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Question coming from Cosmological Perturbation

Welcome to the physics StackExchange (seeing that you are a new user)! For a reference to why we do perturbation theory this way, with the goal in mind to compute (typically) the curvature and scalar ...
MathZilla's user avatar
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1 vote

Distance and luminosity distance

Using the inverse square law we can deduce that the flux F(watts/m^2) received from a source with an intrinsic luminosity L(watts) diminishes as a function of distance squared so $F \propto L/D_e^2$ ...
KDP's user avatar
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Clarification of Weinberg's cosmology book eqns 5.1.44-5.1.47 for scalar perturbation

$B^S_{ij}$ does have terms on the diagonal. It is just that the trace is zero. Here's an example of a $3x3$ matrix with terms on the diagonal but zero trace $$ \begin{bmatrix} 1 & 0 & 0 \\ ...
Andrew's user avatar
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Spatial Curvature of Universe at recombination vs now

We do not use CMB data to directly measure the curvature of the universe. Instead, we use CMB data to estimate the average density (as well as the size of density fluctuations) in the early universe. ...
gandalf61's user avatar
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Are there closed simply connected 2D manifolds that do not require a third dimension?

There are two distinct aspects to the question you are asking about representing a closed simply-connected curved 2D surface like a sphere without requiring extra dimensions. One is topological. The ...
Nullius in Verba's user avatar
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Are there closed simply connected 2D manifolds that do not require a third dimension?

You wrote All the examples of closed 2D manifolds I have found are defined on a 3D surface. From a mathematical perspective, there are several constructions of closed 2D manifolds, and closed ...
Lee Mosher's user avatar
3 votes

Are there closed simply connected 2D manifolds that do not require a third dimension?

Aside from @mmesser314’s nice answer let me mention some other results. From the classification of surfaces, we know that the only (non-empty) 2-dimensional compact boundaryless simply connected ...
peek-a-boo's user avatar
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3 votes

Are there closed simply connected 2D manifolds that do not require a third dimension?

No. A $2$D manifold might fit perfectly well in a $3$D manifold if it exists. But that does not mean that it has to exist. There are a couple ways to approach this. First "exists". Manifold ...
mmesser314's user avatar
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As per Friedmann Equations, is big-bang singularity necessary?

In a closed (unlikely, but still possible within the measurement accuracy) dark energy dominated universe where $\Omega_{\rm K}=1-\Omega_{\Lambda}$ (which it was during inflation if there was any, ...
Yukterez's user avatar
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1 vote

What is the gravitational field of a hole in an infinite perfect crystal?

... overdensity attracts underdensity and underdensity repels overdensity ... Yes, in an empty universe a pair of matched positive and negative masses initially at rest would be self-accelerating ...
A.V.S.'s user avatar
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2 votes

What is the gravitational field of a hole in an infinite perfect crystal?

The equations that govern gravitational fields and matter simply don't make sense when you apply them to an "infinite" volume of matter. Mathematically, we have $\nabla^2 \phi = 4 \pi G \...
Michael Seifert's user avatar
2 votes

What is the gravitational field of a hole in an infinite perfect crystal?

If you put overdensity next to underdensity, overdensity attracts underdensity and underdensity repels overdensity. They chase each other forever with runaway acceleration. Not true. A volume with ...
gandalf61's user avatar
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3 votes
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Will expanding space rupture an empty box floating in outer space

The box would only be ripped apart if dark energy is "phantom energy", which increases in energy density as the Universe expands. In that case, the destruction of the box would occur right ...
Sten's user avatar
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1 vote

Viable values for the $K$ parameter in the FLRW metric

$$c^2 d\tau^2 = c^2 dt^2 - \frac{a(t)^2}{(1-KX^2)} dX^2. $$ For this metric expression, if you set $K=1$, then the spatial surfaces are hyperspheres of radius $a$. Every position on these ...
Sten's user avatar
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2 votes

Viable values for the $K$ parameter in the FLRW metric

First, define a new time coordinate $\eta$ through integrating $$ d\eta = \frac{c dt}{a(t)}.$$ This time coordinate is often called conformal time. With this coordinate, the condition for a radial, ...
11zaq's user avatar
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1 vote

Can a light signal from Earth reach a galaxy outside the Hubble Horizon?

Here are some more reasons a light signal from Earth must be able to reach and pass the Hubble Horizon: The total velocity of a photon is given by its peculiar velocity and its recession velocity as ...
KDP's user avatar
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-1 votes

Can a light signal from Earth reach a galaxy outside the Hubble Horizon?

The details of the answer depend on the specific scenario, so let us assume for simplicity that the horizon is always at a constant distance. The chain of galaxies that your are imagining, assuming ...
Pato Galmarini's user avatar
6 votes

Can a light signal from Earth reach a galaxy outside the Hubble Horizon?

There are various "horizons" in cosmology, and it is important to distinguish between them. This Wikipedia article describes several such cosmological horizons, two of which are the Hubble ...
gandalf61's user avatar
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2 votes

Why is FTL travel impossible if the universe expands FTL?

The expansion of the universe is not associated with any faster-than-light motion. We sometimes speak of faster-than-light recession rates (for example, in the context of Hubble's law), but these ...
Sten's user avatar
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0 votes

Why is FTL travel impossible if the universe expands FTL?

In basic understanding the Universe is not an "Object" by which travels, its' the medium by which Objects Move. The subsequent "speed limit" does not apply to the expansion of ...
LazyReader's user avatar
1 vote

Is the FRW metric, based on spatial homogeneity and isotropy, rotationally and translationally invariant? If so, how?

If what you're confused about is why the metric seems to depend on the coordinates and thus might not be translationally or rotationally invariant, look at the Minkowski metric (flat spacetime) in ...
controlgroup's user avatar
2 votes
Accepted

Is the FRW metric, based on spatial homogeneity and isotropy, rotationally and translationally invariant? If so, how?

Roughly speaking, in addition to Einstein equations, the (spatial) FLRW metric is constructed by assuming that, at fixed time, in the Riemannian manifold defining space: (a) metric properties are ...
Valter Moretti's user avatar
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Understanding expansion of the Universe as things flying apart

Interesting question. If such a hypothetical universe existed without gravitational attraction all objects in motion or at rest would tend to stay that way and most of the particles that exist in our ...
c morrison 973's user avatar
2 votes

Why can we use Bose-Einstein statistics in this expression for number density

This is usually discussed toward the beginning of any solid state textbook. The authors of those cosmology books probably assume that you have a comprehensive education in physics up to that point... ...
T.P. Ho's user avatar
  • 91
1 vote

What causes the 41k obliquity cycle?

Milankovitch cycles cannot be calculated just from Sun-Earth-Moon interactions, but must include the effects of the other planets, especially Venus, Jupiter, and Saturn. These are messy multibody ...
David Bailey's user avatar
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0 votes

Can Bose-Einstein condensates and Fermionic condensates survive for long periods of time in space?

Fermions do not form a condensate, since they can not accumulate in one quantum state. Sometimes people call pairs of fermions condensed, such as in superconducting materials, but I assume that this ...
kai90's user avatar
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Saddle Shaped Universe

It is simply an analogy (and not a very precise one) to illustrate that even in our familiar 3D space there are surfaces with negative curvature at all points. What is drawn in the illustration may be ...
gandalf61's user avatar
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2 votes

Understanding expansion of the Universe as things flying apart

Can the Universe be pretty much seen just as things flying away from each other [...] ? Absolutely! This is precisely what expansion means. That nearby particles are moving apart. (Technically, the ...
Colin MacLaurin's user avatar
3 votes

How can we model the primordial Universe while the interior of a neutron star and comparable states of matter are still mostly unknown?

The centre of a neutron star is at densities of $\sim 10^{18}$ kg/m$^3$, temperatures of $10^8$K-$10^{10}$K. The pressures are $\sim 10^{34}$ Pa and the nucleons are separated by $\sim 10^{-15}$ m. ...
ProfRob's user avatar
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4 votes
Accepted

Beginning of a dark energy (cosmological constant) dominated universe

You have correctly observed that an eternally dark energy dominated universe has no beginning. Another way to see this is to realize that a dark energy dominated universe is a static universe. Its ...
Sten's user avatar
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0 votes
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Friedmann Equation and a contracting universe

The first Friedmann equation says nothing about the sign of $H$. During expansion, it says how fast the universe is expanding. During contraction, it says how rapidly the universe is contracting. ...
Sten's user avatar
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3 votes

Doppler Effect and the concept of relative velocity in GR

For distant objects in a curved spacetime, you can define their velocities by means of parallel transport. Choose a spacetime path between yourself and the object, and drag its 4-velocity vector along ...
Sten's user avatar
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0 votes

Free Neutron Decay

Let's just crank it out with: $$ n \rightarrow H + \bar{\nu_e} $$ I'm going to set $m_{\nu}=0$, for ...reasons. We have: $$ p^{\mu}_n = (m_n, 0) $$ going to: $$ p^{\mu}_H = (E', p') $$ $$ p^{\mu}_{\...
JEB's user avatar
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-3 votes
Accepted

Free Neutron Decay

Definitely, almost obviously. Most free neutron decay would send the electron out, with only a rare subset having bound hydrogen. However, at large enough numbers, then an electron that had already ...
naturallyInconsistent's user avatar
1 vote

Schwarzschild Radius of the Universe

The answer by probably_someone is technically incorrect. He is confusing the Schwarzschild radius with an event horizon. They are not the same thing. the conditions that the author assumed in the ...
KDP's user avatar
  • 6,102
1 vote

Increasing the view of observable universe

This won't work. To increase the observable universe for someone earth-bound, the probe needs to observe events and relay information about that event to earth. The probe may see the event before ...
Brick's user avatar
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1 vote

Homogeneous and Isotropic But not Maximally Symmetric Space

OP's question is closely related to their preceding one. I have actually provided a near-complete answer to the present question there, although it appears that OP's question predates the clarifying ...
Bence Racskó's user avatar
1 vote

Cyclic Universe Problems

If a Big Rip is not part of your plan to rid the expanding universe of mass, you are dependent on particle decay to get the job done right. This is problematic though because while photons are ...
Wookie's user avatar
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2 votes

Homogeneous and Isotropic But not Maximally Symmetric Space

You are correct, a spacetime need not be maximally symmetric to be homogenous and isotropic. Isotropy and homogeneity are restrictions on the spatial structure of the universe, which lead to spatial ...
CompassBearer's user avatar
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Understanding expansion of the Universe as things flying apart

EDIT: I realise this answer falls into the TLDR ("Too Long - Didn't Read") category, so I decided to summarise the main arguments right here in a preface: Let's say there two basic competing ...
KDP's user avatar
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4 votes
Accepted

Understanding expansion of the Universe as things flying apart

The scenario: within a matter-dominated universe, you have prepared two particles with initially constant separation. In this scenario, those two particles will begin to fall toward each other. This ...
Sten's user avatar
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0 votes

Can there be a theoretical synchronised ‘now’ moment at all points across the universe?

A good explanation of what Now means in a relativistic framework. I do have some critical comments. That a light cone defines the area from which information can be received by an observer, this ...
Daniel Jencka's user avatar
1 vote

Could cosmic rays induce a vacuum decay in the future?

The usual line of reasoning about vacuum decay treats it as a random phenomenon: there are events where it might be triggered, and since it has not happened yet we get a bound on how likely it is. ...
Anders Sandberg's user avatar
2 votes

Are there universes where "symmetry breaking" went differently?

In classical physics there is only one history and symmetry breaking is understood in terms of saying that any particular system is very unlikely to perfectly respect the symmetry even if the ...
alanf's user avatar
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1 vote

Are there universes where "symmetry breaking" went differently?

I think, since we can only study the physics of this universe, this question counts as meta-physics. Our current theories allow us to describe the reality that we see, and that's all that we need. As ...
Chaddyfynn's user avatar
1 vote
Accepted

Are there universes where "symmetry breaking" went differently?

I'm not an expert in quantum field theory or string theory, but as I understand it, these "other universes" have different laws of physics (in the low-energy limit). At some very high energy ...
Allure's user avatar
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1 vote

Is it true that $\dot{H}(t)\sim H(t)$, and if so, why?

Wild Feather wrote: "This leads me to believe that, for some reason, $\dot{H}$ can be considered to be of the same order as $H$." In a radiation dominated universe we have $$\rm H=+\frac{1}{...
Yukterez's user avatar
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3 votes
Accepted

Is it true that $\dot{H}(t)\sim H(t)$, and if so, why?

$\dot H$ is not comparable to $H$, since the two have different dimensions. However, for a flat cosmology, you can use the two Friedmann equations to show that $\dot H$ is of order $-H^2$ (except for ...
Sten's user avatar
  • 6,404
2 votes

Black Hole Formation -- How Can an Event Horizon be Observed to Grow?

It's the total mass that counts and we can indirectly observe a black hole growing through changes in the behaviour of objects surrounding it, those affected by its gravity and rotation. Naturally, ...
Wookie's user avatar
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0 votes
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Relation between Earth's and Sun's magnetic axis relative to their paths of travel and relative to each other?

From the Wikipedia article Earth's orbit, The magnetic pole is more complicated. First the magnetic north pole is misnamed from a physics point of view. The north pole of a magnet points north ...
mmesser314's user avatar
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