New answers tagged cosmology
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The cosmic microwave background provides a convenient reference frame for measuring motion (called the co-moving frame). If you are moving relative to the CMB then the doppler shift means the CMB looks slightly hotter in the direction you are moving (the dipole anisotropy), and slightly cooler in the other direction. This motion is called the peculiar ...
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I think I don't quite get your question.
If those galaxies had been moving through space, the redshift from every single object (for instance, the Proxima Centauri and some other star say, Sigma Orionis) would've been the same. It wouldn't have been such a big headache. But, it's not. Hubble's observation itself showed that the whole universe is ...
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Given 2 objects moving at some velocity v relative to one another, is it possible to determine whether they are moving or whether the space between them is expanding?
No, the difference between the two interpretations is fundamentally not testable. It's a verbal distinction that doesn't appear anywhere in the actual mathematical formalism of GR. GR ...
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We know that the CMB is isotropic when viewed outside of the spinning and revolving earth.
As pointed out in Edgar Bonet's comment, this isn't true.
Is it homogeneous?
Realistic cosmological models describe it as approximately, but not exactly, homogeneous. Homogeneity and isotropy can't be perfect, since the universe does have structure.
Can ...
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This is a good question, unfortunately my answer will not be adequate.
The CMB does define a special frame of reference, atleast if one ignores fluctuations, there is a special frame where the CMB spectrum is homogeneous and isotropic. However it will not be in contradiction with the theory of relativity.
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The point of Ted Bunn's answer to the question you link is that when physicists call something a law, e.g. Newton's laws or the Coulomb law, we mean that it is an approximate law. For example Newton's laws are a low speed, low density approximation, and in situations where speeds and densities are high we have to use an improved description i.e. general ...
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For the first question, it is the same quantity, the Minkowski dot product of the four vectors $k$ and $u$ that you may call $A_{O}$ and $A_{S}$, in general
$$g_{\mu\nu}k^{\mu}u^{\nu}=k_{\nu}u^{\nu}\equiv A$$
computed for the source and computed for the observer. So you have
$$1+z=\frac{A_{S}}{A_{O}} $$
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If you want an honest answer, the answer is WE DO NOT KNOW. It could be, it could be not.
Look, just couple of weeks or months ago we find out that our Earth's core is significantly different than we previously thought.
We don't know how many ground water is on our planet. It may be surprising but we really don't know that. Here is more about the fact that ...
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No, light always travels by the speed $c$ when measured as the proper distance over proper time in any frame that locally resembles an inertial one. So in the $c=1$ units, $dx_{\rm proper}/dt_{\rm proper}$ is always equal to one, never two.
The expression $ds$ is the proper time (when a time-like interval is substituted to the formula for $ds^2$) while ...
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You seem to be talking about inflation and expansion as if they were the same thing; they aren't. A Kasner metric has expansion and contraction, but it doesn't have anything like inflation. Inflation is exponential and is driven by a scalar field; the Kasner metrics are vacuum solutions and their behavior isn't exponential.
[...]what exactly would the ...
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I would suggest Stellar Structure and Evolution by Rudolf Kippenhahn and Alfred Weigert. It's a pretty thorough reference guide in terms of describing the processes that occur in stars and how they are used in stellar models. It's also cited quite a bit in astrophysics papers which deal with computational models of stars. It includes a brief chapter on ...
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It has been argued that structure on subgalactic scales is incompatible with standard ideas about cold dark matter, and that instead it indicates "warm dark matter" made of particles with a mass of about 1 keV.
At the same time, there are possible observations of dark matter accumulating at scales of 10 GeV and >100 GeV, which is more in the usual range ...
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It is not so easy to give a definition of time. We see things change when time changes, time is the fundamental measure of the evolution, which could be a local evolution, or the evolution of the entire universe itself.
Say that time is "slowing" would mean that there is another time-like quantity (distinct of time) which would be more fundamental, but, ...
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It looks like as if there was a race between our point running away from those galaxies (with the expansion of universe and space) and the light that was emitted at that time. And only now that light has reached and overtaken us.
That's correct. A photon from a distant source has to overcome the expansion of the universe in order to reach us. I'll ...
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A cosmological horizon isn't the same thing as a black hole horizon--the black hole horizon is an essential feature of the spacetime that is located where it is due to special geometry. A cosmological horizon is an observer-dependent phenomenon that describes when two observers are out of causal contact with each other. The only sense in which white hole ...
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Actually, any uniform expansion of the Universe, no matter how small, will at large enough distances make the galaxies there - which are locally at rest - recede from us faster than the speed of light. No matter how slow the expansion is, as long as it is uniform everywhere in space!
This is not in violation of any laws of physics; Special Relativity ...
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The light from formerly nearby galaxies needed this long time to get here because the Universe – and the distance in between the source galaxy and ours – was expanding as the light was travelling. So when the light got to the middle point, for example, the distance between both galaxies was already just a bit smaller than 1/2 of 13.7 billion light years.
...
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We have a pretty good idea of the expension history of the Universe from numerous measurements etc., and we also know with great precision how fast light travels through empty Space, so if we assume our understanding of the Universe's expansion history is not-too-wrong, we can simply calculate the light travel time, current distance and original distance of ...
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$R=0$ at $t=0$. This is a boundary condition. Then, $K=0$ must be true. Ignore lines 3 and 4, they are wrong. Just take $K=0$ and do what you did with setting $R=R_0$ and $t=t_0$ and you've got the answer.
Edit - Response to comment:
Short answer
I guessed that you were representing the scale-factor by $R(t)$. (I normally use $a(t)$ by the way). We know ...
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The big bang is not an explosion in the conventional sense of the word. The big bang corresponds to an exponential expansion of spacetime and it is this incredible rate of expansion that can be dubbed "explosive".
There are people suggesting that there might be "more universes" and that we can detect their effect in the CMB. I remember however that recent ...
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In an FRW cosmology, we have a scale factor $a$. For example, if $a$ grows by a factor of 2, galaxies have gotten twice as far away as before. The contribution to the stress-energy tensor from radiation goes like $a^{-4}$, from massive particles $a^{-3}$, and from a cosmological constant $a^0$, i.e., it's constant. For these reasons, the universe has gone ...
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The big bang is completely centerless.
The right way to picture where the CMB is coming "from" is to imagine that, long ago, the universe was as hot as the surface of the sun. This means that the universe was filled with a super-hot plasma that looked much like the surface of the sun today. Then, after enough expansion, the gas cooled off, and the ...
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The CMB doesn't come from one place; it comes from everywhere. It also doesn't go in one direction; it goes in every direction. And it didn't happen all at once, but it happened at roughly the same time everywhere. In particular, it's not the light of the Big Bang, but the light of a time roughly 378,000 years after the Big Bang.
The idea is that the ...
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Unbroken supersymmetry at any temperature is not relevant to the motivations for supersymmetry. So, considering that SUSY is usually broken at finite T does not affect the usual rationales given for using SUSY (different question altogether).
Your other question about assuming that $T=0$ for every renormalization scale $Q$ above $Q_c$, the SUSY breaking ...
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One of the Planck papers discusses these anomalies in detail: Planck 2013 results. XXIII. Isotropy and Statistics of the CMB.
How unexpected is this variance from the Standard Model and can it be quantified? How certain is it that the data are accurate? For the recent discovery of the Higgs boson at the LHC, a five sigma result was considered sufficient ...
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This difference is tiny and it's very hard to put any sort of "sigma" confidence to the measurement because we only have one Universe to measure. The anisotropy is small enough that it's hard to say if its statistically significant.
Everything I've read (sorry for some reason I can't find the sources right now) suggest that the Planck CMB measurement ...
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Yes, cosmological horizons are expected to emit Hawking radiation. However, the temperature would be extremely low, about $10^{-30}$ K (Baez 2004 and Hu 2010). This is nothing like a Big Bang. This is all assuming dark energy that behaves like a cosmological constant. For dark energy with certain properties, you can get a Big Rip. See Adams 1997 for a more ...
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Have you read this simple enough article?
Once objects are bound by gravity, they no longer recede from each other. Thus, the Andromeda galaxy, which is bound to the Milky Way galaxy, is actually falling towards us and is not expanding away. Within our Local Group of galaxies, the gravitational interactions have changed the inertial patterns of objects ...
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The real proof of a matter = antimatter universe would be to find antimatter galaxies. If all light is exactly the same this would seem imposable. But if General Relativity works the same in a matter galaxy and in an antimatter galaxy, then it follows that a test of the light from an antimatter galaxy, similar to the test of General Relativity should show ...
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The quantity $k_{NL}$ in their paper isn't a "cutoff scale"; it is a scale at which the nonlinearities (therefore NL) in some quantities become substantial.
There is no cutoff scale $\Lambda$ in dimensional regularization; it's one of the main features and virtues of the dimensional regularization. Instead, $\Lambda\to \infty$ is replaced by the $\epsilon = ...
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Not really my field, but I'll take a crack at it.
I understand that if the region is smaller than the Jeans length, then pressure, which travels at the speed of sound (compared to the speed of gravity c), can build up fast enough to counteract the gravitational collapse. But if the region is bigger than the Jeans length, it collapses. What exactly does ...
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The short answer to this question is that we do not know. The subject of your question is still in early "speculative", theorizing, and researching stages. I can say this because collisions of bubble universes under the eternal inflation theory just happens to be my specific area of work.
Non-colliding bubble universes (and the local potential minima in the ...
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Well , the mass of the visible universe 6e51 kg contradicts this simplification .
The HUP is an inequality. It tells you that the mass must be at least this small number you give, it does not put bounds from above, that is why it is immaterial for our classical existence.
In addition the existing universe is not in a quantum mechanical virtual state, ...
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The radius of the observable universe is about 46 billion light years, which is considerably greater than its age of about 14 billion years. Since the radius of the observable universe is defined by the greatest distance from which light would have had time to reach us since the Big Bang, you might think that it would lie at a distance of only 14 billion ...
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Firstly, the light year measures distance, and not age. But I see your question here: "Can the radius of the universe(in ly) be more than its age (in y)?"
The answer is (surprisingly) yes. In fact, this is indeed the case.
Firstly, a little side story:
This is something that confounded Einstein himself, way before we even knew about the Big Bang. When he ...
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The formula for the Hubble Parameter is
$$ H_{a} = \text{H0}\sqrt{\text{$\Omega $R} \cdot a^{-4}+\text{$\Omega $M} \cdot a^{-3}+\text{$\Omega $K} \cdot a^{-2}+\Omega \Lambda } $$
where only
$$ \Omega \Lambda $$
is for dark Energy. The other Omegas are for radiation, matter and curvature. Reducing to dark energy reduces the formula to
$$ H_{vac} = ...
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If you regard curvature as a form of energy density (see Chris White's post), then yes, you should set $k=0$. If not, then $H$ will only approach a constant value as $a\rightarrow \infty$:
$$
H^2 \rightarrow \frac{8\pi G}{3}\!\rho_\Lambda \qquad \text{for $a\rightarrow \infty$.}
$$
You can check that the general solutions of
$$
\dot{a}^2 - \frac{8\pi ...
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Yes, you should probably assume $k = 0$, since otherwise the statement is not true, as you have shown.
The nature of the Friedmann equations allows us to rewrite the $k/a^2$ term as though it were itself a source of energy density $\rho_k$ varying as $a^{-2}$, complete with its own fraction $\Omega_k$ of the critical density. This is done by writing
$$ ...
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The answer asks for a solution that must encompass a much larger spatial region then the overdense/underdense region of interest. I think that you are aware of the usual answers focused in the overdense situation and I will focus mainly in the underdense case.
The underdense case: in the fig A below we have an intuitive description: in In an isodence ...
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