Motivation behind the theory of cosmological perturbations

  • #1
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Main Question or Discussion Point

What is the main motivation behind the theory of cosmological perturbations? Is it the observational data, observational hints, or perhaps the theory of inflation?
 

Answers and Replies

  • #2
Chalnoth
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What is the main motivation behind the theory of cosmological perturbations? Is it the observational data, observational hints, or perhaps the theory of inflation?
Cosmological perturbations are an attempt to mathematically model the fact that our universe isn't perfectly uniform in density. The idea is to assume you can define an isotropic, homogeneous space-time, and get from that idealized model to our real universe by adding perturbations to that basic space-time.
 
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What physical quantities are perturbed in this analysis? Because I can only see that the density and the temperature of the universe can be perturbed.
 
  • #4
Chalnoth
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The typical perturbation is in the density. The temperature does vary from place to place, and those variations can be estimated from the density perturbations.
 
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Is the theory of cosmological perturbations independent of the theory of inflation, or the are the theories interconnected and one cannot survive without the other?

If the theory of inflation is proven to be wrong, will there still be a need for a theory of cosmological perturbations?
 
  • #6
Chalnoth
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Is the theory of cosmological perturbations independent of the theory of inflation, or the are the theories interconnected and one cannot survive without the other?

If the theory of inflation is proven to be wrong, will there still be a need for a theory of cosmological perturbations?
Yes, they're independent of inflation.

Think of perturbations as a mathematical tool, rather than any kind of strong statement about what the universe is. They're a way of mathematically dividing up the ways in which the density of the universe can vary from place to place into a series of "modes" (each with a different wavelength). Within a certain approximation, each mode varies independently of every other mode. This makes it so that rather than being forced to calculate all of the different possible ways the universe can vary, we can just examine one mode at a time.

This approximation doesn't work very well at small scales: gravitationally-bound systems such as star systems, galaxies, and galaxy clusters break the assumption that each mode is independent of every other mode. So perturbations are mostly just used to describe how structure forms at scales much larger than galaxy clusters. Other approximations can be used to estimate how things operate at smaller scales, but typically the best tools are N-body simulations (which simulate the universe as a collection of individual particles distributed throughout the universe).
 
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Thanks for the detailed answer.

On a different note, why does the Newtonian theory of cosmological perturbations neglect perturbations of the metric?
 
  • #8
Chalnoth
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Thanks for the detailed answer.

On a different note, why does the Newtonian theory of cosmological perturbations neglect perturbations of the metric?
The metric doesn't exist in Newtonian gravity.
 
  • #9
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Well, I guess then, in general relativity, the metric does perturb.

Does the metric perturb on all scales? Or does the metric perturb only on scales larger than the Hubble radius?
 
  • #10
Chalnoth
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Well, I guess then, in general relativity, the metric does perturb.

Does the metric perturb on all scales? Or does the metric perturb only on scales larger than the Hubble radius?
All scales. Perturbations larger than the Hubble radius get "frozen" and tend not to evolve much (they do evolve some, just not much).
 
  • #11
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Why do perturbations larger than the Hubble radius get ''frozen''?
 
  • #12
Chalnoth
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Speed of light limitations: a wave that has a wavelength longer than the horizon, then a ray of light from the peak can't make it to the trough, unless the rate of expansion slows enough that the peak and trough are within one another's horizon.
 
  • #13
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Would it be possible to explain in a bit more detail?

I don't understand what the wave has to do with the perturbation of the metric. Also, how does the slowing down of the rate of expansion help?
 
  • #14
Chalnoth
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Would it be possible to explain in a bit more detail?

I don't understand what the wave has to do with the perturbation of the metric. Also, how does the slowing down of the rate of expansion help?
The "modes" I mentioned before are waves. In a gas, they become pressure (sound) waves.

Slowing down the rate of expansion increases the Hubble radius. If the expansion slows fast enough, modes which were once larger than the horizon become smaller than the horizon.

Hey, it's alright with your explanation.

I was wondering if you can explain the following sentence.

The first step in the analysis of metric fluctuations is to classify them according to their transformation properties under spatial rotations.

I always thought that the metric is a two-tensor covariant object. Where does this business of classifying them according to their transformations under spatial rotations come from?
It is a 2-tensor covariant object. But the covariance of the metric is not the same as invariance: the metric does change under a change in coordinates. The covariance means that it changes in a very specific way under a change in coordinates. This statement doesn't say that the metric itself has any particular symmetries (though it is always possible to find coordinates for which the metric is diagonal). In particular, if the metric is to describe our universe it can't be symmetric under rotations because the universe we observe is different in different directions.

Dividing the metric into components that behave differently under spatial rotations is a way to simplify the math.
 
  • #15
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Slowing down the rate of expansion increases the Hubble radius.
You mention that slowing down the rate of expansion increases the Hubble radius. But isn't the Hubble radius increasing all the time anyway, regardless of the expansion (or not) of the universe? I say this because the Hubble radius in, in crude terms, the size of the observable universe, and the size of the observable universe is increasing because light from beyond the horizon are travelling towards us all the time.

Where's the fault in my logic?

In particular, if the metric is to describe our universe it can't be symmetric under rotations because the universe we observe is different in different directions.
You say that the universe we observe is different in different directions, but isn't is a basic assumption (i.e. the cosmological principle) that the universe is isotropic?
 
  • #16
Chronos
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You fail to grasp the math.
 
  • #17
Chalnoth
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You mention that slowing down the rate of expansion increases the Hubble radius. But isn't the Hubble radius increasing all the time anyway, regardless of the expansion (or not) of the universe?
The Hubble radius is simply ##c/H##.

You say that the universe we observe is different in different directions, but isn't is a basic assumption (i.e. the cosmological principle) that the universe is isotropic?
The assumption of isotropy is approximate, not exact. The existence of the Earth and stars and galaxies prove that it's not exactly the same in every direction. If it were exactly the same in every direction, our universe would be a perfectly-uniform gas with no structure.
 
  • #18
PeterDonis
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it is always possible to find coordinates for which the metric is diagonal
In a sufficiently small neighborhood of a particular event, yes. But it's not always possible to find a single coordinate chart in which the metric is diagonal throughout the entire spacetime.
 
  • #19
Chalnoth
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In a sufficiently small neighborhood of a particular event, yes. But it's not always possible to find a single coordinate chart in which the metric is diagonal throughout the entire spacetime.
True. But it's also usually not possible to find a single coordinate chart which is valid for the entire spacetime.
 
  • #20
PeterDonis
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it's also usually not possible to find a single coordinate chart which is valid for the entire spacetime.
I'm not sure I would say "usually". In all of the FRW spacetimes, and in all of the Kerr-Newman family of black hole spacetimes, it is indeed possible to find a single coordinate chart that is valid for the entire spacetime. In the FRW case, it's the commonly used chart; but in the black hole case, it isn't (either the Kruskal-type charts or the Penrose charts, neither of which are commonly used, at least not pedagogically). Those two families of spacetimes are the most used in actual practical work.
 

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