New answers tagged

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real astronomical objects are embedded in an expanding spatially flat FRW metric Not really. If you think of the cosmos as clumps of matter on top of a FRW background, you're counting the same matter twice: once in a perfectly uniform distribution and then again in its actual clumped location. You can start with FRW if you plan to construct a so-called ...


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In relativity, velocity is direction/angle and elapsed time is distance, so if you have different objects moving at different velocities from a common point for the same amount of time (as measured by their intrinsic evolution), they'll end up on a sphere of constant distance from the starting point, not in a plane as they would in a Newtonian world with ...


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RW assumes that matter is uniformly distributed over the spatial slices, and elliptic, flat, and hyperbolic geometries have very different distributions, in terms of the amount of matter within a given distance of any given point. There's no way you could move the matter around to be homogeneous in a different geometry without violating homogeneity in the ...


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Yes. The Hubble constant refers specifically to the current value of the Hubble parameter, which is not a constant on cosmological time scales. As such, its measurement is indeed restricted to low redshifts. Other measurements are possible, such as measurements on the CMB. Such measurements place restrictions on cosmological parameters generally. But they ...


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Do all measurements of the Hubble Constant have to be made at low redshift values? No. Take for example the cosmic microwave background (CMB) with a redshift of roughly $z \approx 1100$. The Planck satellite has measured the anisotropies from the CMB across the full sky. Assuming a flat $\Lambda$CDM cosmology you can estimate the cosmological parameters ...


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Yes. "The curvature of the universe" is an imprecise term, and describing the curvature of a general four-dimensional spacetime takes 20 numbers at every point. But I'll assume that your phrase should mean the Ricci scalar curvature $R$, which is a single number at each point that is a kind of average curvature of spacetime (where the averaging is ...


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No The curvature parameter $k$ of the universe remains constant throughout its evolution. If the universe is open ($k < 0$) , it will stay open, and if it is closed ($k > 0$), it will stay closed. That's because the amount of matter-energy of the universe is conserved, so if the density is greater than critical now it will forever be greater than ...


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Assuming (as this chart does) that the universe contains only dust and dark energy, the scale factor satisfies $$\dot a^2 = H_0^2 \left( Ω_{Λ,0}\, a^2 + Ω_{k,0} + Ω_{m,0}\, a^{-1} \right)$$ where $Ω_k = 1 - Ω_m - Ω_Λ$. (The exponents are $-1{-}3w$ where $w$ is the equation of state parameter.) The boundary of the $κ=\pm1$ regions is just the line $Ω_{k,0}=0$....


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There is a close analogy: If you jump say 3 feet high, you can calculate your speed as a function of you position between 0 to 3 feet high. Then you wonder what is your speed at 6 feet high during your jump. The good old high school Newtonian mechanics offers an answer: your speed was imaginary at 6 feet high according to the total energy conservation ...


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One way to interpret Friedmann equation with an imaginary Hubble parameter is as arising from some solution with Euclidean metric signature. One class of such solution, termed “Euclidean wormholes” consists of two large asymptotic regions connected by a “throat”. Many FLRW cosmologies analytically continued into an imaginary time become such Euclidean ...


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Yes. The answer is kind of obvious, but if you do not see this, just note that the FLRW metric is: $$g=-c^2dt^2+a(t)^2d\Sigma^2_t,$$ where $d\Sigma^2_t$ is metric induced up to the scale factor on the spacelike hypersurfaces of constant time $t$. This metric is independent of $t$, from which it follows, that if you have a curve living on one of this ...


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As far as we know, it's a coincidence. $1/H = a(t)/a'(t)$. In the early radiation-dominated era, $a(t) \propto t^{1/2}$, so $1/H$ is 2 times the actual age of the universe. In the later (but pre-modern) matter-dominated era, $a(t) \propto t^{2/3}$, so $1/H$ is 1.5 times the actual age. In the future dark-energy-dominated era, $a(t) \propto e^{t/t_0}$ where $...


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Velocities of galaxies When galaxies are gravitationally bound to each other in groups or clusters, they move on more or less elliptical orbits in the common gravitational potential from all the other galaxies (as well as all the dilute intracluster gas which is also a significant part of the total mass). I say "more or less" because galaxies do ...


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In theories with dark energy, an energy density is just a property of space — even “empty” space devoid of matter and radiation. It’s closely related to Einstein’s old idea of a “cosmological constant”. Dark energy isn’t something material that comes from somewhere. As space expands, you simply get more dark energy because there is more volume; a dark energy ...


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Your model works quite well in Newtonian gravity. You can even derive the Friedmann equations describing the rate of expansion of the universe from your model, and they match the equations from real cosmology, in the appropriate $c\to\infty$ limit. If you adapt your model to general relativity, you get the standard cosmological model. If you start with a ...


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You're right that in the case of matter moving away at different velocities from a given point, any observer would be moving away from any other observer at a velocity proportional to the distance between them, i.e. Hubble's law would also be true. If this scenario were true, it would mean that our Universe wouldn't be described by general relativity (GR), ...


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Hubble's formula $v=H.d$ is just an approximation available only for the galaxies in which the recessional velocity is moderated. Indeed, if we observe a galaxy located at $2$ million light-years, we measure today $t_0$ the speed and the distance this galaxy was 2 million years ago, in other words, the time the light took to get from the galaxy to us! If ...


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That's simply because there the dark matter density in galaxies is high, but the density in intergalactic space is low. The reason for that, in turn, is because dark energy is a property of spacetime itself - any spacetime, wherever it is. In contrast, dark matter like normal matter follows gravitational forces and clumps into galaxies and other such ...


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These aspects of astronomy and cosmology are indeed very interesting and very significant, but don't allow the names to get in the way of your understanding. Dark matter is a form of matter made (most likely) of particles which don't interact very much with the matter we are more familiar with (i.e. protons, neutrons, electrons etc.). The evidence for it has ...


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Hubble law states that galaxy receding speed is : $$ v = HD $$ Acceleration by definition is speed change over time, so : $$ a=\frac{dv}{dt}=\frac{d\left(HD\right)}{dt} $$ Applying product rule gives : $$ a = D\frac{dH}{dt} + H\frac{dD}{dt}$$ Usually Hubble parameter change over time is expressed as : $$ \frac{dH}{dt} = -H^2(1+q) $$ Substituting it into ...


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Both experiments report $H_0$, the Hubble parameter at redshift $z=0$, now. The different experiments directly measure $H(z)$ at different $z$'s then use the standard model of cosmology to determine $H_0$. Late time or local measurements, like those using Type Ia supernovae and a cosmic distance ladder, measure $H(z)$ for small $z\sim 1$. Early time ...


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Both sets of experiments are measuring the same quantity: $H_0$, the Hubble parameter today. When people say that Type Ia measure H0 "locally" they mean that the observations are happening at a small redshift. On the other hand, the surface of last scattering was generated at a redshift of about 1100. Nevertheless, both sets of observations are ...


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Your first sentence is already not defined. If by "standpoint of a light ray" you mean "a hypothetical observer travelling at the speed of light", you are already in undefined territory. The Lorentz transformations, which mathematically tell us how to change between inertial reference frames contain a factor: $$\gamma=\frac{1}{\sqrt{1-v^2/...


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The question asks about opposite edges of the observable universe, but it's a special case of a more general question: How can two objects that move away from a common point for a time $Δt$ end up more than $2cΔt$ apart? You don't need to understand general relativity to understand the answer to this question, since it can happen even in special relativity....


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We can't locate the centre of the universe because there isn't one. Every part of the universe was part of the big bang and there are numerous duplicates on Physics SE that address this. e.g. What is our location relative to the Big Bang? Does the universe have a center? Can the coordinate of the big bang point be calculated via observed universe or it is ...


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I think there are two different questions in your question. Or maybe it is better to say that before asking your question, another question should be considered first. Your question: Can we estimate where the center of the universe is? You also ask about the rate of expansion but I think we have to address the question above first. And maybe your question ...


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Yes, gravity can slow down the expansion of the universe. That's why if the universe's average density is greater than critical, it is forecast to collapse into a big crunch (see this section of the Wikipedia article on the Friedmann equations). I don't understand your second paragraph very well, but it seems like a misconception about what actually is ...


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Gravitationaly bound objects, such as the galaxies in the local group (Milky Way and Andromeda for instance) don't destroy the space between them, rather space doesn't expand between them. The expansion of the universe was slowed by the gravity of the objects in the universe until 5 billion years ago when it was dissipated enough that dark energy began to ...


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I think an example should help illustrate what's going on. First, there is no discontinuity when inflation ends: the rate of expansion goes smoothly from accelerating to decelerating, much like a ball rolling down a hill. To start, consider the Friedmann equation, $$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho + 3p),$$ where $p = w\rho$ relates the pressure ...


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As far as we know, the Universe does not have a boundary. But, if it does, then we will need to know what are the boundary conditions in order to answer your question of what a particle will do when it reaches it. For example... The Universe could have reflecting boundary conditions, in which case the photon would bounce back from the boundary. The Universe ...


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So what is the flaw in my reasoning? The flaw is that the FLRW spacetime is not a vacuum solution. Indeed, as you correctly point out the one you wrote down has positive scalar curvature. A vacuum solution has a zero Ricci tensor and therefore a zero scalar curvature. Physically, the FLRW spacetime represents an isotropic and homogenous distribution of ...


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The redshift that is measured is the sum of the redshift due to the expansion of the universe, a Doppler shift caused by that component of the peculiar velocity that is radial and a second order (negligible) Doppler shift due to the tangential component of the peculiar velocity. It is not possible in general to separate out these effects. The exception is ...


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The peculiar velocity of a distant galaxy can affect the overall red shift. For example there are local galaxies like Andromeda that are slightly shifted blue because they are advancing.


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how does the dark energy affect the universe (why we usually related them with universal expansion) To answer your second question, you need to look at the acceleration equation. $$\frac{\ddot a}{a}=-\frac{4\pi G}{3c^2}\sum_i(\varepsilon_i + 3w_i\varepsilon_i)$$ $$\frac{\ddot a}{a}=-\frac{4\pi G}{3c^2}\sum_i\varepsilon_i(1 + 3w_i)$$ Where $\varepsilon_i$ ...


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What appears to be a sufficient answer to the question can be found in this SE answer by @JohnRennie, combined with a few other articles. I had confused "flat space" with "flat spacetime". As John Rennie said in that answer, spacetime is not flat in an expanding universe, but space can be flat. So, indeed, it is necessary to account for ...


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A couple of things up front: first, you seem to be talking about the present-era expansion of the universe, not inflation. Inflation, if it happened at all, ended about 14 billion years ago, long before the formation of large structures like galaxies and very long ropes. Second, a galaxy with a recession velocity of c in the present era is going to be some ...


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The figure 4.2 sigma means that the discrepancy between the two numbers is 4.2 times the estimated standard deviation of the difference between the two numbers, under the assumptions that the measurements are independent and the uncertainties have an approximately normal distribution. If I have two measurements $a \pm b$ and $c \pm d$, where $b$ and $d$ are ...


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You have two data sets $A+\delta A$ and $B+\delta B$. To see that these two measurements agree or not we can calculate their difference. Let us call this value as $D$ such that $D=A-B$ and $\delta D = \sqrt{\delta A^2 + \delta B^2}$. For instance, in some measurement, we find that $D\pm\delta D \equiv 0.03\pm0.05$ In this case, as you can see D ranges ...


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I guess you are looking for angular diamater distance. For different curvature the equation takes different forms. See here https://en.wikipedia.org/wiki/Angular_diameter_distance


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NEWTON the energy is: $$\frac{1}{2} \dot{a}^2+V(a)=E\tag 1$$ where $V(a)$ is the potential energy and $\frac{1}{2} \dot{a}^2$ the kinetic energy. you can describe equation (1) using the Habbel parameter $H=\frac{\dot a}{a}$ $$H^2-\frac{2E}{a^2}=\frac{8\pi\,G}{3}\,\epsilon\tag 2$$ where $\epsilon$ is the energy density with equation (1) in (2) you obtain ...


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