# Tag Info

20

The rough idea is that under the assumptions contained in the cosmological principle, the application of Einstein's equations leads us to the equation $$d(t) = a(t) \chi$$ where $d(t)$ is called the proper distance and $\chi$ is called the comoving distance between two points in space. $a(t)$ is the time-dependent scale factor, which is by convention set to ...

15

I'm not too interested in providing an answer from the cosmological point of view. It is clear that the age of the universe derived in that way is model-dependent. The age thus obtained depends on certain assumptions (e.g. that the dark energy density remains constant). I will just add a couple of additional age determination methods that rely on alternative ...

6

To compute the age of the universe, one must solve the equation: $$\frac{1}{a}\frac{da}{dt} = H_0 \sqrt{\frac{\Omega_{\gamma,0}}{a^4}+\frac{\Omega_{m,0}}{a^3}+\frac{\Omega_{k,0}}{a} +\Omega_{\Lambda,0}}$$ where $\Omega_\gamma$, $\Omega_m$, $\Omega_k$, $\Omega_\Lambda$ are the densities of radiation, matter, curvature, and vacuum energy, and the subscript '0' ...

5

This is not a full answer, but I think it will help if you separate out inflation from the rest of the picture. The age of the universe can be estimated in the first instance as the time elapsed since some very early epoch where the temperature was low enough that the Standard Model of particle physics applies to reasonable approximation. This means you can ...

3

One can answer your question by using slightly odd measures of space and time that make the problem simple. A great (if perhaps technical) paper looking at this is Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe by Davis & Lineweaver (2003). You are right that the distance where galaxies ...

3

What's called the "age of the universe" would more accurately be called the age of the most recent epoch in the universe's history. That epoch began with the end of inflation, or with the end of whatever noninflationary process created the incredibly uniform expanding quark-gluon plasma that eventually clumped into stars, planets and us. We don't ...

2

$\Lambda CDM$'s claim of the Universe being 13.8B years old should be taken with a grain of salt. The Universe (as depicted by $\Lambda CDM$) has hypothetically underwent inflation only for a fraction of a second shortly after the Bang, negligible when compared with its current age. Therefore, you shouldn't be hung up on inflation when it comes to guessing ...

2

I would not say that this is the definition of the Friedmann equation, although it is a way of writing it. It is clearly well defined for $1 + z > 0$, i.e. $z>-1$. Indeed redshift does not make sense for $z \leq -1$, so I don't see where the problem lies.

2

The short answer is yes. The long answer is that Newton's First Law of Motion, as pretty much all of the Newtonian Mechanics concepts, are correct (meaning: they work) as long as you're in the right regime. If you want to measure the motion of a small ball when thrown, Newtonian Mechanics gives you the right answer. What this actually means is that the ...

2

It is not an expanding universe of stationary balls. There is an initial velocity to astronomical objects left over from the collapse of the primordial plasma which coalesced to the present day observable universe. That space is expanding means that the velocity of andromeda towards the milky way is a tiny bit smaller than it would have been if there were ...

2

Einstein's equations with $\Lambda=0$ are perfectly capable of describing an expanding spacetime. The cosmological principle leads to a metric of the form $$ds^2 = -c^2 dt^2 + a(t)^2 d\Sigma^2$$ where $a(t)$ is the so-called scale factor which is conventionally set to $1$ at the present time, and $d\Sigma^2$ is a spatial 3-metric with constant curvature $\... 1 You've changed your coordinates, but not the components of your metric. Remember that under a change of coordinates, $$g_{\mu\nu} \rightarrow g'_{\mu\nu} = \frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^\nu} g_{\alpha\beta}$$ In this case, letting$i,j=1,2,3\$ be spatial indices, $$g'_{ij} = \left(\alpha \delta^\alpha_i\right)\... 1 The equation for the sound horizon is simply the equation for the particle horizon, with the speed of light replaced by the speed of sound, there's nothing more to it. Nevertheless, you have to keep in mind that the speed of sound also changes with time, since the matter density dilutes with a growing scale factor, so you have to treat the speed of sound as ... 1 In de Sitter space the event horizon coincides with the Hubble radius, with the Hubble constant$$\rm H=c\sqrt{\Lambda/3}$$so the line element in terms of \rm H is$${\rm ds^2} = g_{\rm tt} \ \rm dt^2 - {...} = \left( 1 - H^2 r^2 / c^2\right) dt^2 - {...}$$which gives the recessional velocity$${\rm v = c} \ \sqrt{1-1/g^{\rm tt}} = \rm H \ r$$If you ... 1 Yes, it is the average redshift of galaxies that belong to a cluster. There is of course an uncertainty in that, but a typical velocity dispersion among galaxies in a massive galaxy cluster is 1000 km/s. So the redshift error due to the uncertainty in the mean is$$\Delta z \sim \frac{\Delta v}{c} = \frac{\sigma_v}{c\sqrt{n}}= \frac{0.0033}{\sqrt{n}},$$... 1 The answer is yes. Hubble's law relates the "velocity of recession" to the proper distance - which is the distance to the other galaxy now. That is not usually the distance we measure, as you have identified. The difference between the two is very small at low redshifts, but becomes larger at higher redshifts, and depends on the expansion history ... 1 I suppose it is a one dimensional problem, and the rod is just lying on the smooth surface, so that it is free to elongate as being heated. We can not expect any stress from a possible restriction to deformation caused by the surface. Even if the heating process is such that one end of the rod reaches a temperature much greater than the other end, again ... 1 For a rod experiencing both thermal strain and stress loading, the formula is not$$\sigma=Y\epsilon$$. The correct formula is$$\sigma=Y(\epsilon-\alpha \Delta T)$$In this example, since the surface is smooth, there is no friction to allow tensile stress to build up in the rod when it is heated. So it experiences unconstrained expansion, in which case,$$\...

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The gravitational influence of objects, which is how their masses are estimated, already includes the effects you are talking about. In fact this is a very minor issue because most of the masses within the universe are not moving at relativistic velocities with respect to each other. The exception to this is perhaps primordial neutrinos, but even there, the ...

1

These are different ideas. The local speed of light is constant. That is the speed of light as measured locally by an observer. This is unrelated to the increasing distance of the point from which the light was originally emitted. To deal with large regions of space, involving expansion, we have to use maps involving scaling distortions, much as we do when ...

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This became too long for the comments. Before I continue, maybe you should take a look at this answer too. I don't claim the following is a good answer but maybe it gives you ideas... So I think the point is that the speed of light will vary if you are in a reference frame, that is experiencing acceleration/gravity. If you are in an inertial reference frame, ...

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