grav-universe said:
By singularity there, I meant the place where the external coordinate systems become singular, the event horizon at 2 m, not the point at the center.
Ah, ok. Sorry for the mixup on my part.
grav-universe said:
Right, which is why I think that setting h(r) = 1 as an original condition for the derivation of the Schwarzschild metric, such that tangent lengths of rulers are inferred the same for all r, guarantees that the singular radius, or event horizon, will lie at a non-zero radius, at 2 m in this case, because we shouldn't infer the same tangent lengths all the way down to a point. Not that there's anything particularly wrong with that coordinate system, I suppose, but I'm thinking that perhaps we shouldn't set h(r) = 1 originally, but apply some other original condition instead.
Actually, the way you wrote down the "original condition", with three functions of r, f(r), g(r), and h(r), is a better way of doing it than the Wiki page. Then the question becomes, is there any way to reduce the number of functions? It turns out there is, by adopting a particular scaling for the r coordinate. There are at least two ways of doing that that are used in the relativity literature:
(1) The standard Schwarzschild r coordinate, which makes h(r) = 1 and leaves f(r) and g(r) as the two functions;
(2) The isotropic r coordinate, which makes g(r) = h(r), so again there are two functions, f(r) and g(r), but now g(r) multiplies all three space components of the line element, instead of just the r component.
Either of these r coordinates can be adopted without loss of generality, except for one thing, which has been mentioned in other threads: if we use the isotropic r coordinate, we can only cover the region outside the horizon. The standard Schwarzschild r coordinate can be used inside the horizon, although in that chart it is timelike there instead of spacelike.
I think that more or less addresses the part of your previous post that I didn't comment on before, but I'll say a few more things in response to it:
grav-universe said:
The metric is basically just a transformation from the locally measured values to those of the distant observer.
Ok, this clears up how you are interpreting the metric (which is a perfectly good way to interpret it).
grav-universe said:
The distant observer measures dt, dr, and dθ r.
You should probably try to use the word "infers" here, since you have agreed that this is how you mean the word "measures" when you are talking about the distant observer.
grav-universe said:
The local observer measures dt' = sqrt(f(r)) dt, dr' = sqrt(g(r)) dr, and dθ' r' = sqrt(h(r)) dθ r.
Ok, except that you put a prime on dθ'. Strictly speaking, that's not correct; see below.
grav-universe said:
In Schwarzschild, we are given h(r) = 1
Because we adopt the standard Schwarzschild r coordinate, yes.
grav-universe said:
In the link, they give g_22 dθ^2 = dθ^2 r^2. The left side is the local measurement, same as h(r) dθ^2 r^2, and the right hand side is according to the distant observer.
Not really. I can't really speak for whoever wrote the Wiki page, since I don't know what they had in mind, but here's how I read the above. First, there is no difference between "local" and "distant" observers for the angular coordinates; dθ is the same regardless of who is interpreting the measurement. That follows from spherical symmetry; more precisely, given spherical symmetry, we can always *define* angular coordinates so that's true, and that's how \theta and \phi were defined. In other words, the physical circumference of any 2-sphere covers an angle of 2 \pi, and the physical area of any 2-sphere covers a solid angle of 4 \pi. The only thing that can change, depending on how we scale the angular coordinate, is what quantity with units of distance or area multiplies the angle or solid angle to obtain the actual physical circumference or area.
Given that, a better way of expressing what I think the Wiki page was trying to express would be to say that, with spherical symmetry, we can always adopt angular coordinates that make the angular part of the metric look like this:
g_{22} d\theta^2 + g_{33} d\phi^2 = h(r) \left( d\theta^2 + sin^2 \theta d\phi^2 \right)
Then it only remains to show that, if we adopt the standard Schwarzschild r coordinate, we must have h(r) = r^2. That, IMO, better separates out the behavior of the angular coordinates.
(And once again, there is nothing *requiring* us to adopt angular coordinates that behave this way; but given spherical symmetry, it's an obvious thing to do because it makes angles and solid angles behave the way we intuitively think they "should", and it can be done without loss of generality. I can't really fault the Wiki page on this point, since I don't think I've see it discussed even in GR textbooks because everybody just takes it for granted that it can be done. But if we're interested in identifying *all* of the steps of the logic, our treatment of the angular coordinates needs to be stated explicitly too.)
grav-universe said:
For instance, with Eddington's isotropic coordinates, we are only changing radiuses of 2-spheres.
Not really; you are changing how the 2-spheres are *labeled*. In other words, a given 2-sphere, with a given physical radius, circumference, and area, will be labeled by one number, r, with units of length, in standard Schwarzschild coordinates; but with a *different* number, \bar{r}, with units of length, in isotropic coordinates. That obviously has to change the relationship between the number with units of length that labels the 2-sphere, and the physical radius, circumference, and area of that 2-sphere (which do not change when you change its labeling).
This, btw, brings up a possible issue with interpreting the global coordinates (whether we are using Schwarzschild or isotropic) as "what the distant observer infers". Actual physical distances, circumferences, and areas are invariant; they don't change when you change coordinate charts. Say, for example, I pick two 2-spheres, with physical areas A and A + dA, where dA << A. What is the physical radial distance between these two 2-spheres?
First, suppose I'm using the Schwarzschild r coordinate, so I label these two 2-spheres with r = \sqrt{A / 4 \pi} and r + dr. Then, using the Schwarzschild line element, I calculate the physical distance as
ds = \frac{dr}{\sqrt{1 - 2M / r}}
Now suppose I'm using the isotropic \bar{r} coordinate. Then I label the two 2-spheres with \bar{r} = \sqrt{A / 4 \pi} / \left( 1 + M / 2\bar{r} \right)^2 and \bar{r} + d\bar{r}. (Note that, as I've written it, the equation for \bar{r} in terms of A still has to be solved for \bar{r}; but we don't need to do that to write down the expression for ds.) Then, using the isotropic line element, I calculate the physical distance as
ds = d\bar{r} \left( 1 + \frac{M}{2\bar{r}} \right)^2
These two physical distances are the *same*, because they are both expressions for the physical distance between the same two 2-spheres.