I Understanding the Zero Order Universe in Cosmology

  • I
  • Thread starter Thread starter shirin
  • Start date Start date
  • Tags Tags
    Universe Zero
AI Thread Summary
The concept of a "zero order" universe in cosmology refers to a simplified model where the universe is assumed to be smooth and in equilibrium, with constant density throughout. This serves as a foundational framework for understanding more complex models that incorporate deviations from this ideal state, such as inhomogeneities and gravitational influences. The zero order approximation allows cosmologists to grasp the large-scale behavior of the universe before adding complications. It is not a strict term but rather a manner of describing a basic case for comparison with more intricate scenarios. Understanding this zero order model is essential for exploring the richness of the universe's structure and dynamics.
shirin
Messages
47
Reaction score
0
In chapter 2 of Dodelson’s modern cosmology book, it reads:
“Implicit in this discussion will be the notion that the universe is smooth (none of the densities vary in space) and in equilibrium (the consequences of which will be explored in Section 2.3). In succeeding chapters, we will see that the deviations from equilibrium and smoothness are the source of much of the richness in the universe. Nonetheless, if only in order to understand the framework in which these deviations occur, a basic knowledge of the "zero order" universe is a must for any cosmologist. “
What does zero-order in this context mean? Zero order of what quantity? Does it mean a universe with a constant density everywhere?
 
Space news on Phys.org
shirin said:
In chapter 2 of Dodelson’s modern cosmology book, it reads:
“Implicit in this discussion will be the notion that the universe is smooth (none of the densities vary in space) and in equilibrium (the consequences of which will be explored in Section 2.3). In succeeding chapters, we will see that the deviations from equilibrium and smoothness are the source of much of the richness in the universe. Nonetheless, if only in order to understand the framework in which these deviations occur, a basic knowledge of the "zero order" universe is a must for any cosmologist. “
What does zero-order in this context mean? Zero order of what quantity? Does it mean a universe with a constant density everywhere?
Zero order doesn't refer to a number, it's saying that the simple Universe he described is a "base case" that we can compare more complicated models to.

-Dan
 
  • Like
Likes Bandersnatch
It means the most coarse, simplest model. See here for more on the usage of such phrases: https://en.wikipedia.org/wiki/Order_of_approximation
(note that it's more of a mannerism than a well-defined term - zeroth order for one person can be first order for another)

In this context, it means that you assume the cosmological principle to hold perfectly everywhere in the universe. As you say - that the density of all components is the same everywhere. Something that is obviously not true even as you as much as look around you - there's a bunch of rock in one direction, and a whole lot of vacuum in another. There are stars here, and not much of anything there. Etc.
But starting this way lets you understand the underlying large-scale behaviour before you begin to include higher order approximations in the form of e.g. inhomogeneities nucleating filaments and galaxies.

Much like when modelling the motion of the Earth in the Solar System you could start by assuming, as your zeroth order approximation, that the Earth moves solely in the central gravitational field of the Sun and there are no other forces at play. Or maybe just that Kepler's laws hold perfectly. Which would lead you to a pretty good basic understanding of orbital motion. But then you'd want to start including perturbations from other planets and smaller bodies to improve your model, so that you can have precession, planetary migration, tides, and so on.
 
  • Like
Likes Jorrie and topsquark
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