B Cosmological constant for the Universe's expansion

AI Thread Summary
The discussion centers on the cosmological constant's role in explaining the universe's expansion. It critiques the notion that a cosmological constant can balance gravitational forces, arguing that such a model misrepresents the dynamics of cosmic expansion. The conversation emphasizes the inadequacy of using Newtonian gravity in cosmological contexts, advocating for the use of general relativity and FLRW equations instead. It clarifies that Hubble's law describes the velocities of different objects at a single moment rather than providing a straightforward equation of motion. Ultimately, the dialogue underscores the need for a proper understanding of gravitational dynamics and the implications of a cosmological constant in cosmology.
jeremyfiennes
Messages
323
Reaction score
17
TL;DR Summary
Astronomers hold that a cosmological constant can account for thr universe's increasing expansion. This doesn't seem to fit.
physicsforum.JPG


Astronomers1) tell us that a 'cosmological constant' can account for the universe's increasing expansion.

Representing the universe by a symbolic expanding ring, Fig.a, at an instantaneous radius r the inward gravitational force varies as the inverse square of this radius, giving Fg ~ 1/r^2 ('~' = 'proportional to').

A cosmological constant to balance this tendency would need to have the form lambda = const/r^2.

Hubble's law gives a radial expansion rate dr/dt = H0r, Fig.b. Differentiating gives an acceleration d^2r/dt^2 = H0^2*r. Requiring an outward expansionist force Fh proportional to r, Fig.c. And not proportional to the 1/r^2 of a cosmological constant.

Their thesis doesn't seem to fit.

1) For instance Ryden, B. (2006) "Introduction to Cosmology", p.71: "Learning to love lambda".
 
Space news on Phys.org
jeremyfiennes said:
Representing the universe by a symbolic expanding ring
Which it isn't. You are just pulling this claimed "representation" out of thin air, not from any valid model of the universe with a cosmological constant. Garbage in, garbage out.
 
jeremyfiennes said:
A cosmological constant to balance this tendency would need to have the form lambda = const/r^2
The correct way to proceed is not to start with Newtonian gravity, but to arrive at Newtonian gravity by making appropriate approximations after first starting with general relativity. This can done by assuming: 1) a cosmological constant is not large enough to make a contribution; 2) a cosmological constant is large enough to make a contribution.

1) The weak-field limit of Einstein's equation without cosmological constant/dark energy leads to Poisson's equation,
$$\nabla^2 \Phi = - \vec{\nabla} \cdot \vec{g}= 4 \pi G \rho,$$
where ##\Phi## is gravitational potential and ##\vec{g}= - \vec{\nabla} \Phi## is the gravitational acceleration of a small test mass.

2) The weak-field limit of Einstein's equation with cosmological constant/dark energy ##\Lambda## leads to a modified "Poisson" equation,
$$\nabla^2 \Phi = 4 \pi G \rho - \Lambda c^2.$$
For a spherical mass ##M##, the divergence theorem applied to the above gives
$$\vec{g} = \left(-\frac{GM}{r^2} + \frac{c^2 \Lambda}{3} r \right) \hat{r}.$$
The second term is a "springy" repulsive term for positive ##\Lambda##.
 
  • Like
Likes Bandersnatch
There are more basic conceptual issues with what the OP wrote than not using FLRW.
I think it's fine to use Newtonian motion in one dimension to help make sense of what's going on. Ryden's textbook does it in the same chapter the OP refers to (so I'm not sure how helpful it is to tell the OP it's wrong and should learn GR instead - IMHO).
We have a ball and we throw it up in a central gravitational field, and that's meant to be analogous to a galaxy receding away. There are limitations to this approach, but they don't show up here.
Still, it has to be done properly. Here, for example:
jeremyfiennes said:
A cosmological constant to balance this tendency would need to have the form lambda = const/r^2.
This assumes for some reason that the cosmological constant should balance gravity. If it did that, it would nett steady expansion with constant recession velocities. Same as we would have V=const if we canceled out gravity for our ball. We could have some large constant in front of our ##1/r^2## force, such that lambda>Fg. But that would nett acceleration for all time. This is not what is observed or claimed.

Whereas to get outward acceleration >at some point in the evolution of the system<, all we need is the extra force to scale with r slower than gravity's ##r^{-2}##, i.e. anything with the exponent higher than -2. So, e.g. for a force that scales as r, we have something of a form: ##\ddot x=-a/r^2+br##. Which is the same form as the last equation George posted above.
As long as the ball (galaxy) has escape velocity, the second term will start to dominate at some point in the evolution of the system. I.e., we'll have a period of deceleration followed by a period of acceleration.

Further:
jeremyfiennes said:
Hubble's law gives a radial expansion rate dr/dt = H0r, Fig.b. Differentiating gives an acceleration d^2r/dt^2 = H0^2*r. Requiring an outward expansionist force Fh proportional to r, Fig.c. And not proportional to the 1/r^2 of a cosmological constant.
This in turn show misunderstanding of the Hubble law that leads to comparing apples and oranges.
Linear Hubble law with constant ##H_0## is not an equation of motion. It does not mean that if you have an object today at some distance, receding with some velocity, it'll recede with twice velocity once it gets to twice the distance (having thus accelerated). It shows the spread of velocities of >different objects at one point in time<. Only in this sense can H be treated as a constant. In general, it's time-dependent and monotonically decreasing.
In the very far future, when matter dilutes, the Hubble parameter will approach a constant value, and it will be approximately valid to use ##H_\infty## as a constant in the Hubble law. At that point the expansion will approximate exponential, and it will be true that a single object moving to twice the distance accelerates to twice the velocity.
This is equivalent to the first term in:
##\vec{g} = \left(-\frac{GM}{r^2} + \frac{c^2 \Lambda}{3} r \right) \hat{r}##
vanishing for large r.
So there's no disagreement between a) and c) in the OP.
 
https://en.wikipedia.org/wiki/Recombination_(cosmology) Was a matter density right after the decoupling low enough to consider the vacuum as the actual vacuum, and not the medium through which the light propagates with the speed lower than ##({\epsilon_0\mu_0})^{-1/2}##? I'm asking this in context of the calculation of the observable universe radius, where the time integral of the inverse of the scale factor is multiplied by the constant speed of light ##c##.
The formal paper is here. The Rutgers University news has published a story about an image being closely examined at their New Brunswick campus. Here is an excerpt: Computer modeling of the gravitational lens by Keeton and Eid showed that the four visible foreground galaxies causing the gravitational bending couldn’t explain the details of the five-image pattern. Only with the addition of a large, invisible mass, in this case, a dark matter halo, could the model match the observations...
Hi, I’m pretty new to cosmology and I’m trying to get my head around the Big Bang and the potential infinite extent of the universe as a whole. There’s lots of misleading info out there but this forum and a few others have helped me and I just wanted to check I have the right idea. The Big Bang was the creation of space and time. At this instant t=0 space was infinite in size but the scale factor was zero. I’m picturing it (hopefully correctly) like an excel spreadsheet with infinite...
Back
Top