I Demonstration of comoving volume between 2 redshifts

[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.