Podcast #128: We chat with Kent C Dodds about why he loves React and discuss what life was like in the dark days before Git. Listen now.

# Tag Info

2

How large is the non-normalized scale factor (solution of the Friedmann equation) in today‘s universe? Scale factor by itself does not mean anything. The important thing is the ratios of the two scale factors. If we are describing the expansion of the universe in terms of the scale factor, the important question becomes, scale factor relative to what ? ...

1

Take any two particles A and B that have interacted with each other, and they are for all practical purposes forever entangled. If particle C then interacts with B, C is thereafter entangled with both B and A, and so on and on. The entanglement with a particle Z gets more and more diluted as the number of interactions with other particles increases between ...

5

Because we don't know whether the Universe is finite or infinite, it does not make sense to talk about an absolute size, or radius, of the Universe. Instead we use an arbitrary, dimensionless scale called the scale factor, usually denoted $a$. As the Universe expands, all (cosmological) distance scale with $a$. We define this to be equal to unity today, and ...

6

The scale factor $a$ has an arbitrary normalization. Typically people take $a=1$ for the present time.

0

At least for the first question, if I understand correctly what you are doing, the data you have are actually $b(z,i)$ so the biais as a function of the redshift $z$ and the type of the galaxy $i$. Therefore the the fact that the bias is zero is then an information and removing this value from the dataset will remove information on your system because you ...

2

If there were different regions, some with matter and others with antimatter, we would see the telltale signs of annihilation at the boundaries between them, there is none. As to why there's only matter when both matter and antimatter should be produced in equal amounts, don't know, pretty big mystery.

17

The answer touches upon the concept of Big Bang Nucleosynthesis (BBN), which is excellently explained on a graduate level in Baumann's lecture notes. The key idea is the following: In order to form metals (anything heavier than hydrogen and helium), you need deuterium nuclei. But deuterium nuclei are only formed significantly when the temperature of the ...

2

Hubble observations can only produce a local measurement of Hubble constant $H_0$. The cosmic ladders such as cepheids and supernovae are used to arrive at the Hubble parameter $H(t)$ at low redshift. In other words, Hubble observations only tell you the cosmic behavior of the recent epoch. It's a snapshot of the universe in its adulthood, with beards and ...

2

You have partly answered your own question. The Planck measurements are more precise. Moreover, it is generally assumed, perhaps wrongly, that the measurements based on the cosmic microwave background are subject to fewer systematic uncertainties than other methods. You also need to think about how the age is derived. For that you need more than just the ...

1

I think that you have a confusion between redundant information and uniform probability distribution. To answer 1) and 2) The error here is that the probability distribution is not the one you think. You only have one possible outcome (assuming that only microstate $j$ is observed) so you have a Dirac distribution P_i = \left\{ \begin{array}{...

0

Reading the phrase you cited I think you are studing large-scale structure formation, right? If so, you can think those modes as waves randomly moving in the Universe. Each modes corresponds to a wave with a respective frequency. That is, we know the relation between $k$ and $\lambda$. It is possible because we already assumed the fluctuation field can be ...

2

$h$ is a dimensionless version of the Hubble parameter $H_0$: $$H_0=h\times 100\text{ km s}^{-1}\text{ Mpc}^{-1}$$ (See Wikipedia.) For example, $h=0.7$ means that the Hubble parameter is 70 kilometers per second per megaparsec.

3

This is a question that entails the ability to manage QCD at low energies and this is an active field of research yet as we are not able to do it, unless for lattice computations. It can be considered as part of the more general problem of the determination of the phase diagram of QCD (see my answer here). The idea of chiral symmetry breaking in strong ...

0

As a warm-up, let's think about this in the context of a Newtonian cosmology, such as the one that Newton originally envisioned, with a uniform mass distribution that stays in (unstable) equilibrium because all forces cancel by symmetry. You're proposing setting the whole universe in translational motion. However, this is an unobservable thing. Newton can ...

2

The second law essentially just says the most likely thing that can happen will happen. If this law is violated for the universe then there would need to be some sort of external influence that makes this not the case. In other words, treating the universe as an entire system, it couldn't be a closed system so that something external could lower the entropy ...

1

So take a look at the definition of the Fisher Information: $$\mathcal{I}(\theta) = E\left[ \left. \left( \frac{\partial}{\partial \theta} \log f(X;\theta) \right)^2 \right| \theta \right] = \int \left( \frac{\partial}{\partial \theta} \log f(x;\theta) \right)^2 f(x;\theta) dx$$ Nowhere in the above will you find any $p(\theta)$. That is because the ...

4

Short answer: Yes, mostly, with some exceptions. Long answer: Nuclei can only be changed by nuclear processes. So the everyday chemical transformations that go on in our bodies and in the world around us do not affect the nuclei of the atoms involved. However, some nuclei are naturally unstable and will spontaneously decay or transform themselves into ...

0

I think it is just slight misunderstanding. Big bang itself indeed denote the moment of singularity in the universe. Big bang theory is a theory that describe the evolution of the universe assuming there is such moment of singularity in the beginning of its evolution. Compare for example to the steady-state theory for which the universe is expanding but ...

1

Yes, that's how it's done. You can see that it furnishes an estimate rather than a precise answer; that estimate can be improved by other techniques that astrophysicists have in their toolbox.

1

The Big Bang was a singularity. By analogy, the function $f(x)=1/\sqrt{x}$ has a domain $(0,\infty)$, which doesn't include $x=0$ because the function misbehaves there. The Big Bang singularity is technically not considered an "event" because it's not a point or set of points that we include on the spacetime manifold. So time since the big bang means the ...

0

The expansion event we call the "Big Bang" had a beginning. Depending on the context, that term can either refer to its beginning point (or very close to it) in time, or to the first few minutes of it during which all the hydrogen and helium in the early universe was formed.

3

What is referred to as the "Hubble constant", $H_0$, is in fact the value of the Hubble parameter now. The Hubble parameter, defined as $\dot{a}/a$ where $a(t)$ is the scale factor, varies with cosmic epoch. It was larger in the past and it seems to be decreasing now.

1

Inflation ? It doesn't have to be massive structures. As long you have a time dependent energy momentum tensor $T_{\mu \nu}$ that has non zero quadrupole moment, it will generate gravitational waves.

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I think it is a common misconception. Hubble constant is only a constant a fixed time (of course). In fact the whole point of Friedmann equation is to find the evolution of Hubble "constant".

4

The question you ask isn't well-posed (whose time is "flowing backwards?") but there's still a clear answer: no. This is because the expansion of space is unconstrained by the speed of light. No physical thing can travel faster than light, but the universe can still expand faster than light because it's not physical. Since the expansion of the universe isn'...

1

It can easily be seen from the plot that there are at least 3 gaps between objects in the large, there is the gap between the planets and the moons, and between the planets and stars, and between stars and galaxies. First of all, that's just a picture. It does not correctly represent the distance scales. I don't think we can consider the distance between ...

0

After studying the many answers (related or not - but many thanks anyway) to my question, I can now summarize my understanding: As the 3D (= 2D surface + time) past lightcone shell has zero spatial depth, its fractional extent within a 4D (= 3D space + time) spacetime object like the particle horizon "volume" must be also zero or infinitly small. - Cosmic ...

1

Let's start with what is definitely known. Suppose you place an atom in a superposition of the ground state $|g\rangle$ and an excited state $|e\rangle$ in vacuum, and wait for much longer than the decay time. What's the state of the electromagnetic field afterward? We know that $|g \rangle$ will stay in the ground state, so the electromagnetic field will ...

0

In this paper, we attempt to constrain the large scale curvature of the Universe using distances obtained from observations of Type Ia supernovae together with inferred ages of passively evolving galaxies and Hubble parameter estimates from the large scale clustering of galaxies. Current data are consistent with zero spatial curvature, although the ...

2

The primordial abundances tell us about the density of baryons only. Before the discovery of what is now called dark energy, it was widely believed that dark matter was the gap between the baryonic density and the total energy density of the universe. In turn, the total energy density was thought to be close to (though measured to be less than) the critical ...

1

First we should remember that, $P = w\epsilon$ where $\epsilon$ is the energy density. as pressure goes up in the system mass increases its attractive force and collapses inward, Mass does not increase its attractive force as P increases. We know that $w = 0$ for matter. This means that there is no exerted pressure by the non-relativistic matter An ...

1

whereas before the inertial and Rindler observers can also be in the same place, now we should have an observer away from any gravitational field and one near a black hole; moreover, while the latter would see the black hole radiate as if it were a black body at Hawking's temperature, Minkoski's observer would still be immersed in the void and therefore, ...

2

If we assume the universe is electrically neutral (it might not be completely neutral, but it is neutral to a large degree) then there are roughly the same number of protons as there are electrons. Because the mass of protons (and neutrons) is about 1836 times that of electrons, it's safe to assume that the contribution from electrons is negligible.

3

You're certainly on the right track. I'll review the situations as you stated, so that the comparison is clear. 1) In Minkowski spacetime we are interested in comparing two families of observers, the inertial ones and the uniformly accelerated ones. As you clearly stated what the inertial observers consider to be vacuum, i.e. absence of particles, the ...

1

According to Wikipedia, https://en.wikipedia.org/wiki/Observable_universe even though the Universe is only about 14 billion years old, the present distance of the farthest objects whose light reaches us now is as far as 46.5 billion light years or $4.4\ 10^{26}$ meters. This distance is just what is meant by the "particle horizon". The volume of the sphere ...

2

Keep in mind that CCC is a dead theory. It became clear by 2010 that it was not viable because it made predictions about particle physics that were not consistent with what we know about particle physics. The best reference on CCC that I know of, other than the popularization in Penrose's book, is: R. Penrose, Singularities and time-asymmetry, in General ...

0

For Particle Horizon distance you can use the $$r = c\int_{t_e}^{t_o}\frac{dt}{a(t)}$$ But you can write this equation in terms of z and (making limits from $\infty$ to $0$) to make calculations easier. To calculate the conformal time part you have to divide above equation by c. $$\eta = \int_{t_e}^{t_o}\frac{dt}{a(t)}$$ Edit: So we know that 1 + z =...

1

The problem is three dimensional, not on a 2D sphere. The crucial point here is that space is homogeneous. Any point in space can be taken as the origin. So suppose you have an observer. You are perfectly allowed to decide that the position of this observer is the origin of your system of coordinated. So he is sitting at the point $r_0=0$ and there, $\phi$ ...

0

I ll try to answer Once the north and the south poles are fixed, one can imagine and draw the shortest arc-like path connecting two arbitrary points on the sphere that is neither along a latitude nor along any longitude. Such paths are also geodesics with varying θ and ϕ. I ll try to explain it over the 2-sphere where we have only $(r, \theta)$. So the ...

0

I think that the Sagnac effect is due to the relativistic Coriolis force, once you clarify what the relativistic Coriolis force actually is. It's important to understand that fictitious forces in spacetime are rather different from fictitious forces in Newtonian space+time. In the Newtonian case, you can introduce $F=ma$ forces to switch between frames ...

-1

Michelson and Gale measured the CORIOLIS EFFECT, which is proportional to the area of the interferometer, and not the SAGNAC EFFECT, which is proportional to the velocity (and thus to the radius of rotation). Here is the derivation of the Coriolis effect formula featured in the 1925 paper published by A. Michelson: https://www.ias.ac.in/article/fulltext/...

0

Various cyclical cosmological models have been discussed at various times: https://en.wikipedia.org/wiki/Cyclic_model Tolman used thermodynamics to rule out the most straightforward ones in 1934. There is currently no promising-looking model of this type that doesn't involve a lot of very speculative and unproven physics. Various proposals of this type, such ...

0

I am not sure how to prove it mathematically but the answer is yes. Now, we are assuming that the universe is homogeneous and isotropic so when the universe expands this homogeneity and isotropy should remain the same (it cannot be distorted). While we describe such expansion, scale factor cannot have dependence on the position. Actually thats the only ...

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