Why are ds and dω taken as zero in calculating proper distance in cosmology?

[Apashanka]

While studying the proper distance in cosmology I came across the thing as
The FRW metric
ds²=c²dt²-a(t)²[dr²+S_k(r)²dω²]
And S_k(r)=Rsin(r/R)
Where a(t) is the scale factor and dω²= dθ²+sin²θdΦ²,
While calculating the proper distance at the time of emmission of light the term ds and dω are taken zero and thus cdt=a(t)dr
And proper distance is taken to be ∫dr=c∫dt/a(t). The limit being from t_emmt. to t_observ
The explanation given is along spatial geodesic dω=0
Could someone please help me in sort out why dω and ds are taken 0 and proper distance being ∫dr.
Thanks

 

[PeterDonis]

While studying the proper distance in cosmology

From what reference? Please give a specific reference.

 

[Orodruin]

The explanation given is along spatial geodesic dω=0

This is generally not true. However, it is always possible to choose coordinates such that it is the case, i.e., such that your geodesics are radial. The requirement of ds=0 is just telling you that the world line is light-like.

 

[Apashanka]

This is generally not true. However, it is always possible to choose coordinates such that it is the case, i.e., such that your geodesics are radial. The requirement of ds=0 is just telling you that the world line is light-like.

Ok sir that means that ∫dr tells the comoving distance traveled by the light from the time of emmission to the present time (time of observation) if the light moves keeping θ and Φ constant.
That means the actual distance (due to expansion) traveled by the light will be a(t₀)∫dr where t₀ is present time

 

[Orodruin]

That would be the proper distance now to the comoving coordinate where the signal was emitted. I am not sure I would call that the distance travelled.

 

[Apashanka]

That would be the proper distance now to the comoving coordinate where the signal was emitted. I am not sure I would call that the distance travelled.

cdt=a(t)dr that's fine for which dt is the time between emmited and observed light and dr is the comoving distance for light ray following radial geodesics.
But sir what does ∫cdt/a(t) from t_emmited to t_observed implies (e.g ∫dr and what will be the integration limit for this).
I have trouble in understanding this.
Thank you.

 

[Orodruin]

But sir what does ∫cdt/a(t) from temmited to tobserved implies (e.g ∫dr and what will be the integration limit for this).

This is just integrating $c\,dt/a(t) = dr$ between the emission and observation events. The integration limits of $r$ are the $r$ coordinates of the emission and observation events, respectively. For the end result in terms of the cosmological redshift, it really does not matter what exactly those coordinates are. The only thing that matter is that they do not change between the different signals.

 

[Apashanka]

This is just integrating $c\,dt/a(t) = dr$ between the emission and observation events. The integration limits of $r$ are the $r$ coordinates of the emission and observation events, respectively. For the end result in terms of the cosmological redshift, it really does not matter what exactly those coordinates are. The only thing that matter is that they do not change between the different signals.

That means sir r is not the comoving coordinate
Only elementary length dr is the comoving coordinate

 

[Orodruin]

That means sir r is not the comoving coordinate

No, $r$ is definitely the comoving coordinate. Any other claim is wrong. $dr$ is the infinitesimal change in the comoving coordinate. Saying $r$ is not the comoving coordinate is like saying that $x$ is not a coordinate in Euclidean space just because the distance between two objects need not be the $x$ value of one of them.

The comoving distance between two (radially located) comoving positions is the difference in those positions' $r$ coordinates and is what you get out from the integration.

 

[Apashanka]

No, $r$ is definitely the comoving coordinate. Any other claim is wrong. $dr$ is the infinitesimal change in the comoving coordinate. Saying $r$ is not the comoving coordinate is like saying that $x$ is not a coordinate in Euclidean space just because the distance between two objects need not be the $x$ value of one of them.

The comoving distance between two (radially located) comoving positions is the difference in those positions' $r$ coordinates and is what you get out from the integration.

No, $r$ is definitely the comoving coordinate. Any other claim is wrong. $dr$ is the infinitesimal change in the comoving coordinate. Saying $r$ is not the comoving coordinate is like saying that $x$ is not a coordinate in Euclidean space just because the distance between two objects need not be the $x$ value of one of them.

The comoving distance between two (radially located) comoving positions is the difference in those positions' $r$ coordinates and is what you get out from the integration.

Oh sir that means we are trying to find the distance r for those emmited signals which are received at present.

 

[Orodruin]

Oh sir that means we are trying to find the distance r for those emmited signals which are received at present.

No it does not, it means that you are getting an expression for the same comoving distance in terms of two different light signals and that you can use that fact to find the expression for the cosmological redshift.

 

[Apashanka]

No it does not, it means that you are getting an expression for the same comoving distance in terms of two different light signals and that you can use that fact to find the expression for the cosmological redshift.

Sir I thought of an example which may be silly.
If two galaxies A and B separated by comoving distance r and having the same angular coordinates θ,Φ.
At time t₁ the distance between them is a(t₁)r and light from Galaxy A is emmited towards B.
The emmited light reaches Galaxy B at time t₂ and the distance between them now is a(t₂)r ,therefore
c(t₂-t₁)(distance traveled by light between t₂ and t₁)=a(t₂)r from this can't r(comoving distance) be calculated??

 

[Orodruin]

If two galaxies A and B separated by comoving distance r

You are confusing yourself here by calling the comoving distance between the galaxies the same as the radial coordinate.

The emmited light reaches Galaxy B at time t2 and the distance between them now is a(t2)r ,therefore
c(t2-t1)(distance traveled by light between t2 and t1)=a(t2)r from this can't r be calculated??

No, this is incorrect. The proper distance between the galaxies at $t_2$ is not the same as the distance traveled by the light between $t_1$ and $t_2$.

 

[Apashanka]

You are confusing yourself here by calling the comoving distance between the galaxies the same as the radial coordinate.

Sir Galaxy A have coordinates (r₁,θ,Φ) and Galaxy B have coordinates (r₂,θ,Φ) since r,θ,Φ are the comoving coordinates therefore the radial distance between them is r₂-r₁ which is also comoving therefore previous r is taken as r₂-r₁
Am I right??

 

[TEFLing]

This is just integrating $c\,dt/a(t) = dr$ between the emission and observation events. The integration limits of $r$ are the $r$ coordinates of the emission and observation events, respectively. For the end result in terms of the cosmological redshift, it really does not matter what exactly those coordinates are. The only thing that matter is that they do not change between the different signals.

like Orodruin said, that's the right-hand side (RHS)

you just integrated $\int \, dr = \Delta r$, the comoving coordinate difference between emitter & absorber

but to know what that turns out to be, for a photon traveling radially from epoch $t_1$ to epoch $t_2$, you have to integrate the LHS, $\int \,c\,dt/a(t)$. Usually you use $dt = da/\dot a$

 

[Apashanka]

like Orodruin said, that's the right-hand side (RHS)

you just integrated $\int \, dr = \Delta r$, the comoving coordinate difference between emitter & absorber

but to know what that turns out to be, for a photon traveling radially from epoch $t_1$ to epoch $t_2$, you have to integrate the LHS, $\int \,c\,dt/a(t)$. Usually you use $dt = da/\dot a$

Yes I am also saying that galaxy A have coordinates (r₁,θ,Φ) and B have coordinates (r₂,θ,Φ),since r,θ,Φ are the comoving coordinates the coomoving distance between them is r₂-r₁=δ ,and if light at time t₁ is emmited from Galaxy A ,the distance between them is a(t₁)δ similarly the time t₂ at which the emmited light reaches Galaxy B the distance between them is a(t₂)δ
If this emmited light follows only the radial geodesic then c(t₂-t₁)=a(t₂)δ
From it can't the coomoving distance δ between the galaxies A and B be calculated??

 

[Orodruin]

If this emmited light follows only the radial geodesic then c(t2-t1)=a(t2)δ

No. You have already been told in #13 and again in #15 that this is simply wrong.

 

[Apashanka]

No. You have already been told in #13 and again in #15 that this is simply wrong.

Sir actually I am not getting this ,don't mind if you please explain this to me .
Thank you.

 

[Orodruin]

Sir actually I am not getting this ,don't mind if you please explain this to me .
Thank you.

It is just not true. The left-hand side is not $c(t_2-t_1)$. The relation between times and the comoving distance $\delta$ is given by integrating the relation $c\, dt = a(t) dr$ after dividing both sides by $a(t)$. This gives you the integral
$$
\delta = r_2-r_1 = \int_{r_1}^{r_2} dr = c \int_{t_1}^{t_2} \frac{dt}{a(t)}.
$$
The right-hand side in this relation does not integrate to $c(t_2-t_1)/a(t_2)$, which it would need to do for your assumed relation to hold.

 

[Apashanka]

It is just not true. The left-hand side is not $c(t_2-t_1)$. The relation between times and the comoving distance $\delta$ is given by integrating the relation $c\, dt = a(t) dr$ after dividing both sides by $a(t)$. This gives you the integral
$$
\delta = r_2-r_1 = \int_{r_1}^{r_2} dr = c \int_{t_1}^{t_2} \frac{dt}{a(t)}.
$$
The right-hand side in this relation does not integrate to $c(t_2-t_1)/a(t_2)$, which it would need to do for your assumed relation to hold.

Ok sir thanks.

 

[Apashanka]

Ok sir thanks.

But sir this δ is constt. since comoving distance between two galaxies doesn't change with time.
Isn't it??

 

[Orodruin]

But sir this δ is constt. since comoving distance between two galaxies doesn't change with time.
Isn't it??

Yes, so what?

 

[Arman777]

While calculating the proper distance at the time of emmission of light the term ds and dω are taken zero and thus cdt=a(t)dr
And proper distance is taken to be ∫dr=c∫dt/a(t). The limit being from t_emmt. to t_observ
The explanation given is along spatial geodesic dω=0
Could someone please help me in sort out why dω and ds are taken 0 and proper distance being ∫dr.
Thanks

Are you sure that this is true? We know that $r$ is the comoving distance, not the proper distance. By using your integral you cannot find the proper distance, you can only find the relationship between the redshift and scale factor. Since $ds=0$ only for light.

In general, derivation of proper distance goes likes this. Let us say, we want to measure the distance of an object at time $(t=t_0)$ (hence as you noticed we fixed the time). And let's assume the object moves in the only radial direction. Then in the FLRW metric $-c^2dt^2$and $dΩ^2$ terms becomes $0$. S we are left with,

$$ds(t_0)^2=a(t_0)^2dr^2$$ or
$$s_p(t_0)=a(t_0)\int dr $$
$$s_p(t_0)=a(t_0)r $$
Where $s_p(t_0)$ the proper distance of the object at time $t_0$.

This means that at any time $t$ the proper distance of the object is just the proportional to the scale factor for that given time with respect to the choosen comoving coordinate

Now If you want to proper distance between emission and observation then you just have to know the a(t) values at a time. Since r is the comoving coordinate it doesn't change with time. $$(dr/dt=0)$$

So the proper distance at the emission is just $$s_p(t_e)=a(t_e)r$$ and the and proper distance at the observation would be $$s_p(t_o)=a(t_o)r$$

 

[Apashanka]

Are you sure that this is true? We know that $r$ is the comoving distance, not the proper distance. By using your integral you cannot find the proper distance, you can only find the relationship between the redshift and scale factor. Since $ds=0$ only for light.

In general, derivation of proper distance goes likes this. Let us say, we want to measure the distance of an object at time $(t=t_0)$ (hence as you noticed we fixed the time). And let's assume the object moves in the only radial direction. Then in the FLRW metric $-c^2dt^2$and $dΩ^2$ terms becomes $0$. S we are left with,

$$ds(t_0)^2=a(t_0)^2dr^2$$ or
$$s_p(t_0)=a(t_0)\int dr $$
$$s_p(t_0)=a(t_0)r $$
Where $s_p(t_0)$ the proper distance of the object at time $t_0$.

This means that at any time $t$ the proper distance of the object is just the proportional to the scale factor for that given time with respect to the choosen comoving coordinate

Now If you want to proper distance between emission and observation then you just have to know the a(t) values at a time. Since r is the comoving coordinate it doesn't change with time. $$(dr/dt=0)$$

So the proper distance at the emission is just $$s_p(t_e)=a(t_e)r$$ and the and proper distance at the observation would be $$s_p(t_o)=a(t_o)r$$

Yes if the distance between two points be r (e.g comoving distance) then the distance between them grows with time as a(t)r.
Likewise at the emmission time ,the distance between them is a(t_e)r
And at the time of observation the distance between them is a(t_o)r.

 

[Arman777]

Yes if the distance between two points be r (e.g comoving distance) then the distance between them grows with time as a(t)r.
Likewise at the emmission time ,the distance between them is a(t_e)r
And at the time of observation the distance between them is a(t_o)r.

So you understand the concept or your answer to the question ?