Spoiler: I have found a way of understanding it. Not sure is correct, but it's good for me at the moment. Yes. I do believe too that I have overthought this too much.
Well, I have found two main justifications:
- The problem is spherical and in these symmetric conditions, $C=D=1$
- They obviously are equal $(C=D)$ because symmetry and also it turns out their value doesn't matter because we can normalize the metric expression in order to make them equal to 1.
I'm having problems in understanding the first reason because although I understand there is a spherical symmetry, I don't see why terms should be equal when even the previous metric factors are not. I mean, we have $r^2d\theta^2$ and $r^2sin^2\theta d\varphi^2$. That doesn't seem very symmetrical, right?
For a moment, the second reason almost convinced me. But again, something is fishy. If we normalize to convert in one the functions in $C$ and $D$, what happens with the term $ds^2$ in the other part of the equal sign? Is it the same after the normalization? If so, why no one seems to say nothing about it?
Symmetry in angular coordinates
You probably remember that in plane space...
\begin{equation}
ds^2=c^2dt^2-dx^2-dy^2-dz^2 \rightarrow ds^2 = c^2dt^2+dr^2+r^2(d\theta^2+sin^2\theta d\varphi^2)
\end{equation}
ds^2=c^2dt^2-dx^2-dy^2-dz^2 \rightarrow ds^2 = c^2dt^2+dr^2+r^2(d\theta^2+sin^2\theta d\varphi^2)
\end{equation}
So if we consider $C=D=1$, then we are saying that Schwarzschild metric is equivalent to the null metric in its angular coordinates. First things first. How come we have a $sin\theta$ term in the $\varphi$ coordinate? It happens as you probably recall as a consequence of defining a line element: in the $\varphi$ coordinate, a differential line element increments $rsin\theta d\varphi$ times if all the other coordinates remain constant. It's a consequence of the way we define the coordinates and the convention in angles. As summary: we define $\varphi$ as the angle of the projected $r$ over the $XY$ plane; we do that because we need $\theta$ and $\varphi$ to be defined orthogonal: if not, expresions would be a lot more complicated. The definition which makes the line element so weird, it's actually the warranty that symmetry is there.
Ok, so despite metric terms are not exactly equal, there's a symmetry. So, we can believe that $C=D$. Are they equal to one? If so, we are stating that Schwarzschild metric is actually equal to a null metric in its angular coordinates. That's like saying a lot!
When $r$ is not $r$ but actually $r_c$
Well, after digging a little, I have discovered that it turns out that the $r$ in which $C=D=1$ is actually a circunferential $r$ or $r_c$ for short. Now that I think about it, it seems like a silly remark: a radius is defined as the distance between the center of a circumference and its closure, right? How come there can be two different kind of radiuses? Devil is in details, people: in normal EUCLIDEAN space, $r=r_c$. But we are in CURVED space!
In curved space the length of a circumference can be greater (or smaller) thant $2\pi r$. Do you want an example?
Imagine you are at the North Pole. You have a magical rope with a length $r \approx 20.000 km$ and you ask a friend in the equator to take the other extreme of the rope for a walk. Your friend walks all over the equator line by car, ship or feet, without releasing the magical-mystical rope, which guarantees thatyou are he is actually over the equator everytime. After a couple of kidnapps, robberies and attempted murders in those parts of the globe where people doesn't care about human life too much, your friend emails you how much distance he has measured: $l \approx 40075 km.$. He then proceeds to ask himself why in the hell did he had to make the journey in the first place if he could have asked the internet for the numbers. You answer him because science, and he then proceeds to say he is your friend no more (I have heard this joke in somewhere, maybe in Wheeler et al, sorry). It turns out, you discover in your cold new home, that the distance is smaller than $2 \pi r \approx 126000 km$.
In curved space the length of a circumference can be greater (or smaller) thant $2\pi r$. Do you want an example?
Imagine you are at the North Pole. You have a magical rope with a length $r \approx 20.000 km$ and you ask a friend in the equator to take the other extreme of the rope for a walk. Your friend walks all over the equator line by car, ship or feet, without releasing the magical-mystical rope, which guarantees that
So, TL;DR:
But, we are going off track!! Let's come back to the reason $C=D=1$. Using a $r_c$ coordinate instead of a purely $r$ coordinate, assures that $C=D=1$, which is a good thing for simplification and stuff. Why can we use $r_c$? Doesn't it change the definition of coordinates? What about the proper distance $ds$? Is it the same? Is it the same with a weird constant?
\begin{equation}
ds^2=c^2A'(r)dt^2+B'(r)dr^2+C(r)r^2(d\theta^2+sin^2\theta )\rightarrow \\
\frac{ds^2}{C(r)} = c^2 A(r)dt^2+B(r)dr^2+r^2(d\theta^2+sin^2\theta )~~~ \text{with } ~~~A(r) = \frac{A'(r)}{C(r)}, B(r)=\frac{B'(r)}{C(r)}
\end{equation}
Sooo... What does $\frac{ds^2}{C(r)}$ mean? Well... Funny thing ahead. Since $C(r)$ has no dimensions (none of the unknown functions have, all of them should be dimensionless), it doesn't really matter what is the exact value for $ds$! When we solve the problem for this new $ds'^2=\frac{ds^2}{C(r)}$, we will have a different exact dimensionless value for $ds$, a different exact value for the non-dimensional equation for proper distance, but there will be no change in physics involved. Instead of saying "for a value $r$, $ds$ is equal to something", we will say "$ds$ is equal to other thing" and we will be able to know how to convert this "other thing" into units. There is no lose of generality, as some of the demostrations I have read about the subject, tell. Something, by the way, I didn't grasp at the moment. I'm not a smart dude, you know.
- In a positive curvature space (like the surface of the Earth) $l<2\pi r$
- In a negative curvature space (like hyperbolic space) $l>2 \pi r$
- In a zero curvature space (like
your moman Euclidean plane) $l=2 \pi r$.
But, we are going off track!! Let's come back to the reason $C=D=1$. Using a $r_c$ coordinate instead of a purely $r$ coordinate, assures that $C=D=1$, which is a good thing for simplification and stuff. Why can we use $r_c$? Doesn't it change the definition of coordinates? What about the proper distance $ds$? Is it the same? Is it the same with a weird constant?
Why normalization doesn't actually changes something we should really care about
Normalization in this context means...\begin{equation}
ds^2=c^2A'(r)dt^2+B'(r)dr^2+C(r)r^2(d\theta^2+sin^2\theta )\rightarrow \\
\frac{ds^2}{C(r)} = c^2 A(r)dt^2+B(r)dr^2+r^2(d\theta^2+sin^2\theta )~~~ \text{with } ~~~A(r) = \frac{A'(r)}{C(r)}, B(r)=\frac{B'(r)}{C(r)}
\end{equation}
Sooo... What does $\frac{ds^2}{C(r)}$ mean? Well... Funny thing ahead. Since $C(r)$ has no dimensions (none of the unknown functions have, all of them should be dimensionless), it doesn't really matter what is the exact value for $ds$! When we solve the problem for this new $ds'^2=\frac{ds^2}{C(r)}$, we will have a different exact dimensionless value for $ds$, a different exact value for the non-dimensional equation for proper distance, but there will be no change in physics involved. Instead of saying "for a value $r$, $ds$ is equal to something", we will say "$ds$ is equal to other thing" and we will be able to know how to convert this "other thing" into units. There is no lose of generality, as some of the demostrations I have read about the subject, tell. Something, by the way, I didn't grasp at the moment. I'm not a smart dude, you know.
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