New answers tagged

0

Can anyone tell me from where Eq. (1) comes? The curvature come from the right-hand side ($U$) of your first equation (modified a bit, merged $a$ and $x$ into a single $a$, since $x$ in your equation is apparently a fixed constant which can be absorbed into $a$ or set to $x=1$ in the chosen unit): $$ U=\frac{1}{2}m\dot{a}^2-\frac{4\pi}{3}G\rho a^2m $$ In ...


-1

I've seen this "derivation" before, even though I don't like it. If you start with $$\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}$$ and multiply both sides by $a^2x^2/2$ you get $$\frac{1}{2}\dot{a}^2x^2=\frac{4\pi G}{3}\rho a^2x^2-\frac{1}{2}kc^2x^2.$$ Now since $v=x\dot{a}$ and $\rho=\displaystyle\frac{M}{\displaystyle\...


1

I'm interpreting this question as "what lies beyond the observable universe?", asked in different words. The answer to that is "we don't know", because things that are beyond the observable universe are by definition not observable, so there could be invisible pink unicorns out there which we cannot prove doesn't exist. Any possible ...


1

$$\Gamma_{ab}{}^{c} = \frac{1}{2}g^{cd}\left(g_{ad, b} + g_{bd,a} - g_{ab,d}\right)$$ $$R_{ab} = \partial_{c}\Gamma_{ab}{}^{c} - \partial_{a}\Gamma_{bc}{}^{c} + \Gamma_{ab}{}^{c}\Gamma_{ce}{}^{e} - \Gamma_{ad}{}^{c}\Gamma_{bc}{}^{d}$$ So, given a metric, you can compute any christoffel symbol, and given any christoffel symbol, you can calculate the ricci ...


3

This is a typical example where a Newtonian derivation is much simpler and quicker, and gives the same answer. Which you can easily find online. But if you want to do this from within GR, then you have to work out the Ricci tensor entry $R_{00}$, the Ricci scalar $R$, and the metric entry $g_{00}$: $g_{00} = 1$; $R_{00}$: $$ R_{00} = R^m_{tmt} = R^r_{rtr} +...


1

In cosmology, the perturbations are usually treated as adiabatic or isothermal, and the real case is usually a combination of the two. Note that your definition of $c_s$ isn't precisely correct: as you can see, the speed of sound is defined as $c_s^2=\left(\frac{\partial P}{\partial\rho}\right)_S$, and that little S is important. It means "measured at ...


0

In short yes. I believe this idea arose when people were thinking of the symmetry breaking from a bigger symmetry group (for example some hypothetical GUT scale group), down to the standard model as the universe cooled down. Morally speaking, the same story of the Higgs mechanism applies. That is we have a theory with symmetry group $G$ in which the vacuum ...


1

You've changed your coordinates, but not the components of your metric. Remember that under a change of coordinates, $$g_{\mu\nu} \rightarrow g'_{\mu\nu} = \frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^\nu} g_{\alpha\beta}$$ In this case, letting $i,j=1,2,3$ be spatial indices, $$g'_{ij} = \left(\alpha \delta^\alpha_i\right)\...


1

The equation for the sound horizon is simply the equation for the particle horizon, with the speed of light replaced by the speed of sound, there's nothing more to it. Nevertheless, you have to keep in mind that the speed of sound also changes with time, since the matter density dilutes with a growing scale factor, so you have to treat the speed of sound as ...


3

The strongest motivation for initially equal amounts of matter and anti-matter actually comes not from the standard model but from cosmology - there is a staggeringly large number of photons in the Universe, relative to baryons$^1$. It's possible to arrive at this conclusion a couple of different ways. One is to look at the cosmic microwave background, which ...


1

The mathematical meaning of a singularity is that we cannot define mathematical quantities. Since we cannot define mathematical quantities, we have no idea of the physics in the initial singularity. Anything we assume is therefore a speculation. We cannot know whether the universe was created out of nothing, and, if so, whether it was created with a matter-...


5

What is the energy spectrum of all photons in the observable universe? All photons means: photons from the cosmic microwave background. It is the best black body radiation spectrum known at very low temperature. 2)photons radiated by stars in galaxies. The approximate black body spectrum of stars is in the thousands of kelvin, similar to our sun, the ...


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I know you want a careful treatment, but if you already have Kolbe and Turner then they have the most careful treatment I've seen at a textbook level. They essentially assume the boltzmann equation is familiar, however, so I recommend something like Kardar volume 1 chapter 3 or other statistical physics book. The actual cross section formula is complicated, ...


1

It is true, but the matter to radiation ratio is not important, only the fact that it is closed and first expands and then recollapses is important, that alone is enough to discover that the statement is correct for purely geometric reasons. If the circumference (2π times the radius of curvature) and/or Hubble parameter is small enough though light might ...


0

The power spectrum of primordial density perturbations is often modeled as a power law, $$P(k) = A \left(\frac{k}{k_0}\right)^{n-1}.$$ This is the form predicted by slow roll inflation and it fits current CMB and large scale structure data well. Here, $k_0$ is the pivot scale. It is the wavenumber (scale) were the amplitude, $A$, is measured. It is ...


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You can have field fluctuations on any scale, but in a non-inflationary spacetime they just return to the vacuum. You need inflation to convert the fluctuation to a classical curvature perturbation.


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Short answer: it depends. Edit: while I still disagree with Rob Jeffries on the conclusion, it's true that the main focus on your question shouldn't be on Shall we ever stop receiving these CMB photons here on earth? but indeed on If this moment can be compared with the switching on of a cosmic lamp because it shows a fundamental misconception. The ...


2

Einstein's equations with $\Lambda=0$ are perfectly capable of describing an expanding spacetime. The cosmological principle leads to a metric of the form $$ds^2 = -c^2 dt^2 + a(t)^2 d\Sigma^2$$ where $a(t)$ is the so-called scale factor which is conventionally set to $1$ at the present time, and $d\Sigma^2$ is a spatial 3-metric with constant curvature $\...


2

Wikipedia calls it the “radiation constant” or the “radiation density constant” (and omits the $B$ subscript). It is related to the Stefan-Boltzmann constant $\sigma$ by $a_B=4\sigma/c$. There is no standards body that gives official names to physical constants, as far as I know.


1

Yes, it is the average redshift of galaxies that belong to a cluster. There is of course an uncertainty in that, but a typical velocity dispersion among galaxies in a massive galaxy cluster is 1000 km/s. So the redshift error due to the uncertainty in the mean is $$\Delta z \sim \frac{\Delta v}{c} = \frac{\sigma_v}{c\sqrt{n}}= \frac{0.0033}{\sqrt{n}},$$ ...


0

I will answer your question 2 in full, and give a pointer to the answer to question 1. Someone else may like to answer question 1 more fully. You do often see somewhat sloppy language in which the spectrum at low $l$ is said to be "caused by" or "owing to" the Sachs-Wolfe effect. This is not quite right. The fluctuations, whether at low $...


3

I think you have the main lines of evidence right. You didn't mention the overall age of the universe, but arguably this is part of your item 5. (The point being that if you don't include $\Lambda$ in a $\Lambda$CDM cosmology then even rough observations of the amount of matter and the amount of curvature cannot square with the age of the oldest stars.) From ...


1

Since no-one qualified would answer, I'll a give it a shot: Talking quantum mechanics, these bosons have no identity, and their overall state does not even need to have a well-defined particle number in Fock space. This means, mathematically, the cosmic state of e.g. the photon field would be a superposition of one-, two-, three- ... gazillion-photon states,...


1

The answer is yes. Hubble's law relates the "velocity of recession" to the proper distance - which is the distance to the other galaxy now. That is not usually the distance we measure, as you have identified. The difference between the two is very small at low redshifts, but becomes larger at higher redshifts, and depends on the expansion history ...


0

In eqn. $16$ only magnitude of gravitational acceleration is equated to receding acceleration. And it is not Newton’s gravitational acceleration but (cosmological) time dependent Susskind’s gravitational acceleration. Second derivative of $D$ in the eqn. $16$ is equivalent to second derivative of $r$ in Newton’s inverse square law. Equation $15$ is more ...


2

It is not an expanding universe of stationary balls. There is an initial velocity to astronomical objects left over from the collapse of the primordial plasma which coalesced to the present day observable universe. That space is expanding means that the velocity of andromeda towards the milky way is a tiny bit smaller than it would have been if there were ...


0

Imagine a moving walkway as is common in airports. Step on to it while your friend does not. If you do nothing then you will recede from him. The space between you will expand. However, unless it is an unusually fast walkway, you could walk against it and approach him. The analogy is not perfect as the speed of recession would be constant but I think it ...


0

Relativistic mass becomes important as an increased inertial mass at speeds commensurate with the speed of light. Dark matter was postulated in order to reconcile observed rotational curves of galaxies with the calculation of expected curves. Rotation curve of spiral galaxy Messier 33 (yellow and blue points with error bars), and a predicted one from ...


1

The gravitational influence of objects, which is how their masses are estimated, already includes the effects you are talking about. In fact this is a very minor issue because most of the masses within the universe are not moving at relativistic velocities with respect to each other. The exception to this is perhaps primordial neutrinos, but even there, the ...


1

This is just the first Friedmann equation (including the cosmological constant within $\rho$): $$H^2 + \frac{k}{a^2} = \frac{8\pi G \rho}{3},$$ which you can rearrange and evaluate at the present time (so $a=1$) to get $$k = H^2 \left( \frac{\rho}{\rho_c} - 1 \right) .$$ This gives you the present curvature as a function of the quantities you have. Note ...


2

$\Lambda CDM$'s claim of the Universe being 13.8B years old should be taken with a grain of salt. The Universe (as depicted by $\Lambda CDM$) has hypothetically underwent inflation only for a fraction of a second shortly after the Bang, negligible when compared with its current age. Therefore, you shouldn't be hung up on inflation when it comes to guessing ...


0

The future cosmic Event horizon is the source of de Sitter (aka cosmic Hawking) radiation, also characterised by a specific temperature, the de Sitter temperature $T$ (as per the OP). It is the minimum possible temperature of the universe. To an observer in our universe, a de Sitter Universe is in their infinite future, i.e. when the Hubble sphere and Event ...


0

The cosmological multiverse does what you ask. There are several kinds of cosmological multiverse, the most common being eternal inflation, but first we do not think our universe will end in a Big Crunch: If the cosmological constant were actually zero, the critical density would also mark the dividing line between eventual recollapse of the universe to a ...


3

What's called the "age of the universe" would more accurately be called the age of the most recent epoch in the universe's history. That epoch began with the end of inflation, or with the end of whatever noninflationary process created the incredibly uniform expanding quark-gluon plasma that eventually clumped into stars, planets and us. We don't ...


15

I'm not too interested in providing an answer from the cosmological point of view. It is clear that the age of the universe derived in that way is model-dependent. The age thus obtained depends on certain assumptions (e.g. that the dark energy density remains constant). I will just add a couple of additional age determination methods that rely on alternative ...


0

It would appear that dislexia got in the way of interpreting the $\Omega_B h^2=0.0224$ notation properly. As the conversation above goes, the answer - based on the Planck 2018 results - is: $$\Omega_Bh^2=0.0224$$ $$\Omega_B=\frac{0.0224}{h^2}$$ $$\Omega_B=0.0224h^{-2}$$ $$\rho_B=\Omega_B\cdot \rho_c=0.0224\cdot 0.674^{-2}\cdot 8.53\times 10^{-27}\space kg\...


6

To compute the age of the universe, one must solve the equation: $$\frac{1}{a}\frac{da}{dt} = H_0 \sqrt{\frac{\Omega_{\gamma,0}}{a^4}+\frac{\Omega_{m,0}}{a^3}+\frac{\Omega_{k,0}}{a} +\Omega_{\Lambda,0}}$$ where $\Omega_\gamma$, $\Omega_m$, $\Omega_k$, $\Omega_\Lambda$ are the densities of radiation, matter, curvature, and vacuum energy, and the subscript '0' ...


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The rough idea is that under the assumptions contained in the cosmological principle, the application of Einstein's equations leads us to the equation $$d(t) = a(t) \chi$$ where $d(t)$ is called the proper distance and $\chi$ is called the comoving distance between two points in space. $a(t)$ is the time-dependent scale factor, which is by convention set to ...


5

This is not a full answer, but I think it will help if you separate out inflation from the rest of the picture. The age of the universe can be estimated in the first instance as the time elapsed since some very early epoch where the temperature was low enough that the Standard Model of particle physics applies to reasonable approximation. This means you can ...


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