Demonstration of comoving volume between 2 redshifts

[fab13]

TL;DR Summary

I would like to get help about an expression of comoving volume between 2 redshifts

1) I can't manage to find/justify the relation ##(1)## below, from the common relation ##(2)## of a volume.

2) It seems the variable ##r## is actually the comoving distance and not comoving coordinates (with scale factor ##R(t)## between both).

The comoving volume of a region covering a solid angle ##\Omega## between two redshifts ##z_{\mathrm{i}}## and ##z_{\mathrm{f}},## to find is :

##
V\left(z_{\mathrm{i}}, z_{\mathrm{f}}\right)=\Omega \int_{z_{\mathrm{i}}}^{z_{\mathrm{f}}} \frac{r^{2}(z)}{\sqrt{1-\kappa r^{2}(z)}} \frac{c \mathrm{d} z}{H(z)}\quad\quad(1)
##

for a spatially flat universe ##\kappa=0)## the latter becomes

##V\left(z_{\mathrm{i}}, z_{\mathrm{f}}\right)=\Omega \int_{r\left(z_{\mathrm{i}}\right)}^{r\left(z_{\mathrm{f}}\right)} r^{2} \mathrm{d} r=\frac{\Omega}{3}\left[r^{3}\left(z_{\mathrm{f}}\right)-r^{3}\left(z_{\mathrm{i}}\right)\right]
\quad\quad(2)##

I would like to demonstrate it from the comoving distance with :

##D_{\mathrm{A}}(z)=\left\{\begin{array}{ll}
{(1+z)^{-1} \frac{c}{H_{0}} \frac{1}{\sqrt{\left|\Omega_{\mathrm{K}, 0}\right|}} \sin \left[\sqrt{\left|\Omega_{\mathrm{K}, 0}\right|} \frac{H_{0}}{c} r(z)\right],} & {\text { if } \Omega_{\mathrm{K}, 0}<0} \\
{(1+z)^{-1} r(z),} & {\text { if } \Omega_{\mathrm{K}, 0}=0} \\
{(1+z)^{-1} \frac{c}{H_{0}} \frac{1}{\sqrt{\Omega_{\mathrm{K}, 0}}} \sinh \left[\sqrt{\Omega_{\mathrm{K}, 0}} \frac{H_{0}}{c} r(z)\right]} & {\text { if } \Omega_{\mathrm{K}, 0}>0}
\end{array}\right.
##

Anyone could give me some clues/tracks/suggestions to get it ?

Regards

 

[fab13]

Is it right to write :

##\dfrac{1}{\sqrt{1-\kappa r^{2}(z)}} \dfrac{c \mathrm{d} z}{H(z)}=\mathrm{d} r## ??

I can't infer this relation, so if someone could help me, this would be fine. The only thing I have found is :

##\dfrac{c\text{d}z}{H(z)}\dfrac{a(t)}{\sqrt{1-\kappa r^{2}(z)}} =-c\,\mathrm{d} t##

Regards

 

[PeterDonis]

The comoving volume of a region covering a solid angle ##\Omega## between two redshifts ##z_{\mathrm{i}}## and ##z_{\mathrm{f}}##

There are several possible issues here that need to be resolved before we can even do a calculation.

First, the "volume" you are talking about is a 4-dimensional spacetime volume, not a 3-dimensional spatial volume. Do you realize that?

Second, what does "a region covering a solid angle" mean? A solid angle alone isn't enough to bound a finite region; pick a solid angle in the sky and you can look in any direction within that solid angle out to infinity. (Unless you are in a closed universe, but you mention a spatially flat universe, which is not closed, so whatever calculation you are going to make has to cover that case.)

Third, what do you mean by "between two redshifts"? Do you mean "between two spacelike hypersurfaces of constant cosmological time at these two redshifts"? That would seem to be the most natural interpretation, but you should be explicit.

 

[PAllen]

Do you, @fab13 , mean reshift of galaxies as observed by us, or redshift of the CMB from when the universe became transparent? Two completely different things. I am guessing you meant the first.

 

[fab13]

(1) Sorry, I have a naive point of view of the comoving volume notion, i.e I thought that it was the difference between the volume of universe at ##z=z_f## and the volume of universe at ##z=z_i## with ##z_i<z_f##, wasn't it ?

By 4-dimensional spacetime volume, you mean that we have to mix the time with 3D spatial coordinates, like I said above ?. Redshift has the special property which is duality : it represents both time and distance, so this is where my confusions appear.

So from my point of view, I think I have to do calculations taking into account this duality and the difference of physical volume (classical volume of a sphere) that I have cited into (1) sentence above. Concerning the solid angle, I don't know exactly its expression in this case (I only know the classical ##\text{d}\Omega=\text{d}\theta\,\text{sin(}\theta\text{)}\text{d}\phi##).

What do you think about it ?

Regards

 

[PeterDonis]

I have a naive point of view of the comoving volume notion, i.e I thought that it was the difference between the volume of universe at ##z=z_f## and the volume of universe at ##z=z_i## with ##z_i<z_f##

The spatial volume of the universe is infinite, so this makes no sense.

By 4-dimensional spacetime volume, you mean that we have to mix the time with 3D spatial coordinates, like I said above ?

It's not a matter of "mixing". Spacetime is 4-dimensional. So a subset of it will in general be 4-dimensional as well. It is possible to pick out subsets that have fewer dimensions, but it's not clear to me whether that's what you're trying to do.

Redshift has the special property which is duality : it represents both time and distance

No, by itself it represents neither. It can be correlated with either one, if you have other data as well as the redshift.

What do you think about it ?

I still can't tell what you're trying to calculate. I suspect you don't fully understand what you're trying to calculate. So I think we need to take a step back: where are you getting this "comoving volume between 2 redshifts" thing from in the first place? Why do you care about it? What higher level question are you trying to answer?

 

[fab13]

@PeterDonis

here's the source and the context of my initial post where equation ##(1)## appears :

2.2. Distance measurements
The comoving distance to an object at redshift ##z## can be computed as
##r(z)=\frac{c}{H_{0}} \int_{0}^{z} \frac{\mathrm{d} z}{E(z)}##
Although this quantity is not a direct observable, it is closely related to other distance definitions that are directly linked with cosmological observations. A distance that is relevant for our forecasts is the angular diameter distance, whose definition is based on the relation between the apparent angular size of an object and its true physical size in Euclidean space, and is related to the comoving distance by
##
D_{A}(z)=\left\{\begin{array}{ll}
(1+z)^{-1} \frac{c}{H_{0}} \frac{1}{\sqrt{\left|\Omega_{k 0}\right|}} \sin \left[\sqrt{\left|\Omega_{k, 0}\right|} \frac{H_{0}}{c} r(z)\right] & \text { if } \Omega_{x, 0}<0 \\
(1+z)^{-1} r(z) & \text { if } \Omega_{K, 0}=0 \\
(1+z)^{-1} \frac{c}{H_{0}} \frac{1}{\sqrt{\Omega_{K}, 0}} \sinh \left[\sqrt{\Omega_{K}, D_{0}} \frac{H_{0}}{c} r(z)\right], & \text { if } \Omega_{K, 0}>0
\end{array}\right.
##
Also relevant for our forecasts is the comoving volume of a region covering a solid angle ##\Omega## between two redshifts ##z_{i}## and ##z_{f},## which is given by ##V\left(z_{i}, z_{i}\right)=\Omega \int_{a_{i}}^{a} \frac{r^{2}(z)}{\sqrt{1-K r^{\prime}(z)}} \frac{c d z}{H(z)}##
(14)
for a spatially flat universe ##(K=0),## the latter becomes ##V\left(z_{i, z_{i}}\right)=\Omega \int_{r_{(i)}}^{\left(i_{i}\right)} r^{2} \mathrm{d} r=\frac{\Omega}{3}\left[r^{3}\left(z_{i}\right)-r^{3}\left(z_{i}\right)\right]##
(15)
These expressions allow us to compute the volume probed by Euclid within a given redshift interval.

I hope that you will understand better the context of my initial question and so help me.

Regards

 

[PAllen]

So this says my guess in post #4 is correct. Your recent post (showing why we emphasize a reference for discussion so much) immediately points up a misunderstanding. You say you want to see how to compute a given formula for comoving volume for a solid angle and red shift range from us, in terms of comoving distance, but the formulas you then quote are for angular size distance. Why on Earth would you want to compute a comoving volume from angular size distance formulas?

 

[fab13]

I don't want to make confusions : the only thing that I wanted is just to get the demonstration which allows to find :

##V\left(z_{i}, z_{i}\right)=\Omega \int_{a_{i}}^{a} \frac{r^{2}(z)}{\sqrt{1-K r^{\prime}(z)}} \frac{c d z}{H(z)}##

nothing else.

 

[PeterDonis]

here's the source

What book is this from? Just cutting and pasting doesn't tell us the source.

 

[fab13]

I am doing bibliographic researches on this paper , you will see the formulas that I have cited.

Thanks for your help

 

[PAllen]

Actually, looking at your post #7, I immediately noticed some inconsistencies in your volume formulas, and guessed what they ought to be. Please compare yours to the paper carefully. There are important errors in your rendition of them.

[edit: actually, in your OP you have the volume formulas correct, the error is only in post #7. So you just want to know how these are derived. You are not insisting they be derived from the irrelevant angular distance formulas. I believe the derivation is pretty straightforward, but I will not have the time to write it up any time soon.]

 

[fab13]

@PAllen . Thanks, I realized that I have written bad bounds for integral :

it is not : ##V\left(z_{\mathrm{i}}, z_{\mathrm{f}}\right)=\Omega \int_{z_{\mathrm{i}}}^{z_{\mathrm{f}}} \frac{r^{2}(z)}{\sqrt{1-\kappa r^{2}(z)}} \frac{c \mathrm{d} z}{H(z)}\quad\quad(1)##

but rather :

##V\left(z_{i}, z_{i}\right)=\Omega \int_{a_{i}}^{a} \frac{r^{2}(z)}{\sqrt{1-K r^{\prime}(z)}} \frac{c d z}{H(z)}##

with bounds which refer to the scale factor, doesn't it ?

Do you agree ? However, I can't still get to demonstrate this relation. A little help wouldn't be too much.

Regards

 

[PAllen]

@PAllen . Thanks, I realized that I have written bad bounds for integral :

it is not : ##V\left(z_{\mathrm{i}}, z_{\mathrm{f}}\right)=\Omega \int_{z_{\mathrm{i}}}^{z_{\mathrm{f}}} \frac{r^{2}(z)}{\sqrt{1-\kappa r^{2}(z)}} \frac{c \mathrm{d} z}{H(z)}\quad\quad(1)##

but rather :

##V\left(z_{i}, z_{i}\right)=\Omega \int_{a_{i}}^{a} \frac{r^{2}(z)}{\sqrt{1-K r^{\prime}(z)}} \frac{c d z}{H(z)}##

with bounds which refer to the scale factor, doesn't it ?

Do you agree ? However, I can't still get to demonstrate this relation. A little help wouldn't be too much.

Regards

NO. The first is correct, the second makes no sense. Look at the paper you referenced. As to the general method, in GR, a volume element is given by the square root of the metric determinant times a raw coordinate volume element. I would guess all they have done is to integrate the volume of a comoving constant time slice over a solid angle between two red shifts using the determinant of the induced 3 metric. All of this starting from the FLRW metric.

 

[fab13]

NO. The first is correct, the second makes no sense. Look at the paper you referenced. As to the general method, in GR, a volume element is given by the square root of the metric determinant times a raw coordinate volume element. I would guess all they have done is to integrate the volume of a comoving constant time slice over a solid angle between two red shifts using the determinant of the induced 3 metric. All of this starting from the FLRW metric.

1) I don't understand how can I demonstrate the relation (1) from the metric determinant times a raw coordinate volume element. If you could tell me more about this.

2) In general, I believed that big papers like this one were examined perfectly before published and I realize that's not the case : it might be many errors and however, the paper would be accepted : strange, isn't it ? Surely, I idealize too much the research domain.

3) @PAllen : if you had any starting point for my demonstration (equation 1), this would be nice to show it.

Regards

 

[PAllen]

They are doing what I described, with a few tricks.

Note, from definitions earlier in the paper, that the collection of terms near dz makes it effectively dr. Then, note that restricted to a constant cosmological time, you have one of the 3 homogeneous, isotropic 3 metrics (this is the induced 3-metric I was referring to). See, for example, the reduced circumference coordinate section of:
https://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric

Then, instead of working from a complete volume element, they note that holding r constant, this 3-metric gives area as just solid angle times radius squared. Solid angle is constant for all r, by specification, so it comes out as a constant. Then the relation of dr to comoving distance to complete an infinitesimal shell volume is given by the square root of the line element above holding angle constant. With these steps you arrive exactly at the formula they give. There is no mistake, and further, all of this is expected to be straightforward to the intended audience of the paper.

 

[fab13]

Hello,

I have just taken over this issue and I tried to make progress. Following PAllen's advises, I did the following small calculation :

The FLRW metric can be expressed under following (0,2) tensor form :

##\left[\begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & -\frac{R^{2}(t)}{1-k r^{2}} & 0 & 0 \\ 0 & 0 & -R^{2}(t) r^{2} & 0 \\ 0 & 0 & 0 & -R^{2}(t) r^{2} \sin ^{2} \theta\end{array}\right]##

If I consider only slice times constant, my goal is to compute the volume probeb by a satellite between 2 redshifts.

1) We can easily find that :

$$\int_{0}^{z_0}\frac{cdz}{H(z)} = \int_{0}^{t_0}\frac{cdt}{R(t)}$$

2) Then, If I consider a volume with ##\text{d}r##, ##\text{d}\theta## and ##\text{d}\phi## coordinates, I have the following expression for determinant :

##g=\text{det}[g_{ij}] = -\dfrac{R(t)^6}{1-kr^2}\,r^4\,\text{sin}^2\theta##

Which means that I have :

##\text{d}V=\sqrt{-g}\text{d}^3x = \dfrac{R(t)^3}{\sqrt{1-kr^2}}\,r^2\,\text{d}r\,\text{sin}\theta\,\text{d}\theta\, \text{d}\phi##

##V=\int\text{d}V= \int \dfrac{R(t)^3}{\sqrt{1-kr^2}}\,r^2\,\text{d}r\,\text{sin}\theta\,\text{d}\theta\, \text{d}\phi##

$$V = \int \text{d}\Omega \int \dfrac{R(t)^3}{\sqrt{1-kr^2}}\,r^2\,\text{d}r$$

$$\rightarrow\quad V = \Omega \int \dfrac{R(t)^3}{\sqrt{1-kr^2}}\,r^2\,\text{d}r$$with ##\Omega## the solid angle considered.

But as you can see, I am far away from the expression ##(1)## that I would like to find, i.e :

##V\left(z_{\mathrm{i}}, z_{\mathrm{f}}\right)=\Omega \int_{z_{\mathrm{i}}}^{z_{\mathrm{f}}} \frac{r^{2}(z)}{\sqrt{1-\kappa r^{2}(z)}} \frac{c \mathrm{d} z}{H(z)}\quad\quad(1)##

Where is my error ?

Any help would be fine, I am stucked for the moment.

 

[fab13]

@PAllen

a volume element is given by the square root of the metric determinant times a raw coordinate volume element. I would guess all they have done is to integrate the volume of a comoving constant time slice over a solid angle between two red shifts using the determinant of the induced 3 metric. All of this starting from the FLRW metric.

What do you mean please by "a raw coordinate volume element" ?

 

[fab13]

Isn't really anyone that could help me to prove the relation ##(1)## above from my attempt in post #17 ?

 

[PeterDonis]

If I consider only slice times constant

You don't seem to be, since you start out with an integral from ##0## to ##t_0##, which is an integral over multiple time slices.

Also, the expression you appear to be trying to find appears to be integrating between multiple time slices, since it is treating the Hubble value ##H(z)## as a function of redshift. But the Hubble value is the same everywhere on a slice of constant time; it's not a function of anything unless you are considering multiple time slices.

I still don't think you fully understand what you are trying to calculate.

 

[PAllen]

I don't have time to say more now. However, I calculated exactly as described in my post #16, and got exactly the equation in the paper. Please try to follow the recipe in post #16.

 

[PeterDonis]

I calculated exactly as described in my post #16, and got exactly the equation in the paper

To be sure I understand, is the following description of what your recipe is calculating correct?

(1) We have two chosen redshifts, ##z_1## and ##z_2##.

(2) We have a chosen solid angle.

(3) We have a chosen instant of comoving time (understood to be the instant "now"), which defines a spacelike 3-surface. We use spherical coordinates on this spacelike 3-surface.

(4) We convert the chosen redshifts into values of ##r## on the spacelike 3-surface as follows: for each redshift, we find the ##r## that corresponds to a comoving worldline that emitted light, at some past time, which is just now reaching the observer at ##r = 0## and is observed to have the chosen redshift.

 

[fab13]

@PeterDonis @PAllen

I don't know yet how to do the demonstration and I don't want to be boring, so when you will have some time, I would be very grateful to help me and give me the demonstration of this "little calculation" (not such"little" for me in any case as you can see). The only calculation I could have done is in my post #17 : I tried to follow your advices but without success.

I begin to desperate, any support is welcome.

Regards

 

[fab13]

Maybe I have found a partial explanation to my issue to determine the expresion of this volume between 2 redshifts. Here below a formula :

The main expression to keep in mind is :

##d V_{C}=D_{H} \frac{(1+z)^{2} D_{A}^{2}}{E(z)} d \Omega d z\quad(3)##

1) Could anyone explain me please the different justifications to introduce all the factors implied in this expression ?

2) I have not yet with this expression the same expression (1) at the beginning of my post, so could anyone manage to find (1) from (3) ? Any help would be fine, I am stucked for the moment.

 

[PeterDonis]

Here below a formula

Where is this from? Please give a reference. It's impossible to respond to your questions without knowing what reference this is from.

 

[fab13]

Sorry , the formula comes from : https://ned.ipac.caltech.edu/level5/Hogg/Hogg9.html

 

[PeterDonis]

Ok, having taken a look at both references (the Hogg web page and the Euclid paper) and compared them, I think I can answer a few questions.

Is it right to write :

##\dfrac{1}{\sqrt{1-\kappa r^{2}(z)}} \dfrac{c \mathrm{d} z}{H(z)}=\mathrm{d} r## ??

No. Note that in the Euclid paper, when they go from equation (14) to equation (15), they say that step is only valid for a spatially flat universe, ##K = 0##.

If you look back at equation (12) of the Euclid paper, you see:

$$
r(z) = \frac{c}{H_0} \int_0^z \frac{dz}{E(z)}
$$

which, using an earlier equation, can be rewritten slightly to

$$
r(z) = \int_0^z \frac{c dz}{H(z)}
$$

And since by definition ##r = \int dr##, we have

$$
dr = \frac{c dz}{H(z)}
$$

This is valid regardless of whether the universe is spatially flat, open, or closed; so the counterintuitive part here is that the curvature factor ##K## does not appear explicitly. But if you look back at the equation for ##E(z)## (which is included in ##H(z)##), you will see that it has a term under the square root that involves ##K##, so the curvature factor is still being taken into account.

Now, what does ##r (z)## represent? It is the "comoving distance" to an object with redshift ##z##. Let's unpack what that means:

(1) "Comoving" means that this is not a proper distance; it is not what you would read off, say, a very long tape measure stretched between you and the object. It is the difference in the comoving coordinate ##r## between you and the object. (You are assumed to be at ##r = 0##.) This assumes that both you and the object are comoving objects, i.e., you are both moving along with the Hubble flow, which means both of your comoving ##r## coordinates are constant with time. So the comoving distance between you is also constant with time; it does not increase as the universe expands. (The usual term for the distance that does increase as the universe expands is "proper distance".)

(2) So what ##r(z)## is actually doing is "translating" a redshift ##z##, which is what we directly observe, into a difference in comoving coordinates, which is what can easily be plugged into the mathematical model we use to describe the universe. Note that we can do this between any two objects with different redshifts; one of the objects does not have to be us at ##r = 0##.

Now, given that, what is a "comoving volume"? As I read the references, what they mean is basically what I described in post #22. To recap briefly: we pick two different redshifts, ##z_1## and ##z_2##; we obtain the difference in comoving distances ##r(z_1) - r(z_2)##; we integrate that over some solid angle ##\Omega## to get a volume, using some relationship between radial comoving distance and transverse comoving distance (the specific relationship depends on whether the universe is spatially flat, open, or closed) to convert the solid angle to a comoving area. Note the key phrase in the Hogg article: "the volume measure in which number densities of non-evolving objects locked into Hubble flow are constant with redshift". This can't be true of a volume that increases as the universe expands, because number densities for such a volume decrease as the volume increases. Basically, the purpose of comoving distance and comoving volume is to remove the effects of the expansion of the universe in order to make certain kinds of analysis, the kind that these cosmologists are trying to do, simpler.

The main expression to keep in mind is :

##d V_{C}=D_{H} \frac{(1+z)^{2} D_{A}^{2}}{E(z)} d \Omega d z\quad(3)##

1) Could anyone explain me please the different justifications to introduce all the factors implied in this expression ?

2) I have not yet with this expression the same expression (1) at the beginning of my post, so could anyone manage to find (1) from (3) ?

Let's compare this expression with the one in the Euclid paper:

$$
V_C = \Omega \int \frac{r^2(z)}{\sqrt{1 - K r^2(z)}} \frac{c dz}{H(z)}
$$

We can rewrite this in terms of ##E(z)## instead of ##H(z)## (and also pull the constant ##c## outside the integral):

$$
V_C = \frac{c}{H_0} \Omega \int \frac{r^2(z)}{\sqrt{1 - K r^2(z)}} \frac{dz}{E(z)}
$$

We can substitute ##D_H = c / H_0## (as defined in an earlier page of the Hogg article):

$$
V_C = D_H \Omega \int \frac{r^2(z)}{\sqrt{1 - K r^2(z)}} \frac{dz}{E(z)}
$$

But remember that above we said we wanted to integrate over solid angle; we want to know how comoving volume varies with solid angle as well as with radial comoving distance. So we need to pull ##\Omega## inside the integral:

$$
V_C = D_H \int \frac{r^2(z)}{\sqrt{1 - K r^2(z)}} \frac{dz}{E(z)} d\Omega
$$

We can rewrite this as an equation between differentials (i.e., undoing the integral so we have an equation for the differential of ##V_C##:

$$
dV_C = D_H \frac{r^2(z)}{\sqrt{1 - K r^2(z)}} \frac{1}{E(z)} d\Omega dz
$$

And, finally, we can observe that

$$
\frac{r^2(z)}{\sqrt{1 - K r^2(z)}} = \left( 1 + z^2 \right) D_A^2
$$

And that gives us the equation in the Hogg article.

That last observation is the difficult part, but hopefully you can see why it's true by reading through all the pages of the Hogg article (pages 5 and 6 in particular might be helpful).

 

[fab13]

Hello PeterDonis,

thanks for your answer. Everyting is right except the demonstration of the last equation, i.e :

$$\frac{r^2(z)}{\sqrt{1 - K r^2(z)}} = \left( 1 + z^2 \right) D_A^2$$

I know that ##D_A(z) = \dfrac{D_C}{1+z}## but I can't conclude with the presence of ##\dfrac{1}{\sqrt{1-Kr^2}}.##

Could tell me more please about the trick to apply to find :

##\dfrac{r^2(z)}{\sqrt{1 - K r^2(z)}} = \left( 1 + z^2 \right) D_A^2##

Thanks

 

[PeterDonis]

Everyting is right except the demonstration of the last equation

No, everything is right including the last equation.

Could tell me more please about the trick to apply to find :

##\dfrac{r^2(z)}{\sqrt{1 - K r^2(z)}} = \left( 1 + z^2 \right) D_A^2##

They are both equivalent to the square of the transverse comoving distance ##D_M##, which is described in section 5 of the Hogg article you linked to. The equivalence ##D_M = \left( 1 + z \right) D_A## is from section 6 of that article. The equivalence ##D_M = \dfrac{r^2(z)}{\sqrt{1 - K r^2(z)}}## follows from the definition of ##D_M## given in section 5.

 

[fab13]

Sorry, there was a typo from recent posts, the expression to demonstrate is :

##\dfrac{r^2(z)}{\sqrt{1 - K r^2(z)}} = \left( 1 + z \right)^{2} D_A^2##

and not ##\dfrac{r^2(z)}{\sqrt{1 - K r^2(z)}} = \left( 1 + z^2 \right) D_A^2##

 

[fab13]

@PeterDonis

The equivalence ##D_M = \dfrac{r^2(z)}{\sqrt{1 - K r^2(z)}}## follows from the definition of ##D_M## given in section 5.

I can't manage to get this expression of ##D_M##. In section 5 of Hogg's paper, there is only the following expression :

##D_{M}=\left\{\begin{array}{ll}D_{H} \frac{1}{\sqrt{\Omega_{k}}} \sinh \left[\sqrt{\Omega_{k}} D_{C} / D_{H}\right] & \text { for } \Omega_{k}>0 \\ D_{C} & \text { for } \Omega_{k}=0 \\ D_{H} \frac{1}{\sqrt{\left|\Omega_{k}\right|}} \sin \left[\sqrt{\left|\Omega_{k}\right|} D_{C} / D_{H}\right] & \text { for } \Omega_{k}<0\end{array}\right.##

How to make the link with ##D_M = \dfrac{r^2(z)}{\sqrt{1 - K r^2(z)}}## ?

Sorry, I don't want to be boring.

 

[PeterDonis]

Sorry, there was a typo from recent posts, the expression to demonstrate is :

##\dfrac{r^2(z)}{\sqrt{1 - K r^2(z)}} = \left( 1 + z \right)^{2} D_A^2##

and not ##\dfrac{r^2(z)}{\sqrt{1 - K r^2(z)}} = \left( 1 + z^2 \right) D_A^2##

I think the square root in the denominator on the LHS also doesn't belong there. It would if we took the square root of both sides.

 

[PeterDonis]

In section 5 of Hogg's paper, there is only the following expression :

##D_{M}=\left\{\begin{array}{ll}D_{H} \frac{1}{\sqrt{\Omega_{k}}} \sinh \left[\sqrt{\Omega_{k}} D_{C} / D_{H}\right] & \text { for } \Omega_{k}>0 \\ D_{C} & \text { for } \Omega_{k}=0 \\ D_{H} \frac{1}{\sqrt{\left|\Omega_{k}\right|}} \sin \left[\sqrt{\left|\Omega_{k}\right|} D_{C} / D_{H}\right] & \text { for } \Omega_{k}<0\end{array}\right.##

Do you understand what that expression means? And do you understand what the ##K## in the expression you are trying to match with it means?

Try writing down three different versions of the expression that has ##r## in it: one for each of the three different possible values of ##K##. Then consider what coordinates those expressions are in, and what coordinates the expressions in section 5 of the Hogg article are in. How are those two different coordinate systems related? (Hint: the flat ##K = 0## case is the easy one.)

 

[fab13]

Ok, having taken a look at both references (the Hogg web page and the Euclid paper) and compared them, I think I can answer a few questions.
No. Note that in the Euclid paper, when they go from equation (14) to equation (15), they say that step is only valid for a spatially flat universe, ##K = 0##.

If you look back at equation (12) of the Euclid paper, you see:

$$
r(z) = \frac{c}{H_0} \int_0^z \frac{dz}{E(z)}
$$

which, using an earlier equation, can be rewritten slightly to

$$
r(z) = \int_0^z \frac{c dz}{H(z)}
$$

And since by definition ##r = \int dr##, we have

$$
dr = \frac{c dz}{H(z)}
$$

This is valid regardless of whether the universe is spatially flat, open, or closed; so the counterintuitive part here is that the curvature factor ##K## does not appear explicitly. But if you look back at the equation for ##E(z)## (which is included in ##H(z)##), you will see that it has a term under the square root that involves ##K##, so the curvature factor is still being taken into account.

Now, what does ##r (z)## represent? It is the "comoving distance" to an object with redshift ##z##. Let's unpack what that means:

(1) "Comoving" means that this is not a proper distance; it is not what you would read off, say, a very long tape measure stretched between you and the object. It is the difference in the comoving coordinate ##r## between you and the object. (You are assumed to be at ##r = 0##.) This assumes that both you and the object are comoving objects, i.e., you are both moving along with the Hubble flow, which means both of your comoving ##r## coordinates are constant with time. So the comoving distance between you is also constant with time; it does not increase as the universe expands. (The usual term for the distance that does increase as the universe expands is "proper distance".)

(2) So what ##r(z)## is actually doing is "translating" a redshift ##z##, which is what we directly observe, into a difference in comoving coordinates, which is what can easily be plugged into the mathematical model we use to describe the universe. Note that we can do this between any two objects with different redshifts; one of the objects does not have to be us at ##r = 0##.

Now, given that, what is a "comoving volume"? As I read the references, what they mean is basically what I described in post #22. To recap briefly: we pick two different redshifts, ##z_1## and ##z_2##; we obtain the difference in comoving distances ##r(z_1) - r(z_2)##; we integrate that over some solid angle ##\Omega## to get a volume, using some relationship between radial comoving distance and transverse comoving distance (the specific relationship depends on whether the universe is spatially flat, open, or closed) to convert the solid angle to a comoving area. Note the key phrase in the Hogg article: "the volume measure in which number densities of non-evolving objects locked into Hubble flow are constant with redshift". This can't be true of a volume that increases as the universe expands, because number densities for such a volume decrease as the volume increases. Basically, the purpose of comoving distance and comoving volume is to remove the effects of the expansion of the universe in order to make certain kinds of analysis, the kind that these cosmologists are trying to do, simpler.
Let's compare this expression with the one in the Euclid paper:

$$
V_C = \Omega \int \frac{r^2(z)}{\sqrt{1 - K r^2(z)}} \frac{c dz}{H(z)}
$$

We can rewrite this in terms of ##E(z)## instead of ##H(z)## (and also pull the constant ##c## outside the integral):

$$
V_C = \frac{c}{H_0} \Omega \int \frac{r^2(z)}{\sqrt{1 - K r^2(z)}} \frac{dz}{E(z)}
$$

We can substitute ##D_H = c / H_0## (as defined in an earlier page of the Hogg article):

$$
V_C = D_H \Omega \int \frac{r^2(z)}{\sqrt{1 - K r^2(z)}} \frac{dz}{E(z)}
$$

But remember that above we said we wanted to integrate over solid angle; we want to know how comoving volume varies with solid angle as well as with radial comoving distance. So we need to pull ##\Omega## inside the integral:

$$
V_C = D_H \int \frac{r^2(z)}{\sqrt{1 - K r^2(z)}} \frac{dz}{E(z)} d\Omega
$$

We can rewrite this as an equation between differentials (i.e., undoing the integral so we have an equation for the differential of ##V_C##:

$$
dV_C = D_H \frac{r^2(z)}{\sqrt{1 - K r^2(z)}} \frac{1}{E(z)} d\Omega dz
$$

And, finally, we can observe that

$$
\frac{r^2(z)}{\sqrt{1 - K r^2(z)}} = \left( 1 + z^2 \right) D_A^2
$$

And that gives us the equation in the Hogg article.

That last observation is the difficult part, but hopefully you can see why it's true by reading through all the pages of the Hogg article (pages 5 and 6 in particular might be helpful).

Thanks for your patience.

I have still difficulties to get the equality :

$$\frac{r^2(z)}{1 - K r^2(z)} = \left( 1 + z \right)^2 D_A^2\quad(1)$$

Indeed, we can write by definition of Angular diameter distance :

$$D_c^2=\left( 1 + z \right)^2 D_A^2\quad(2)$$

where ##D_c## is the comoving distance.

1) Apparently, ##r=r(z)## in eq(1) is assimilated to comoving distance and not comoving coordinate "##r##"
Could anyone confirm it ?

2) Actually, there are 3 ways to compute the comoving coordinate "##r_1##" (caution : not "##r(z)##" above !) :

$${\large\int}_{t_{1}}^{t_{0}}\dfrac{c\text{d}t}{R(t)}={\large\int}_{0}^{r_{1}}\dfrac{\text{d}r}{\sqrt{1-kr^{2}}}=S_{k}^{-1}(r_{1})$$

with :

\begin{eqnarray}
S_{k}^{-1}(r_{1})=
\left\{
\begin{aligned}
&\,\,\,\, \text{arcsin}(r_{1})\,\,\,\, \text{si}\,\, k=+1 \\
\\
&\,\,\,\, r_{1}\,\,\,\, \text{si}\,\, k=0 \\
\\
&\,\,\,\, \text{argsh}(r_{1})\,\,\,\, \text{si}\,\,k=-1
\end{aligned}
\right.
\end{eqnarray}

What I don't know is how to make appear the curvature parameter under the form ##\sqrt{1-kr^2}## in the
expression of comoving coordinate above.

What I can only write in flat space is :

$${\large\int}_{0}^{r_{1}}\dfrac{\text{d}r}{\sqrt{1-kr^{2}}}=S_{k}^{-1}(r_{1})=r_1$$

and so having as in the paper cited : $$R_0^2\,r_1^2(z) = D_c^2=\left( 1 + z^2 \right) D_A^2$$but how to show the factor ##\sqrt{1-kr^2}## for the 2 other cases.

SUGGESTION : Maybe if I differentiate the expression :

$${\large\int}_{0}^{r_{1}}\dfrac{\text{d}r}{\sqrt{1-kr^{2}}}=S_{k}^{-1}(r_{1})=r_1$$

$$\Rightarrow$$ $$\dfrac{\text{d}r}{\sqrt{1-kr^{2}}}=\text{d}r_1$$ but then, ##r## and ##r_1## represents both comoving coordinates and not comoving distance, don't they ?

But then, what's the utility of differentiate in this case since I can only take ##S_{k}^{-1}(r_{1})=r_1## only in flat case (k=0) ? I mean, this way, I make appear curvature parameter "##k##" whereas I consider only ##k=0##, do you understand ?

Sorry if I insist, I would really like to conclude, thanks for your patience.

 

[PeterDonis]

@fab13 please don't quote an entire post when responding. Just quote the part you're responding to.