# Tag Info

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I believe a great answer to your question is: We don't know We still can't resolve the time before electroweak interactions, so how can we even come close to answering this question? You might get answers from some theories (or I prefer to call them theorems because they're only math until today) like string theory or loop quantum gravity or M-theory or ...

8

SN 1987A is (was?) in the Large Magellanic Cloud, which is gravitationally bound to the Milky Way. This means that its motion relative to us is only minimally affected by cosmological expansion, and talking about it in terms of a $z$ parameter is misleading at best. The best estimates of the distance to SN 1987A are about 168,000 light-years. If you ...

6

Short answer: The ratios have changed over time... drastically. This is a consequence of the expansion of our universe. Initially (and by that I mean after the conjectured inflationary epoch, which I will not consider here), radiation dominated all other forms of energy by far. However, as the universe expands---as measured by the increase of the ''scale ...

6

This is a metaphysical/philosophical question, imo. There is the platonic ideals school, in this case read for ideals=mathematics, which postulated that ideals existed and nature fell into their form. I have seen a number of theoretically inclined people who are really of that school. One does not have to think of the beginning of the universe to start ...

4

There is a limited sense in which this correct. When inflation ended there was a temperature rise known as reheating, and we believe it was at this stage that the standard model particles were first created, or at least the majority of them. If you are measuring temperature by the energies of the standard model particles then this would have been the hottest ...

4

You understand that $\mathcal C$ violation is required, as if it weren't, processes related by $\mathcal C$ that violated baryon number conservation would balance, i.e. $$P \to Q B \qquad \mathcal C:\qquad \overline P \to \overline Q \, \overline B$$ would result in no net baryon number violation. In these expressions, $B$ is a fermion carrying baryon ...

3

I think there are two ways to approach the question. If you are coming from the point of view of a theoretician trying to come up with a working model of the universe, you would definitely like to make the assumption of isotropy and homogeneity. This is usually what we do, as least to first order approximation. One of the reasons for that is that the ...

3

Gravity acts on all matter, not just water (it just so happens that water flows with less resistance than rock) which is why we get noticeable water tides but not very noticeable earth tides. However, if you were to bring a very large gravitating body too close to earth, you would find that the earth isn't quite as solid as it feels. The answer to your ...

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The Magellanic clouds are satellite galaxies of the Milky Way. They are right next door. Google says 61 kPc to the LMC which means trivial cosmological redshift.

3

It depends on the scales which you are interested in. During the deSitter evolution of the universe (let's assume exact deSitter and completely flat potential for simlicity) the fluctuations of the inflaton field exit the horizon and freeze in with a power spectrum of $H_*^2/(2\pi)^2$ where $*$ denotes that the value of the Hubble parameter is taken at the ...

3

In a static universe it would indeed be true that if you looked at an object, say, 10 billion light years away you would be looking at it as it had been 10 billion years ago. This isn't really an application of special relativity and is merely a consequence of a finite speed of light. Our universe, however, is expanding and so you can actually see across ...

3

As far as I understand, Physics is not able to awnser this question, because the physical laws we use to describe the Universe are not valid up to the exact event of the Big Bang (the Big Bang is said to be a singularity of spacetime). Physics attempts to describe the Universe at a moment when it already existed, but does not states causes for its existence ...

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Cosmic microwave background radiation is very "cold", i.e. the average energy is very low. It started in the region of atomic levels, order of electron volts, but is now order of magnitude lower and the only interactions it can have with atoms in the atmosphere are elastic scatters. The origin of the CMB radiation at 380.000 years after the Big Bang is ...

2

So the 'old' model of inflation was based on the idea that a scalar field (the inflaton) was trapped initially in some meta-stable vacuum. If it's energy density dominates the Universe at this time then the Universe inflates. This continues until the field tunnels out of this vacuum and inflation ends. This leads to bubble nucleation as different patches of ...

2

We cannot see anything closer than 380,000 years after the big bang because that is when radiation and matter decoupled. The CMB is a picture of what the universe looked like at that point. All clumping of matter into stars, galaxies, etc has occurred since then. If we had looked 1 billion years ago, we would see the same except that the CMB temperature ...

2

The main point to grasp is that the tiny inhomogeneities seen in the CMB are too small (by a factor of 100) to grow into the structures we see today without something like "cold dark matter". The CMB was formed at the epoch when normal (baryonic) matter and the radiation filling the universe decoupled. Only at this point was normal matter free to start ...

2

The Christoffel symbols for your metric are $$\begin{split} &\Gamma^0_{ij}=a\dot{a}\gamma_{ij}\\ &\Gamma^i_{0j}=\frac{\dot{a}}{a}\delta^i_j\\ &\Gamma_{jl}^{i}=\tilde\Gamma^i_{jl} \end{split}$$ then $$... 2 In a static spacetime, there is (by definition) a timelike Killing vector field \xi^\mu, which implies that geodesics with four-velocity u^\nu have a conserved quantity \epsilon = -g_{\mu\nu}\xi^\mu u^\nu. For example, in Schwarzschild spacetime, this is$$\epsilon = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\lambda}\text{,}$$where ... 2 the cutting by universes is a way : to introduce possible new physics for each of these universes without leaving the homogeneity and isotropy cosmological principles, the known constants and the known physics of "our" universe to defer the infinity issue from our universe to a parent structure : the multiverse Homogeneity and isotropy are the main ... 2 Very often we don't. We just stick to k=0, because for most practical purposes, this is fine. But all measurements contain uncertainties. The latest Planck results (Parade et al. 2015) combined with observations of the baryonic acoustic oscillations yield^\dagger \Omega_k = 0.000\pm0.005. We don't know whether the curvature on much larger scales than ... 2 A comoving observer and an observer that has been moving at 0.866c since Big Bang will disagree on their measured age of the Universe by a factor of 2. While both measurements are correct, we can say that the comoving observer measures a more "natural" age of the Universe. For instance, the comoving observer is the only observer who will measure the ... 2 Suppose two observers, Alice and Bob, are moving relative to each other since the beginning of the universe. While they do it, they construct the chronologies of all the events of the universe, as they record them in their frame of reference. They will construct different chronologies. However, and this is key, each can reconstruct the other's chronology. ... 2 From the theory of Thompson scattering (see http://farside.ph.utexas.edu/teaching/em/lectures/node96.html ) we know that a charged particle of mass m interacting with a plane wave electromagnetic field of Strength E_0 and frequency \omega has an effective dipole moment of magnitude$$d=\frac{e^2E_0}{m\omega^2} Note that the dipole moment scales ...

2

Actually there is a geometry that describes something like the naive idea of the Big Bang. But it's a bit of a cheat because it's really just a piece of the usual expanding universe metric. The first metric suggested to describe a collapsing star was the Oppenheimer-Snyder metric, which describes a spherical ball of dust collapsing under its own gravity. ...

1

Pulling together what's been said in various comments: 1) General relativity admits models where spacetime is foliated by spacelike leaves, all of which are indexed by a global time coordinate. The simplest of these models is Minkowski space. All of your observations about models with comoving observers apply equally well to Minkowski space, so if you ...

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You have$\frac E{m_0}$, the energy divided by the rest mass is $\gamma=\sqrt{1-\frac{v^2}{c^2}}$. The proper time is lab time divided by $\gamma$. Since you have a fixed $E$, as $m_0 \to 0, \gamma \to \infty$ and the proper time goes to $0$. For the last part, you are supposed to assume that an $11$ MeV neutrino arrived $7$ seconds before a $7$ MeV ...

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Of course you won't find it anywhere - SN 1987A is in the Large Magellanic Cloud, just 168000 ly away. At this scales cosmological expansion is negligible compared to other processes so measuring its redshift is says little useful about its distance. $z=0.1$ corresponds roughly to 1 Gly. The universe is HUGE. Here's plot from wikipedia

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What would happen if you were to release the energies of the big bang in our universe a second time? Have a look at this standard history of the universe, History of the Universe - gravitational waves are hypothesized to arise from cosmic inflation, an expansion just after the Big Bang Our universe is now at the far right. Note the beginning ...

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The term Big Bang does not have rigorous physical definition. And if you mean specifically inflation, then the modern inflationary models (chaotic inflation with various scalar field potentials) predict that inflation is always present at certain regions of space. Inflation is driven by a scalar field, the value of which lowers with time. When it drops ...

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As stated in the answer you linked, the density of a black hole is defined by the ratio of its mass over the volume spanned by its Schwarzschild radius. That does not mean that there is actually uniformly distributed matter inside the Schwarzschild radius. All of the matter is packed very densely (in something with the characteristic length of the Planck ...

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