Cosmological constant for the Universe's expansion

[jeremyfiennes]

TL;DR

Astronomers hold that a cosmological constant can account for thr universe's increasing expansion. This doesn't seem to fit.

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".

 

[Ibix]

gravitational force

When you say these words in the context of cosmology, stop. You are using Newtonian gravity. It won't work.

See also this thread, currently third in the forum list.

 

[PeterDonis]

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.

 

[anorlunda]

Rather than a ring, beginners in cosmology should use the FLRW equations as a simple model.

https://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric

Or you can study it right here on PF. Here is the first of a series of PF Insights articles on the subject.
https://www.physicsforums.com/insights/journey-cosmos-friedmann-equation/

 

[George Jones]

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 $\Lambda$.

 

[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:

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:

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.