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Newtons theory of gravity involves a gravitational constant $G$, however one does not refer to it directly, we speak instead of gravity or the force of gravity.

Now Einsteins introduced a cosmological constant in his equation for GR, and ever since it's been referred to as such; however, this usage seems strange to me, given the above parallel; it seems to me we ought to be referring to something physical, which can then be quantified and whose quantification involves the cosmological constant.

Does the cosmological constant then refer to a cosmological force? A force that is, contra gravity, repulsive? And which is even weaker than gravity, kicking in on the scale of galaxies rather than solar systems?

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  • $\begingroup$ It's more fashionably known as 'dark energy' (disclaimer: only mat be the cosmological constant I think). $\endgroup$ – tfb Jan 16 at 23:13
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The cosmological constant is the answer to the question: what curvature does the universe have in the absence of matter and other forms of energy?

Starting from the Einstein field equations $$R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu}+\Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$ where $R_{\mu\nu}$ is the Ricci curvature tensor, $R=g^{\mu\nu}R_{\mu\nu}$ is the Ricci scalar, $g_{\mu\nu}$ is the metric, and $T_{\mu\nu}$ is the stress-energy tensor. Just drop in $T_{\mu\nu}=0$ and take the trace of both sides using $g^{\mu\nu}$ to get $$R=4\Lambda.$$

We have somewhat of a bias toward assuming that the curvature of space-time without matter and energy in it should be zero, and so we want to interpret $\Lambda$ as being a "dark energy." In reality, it could just be that the default curvature of space-time is not flat, for whatever reason. This should not be confused with the observed flatness of space, parameterized by $\Omega_k$ in most variations of the $\Lambda$CDM cosmologies, since $R$ is the overall flatness of space-time, not just space.

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  • $\begingroup$ When you say, "what curvature does the universe have in the absence of matter and other forms of energy? " does that not also mean you set $\Lambda = 0$? Why are you choosing to hold $\Lambda$ separately from "other forms of energy" ? $\endgroup$ – N. Steinle Jan 17 at 5:37
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    $\begingroup$ The question of energy in GR does not have a simple answer. Regarding $\Lambda$ as being a type of energy is not something that is automatic in the equations, but a choice we make that is one possible interpretation. $\endgroup$ – Sean E. Lake Jan 17 at 7:49
  • $\begingroup$ but the curvature parameter Ωk falls off with 1/a², while the dark energy parameter Ωλ stays constant, how does that fit together? $\endgroup$ – Yukterez Mar 28 at 4:00
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    $\begingroup$ @Yukterez First, the density parameters are defined to be constants independent of the scale factor, but you obviously meant their related energy densities. For that, you're confusing the curvature of space today (given by $k=H_0^2c^2 \Omega_k$) with the curvature of space-time as a whole (given by the $R=4\Lambda + \frac{8\pi G}{c^2}\left(\rho - \frac{3p}{c^2}\right)$ in a FLRW universe). $\endgroup$ – Sean E. Lake Mar 28 at 5:01
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Newtons theory of gravity involves a gravitational constant G, however one does not refer to it directly, we speak instead of gravity or the force of gravity.

Indeed! $G$ is a constant of proportionality that is (from Newton's gravitation law) determined experimentally by measuring accelerations and masses.

Here's a brief history of the cosmological constant problem, and an even briefer one here.

Does the cosmological constant then refer to a cosmological force? A force that is, contra gravity, repulsive?

Yes, however in GR most physicists don't use the word "force." Rather, as you said, the constant is gravity due to a negative pressure that couples to the metric. Einstein introduced the cosmological constant, $\Lambda$, in his steady-state cosmological model in order to counteract the gravitational collapse of the universe - at the time we hadn't yet observed outside of our own galaxy, so the universe did appear to be static, but Hubble's (and subsequent) discoveries proved the universe is much bigger. This made the cosmological constant "Einstein's biggest blunder," however it re-emerged in the 1990's with the discovery that the universe's expansion is accelerating. This is why it is fashionable these days to associate the cosmological constant with the concept of dark energy - perhaps the acceleration of the universe's expansion is due to a cosmological constant?

This all results in a conundrum called the cosmological constant problem (also called the vacuum catastrophe), in which the astrophysical value of $\Lambda$ differs from the quantum mechanical value by at least 120 orders of magnitude.

And which is even weaker than gravity, kicking in on the scale of galaxies rather than solar systems?

So, if you interpret the cosmological constant as the mechanism driving the universe's accelerated expansion, then one arrives at the following graph: enter image description here (image source)

This figure shows the main contributions to the energy density of the universe, where you can see that the energy in dark energy is a constant, which is small compared to the energy in matter and in radiation at small times. But as time increases, the energy density of matter and of radiation decreases below that of dark energy, meaning the acceleration of the universe's expansion dominates the growth of the universe.

Lastly, here's a nice article about how the cosmological constant acts on large scales, as you suggest, since relativistic GR treats each point of the manifold as a galaxy in space, from which I quote: "We have two competing effects: on the one hand, the overall mass of the universe tends to slow the expansion over time. On the other hand, the mysterious dark energy tends to accelerate it. Actually, the term dark energy is just a placeholder for a theory to come: it's a way of talking about the fact that the expansion is accelerating, an observational discovery. We don't (yet) understand why the expansion is accelerating, but it is."

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Now Einsteins introduced a cosmological constant in his equation for GR, and ever since it's been referred to as such; however, this usage seems strange to me, given the above parallel; it seems to me we ought to be referring to something physical, which can then be quantified and whose quantification involves the cosmological constant.

The Cosmological Constant first appeared, when Einstein (and other scientists in that period of time, before hubble law discovered) thought that the universe was static.

In the Newtonian context, We can write the relationship between mass density and gravitational potential by the Poisson's Equation,

$$\nabla^2 \Phi =4\pi G\rho\,\,\, (Eqn.1)$$

And gravitational acceleration can be written as,

$$\vec {a(t})=-\nabla \Phi\,\,\,$$

For static universe $a(t)$ has must be zero. But this means that $\nabla \Phi=0$, and from (Eqn.1) we see that $\rho=0$. Hence, the only allowable static universe is an empty universe.

To solve this problem Einstein added a term called cosmological constant in his equations. In Newtonian terms what we did was adding a constant to the Poisson's Equation and write it like,

$$\nabla^2 \Phi + \Lambda=4\pi G\rho\,\,\, (Eqn.2)$$ for $\Lambda=4\pi G\rho$ we see that universe becomes static with non-zero matter density.

In today's cosmology, cosmological constant described as a sort of constant energy density in the universe, which has constant negative pressure. Matter and radiation density decreases while the universe expands (by $a(t)^{-3}$ $a(t)^{-4}$ respectively). However, $\Lambda$ remains constant while the universe expands.

Does the cosmological constant then refer to a cosmological force? A force that is, contra gravity, repulsive? And which is even weaker than gravity, kicking in on the scale of galaxies rather than solar systems?

Cosmological constant is not a force, but it can be represented as a force, I solved a question like this in here couple days ago. If we modified the Newton's law of universal gravitation according to the cosmological constant, we get

$$F=-k/R^2+\Lambda mR/3$$ for $k=GMm$

For the question you can look here

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  • $\begingroup$ It's not just that Einstein thought the universe was static - before Hubble that's what all the evidence suggested was true. $\endgroup$ – N. Steinle Jan 17 at 5:41
  • $\begingroup$ Yes indeed, I meant that einstein introduced the idea of cosmological constant. I edited my post. $\endgroup$ – Reign Jan 17 at 6:35
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Does the cosmological constant then refer to a cosmological force?

The Cosmological Constant acts as “repelling” gravity. This can be seen from the Friedmann equations, here it works opposite to matter and radiation density due to the negative pressure it exerts.

This just means that the universe expands accelerated in case the CC dominates matter and radiation density (which is negligible today) in accordance to current observational data. But we still don’t talk about forces because General Relativity concerns geodesics and hence objects in free fall.

Forces come into play if we regard extended bodies and are called tidal forces then, known as “spaghettification” in the context of black holes. In our accelerated expanding universe (due to the CC) an extremely long rubber band would be stretched due to tidal forces which can be seen from second Friedmann equation also called acceleration equation.

In the ideal case, the perfect fluid assumption on which the FRL modell is based tidal acceleration is scale independent.

I’m quite late, not sure if you read that.

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