This brought to mind that if in Newton's times the concept of manifold would be sufficiently understood, the same could be achieved with a finite (even flat) universe - for example, a three-torus like $\mathbb R^3/\mathbb Z^3$ – მამუკა ჯიბლაძე Jan 19 '15 at 4:57

Newton also assumed that the speed of gravity is infinite. This led him to further support his idea of an infinite universe. If he had somehow known that the "c" is finite, then he would have eventually reasoned that the universe can not be static.

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