Demonstration of comoving volume between 2 redshifts

[PeterDonis]

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.

Yes. Which means that we can write:

$$
\frac{r^2}{1 - K r^2} = D_c^2
$$

or, taking the square root,

$$
\frac{r}{\sqrt{1 - K r^2}} = D_c
$$

Apparently, $r=r(z)$ in eq(1) is assimilated to comoving distance and not comoving coordinate "$r$"

No. $r(z)$ is the comoving coordinate. But it has a well-known relationship to comoving distance. Do you know what that relationship is? (Hint: is it true that the equation I just wrote above is that relationship?)

 

[fab13]

Yes. Which means that we can write:

$$
\frac{r^2}{1 - K r^2} = D_c^2
$$

Okay but $r$ corespond to the comoving distance from :

$r(z)=\int\dfrac{c\text{d}z}{H(z)}$ which represents the comoving distance and not from
$r=\int\dfrac{c\text{d}t}{R(t)}$ which represents the comoving coordinate. Indeed, there is a $R_0$ missing between both.

or, taking the square root,
$$
\frac{r}{\sqrt{1 - K r^2}} = D_c
$$

Here, you assimilate "$r$" to a comoving distance, don't you ? whereas it is a comoving coordinate coming from

$$r=\int\dfrac{c\text{d}t}{R(t)}$$

 

[PeterDonis]

$r$ corespond to the comoving distance

I think I see the problem; the Hogg article defines two "comoving distances", not just one.

The comoving distance $D_M$ is the transverse comoving distance. $r$ is equal to $D_M$.

The comoving distance $D_C$ is the line of sight comoving distance. $r$ is not equal to $D_C$.

Try re-reading the previous posts with that in mind.

 

[fab13]

Yes. Which means that we can write:

$$
\frac{r^2}{1 - K r^2} = D_c^2
$$

or, taking the square root,

$$
\frac{r}{\sqrt{1 - K r^2}} = D_c
$$

1) It seems here that you go too fast for $D_c$. I think we should write for this comoving distance.

$$
r=D_c=\int\dfrac{c\text{d}z}{H(z)}=R_0\int\dfrac{c\text{d}t}{R(t)}=R_0{\large\int}_{0}^{r}\dfrac{\text{d}r}{\sqrt{1-kr^{2}}}
$$

2) What is surprising is that in EUclide paper, they keep the factor $\dfrac{1}{\sqrt{1-kr^{2}}}$ whereas they keep using the euclidean case by write : $r=D_c$ and not $r=D_c=sin()$ or $r=D_c = sinh()$.

I know that forecasts in Euclid consider a space almost flat : maybe that's why they do this approximation, i.e use

$$r=D_c=\int_{0}^{r}\dfrac{\text{d}r}{\sqrt{1-kr^{2}}}$$

since $\Omega_k$ is very closed to 0. That would explain the quantity "$r=D_c\,\neq sin()$ or $\neq sinh()$" always present even if we are not in flat case.

What do you think about ?

 

[PeterDonis]

What do you think about ?

Read my post #38. There are two comoving distances, not one.

 

[Arman777]

I just want to write some expressions, but I hope it will clear things up. There are indeed two co-moving coordinates. The first one is $\chi$, and the second one is $r$.

Now $r$ represents the transverse co-moving coordinate. The FLRW metric (in terms of the $r$) can be written as

$$ ds^2 = -c^2dt^2 + a^2(t)[\frac{dr^2}{1-kr^2} + r^2d\Omega^2]~~~~(1)$$

where

$$k =
\begin{cases}
-1, & \text{Hyperspherical} \\
0, & \text{Euclidean} \\
+1, & \text{Spherical}
\end{cases} ~~\text{and}~~d\Omega^2 = d\theta^2 + sin^2(\theta)d\phi^2$$.

There is also another co-moving coordinate, $\chi$, which represents the line of sight co-moving coordinate. Line of sight co-moving coordinate defined as

$$d\chi = \frac{dr}{\sqrt{1-kr^2}}$$By taking the integral of both sides, we obtain$$\chi \equiv \int_0^r \frac{dr}{\sqrt{1-kr^2}} = \begin{cases}
sinh^{-1}(r) & k= -1 \\
r & k = 0 \\
sin^{-1}(r) & k = +1
\end{cases}$$Or we can take the inverse of the above equality and write$$ r \equiv S_k(\chi) =
\begin{cases}
sinh(\chi) & k= -1 \\
\chi & k = 0 \\
sin(\chi) & k = +1
\end{cases}$$In this case, equation (1) can be written as,$$ ds^2 = -c^2dt^2 + a^2(t)[d\chi^2 + S_k^2(\chi)d\Omega^2]$$

The most important part is actually realizing that, Hoggs is using $D_C$ and $D_M$ instead of $\chi$ and $r$.

$$r \equiv D_M \equiv S_k(\chi)$$

and

$$\chi \equiv D_C$$

So, Hoggs is using a different kind of notation that's all.

In this case, you will also realize that,

$$
D_A = \frac{\Delta S}{\Delta\theta} = \frac{r}{1+z} \equiv \frac {S_k(\chi)}{1+z}
\equiv \frac{D_M}{1+z}$$

 

[PeterDonis]

I just want to write some expressions, but I hope it will clear things up. There are indeed two co-moving coordinates. The first one is $\chi$, and the second one is $r$.

This is not correct as you state it. It is correct that there is a coordinate chart in which $\chi$ is the "radial" comoving coordinate and $r$ is the "transverse" one. But there is also a coordinate chart in which $r$ is both the "radial" and the "transverse" coordinate; it's just that $r$ only represents comoving distance in the transverse direction.

In fact, in your very next paragraph, you admit this and contradict yourself:

Now $r$ represents the transverse co-moving coordinate. The FLRW metric (in terms of the $r$) can be written as

$$ ds^2 = -c^2dt^2 + a^2(t)[\frac{dr^2}{1-kr^2} + r^2d\Omega^2]~~~~(1)$$

In this metric, $r$ is both a radial and a transverse coordinate. There is no $\chi$ anywhere. But, as is obvious from the metric, $r$ only directly represents distance in the transverse (angular) direction; it does not directly represent distance in the line of sight (radial) direction, because of the extra factor multiplying $dr^2$ in the metric.

There is also another co-moving coordinate, $\chi$, which represents the line of sight co-moving coordinate. Line of sight co-moving coordinate defined as

$$d\chi = \frac{dr}{\sqrt{1-kr^2}}$$

This is a coordinate transformation to a different coordinate chart, in which now there are two "distance" coordinates, the transverse $r$ and the radial $\chi$. And, in fact, I have never actually seen this coordinate chart used, because every source I have seen that uses $\chi$ eliminates $r$ altogether by substituting the appropriate function of $\chi$ (you discuss those functions later in your post). So in these coordinate charts, there is still only one comoving coordinate, but now that coordinate only directly represents radial (line of sight) distance, not transverse distance.

All that said, there is nothing that requires anyone to use this coordinate chart in terms of $\chi$ instead of the one above in terms of $r$, and, in fact, the references under discussion, as far as I can tell, don't.

The most important part is actually realizing that, Hoggs is using $D_C$ and $D_M$ instead of $\chi$ and $r$.

No, he's not. He's not using $D_C$ and $D_M$ as coordinates at all. He derives expressions for them in terms of coordinates, but that does not mean he is using them as coordinates.

 

[Arman777]

I think you are right. Theeχ $\chi$ actually should not be thought as a real coordinate, but maybe more like a function. Its just better to call it $f_K(r)$ (i.e, $\chi \equiv f_K(r)$). All of the distances can be expressed in terms of the $r$ and that is the important point.

No, he's not. He's not using DC and DM as coordinates at all. He derives expressions for them in terms of coordinates, but that does not mean he is using them as coordinates.

Of course, he is not using them as coordinates (I realized that I used a sloppy language). I just mean that $D_M$ is the same as the co-moving distance $r$.

See this from Weinberg 1972

and this is actually also referred in the Hogg's Paper.

Let us refer, basic, old comoving radial coordinate $r$. In this case the co-moving separation between to co-moving observers is

$$c \int_{t_e}^{t_0} \frac{dt}{a(t)} = \int_0^r \frac{dr}{\sqrt{1-kr^2}}\equiv D_C$$

And the transverse comoving distance is actually just $r \equiv D_M$ (see above pictures)

 

[PeterDonis]

I think you are right. The $\chi$ actually should not be thought as a real coordinate, but maybe more like a function.

It can be one or the other, depending on which coordinate chart you choose. There are certainly coordinate charts in which $\chi$ is a coordinate, not a function of coordinates.

All of the distances can be expressed in terms of the $r$

They can also all be expressed in terms of $\chi$. It all depends on the particular author's preference.

Of course, he is not using them as coordinates (I realized that I used a sloppy language). I just mean that $D_M$ is the same as the co-moving distance $r$.

More precisely, in the case under discussion and in the coordinates chosen in that case, the transverse comoving distance $D_M$ is directly represented by the coordinate $r$. The line of sight comoving distance $D_C$ is not.

 

[Arman777]

I agree

 

[fab13]

1) I can see these 2 definitions of comoving distance (line of sight and transverse) lead to different subtilities.

Up to know, I have always considered the variable "r" as a coordinate in the metric :

$$ds^2 = -c^2dt^2 + a^2(t)[\frac{dr^2}{1-kr^2} + r^2d\Omega^2]\quad(1)$$But you will surely tell me this expression is just a particular case, the flat one.

I don't agree for the definition about Hogg's paper when he says that :

5 Comoving distance (transverse)
The comoving distance between two events at the same redshift or distance but separated on the sky by some angle δθ is DMδθ and the transverse comoving distance DM (so-denoted for a reason explained below ) is simply related to the line-of-sight comoving distance DC :
$$
D_{\mathrm{M}}=\left\{\begin{array}{ll}
D_{\mathrm{H}} \frac{1}{\sqrt{\Omega_{k}}} \sinh \left[\sqrt{\Omega_{k}} D_{\mathrm{C}} / D_{\mathrm{H}}\right] & \text { for } \Omega_{k}>0 \
D_{\mathrm{C}} & \text { for } \Omega_{k}=0 \
D_{\mathrm{H}} \frac{1}{\sqrt{\left|\Omega_{k}\right|}} \sin \left[\sqrt{\left|\Omega_{k}\right|} D_{\mathrm{C}} / D_{\mathrm{H}}\right] & \text { for } \Omega_{k}<0
\end{array}\right.
$$
where the trigonometric functions sinh and sin account for what is called "the curvature of space." (Space curvature is not coordinate-free; a change of coordinates makes space flat; the only coordinate-free curvature is space-time curvature, which is related to the local massenergy density or really stress-energy tensor.) For $\Omega_{\Lambda}=0,$ there is an analytic solution to the equations

especially :

The comoving distance between two events at the same redshift or distance but separated on the sky by some angle δθ is DMδθ and the transverse comoving distance

Indeed, by definition, the physical distance between 2 objects of same redshift is like the physical distance between 2 points on a circle, so we should have instead :

$$D_\text{physical}=D_C\delta\theta$$

and not

$$D_\text{physical}=D_{M}\delta\theta$$

where DC is the radial comoving distance (or yet line-of-sight comoving distance).

Why did he use DM instead of DC ?

2) Same thing happends with the Angular Diameter distance :

we should have $D_A=D_C/(1+z)$ where $D_C$ represents the line-of-sight comoving distance.
This is by definition the distance of observer from the object began to emmit.

Hogg defined it with : $DA=D_M/(1+z)$

It should be : $D_A=D_C/(1+z)$

As a conclusion, why give 2 definitions for comoving distance ? we can deduce the transverse comoving distance from line-of-sight comoving distance by just multiplying with a little angle δθ.

Why couldn't we write directly only one comoving distance $D_C$ (the most common I would say) defined like this below :

Hyperbolic Universe :$\Omega_{k}^{0}>0$

$$
D_{h}=\frac{c}{H_{0} \sqrt{\Omega_{k}^{0}}} \sinh \left[\sqrt{\Omega_{k}^{0}} \int_{0}^{z_{e}} \frac{\mathrm{d} z}{\left[\Omega_{m}^{0}(1+z)^{3}+\Omega_{r}^{0}(1+z)^{4}+\Omega_{\Lambda}^{0}+\Omega_{k}^{0}(1+z)^{2}\right]^{1 / 2}}\right]
$$

Euclidean Universe :$\Omega_{k}^{0}=0$

$$
D_{h}=\frac{c}{H_{0}} \int_{0}^{z_{e}} \frac{\mathrm{d} z}{\left[\Omega_{m}^{0}(1+z)^{3}+\Omega_{r}^{0}(1+z)^{4}+\Omega_{\Lambda}^{0}\right]^{1 / 2}}
$$

Spherical Universe : $\Omega_{k}^{0}<0$

$$
D_{h}=\frac{c}{H_{0} \sqrt{\left|\Omega_{k}^{0}\right|}} \sin \left[\sqrt{\left|\Omega_{k}^{0}\right|} \int_{0}^{z_{e}} \frac{\mathrm{d} z}{\left[\Omega_{m}^{0}(1+z)^{3}+\Omega_{r}^{0}(1+z)^{4}+\Omega_{\Lambda}^{0}+\Omega_{k}^{0}(1+z)^{2}\right]^{1 / 2}}\right]
$$

ps : if someone could see what's wrong, it tells me that I can't us macro '#' but I don't use it in these 3 equations above.

Thanks for your support and don't hesitate to give me remarks, I would be glad to break my old certainties.

 

[PeterDonis]

you will surely tell me this expression is just a particular case, the flat one.

I have said no such thing. Obviously this metric is not spatially flat if $k \neq 0$.

Why did he use DM instead of DC ?

Because, as I have said multiple times now, there are two different comoving distances, not one. For "distance along a circle", you use $D_M$, not $D_C$, since the "distance along a circle" is transverse, not line of sight.

 

[fab13]

I agree of course that in one case, we compute along the line-of-sight ($D_C$) but for the distance "along a circle", we just have, given the angle $\delta\theta$ :

$$D_\text{physical,circle}=D_M=D_C\delta\theta$$

Is this relation valid ? Otherwise, I don't understand nothing at all.

 

[PeterDonis]

I agree of course that in one case, we compute along the line-of-sight ($D_C$) but for the distance "along a circle", we just have, given the angle $\delta\theta$ :

$$D_\text{physical,circle}=D_M=D_C\delta\theta$$

Is this relation valid ?

No. That's the whole point: transverse comoving distances are not just "radial distance times angle", as you would expect them to be in ordinary Euclidean space. That's why Hogg goes to all that trouble to distinguish the two comoving distances.

Actually, for the spatially flat case, the space is ordinary Euclidean space, so for that case in particular, we have $D_M = D_C$. But that is only true for that particular case.

 

[fab13]

1) With :

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.

2) and with :

Yes. Which means that we can write:

$$
\frac{r^2}{1 - K r^2} = D_c^2
$$

or, taking the square root,

$$
\frac{r}{\sqrt{1 - K r^2}} = D_c
$$

No. $r(z)$ is the comoving coordinate. But it has a well-known relationship to comoving distance. Do you know what that relationship is? (Hint: is it true that the equation I just wrote above is that relationship?)

How do you infer that :

$$
\frac{r}{\sqrt{1 - K r^2}} = D_c
$$

from the eq(2) above, i.e :

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

?
Sorry the introduction of another comoving distance (transverse : $D_M$) has broken my initial perception.

Thanks for your patience

 

[PeterDonis]

@fab13, we've gotten way too tangled up at this point. Let's start from scratch, giving the definitions of each of the distances as they appear in Hogg's article; we'll then compare with the definitions given in the Euclid paper.

First is the Hubble distance:

$$
D_H = \frac{c}{H_0}
$$

Next is the line of sight comoving distance:

$$
D_C = D_H \int_0^z \frac{dz}{E(z)}
$$

Next is the transverse comoving distance, which I will write in abbreviated form as:

$$
D_M = F(D_C)
$$

where $F(D_C)$ is whichever function of $D_C$ is the correct one for the curvature parameter $K$ we are considering. For $K = 0$ (spatially flat), it is just $F(D_C) = D_C$. I won't write out the other functions explicitly, since that has already been done earlier in this thread.

Finally, there is the angular diameter distance:

$$
D_A = \frac{D_M}{1 + z}
$$

This formula can of course be easily inverted to give:

$$
D_M = D_A \left( 1 + z \right)
$$

And then we can combine this with the other formula for $D_M$ to obtain:

$$
D_A \left( 1 + z \right) = F(D_C)
$$

(Hogg also discusses luminosity distance, but we haven't used it in this thread so I'll leave it out here.)

Now, Hogg does not use $r$ at all. The Euclid paper does. So let's look at how they define $r$:

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

This should look familiar: it' s just the definition of $D_C$ (not $D_M$) from Hogg, with the definition of $D_H$ substituted for it. So this $r(z)$ is not the same as the radial coordinate $r$ in spherical coordinates! The latter is equal to $D_M$, not $D_C$. And the Euclid paper calls $r(z)$ "comoving distance", without any qualification, which completely obscures the fact that there are two comoving distances, not one, as Hogg clearly states. So the Euclid paper's terminology here is quite confusing in comparison with other literature, and I think it was confusing us all previously in this thread.

Having got that resolved, let's now compare the formulas for comoving volume from Hogg and the Euclid paper. I'll just look at the differentials since that is sufficient to see the comparison. Hogg's differential is:

$$
dV_C = D_H \frac{\left( 1 + z \right)^2 D_A^2}{E(z)} \, d\Omega \, dz
$$

This can obviously be rewritten in terms of $D_C$ using our formula above:

$$
dV_C = D_H F(D_C)^2 \frac{1}{E(z)} \, d\Omega \, dz
$$

The Euclid paper's differential is:

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

We can use the earlier definition $H(z) = H_0 E(z)$ from the Euclid paper and Hogg's definition of $D_H$ to rewrite this as:

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

These two formulas are formulas for the same thing: the differential in comoving volume as a function of the differentials in solid angle $\Omega$ and redshift $z$. So they must be equal. And that means we must have (factoring out $D_H$, $1 / E(z)$, $d\Omega$, and $dz$ since all of those appear the same in both formulas):

$$
F(D_C)^2 = \frac{r^2(z)}{\sqrt{1 - K r^2(z)}}
$$

Since the Euclid paper defines $r(z)$ to be equal to Hogg's $D_C$, as we noted above, this means:

$$
F(D_C)^2 = \frac{D_C^2}{\sqrt{1 - K D_C^2}}
$$

And now, two final touches: first, we can switch back to $D_M$ on the LHS of the above to obtain:

$$
D_M^2 = \frac{D_C^2}{\sqrt{1 - K D_C^2}}
$$

Second, we can observe that, based on Hogg's formula for $D_C$ in terms of $D_H$, we have $D_H dz / E(z) = d D_C$ (the differential of comoving distance--i.e., we can eliminate redshift $z$ in favor of comoving distance). And then we can substitute back into the differential comoving volume formula to obtain:

$$
dV_C = D_M^2 \, d\Omega \, d D_C
$$

And now we can actually unpack what this means. Let's take it in steps:

(1) $D_C$ is a function of redshift $z$. What this means, physically, is that, as we look along a given line of sight, we see objects now whose light that we see now has various redshifts. The larger the redshift $z$ of the light we see from an object, the larger its line of sight comoving distance from us. This is simply because, the larger the redshift, the further in the past the light was emitted, so the larger the comoving distance from us has to be for us to be receiving the light with that redshift now.

(2) As we noted above, for the spatially flat case, $K = 0$, we have $D_M = D_C$. For the case of positive curvature, $K = 1$, we have $\Omega_k < 0$, and $D_M < D_C$. For the case of negative curvature, $K = -1$, we have $\Omega_k > 0$, and $D_M > D_C$. (These relationships follow from the properties of the $\sin$ and $\sinh$ functions, respectively.) What this means, physically, is that in a universe with positive spatial curvature, there is less comoving volume at a particular comoving distance from us than there would be in a spatially flat universe, whereas in a universe with negative spatial curvature, there is more.

This might be easier to understand if I invert it: in a universe with positive spatial curvature, there is more comoving distance between us and a 2-sphere with a given surface area (and hence a small "slice" of comoving volume equal to that surface area times the differential of comoving distance) than there would be in flat Euclidean space, whereas in a universe with negative spatial curvature, there is less. This can be understood by analogy with the case of a 2-surface with positive (a 2-sphere) or negative (a "saddle" type shape) curvature, by looking at how distance from a central point (such as the North Pole of the sphere) along the surface relates to the circumference of a circle at that distance; for positive curvature, the distance is larger for a given circumference than it would be in Euclidean space, whereas for negative curvature, it is smaller.

In short: the spatial curvature affects the relationship between transverse and line of sight comoving distances.

(3) So what the formula for the differential of comoving volume is telling us is that that differential is the product of:

- the differential in comoving distance ($d D_C$), times

- the differential of surface area at the given comoving distance, as a function of the differential in solid angle ($D_M^2 d\Omega$, since the distance in this case is transverse, not line of sight).

The two different sources--Hogg and the Euclid paper--were just choosing different ways of expressing both the differential in comoving distance and the transverse comoving area $D_M^2$ in terms of other parameters.

Hopefully this long post helps to clarify what is going on, and clears up confusions from previous posts (including mine, since the differences in notation between the two sources were leading me to make mistakes).

 

[fab13]

Dear PeterDnis,

Big thanks for your detailed answer, you took the time to give me all clarifications and explanations and it is fine from your part.

Just a little remark, maybe I am wrong : when you say :

This might be easier to understand if I invert it: in a universe with positive spatial curvature, there is more comoving distance between us and a 2-sphere with a given surface area (and hence a small "slice" of comoving volume equal to that surface area times the differential of comoving distance) than there would be in flat Euclidean space, whereas in a universe with negative spatial curvature, there is less

Would you want rather say that :

for positive spatial curvature (omega_k =+1 => sinus) there is less comoving distance between us and a 2-sphere with a given surface area...
whereas in a universe with negative spatial curvature (k=-1 => sinh), there is more

?

I say this since as you said,

"comoving volume equal to that surface area times the differential of comoving distance"

.

Sorry if it is a mistake.

Best regards

 

[PeterDonis]

Would you want rather say that

No. What I said is correct as I stated it.

I say this since as you said

What you quoted there is about the differential of comoving volume and how it depends on the differential of (line of sight) comoving distance and the surface area of a 2-sphere (which depends on the transverse comoving distance).

The other quote you gave is about the total (line of sight) comoving distance between us and a 2-sphere with a given surface area. That depends on global properties, not just local ones, which is what the differentials depend on.

Another way of viewing those global properties is that, if we use the transverse comoving distance to estimate how much line of sight comoving distance there is between us and a given object (using the relationships of Euclidean geometry between the surface area of a sphere and its radius), our estimate will be too low (more line of sight comoving distance than the Euclidean estimate) in a positively curved universe, too high (less line of sight comoving distance than the Euclidean estimate) in a negatively curved universe, and just right (exactly as much line of sight comoving distance as the Euclidean estimate) in a spatially flat universe.

 

[fab13]

Dear PeterDonis,

I am not here to bore anyone but just to understand. Your long post above is very important in my attempt of grasping the subtilities for my issue.

I can't yet understand how you can make appear the factor $r(z)^2$ into $\dfrac{r^2}{\sqrt{1-kr^{2}}}$, i.e when you write :

$$F(D_C)^2 = \frac{r^2(z)}{\sqrt{1 - K r^2(z)}}\quad(1)$$

Up to now, the only thing I know is :

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

with :

$$

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.
\quad(3)
$$

How to you make the link between eq(1) and eq(2) ? I mean, could you precise how you justify this function $F$ that seems to be defined like :

$$F(D_C)^2 = \frac{r^2}{\sqrt{1 - K r^2}}\quad(4)$$

which can also be written as : $$F(D_C)^2 = \frac{D_C^2}{\sqrt{1 - K D_C^2}}$$

There is a short-cut in obtaining eq(4) that I can't still to solve, i.e passing from eq(2) to eq(4).

Best Regards

 

[PeterDonis]

I can't yet understand how you can make appear the factor $r(z)^2$ into $\dfrac{r^2}{\sqrt{1-kr^{2}}}$, i.e when you write :

$$F(D_C)^2 = \frac{r^2(z)}{\sqrt{1 - K r^2(z)}}\quad(1)$$

I didn't "make it appear". I just took two formulas, one from Hogg and one from the Euclid paper, pointed out that they are both formulas for the same quantity, namely $dV_C$, and therefore must be equal, wrote down the equation that says they are equal, and canceled common factors.

How to you make the link between eq(1) and eq(2) ?

See above.

I mean, could you precise how you justify this function $F$ that seems to be defined like :

$$F(D_C)^2 = \frac{r^2}{\sqrt{1 - K r^2}}\quad(4)$$

No, that's not how $F(D_C)$ is defined. It's defined based on Hogg's formulas for $D_M$ in section 5 of his article. Each possible case for spatial curvature (negative, zero, positive) gives a formula for $D_M$ as a function of $D_C$ (for the zero curvature, flat case, it is just $D_M = D_C$, as I said). Those formulas are what I am calling $F(D_C)$.

What I am then doing, as I described above, is showing how, in order for Hogg's formula for $d V_C$, the differential of comoving volume, to be equal to the Euclid paper's formula for the same quantity (the integrand in their integral for comoving volume, which is equation 14 in that paper), $F(D_C)^2$ must be equal to $r^2 / \sqrt{1 - K r^2}$. That is done, as I said above, by simply setting the two formulas equal and canceling common factors.

 

[fab13]

No, that's not how $F(D_C)$ is defined.

Sorry, It is not correct, read your post #51, you wrote and infer that :

$$F(D_C)^2 = \frac{r^2(z)}{\sqrt{1 - K r^2(z)}}= \frac{D_C^2}{\sqrt{1 - K D_C^2}}$$

But you don't give the demonstration of this equality (by other way than taking the equaltily between both $dV_C$);

That's why I reformulate my message :

How can you proove that independently from the 2 papers (Hogg and Euclid), i.e without taking the 2 papers, we can write (from your post #51 above such that written) :

$$F(D_C)^2 = \frac{r^2(z)}{\sqrt{1 - K r^2(z)}}\quad(1)$$

Thanks

 

[PeterDonis]

It is not correct

Wrong. What I say is correct. You are simply not understanding what I say.

you don't give the demonstration of this equality (by other way than taking the equaltily between both );

That's because setting both expressions for $d V_C$ equal is the only way to demonstrate the equality you are asking for a demonstration of.

How can you proove that independently from the 2 papers

As far as I know, you can't. I proved it the way I did because that's the only way I can see to prove it.

My question in return is, so what? Why should I have to prove it some other way, when I've already proved it this way?

 

[fab13]

As far as I know, you can't. I proved it the way I did because that's the only way I can see to prove it.

Good, finally, you admit that you can't see another way to prove it. Imagine if you had not Eulcid and Hogg's papers, how could have you been to demonstrate that :

$$F(D_C)^2 = \frac{r^2(z)}{\sqrt{1 - K r^2(z)}}\quad(1)$$

??

So, I am keeping to look for this another way, hoping someone will help me as you tried kindly since many posts.

Thanks

 

[PeterDonis]

you admit that you can't see another way to prove it

When did I ever claim otherwise?

Imagine if you had not Eulcid and Hogg's papers, how could have you been to demonstrate that

The only reason this was an issue in the first place was because of the difference in the formulas between the Euclid paper and the Hogg article. The Euclid paper's notation, as I think I already commented, does not seem to be standard (and, as I said, that confused me and caused me to make mistakes in earlier posts), and the paper does not recognize, as the Hogg article does, that there are two comoving distances, not one, and its non-standard notation invites confusion between the two.

So the only point I see to this exercise at all is to try to make some sense out of what the Euclid paper is saying. I don't think the equation you are looking for another way to demonstrate has any relevance at all except in that context, and then only because of the Euclid paper's non-standard notation, in particular its non-standard usage of $r$.

 

[PeterDonis]

I am keeping to look for this another way

I think you need to stop and think about exactly what you are looking for.

If you are looking for a way to think about "the comoving volume between two redshifts" that is reasonably intuitive physically, I think your best bet is this equation from my earlier post:

$$
d V_C = D_M^2 \, d\Omega \ d D_C
$$

Here $D_C$ is the only function of $z$, so you can just use the definition of $D_C$ in terms of $z$ (which is given in Hogg's paper) to obtain a formula with $dz$ instead of $d D_C$, and integrate that between the two redshifts of interest. Then you just integrate the result over a full 2-sphere (i.e., over the full range of $\Omega$, which just results in multiplying your answer by a factor of $4 \pi$ for the area of the 2-sphere), and you're done. This is all straightforward in terms of the definitions given in Hogg's article, which are, as far as I know, standard in the cosmology literature.

I don't know why the Euclid paper chose a different notation and a different way of doing the integral; presumably they had some reason particular to the specific problem they were working on. Their paper looks like it is intended for an audience of experts in that particular subfield of cosmology, not for a broader readership, so they might have expected that their readers would be familiar with their notation and it wouldn't be an issue.