Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Motivation behind the theory of cosmological perturbations

  1. Sep 26, 2016 #1
    What is the main motivation behind the theory of cosmological perturbations? Is it the observational data, observational hints, or perhaps the theory of inflation?
     
  2. jcsd
  3. Sep 26, 2016 #2

    Chalnoth

    User Avatar
    Science Advisor

    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.
     
  4. Sep 26, 2016 #3
    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.
     
  5. Sep 26, 2016 #4

    Chalnoth

    User Avatar
    Science Advisor

    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.
     
  6. Sep 26, 2016 #5
    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?
     
  7. Sep 26, 2016 #6

    Chalnoth

    User Avatar
    Science Advisor

    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).
     
  8. Sep 26, 2016 #7
    Thanks for the detailed answer.

    On a different note, why does the Newtonian theory of cosmological perturbations neglect perturbations of the metric?
     
  9. Sep 26, 2016 #8

    Chalnoth

    User Avatar
    Science Advisor

    The metric doesn't exist in Newtonian gravity.
     
  10. Sep 26, 2016 #9
    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?
     
  11. Sep 26, 2016 #10

    Chalnoth

    User Avatar
    Science Advisor

    All scales. Perturbations larger than the Hubble radius get "frozen" and tend not to evolve much (they do evolve some, just not much).
     
  12. Sep 26, 2016 #11
    Why do perturbations larger than the Hubble radius get ''frozen''?
     
  13. Sep 26, 2016 #12

    Chalnoth

    User Avatar
    Science Advisor

    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.
     
  14. Sep 26, 2016 #13
    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?
     
  15. Sep 26, 2016 #14

    Chalnoth

    User Avatar
    Science Advisor

    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.

    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.
     
  16. Sep 26, 2016 #15
    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?

    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?
     
  17. Sep 26, 2016 #16

    Chronos

    User Avatar
    Science Advisor
    Gold Member

    You fail to grasp the math.
     
  18. Sep 27, 2016 #17

    Chalnoth

    User Avatar
    Science Advisor

    The Hubble radius is simply ##c/H##.

    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.
     
  19. Sep 27, 2016 #18

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    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.
     
  20. Sep 27, 2016 #19

    Chalnoth

    User Avatar
    Science Advisor

    True. But it's also usually not possible to find a single coordinate chart which is valid for the entire spacetime.
     
  21. Sep 27, 2016 #20

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted