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There exists a huge gap in the strength of the four forces that we have observed in nature between gravity and the other three: In the following image we see that the radiation decouples from the "soup" at energy densities of 0.25eV. That is the snapshot of CMB, Cosmic Microwave Background radiation . CBR in this plot is the CMB, when we have a ...


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Well, the argument is something like this. Matter and radiation are tightly coupled until the time of last scattering. The photons are in thermal equilibrium with the plasma surrounding them. This means the state of the Universe at the time of last scattering is very precisely imprinted on the photons, and the photons travel from last scattering to us more ...


4

In curved spacetime, you can no longer compare velocities at different points in the straight-forward manner we use in flat spacetime. Thus the claim that recession velocities should not be considerer 'real' (as in relative) velocities, but rather rates of expansion of space. If you want to get at the former, you need to parallel transport the source's ...


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I read this question but I didn't understand the physics equations used in the answer. Let me offer a simplified explanation. The origional Big Bang cosmology asserts that the universe is expanding, which if true means that the proper distance $D$ between any 2 points is increasing. Since you can think of this expansion as "space itself expanding," then ...


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The size of the universe is given by a scale factor, normally written as $a(t)$, that is a function of time and is calculated by solving Einstein's equation for an isotropic homogeneous universe. Once we've calculated $a(t)$ we can differentiate it wrt time to get $\dot{a}(t)$ and use this to calculate the recession velocities. The scale factor is a ...


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In the standard model of cosmology, we say that the universe we live in is an FRW-universe. The FRW part just refers to the initials of the guys who first wrote down the description. The equation that describes such a universe is: $$ds^2=a(t)^2\left[-d\tau^2+d\vec x^2\right]$$ Note: this is for a flat, homogeneous, isotropic FRW-universe. In the above ...


3

You are perfectly right: neutrinos hold the promise of providing a window that gives us views much deeper into the big bang than the window conventionally provided by photons. The Hubble Space Telescope gives us snapshots of galaxies in a universe that is only 600 million years old. Although this feat is brought to the wider public as big news, the Hubble ...


3

You've overlooked gravitational entropy. The entropy of a black hole horizon is given by: $$ S = \frac{kA}{4 \ell_p^2} $$ where $A$ is the area of the horizon, $k$ is Boltzmann's constant and $\ell_p$ is Planck's constant. This entropy is absolutely huge, so if you take a uniformly distributed gas in thermal equilibrium and concentrate it into a black ...


3

The dividing line meets at $(\Omega_m,\Omega_\Lambda)=(0,1)$. From the Friedmann equations, it follows that the scale factor $a(t)$ satisfies the relation $$ \frac{\dot{a}^2}{H_0^2} = \Omega_m a^{-1} + (1 - \Omega_m - \Omega_\Lambda) + \Omega_\Lambda a^2. $$ The universe has no big bang singularity if the above expression is negative (or zero) for some ...


3

It is only in the absence of dark energy that the correspondence between geometrical curvature and the ultimate fate of the universe is as straightforward as you describe. Measurements (primarily of the cosmic microwave background) indicate that our universe is flat or very nearly so, which should be interpreted geometrically (i.e. in terms of the sum of ...


3

There are several scenarios that point out cyclic cosmologies. The first one is called Matter bounce cosmology which is based on the idea that that the universe originated from a cosmological bounce in which quantum fluctuations develop into a scale-invariant spectrum of curvature perturbations. The bounce is realized beyond Einstein's General Relativity and ...


3

The most elegant way I've seen to describe this is described in the paper The river model of black holes. If we write the Schwarzschild metric in Gullstrand-Painlevé coordinates we get (in units where $c = G = 1$): $$ ds^2 = -dt_{ff}^2 + \left(dr + \beta dt_{ff} \right)^2 + r^2 d\Omega^2 $$ where: $$ \beta = \frac{2M}{r} $$ This looks like the Minkowski ...


2

General relativity describes gravity as the curvature of space-time. So to understand how black holes work, we must have a basic idea of how space-time works. In regular 3D space, the length $\Delta s$ between two points is $\Delta s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$. This is just the pythagorean theorem applied in 3D. But in space-time, where we ...


2

Reheating is the decaying of the inflaton into the particles that we are currently observing. In the context of quantum field theory this happens simply because there is a coupling of the inflaton field to either the Standard Model (and possibly other, yet unobserved, particles) directly, or to a field $\chi$ which then couples to the Standard Model and ...


1

There is a very nice paper exactly on this topic, where the expressions describing the fit curves are derived: http://arxiv.org/abs/hep-ph/9906447v1 If you have a look at the final expression (formula (6.5) on page 8), you will agree, that the relation between $\Omega_\Lambda$ and the luminosity is hard to describe by words. However, you can try to think ...


1

Detecting cosmic neutrino background (~1.95K) is extremely difficult (compared with cosmic microwave background) and never performed directly so far. That's because neutrinos interacts with matter very weakly, unlike photons. We have to build very large detectors. (But if they behave like photons we can't use them to observe early universe.) I also found ...


1

The above data is the is the anisotropy of temperature of the Cosmic Microwave Background (CMB) as measured by NASA U2 airplanes in the 1970s. The anisotropy is due to the redshift and blueshift of the Earth moving 300 kilometers per second or 1,080,000 kilometers per hour relative to the frame of the CMB, in the direction of the + at the center of the ...


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1 - It is false! If $E = mc^{2}$ is true only for an object that isn’t moving, the mass never changes (is a "Lorentz invariant"). 2 - Can you rephrase it, please? 3 - Energy and mass are not at all the same thing; an object’s energy can change when its motion changes, but its mass remains the same. 4 - In Special Relativity, time can be variable, its ...


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I think, you mix two or even three kinds of redshift effects. First, the classical Doppler shift due to which light emitted from a moving object is shifted to the red or blue. This is dependent on relative movement of sender and receiver only. For transverse motion we still get a redshift (see Wikipedia URL below), but it is very small for low speeds. ...


1

You might find the following paper useful: http://arxiv.org/abs/astro-ph/0310808v2 In the figure 2 on page 7 you see the plots of different $v(z)$ relations, among them the classical and the special relativistic ones that you have used in your calculations. You can see that the classical $v(z)$ relation intersects the $v=1c$ line for some redshift $z$, this ...


1

so what if a light beam was pointed somewhere behind the event horizon to the outside, Inside the horizon, the curvature of spacetime is such that the direction "to the outside" is the past time direction. In other words, to go 'back' towards the horizon is as impossible as it is to go 'back' in time. Indeed, for the same reason that we inexorably ...


1

Carbon has to be produced by the triple-alpha process because there is no stable nucleus with 8 or 5 nucleons. The probability of this is very low, because it requires three different particles to be in the same place at the same time. You'll note that the Wikipedia article says: One consequence of this is that no significant amount of carbon was ...


1

As you said, since nothing can go out from the event horizon, if light is sent from within the event horizon it does not escape. However, it is not true, that light will slow down and go back. In fact, inside the event horizon all accessible paths point towards the center of the Black Hole. This means, that, once inside a Black Hole, you will be pushed ...


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Your question cannot be answered because ultimately it depends on experiment. In general relativity the gravitational constant, $G$, is assumed constant and the geometry of spacetime is derived on this basis. At the moment observation suggests the metric we obtain using GR is a good description of the universe (provided you believe in dark energy), so the ...


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The Milky Way is receding from the members of the Hydra-Centaurus Supercluster. The Hydra cluster has a red shift of 0.0548. The Centaurus cluster has a red shift of 0.0114. The Norma cluster has a red shift of 0.0157. The local group is and will continue moving away from the Hydra-Centaurus Supercluster.


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Space itself was once concentrated in an infinitesimally small point. During the Bang of the Big Bang all distances between points got bigger. If you try to measure the expansion of the universe from any point you will draw the conclusion that the expansion started from that point. It seems that the expansion happened everywhere, and nowhere at the same ...


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Rather than work out the answer for you, I'll work out the equivalent answer in Minkowski space, and leave it to you to develop the rest for the case of the Einstein static universe. In Minkowski space, you have: $$ds^{2} = - dt^{2} + dx^{2} + dy^{2} + dz^{2}$$ If one wants to timelike geodesics, one must find the path that is an extremum of: $$ I = ...


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$\phi$ in this context is typically known as the "inflaton" (a somewhat silly name, I know, but we already have quarks), it is the scalar field that drives inflation. Any field may have a potential component, typically written as $V$. Then the Lagrangian can be written as the sum of kinetic term(s) with (the relevant covariant) derivatives and the potential ...


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This is a great question. I feel it necessary to point out the level of study and understanding that go behind asking this question. Well done! Here's the way I understand it. You analysis is flawless; in a radiation dominated universe, $a\propto\sqrt t$. That said, it is not correct to interpret this as the photons exerting some sort of pressure that ...


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I am not exactly sure which low energy cutoff you refer to; however, there is a low-energy cutoff for photons that I am aware of. Photons with energies on the order of $H_0\sim10^{-33}\text{eV}$ would be super-horizon modes. That is, their wavelengths would be on the order of the Hubble radius, $H_0^{-1}=14.6~Gly$. Larger than this would mean that the ...



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