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I am looking for a way to get, by a simple numerical computation, the 3 curves on the following figure:

size of anisotropies vs Distance

For this, I don't know what considering as abcissa (comoving distance ?, i.e

$$D_{comoving} = R(t)r$$

with $R(t)$ scale factor and $r$ the coordinate which appears into FLRW metric).

Previously, I did a little project about the trajectory of light geodesics in an expanding universe; here below a figure illustrating the expected results with FLRW metric:

light ray trajectory

As you can see, the curve of light is bended since $\Lambda\text{CDM}$ current model produces acceleration of expansion, so light has more and more difficulties to reach our galaxy.

In a first version of this project, I have computed a light geodesic in Einstein-de-Sitter universe : so in this case, the distance between the 2 galaxies (emitting and our galaxy) which is equal to :

$$D_{{\varphi}}=\text{Distance}_{init}\,\bigg[\frac{3\,H_{0}}{2c}\,ct\bigg]^{2/3}$$

I did also a simple computation on the angular diameter distance versus the redshift. I get this figure which is the expected results for the 3 models (k=-1,0,1) :

enter image description here

I would have thought that angle of anistropies was constant during its travel : that is not the case on the first figures at the top of post, which illustrates a bending, producing then a different angle between the initial emission and the final reception by our eyes : is it actually right ?

I think my main issue, to produce this first figure, is to know what I have got to take as abscissa ? the comoving distance, the proper distance, the angular diameter distance ?

Firstly, I believed that I should take the comoving distance but I don't know how to make converge 2 light rays (extremities of one anisotropy) in oberver eyes.

Any help is welcome to reproduce the 3 curves on the first figure of this post : these curves illustrate very well the notion of smaller/bigger value of apparent anisotropies as a function of curvature parameter.

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  • $\begingroup$ -@safesphere ok, I took a look again at your first question on physics.stackexchange.com/questions/357830/… and I didn't undertsand exactly your original issue : you should maybe consider the size of observable universe in your reasoning. Why do you talk about the circumference : what matters is the ratio between the current cosmological distance of object and the term $(1+z)$ to get the angular diameter distance. Could you help me please to clarify the interpretation of your first question, since my english is not perfect ? $\endgroup$ – youpilat13 Aug 7 '18 at 23:34
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    $\begingroup$ In my original question I was trying to describe exactly what is written in Wikipedia here: "An object 'behind' another of the same size, beyond a certain redshift (roughly z=1.5), appears larger on the sky, and would therefore have a smaller "angular diameter distance": en.m.wikipedia.org/wiki/Angular_diameter_distance - I was not familiar with the concept of angular diameter distance (English is not my native either). I discovered this effect on my own and posted a question, but shamefully some very senior people on this site were unable to understand the concept. $\endgroup$ – safesphere Aug 12 '18 at 6:11
  • $\begingroup$ -@safesphere all right, your wikipedia source helped me to understand your issue. The angular diameter distance is just the physical distance (not an "angle") between the object and a comoving observer, at the time (epoch) determined by the redshift you choose. $\endgroup$ – youpilat13 Aug 12 '18 at 19:58
  • $\begingroup$ -@safesphere For example, at the time $z=2$, the angular diameter distance is equal to : $D_{a}(z=2)=R(z=2) r(z=2)$ with $R(z=2)$ the scale factor at $z=2$ and $r(z=2)$ the radial coordinate. For scale factor, you have the relation $1+z=\dfrac{R_{0}}{R(z)}$ and for radial coordinate, you have $r(z)=\int_{0}^{t(z)}\dfrac{c\,\text{d}t}{R(t)}$. With FLRW metric, the physical distance at a time wanted is the product of scale factor $R(t)$ with the radial coordinate $r$ (also often written $\chi$). Hoping this will help you. $\endgroup$ – youpilat13 Aug 12 '18 at 20:23

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