Gravitational Instability and accelerated explanation of Universe?

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  • #1
say_cheese
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TL;DR Summary
The time constant of instability corresponding to the Hubble derived mass density is 10^11 years (without a Cosmological Constant). Why is this not an explanation for the observed acceleration of the expansion of the universe?
Ref: Principles of Physical Cosmology by P.J.E. Peoples, Princeton Univ. Press

This question is likely fully answered somewhere, but I cant find it.

Hubble obtained (1926) the mean mass density of the Universe as 10^-31 g/c.c. The gravitational stability equation gives (McCrea, Milne (1934)) with no dark energy cosmological constant,
equation

later derivation without division by 3.
The resulting exponential time constant for the Gravitational Instability from this equation and the Hubble density is 10^11 years, much longer than the current age of the Universe.
The instability can result in increasing expansion, if the density perturbation is negative. Why is this not an explanation for the observed acceleration in the expansion of the Universe?
 
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  • #2
say_cheese said:
The resulting exponential time constant for the Gravitational Instability
What exponential time constant? Without dark energy the expansion is not exponential.
 
  • #3
The solution to the above Gravitational Stability equation (initial perturbation is e) gives the growth of perturbation

equation


The time constant then is
equation
 
  • #4
Also, in words, Einstein had to introduce the Cosmological Constant to adjust for a static Universe. He abandoned it because that would lead to a Gravitational Instability. So why cant we take the CC to be zero and explain the accelerated expansion as the instability?
 
  • #5
say_cheese said:
The solution to the above Gravitational Stability equation (initial perturbation is e) gives the growth of perturbation

equation


The time constant then is
equation
No, it doesn't. It gives a sinusoid, not an exponential. Check your work.

Also, even that assumes that ##\rho_b## is constant in time. But it isn't. The density of matter decreases with time as the universe expands.
 
  • #6
say_cheese said:
Also, in words, Einstein had to introduce the Cosmological Constant to adjust for a static Universe. He abandoned it because that would lead to a Gravitational Instability. So why cant we take the CC to be zero and explain the accelerated expansion as the instability?
The expansion that Einstein factored out would be a decelerating expansion - as matter and radiation becomes less dense. It requires a vacuum (dark) energy to explain accelerated expansion. That's using the Friedmann equation.
 
  • #7
say_cheese said:
Einstein had to introduce the Cosmological Constant to adjust for a static Universe.
That was why Einstein originally introduced it, but it is not why we currently include it. We currently include it because (a) we have evidence for accelerated expansion of the universe, which requires it, and (b) the term corresponding to it in the Lagrangian should be there by the same reasoning that gives the standard Ricci tensor term.

say_cheese said:
He abandoned it because that would lead to a Gravitational Instability.
He was disappointed to discover that his static universe was unstable, yes.

say_cheese said:
So why cant we take the CC to be zero and explain the accelerated expansion as the instability?
The instability in the static universe is in response to small perturbations about an equilibrium between the density of ordinary matter and the cosmological constant. If you remove the cosmological constant, there is no equilibrium at all and the concept of "instability" as you are using it doesn't even make sense.

In any case, this kind of hand-waving reasoning is not how physics is done. You need to look at the actual math.
 
  • #10
PeterDonis said:
That was why Einstein originally introduced it, but it is not why we currently include it. We currently include it because (a) we have evidence for accelerated expansion of the universe, which requires it, and (b) the term corresponding to it in the Lagrangian should be there by the same reasoning that gives the standard Ricci tensor term.He was disappointed to discover that his static universe was unstable, yes.The instability in the static universe is in response to small perturbations about an equilibrium between the density of ordinary matter and the cosmological constant. If you remove the cosmological constant, there is no equilibrium at all and the concept of "instability" as you are using it doesn't even make sense.

In any case, this kind of hand-waving reasoning is not how physics is done. You need to look at the actual math.
My question does not qualify as a hand waving reasoning. Certainly not a dummy here. I quoted the math behind with references therein and asked a hard question.(See above). It takes a bit of time and not answer from the hip. Please pick up Peebles book and go through it. It is amazing in its comprehensiveness even for its age.
 
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  • #11
say_cheese said:
Certainly not a dummy here. I quoted the math
You claimed that the solution of the equation ##d^2 l / dt^2 = - k l## is an exponential. It's not. It's a sinusoid.

If ##l(t) = e^{\pm \sqrt{k} t}##, then ##dl / dt = \pm \sqrt{k} l## and ##d^2 l / dt^2 = k l## (notice the lack of the minus sign, since taking two derivatives cancels any minus sign in the exponent).

If ##l(t) = \sin \sqrt{k} t##, then ##dl / dt = \pm \sqrt{k} \cos \sqrt{k} t## and ##d^2 l / dt^2 = - k \sin \sqrt{k} t = - k l##. So that solves the equation (and so would ##\cos \sqrt{k} t##).

Given this obvious error on your part, you are in no position to be claiming that others are shooting from the hip. (And, as I have already pointed out, even all of the above assumes that the ##k## in the equation is actually a constant, but it isn't; it contains ##\rho_b##, which is a function of time.)
 
  • #12
say_cheese said:
Please pick up Peebles book
I have read it, and I don't recall any claims about "gravitational instability" along the lines you are making. Please be specific about exactly where in Peebles' book the claims you describe are being made.
 
  • #13
say_cheese said:
There is a current theory that uses this conjecture to explain the accelerated expansion without dark energy.
I see nothing in the paper that remotely resembles the claim you posted in the OP of this thread. The paper does not even reference Peebles' book.
 
  • #14
PeterDonis said:
I have read it, and I don't recall any claims about "gravitational instability" along the lines you are making. Please be specific about exactly where in Peebles' book the claims you describe are being made.
page 67, Peebles book - gravitational instabilty subsection in book, 1993.

"if the mass density were lower than the critical value, the Universe would expand". The equations aare also from the same subsection.

Also, I feel vindicated. There is actually a theoretical paper proposing this which I have given in the later reply. I found it after I posted the thread.
 
  • #16
say_cheese said:
"if the mass density were lower than the critical value, the Universe would expand".
I don't have my copy handy to check the context, but I suspect you are quoting this out of context. As an isolated statement, it makes no sense. There are expanding models with density lower than critical, equal to critical, and higher than critical.

say_cheese said:
The equations aare also from the same subsection.
The equation you posted in the OP is fine in itself, it's just the second Friedmann equation with zero cosmological constant.

The problem is your claim that the solution of this equation is an exponential. It isn't. In fact, once we take into account that ##\rho_b## is a function of time, the solution is a decelerating expansion, which might or might not turn into contraction depending on the density relative to critical. This has been well known basic cosmology for decades.
 
  • #17
PeterDonis said:
You claimed that the solution of the equation ##d^2 l / dt^2 = - k l## is an exponential. It's not. It's a sinusoid.

If ##l(t) = e^{\pm \sqrt{k} t}##, then ##dl / dt = \pm \sqrt{k} l## and ##d^2 l / dt^2 = k l## (notice the lack of the minus sign, since taking two derivatives cancels any minus sign in the exponent).

If ##l(t) = \sin \sqrt{k} t##, then ##dl / dt = \pm \sqrt{k} \cos \sqrt{k} t## and ##d^2 l / dt^2 = - k \sin \sqrt{k} t = - k l##. So that solves the equation (and so would ##\cos \sqrt{k} t##).

Given this obvious error on your part, you are in no position to be claiming that others are shooting from the hip. (And, as I have already pointed out, even all of the above assumes that the ##k## in the equation is actually a constant, but it isn't; it contains ##\rho_b##, which is a function of time.)
Very strange that someone would think I would not know the solution to the simple harmonic motion equation. You have not watched the physics behind it, the explanation for this is beautifully described in Peebles book, This has to do with the fact that if the Universe is a balloon, we are inside it. Please go through Peebles book.
 
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  • #18
say_cheese said:
Very strange that someone would think I would not know the solution to the simple harmonic motion equation.
Why is it strange? You are claiming that the solution to this equation is an exponential. That's what you said in post #3 of this thread.
 
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