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19

Let's start partway through the expansion of the Universe in the matter dominated epoch. At this time the energy density is dominated by matter, but the dark energy and radiation components are still present, just relatively small. The Universe is expanding, but the expansion is gradually slowing down. As the Universe expands, the density of matter scales ...


16

The title and the text actually ask two different questions. While Kyle Oman and Thriveth answer the title excellently, I'll address the question in the text which asks "Why did the Universe expand in the first place, before dark energy (DE) started to dominate". The answer to this is inflation (we think). The first fraction of a second after the creation ...


9

The short version: The amount of matter in the Universe is fixed, so as the Universe expands, matter density will drop because the same amount of matter will be spread out on more space. Dark Energy, on the other hand, is (by definition) constant or almost constant in density. This means that no matter how dilute the Dark Energy is, if it waits long enough, ...


9

Your question is not specific to inflation, and really applies to any case where a bosonic quantum field behaves semiclassically due to macroscopically large occupation numbers. One very simple example of this is the Stark effect in quantum mechanics, where a Hyrodgen atom is placed in a uniform electric field. The atom is treated as a quantum mechanical ...


8

The most distant object that light we emit today can reach in the distant future is at the event horizon $$eH(t) = a(t)\cdot \int_{t}^{t_{max}} \frac{c\cdot \text{d}t'}{a(t')}$$ which is now approximately 17 billion lightyears away, see the future light cone in comoving coordinates which converges to this distance: If the light was emitted at the big ...


8

By your assumption, we are talking about the FLRW Universe. Such a universe is by definition isotropic and homogeneous so there can't be any preferred direction at any point. If there were a difference between the CMB frame and the co-moving frame, it would indeed produce a preferred direction at some/most points (the direction of motion of one frame ...


7

We want the Newtonian limit of the Einstein Field equations for nonzero vacuum energy(=cosmological constant). As $\rho_\mathrm{vac}=\Lambda/4\pi G$ is a mass(=energy) density, Poisson equation is $$ \Delta\Phi=4\pi G\rho(\boldsymbol r)-\Lambda \tag{1} $$ If we assume spherical symmetry, and point-like source $\rho\sim\delta(\boldsymbol r)$, the ...


6

I though I would discuss the transition from radiation to matter dominated phases and from there to the dark energy phase. A fair amount of this can be discussed with just Newtonian mechanics. General relativity changes this by some subtle means, but as a coarse grained view, to borrow a stat mechanics term, Newtonian mechanics captures a lot of this. We ...


5

Cosmic Rays are most often high-energy particles, mostly protons and alpha particles accelerated to high velocities by cosmic magnetic fields. They do not show up in the microwave wavelength range that comprise the CMB. As @ACuriousMind says in the comment, there is contamination in the CMB, but this is mainly due to Galactic dust and Bremsstrahlung from ...


4

If the universe is open, there's obviously more universe that you haven't included in your system. The universe, by definition, contains all energy and matter. An open system, by definition, has an outside system to exchange energy and matter with. If that outside system isn't part of the universe, then where is it?


4

Ok, lets look at how we determine $\mu$ in a cosmological setting. In order to determine $\mu_i$, we can use the fact that, in equilibrium, $\mu$ is conserved in all reactions. This means that if we have a scattering process $i + j \rightarrow a+b$, then we know that $\mu_i + \mu_j = \mu_a + \mu_b$. Fermions in equilibrium, like electrons and neutrinos ...


4

The Gödel universe is homogeneous and every observer anywhere in the universe observes the universe to be rotating around them. So a Gödel universe has no centre.


4

A fine tuning problem is only a problem if we require that the considered model is a good or complete model of how we think the universe behaves. In this case, it appears that we require the density of the universe to be as close as 1 part in $10^{64}$ to the exact density that will make it a flat universe that will expand forever, asymptotically coming to ...


3

The argument is that (mixing happens) => (inflation happens) => (mixed regions are out of causal contact, but have no way to change their local temperature) In this scenario, it doesn't matter when the photons are emitted. Their apparent homogenity is an effect of the left most step.


3

The process by which particles are created after inflation is known as "reheating". One needs a coupling between the inflaton and another particle's field for this to happen. Generally it occurs at the end of inflation when the inflaton is oscillating in the well of its potential and the expansion rate falls below the interaction rate between the inflaton ...


3

EDIT: My first answer seemed to imply that radiation is at rest in the Cosmic Rest Frame. Radiation is not in rest in any frame. See below. The sentence shouldn't be read as "[velocity of energy] forms", but "velocity of [energy forms]"$^\dagger$. The sentence refers to "energy forms", i.e. the different forms in which energy can manifest itself. These ...


3

You get an extra term that increases with r: $$a = -\frac{G\cdot M}{r^2} + j\cdot r$$ with j as the repulsive component.


3

Although the radiation-dominated (RD) era is long in comparison to the matter-dominated (MD) and $\Lambda$-dominated ($\Lambda$D) eras, it is nice to have an answer which can be adapted easily for any cosmological era. If we assume that the Universe is permeated by a perfect fluid we may use the equation of state \begin{equation} w = \frac{P}{\rho}, ...


3

In general relativity a free particle moves on a trajectory called a geodesic and to make it diverge from that geodesic you need to apply a force to it. To take an everyday example, an object momentarily at rest at the surface of the Earth would normally follow a geodesic that leads radially towards the centre of the Earth with an acceleration relative to ...


2

Therefore you need to calculate the future light cone $$LC_{proper}=\int_{a(t_0)}^{a(t_1)} \frac{c\cdot a(t_1)}{\alpha^2\cdot H(\alpha )} \, \text{d}\alpha$$ In comoving coordinates you divide that by the scale factor of the time at absorption $$LC_{comoving}=\frac{LC_{proper}}{a(t_1)}$$ with H as the Hubble parameter $$H(a)=H_0\cdot ...


2

In the geometrical optics approximation light ray is represented by a null geodesic. Therefore you only need to find a null geodesic connecting points $(t_0,0,0,0)$ and $(t_1,x,0,0)$ for some $t_1$ (and this condition will determine $t_1$ uniquely). This is probably quite easy to do directly in this case, but in general for investigation of null curves in ...


2

After answering several similar questions on Physics SE I have realised I can answer this question myself using the concepts of the (comoving) Particle Horizon, the (comoving) Event Horizon and the Comoving Hubble Sphere. If I want to know the maximum distance I can communicate in an any amount of future time, then I should consider the Event Horizon, which ...


2

The inflationary theory you mention is probably eternal inflation. In this theory there is just one universe but different parts of it are causally disconnected i.e. the different parts cannot affect each other in any way. Whether these constitute a multiverse comes down to terminology. In principle there is a continuous spacelike straight line that links ...


2

In the Standard Model, the baryon and lepton number are accidental global symmetries. However, they are conserved only at the classical level: quantum corrections do not respect them, i.e., they are anomalous. The interesting thing is that they are violated by exactly the same amount. In terms of currents we write: $$\partial_\mu J_B^\mu=\partial_\mu ...


2

To answer your specific question: absolutely none. The Millenium run is a "dark matter-only" simulation. In this sort of simulation gas physics is taken to play a negligible role. All the gas (and stars, indeed all "baryonic matter" as it's called in the jargon) is removed and replaced with additional dark matter. The extra dark matter is added just to keep ...


2

This sort of theory is still an active research topic as can be seen by running this search, which turns up plenty of recent papers related to the topic, including several in credible refereed journals. It is not widely regarded as complete/correct/accepted, but it is grounded in sensible physics... though to me some parts of it are a bit strange.


2

The preferred frame of reference is that of the co-moving reference frame that defines the Hubble flow. In practical terms that can be defined by correcting any velocity for the observer's motion with respect to the cosmic microwave background. Individual peculiar velocities for galaxies (including our own) are measured in hundreds to thousands of km/s. This ...


1

The question isn't entirely clear, but I suspect that you're being asked to prove that the given parametric equation for $R(\theta)$ and $t(\theta)$ satisfies the Friedman equation. If so, you shouldn't try to get rid of $\theta$ entirely. Instead, show that the left-hand side (as a function of $\theta$) is equal to the right-hand side (as a function of ...


1

As already explained in other answers and comments in General Relativity (GR) energy is not conserved. Some people and physicists say it is, it simply gets lost by the matter-energy and gained by the gravitational field, or viceversa; this is more pleasing to our sense of conservation of something, but it has problems in that the gravitational energy, is not ...


1

The metric for the de Sitter spacetime, which approximates the observable universe in stationary coordinates is $$ ds^2~=~-\left(1~-~\frac{r^2\Lambda}{3}\right)dt^2~+~\left(1~-~\frac{r^2\Lambda}{3}\right)^{-1}dr^2~+~r^2d\Omega^2 $$ The important term is $$ \left(1~-~\frac{r^2\Lambda}{3}\right), $$ that looks a bit like the Schwarzschild factor. This ...



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