What does $R$ denote in this spherically symmetric metric?

Question

I was going through the book Gravitation by Misner, Throne and Wheeler (1973). On page 595, it is written that

$$ds^2 = -e^{2 \Phi} \ dt^2 + e^{2 \Lambda} \ dr^2 + R^2 \ d \Omega^2 ,$$
where $d \Omega^2 = d \theta^2 + \sin \theta \ d\phi^2$ and $R, \Phi$ and $\Lambda$ are arbitrary functions of $r$ and $t$. But since this is static metric, they are function of $r $ only.

My question here is what does $R$ represent here ? Is it Ricci Scalar ? I am very beginner in this subject, but it doesn't seem to me Ricci Scalar for sure. But then what is the reason for this kind of notation here ? Does it have any kind of relation with Ricci Scalar ? Thanks.

Answer 1

Like the text says, it's just a function, no particular relation to the Ricci scalar. So why did they choose to call it $R$, if there could be confusion? Because it turns out that it can be given a geometric interpretation. Say you have a sphere at a fixed time and with a coordinate radius $r$. Setting $t$ and $r$ constant, the metric on the sphere is

$$ds^2 = R(r)^2 (d\theta^2 + \sin^2 \theta\, d\varphi^2).$$

If we then set $\theta = \pi/2$, we're left with the equator of this sphere, on which we have $ds = R(r)\, d\varphi$. And if we integrate proper length $ds$ along the equator, we find that $s = 2\pi R(r)$ is the circumference of the equator. You can also calculate the determinant of the metric on the sphere, finding

$$\sqrt{\det g} = R(r)^2 \sin^2\theta.$$

We can then integrate this over the sphere to find its surface area $A = 4\pi R(r)^2$.

What does this tell us? Remember that $r$ is in principle just a coordinate: a number identifying different spheres. These results say that the function $R$ is the radius of each sphere as measured by someone on its surface: that is, the area of a sphere of coordinate radius $r$ is not $4\pi r^2$ but $4\pi R^2$, and similarly for the circumference. This is why it makes sense to name this function $R$.