# Problem with Schwarzschild metric derivation

1. Sep 2, 2012

### grav-universe

In this Wiki link for the derivation of the Schwarzschild metric, in the section "simplifying the components", g_22 and g_33 are derived. The problem is that upon deriving them, they first set those local measurements of the components for the metric upon a 2_sphere (on the left side) equal to measurements made by a distant observer (on the right side), rather than between two local observers as they claim. Since these components are the tangent distance, this automatically sets tangent distances the same for all static observers, not actually derived as far as I can tell, since it is originally supposed to only be between two local observers. From what I can see, all we can really say about the measurements of two local observers upon the 2-sphere is that since they are angle and rotation independent, is that g_22 and g_33 will both carry the same co-efficient, but we have not determined what that co-effient is yet, so we cannot just automatically assume it to be unity for all static observers, right? What am I missing or is there something missing in the link?

2. Sep 2, 2012

3. Sep 2, 2012

### Staff: Mentor

The Wikipedia page is wrong right at the start, when it describes the assumptions: the fact that Schwarzschild spacetime is static is *not* an assumption, it is part of what is derived from the assumptions that the spacetime is spherically symmetric and vacuum. The latter two assumptions alone are sufficient to derive the Schwarzschild solution, because of Birkhoff's Theorem. I hate to give another Wikipedia page reference after bashing this one, but at least it's a start, and this page states the conclusion of the theorem correctly and gives references:

http://en.wikipedia.org/wiki/Birkhoff's_theorem_(relativity [Broken])

Can you be more specific about which equation you are referring to? As far as I can see, they are picking a particular 2-sphere with radial coordinate $r_0$ and doing everything in terms of that, which means everything in terms of local measurements on that 2-sphere. They only generalize at the very end.

Not quite; one coefficient is $r^2$ and the other is $r^2 sin^2 \theta$. But I think you are referring to possible extra factors in these coefficients; see below.

The Wiki page does appear to leave out a key step; you have to define the scaling of the radial coordinate $r$. The page simply assumes that that scaling is such that the two angular metric coefficients are what I wrote above, which is equivalent to saying that the physical area of a 2-sphere at radial coordinate $r$ is $4 \pi r^2$. But there is nothing requiring that scaling for the radial coordinate; there are other possible scalings. The Wiki page should have shown how, without loss of generality, you can *adopt* that scaling for the radial coordinate; and doing that is the key step that ensures that the angular metric coefficients are what I gave above, i.e., that they take their simplest possible form (no extra factors in either of them, which I think is what you were wondering about). With any other scaling for the radial coordinate, there *will* be extra factors in the angular metric coefficients.

Last edited by a moderator: May 6, 2017
4. Sep 2, 2012

### grav-universe

Thanks for the links, guys.

Right. Those are what a distant observer measures (infers) for the tangent distance, and they are set equal to what the local observer measures for g_22 and g_33, while the text reads like we should be transforming only between two local observers at different angles along the same 2-sphere. Each of those two factors should contain an extra coefficient, some function of r, which would be the same for both factors, representing a transformation between the measurements of static observers at different r. Without it, the equation says that a distant observer measures the same tangent distance as a local observer, but it is not actually derived, only stated that way, and then substituted directly into the metric without any explanation as to why that should be the case to begin with.

Okay, thank you. I figured that might be the case, that it is an arbitrary condition for the coordinatization, that tangent distances are measured the same by all static observers. That is what you are saying, right? I don't see that stated anywhere in the link though. Seems to me that would be an important thing to mention. I think that the distant observer would always measure 4 pi r^2 anyway though (or 2 pi r for the circumference), with only local observers measuring it differently.

If this is an original arbitrary condition set to make the derivation easier, however, then I can see why the singularity could not lie at r = 0 in Schwarzschild, but at some non-zero radius, as it is difficult to imagine that a distant observer measures the tangent length of a local observer's rod at a singularity at r = 0 the same as the tangent length of the distant observer's own rod. So setting such a condition for all r pretty much guarantees that the singularity lies at a non-zero radius for that coordinatization, although it's somewhat surprising it can even still be finite in that case.

Last edited: Sep 2, 2012
5. Sep 2, 2012

### Staff: Mentor

No, they aren't. They give what a *local* observer measures for tangent distance. For simplicity, consider a tangential path at radius $r_0$ where $\theta = \pi / 2$ and the only nonzero coordinate differential is $d \phi$. Then the tangential distance along the path, *as measured by the local observer at that path*, is $r_0 d \phi$ (the square root of the metric line element).

The distant observer can't directly measure tangent distance locally. See below.

You're right, it isn't.

If it weren't Wikipedia, I would agree with you.

Why do you think this? As I said above, the distant observer can't directly measure distances "locally" (meaning at a different radial coordinate than he is at). Also, I think you are thinking of it backwards. It's not a question of figuring out what distances observers measure, when you know the coordinates; it's a question of figuring out how best to *represent* the physical distances observers measure, by *choosing* coordinates.

The standard Schwarzschild coordinates *choose* a scaling for the radial coordinate such that the physically measured area of a 2-sphere at radial coordinate $r$ is $4 \pi r^2$. But a better way to put it would be that the standard Schwarzschild coordinates *define* the radial coordinate $r$ by labeling each 2-sphere with $r = \sqrt{A / 4 \pi}$, where $A$ is the physical area of the 2-sphere.

6. Sep 2, 2012

### grav-universe

I added more to my last post, by the way. You posted quickly. :)

Right, it is only inferred by the metric. That is why we are allowed to use different coordinates systems, since distances are not directly measured, only inferred, unless we use the ruler distance by placing rulers end to end along a path. By "measure" here, I really mean inferred.

Those are the inferred measurements according to a distant observer that says the local observer lies at r_0. The metric is basically just a transformation from the locally measured values to those of the distant observer. For instance, for a photon for simplicity, we can start with

c^2 = v_r'^2 + v_t'^2

where v_r' and v_t' are the locally measured radial and tangent photon speeds, which becomes

c^2 = (dr' / dt')^2 + (dθ' r' / dt')^2

c^2 dt'^2 = dr'^2 + dθ'^2 r'^2

c^2 [f(r) dt^2] - [g(r) dr^2] - [h(r) dθ^2 r^2] = 0

where f(r), g(r), and h(r) are functions of r used for transforming to the coordinates of the distant observer. The distant observer measures dt, dr, and dθ r. The local observer measures dt' = sqrt(f(r)) dt, dr' = sqrt(g(r)) dr, and dθ' r' = sqrt(h(r)) dθ r. In Schwarzschild, we are given h(r) = 1 and then derive f(r) = 1 - 2 m / r and g(r) = 1 / (1 - 2 m / r). In other words, the values used in the metric are according to the coordinate system of a distant observer. 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. They are setting tangent measurements between local and distant observers equal rather than simply changing angles between two local observers on the 2-sphere as claimed. Changing angles between two observers upon the same 2-sphere really only implies that both of the factors for g_22 and g_33 should carry the same co-efficient h(r) for both, otherwise having nothing to do with how they worked it out as far as I can tell.

According to the distant observer's coordinate system, what it infers, if we place local observers all around the 2-sphere at r, it would form a sphere, and it is spherically symmetrical. For any coordinate system we choose, we are just moving those same local observers inward or outward to different radiuses, but they will still form spheres according to the distant observer. For instance, with Eddington's isotropic coordinates, we are only changing radiuses of 2-spheres. The tangent lengths are then different because we are keeping the same locally measured circumference length, so if we decrease the radius, say, while keeping the same local measure around the circumference, then local rods placed tangent to the circumference must now also be smaller in direct proportion to the decrease in radius according to what the distant observer infers.

Last edited: Sep 2, 2012
7. Sep 2, 2012

### Staff: Mentor

20 minutes isn't *that* quick, is it? Let me comment on that addition first:

It's "arbitrary" only to the extent that it adopts a particular scaling for the radial coordinate, out of many possible scalings. It doesn't affect the fact that the radial coordinate is monotonically increasing as the areas of the 2-spheres increase. Which means that this...

...is incorrect, because the physical area of the "2-sphere" at the singularity is zero--which is to say, the singularity is at the geometric center of the family of nested 2-spheres, and that center itself is a point. One could cook up a radial coordinate that assigned a non-zero r to that point, I suppose, but that would be somewhat perverse. In any case, the standard radial coordinate, which assigns r based on the physical area of the 2-sphere, obviously assigns r = 0 to the singularity.

As above, the "2-sphere" at r = 0 is actually a point, so there is no "tangential length" there.

Ah, ok, that clarifies things. I don't have time to comment further now, but in any case I need to digest the rest of your latest post a little more.

8. Sep 2, 2012

### grav-universe

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.

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.

Last edited: Sep 3, 2012
9. Sep 3, 2012

### Staff: Mentor

Ah, ok. Sorry for the mixup on my part.

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:

Ok, this clears up how you are interpreting the metric (which is a perfectly good way to interpret it).

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.

Ok, except that you put a prime on dθ'. Strictly speaking, that's not correct; see below.

Because we adopt the standard Schwarzschild r coordinate, yes.

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.)

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.

10. Sep 3, 2012

### grav-universe

Thanks PeterDonis. I agree with everything you stated, except for one minor detail having to do with dθ and dθ'. In a recent thread where we worked through the relationship between the Schwarzschild metric and the equivalence principle, it was found that in order for the equivalence principle to hold, if a local observer holds a tangent rod that is straight according to that observer, the distant observer must infer it to be slightly bent downward on each end. The resulting relationship found from that between the local and distant angles inferred between the ends of the rod was dθ' = dθ sqrt(1 - 2 m / r) for Schwarzschild, which leaves r' = r / sqrt(1 - 2 m / r) since dθ' r' = dθ r, so basically just taking the inferred radial distance by extending the local ruler in that direction. The local observer, then, doesn't measure a sphere using that inferred distance, nor a 2 pi r' circumference, but only works toward it with greater radius. I even worked out a proof for the angles from the relationship found, although perhaps not rigorous enough since I only used first order approximations, although I may try to work through it again more precisely, but in the meantime, I generally just try to use dθ r and dθ' r' wherever possible, rather than separate them.

11. Sep 3, 2012

### Staff: Mentor

Can you give a link to the thread? I don't think I participated in it. I'd have to read it before commenting on what you say here; at a glance it doesn't look right, but I may be missing relevant context from the thread.

12. Sep 3, 2012

### grav-universe

Sure. It is the thread gravitational lensing using equivalence principle. I didn't include a "proof" for that in the thread, though, but it's the only way the mathematics works out right. I'll go ahead and post that here, though, so you can look it over and see what you think. As I said, I only used first order approximations, but I think that might actually work out in favor of the proof in a way because I didn't use any factors of sqrt(1 - 2 m / r) for the first order approximations in the proof, yet that is what still falls out in the end.

In post #6 of the thread, I used the exact solution to the metric that BillK gave to find the coordinate radial acceleration of a photon at y = r that a distant observer would measure when the photon initially travels tangently, which is 3 times the coordinate acceleration of a rod freefalling from rest at y = r to first order. According to the distant observer, the rod falls from rest with a radial acceleration of G M / r^2 to first order while the photon has a radial acceleration of 3 G M / r^2 to first order. If the photon initially travels along the rod at the moment it begins freefalling, then since the rod freefalls inertially, the equivalence principle says the photon will continue to travel along the rod as it falls (with its length along x) in order for both to coincide with the end of the rod. But since the radial acceleration of the photon is 3 times greater, that means that according to the distant observer, the end of the rod must be bent downward from y = r a distance of

dy = (a_photon - a_rod) dt^2 /2

with the time along the length of the rod being dt = dx / c to first order, so

dy = (3 G M / r^2 - G M / r^2) (dx^2 / c^2) / 2

dy = dx^2 (G M / (r^2 c^2))

dy = dx^2 m / r^2

Now let's transform that to what the local observer at y = r measures, at the place where the rod and photon start. The local observer just measures a straight rod with no bend. The angle between the line from the starting point on the rod to the center of the gravitating body and the line from the center of the body to the end of the rod according to the distant observer is

sin dθ = dθ = dx / sqrt((r - dy)^2 + dx^2)

The angle according to the local observer is

sin dθ' = dθ' = dx' / sqrt(r'^2 + dx'^2)

Since the Schwarzschild metric keeps the relation dθ' r' = dθ r in the transform, we can re-arrange to get

dθ' / dθ = a and r' / r = 1/a, where a is some so far unknown value that cancels out in the metric. So from what we found for the angles, now we have

dθ' / dθ = a = [dx' / sqrt(r'^2 + dx'^2)] / [dx / sqrt((r - dy)^2 + dx^2)]

where dx' = dx for the tangent distance, and r' = r / a, so

a^2 = ((r - dy)^2 + dx^2) / ((r / a)^2 + dx^2)

r^2 + a^2 dx^2 = r^2 - 2 r dy + dy^2 + dx^2

a^2 = (- 2 r dy + dy^2 + dx^2) / dx^2

and dy = dx^2 m / r^2, so

a^2 = (- 2 r (dx^2 m / r^2) + (dx^2 m / r^2)^2 + dx^2) / dx^2

a^2 = (- 2 m / r + dx^2 m^2 / r^4 + 1)

and dropping the infinitesimal, we have

a^2 = (1 - 2m/r)

a = sqrt(1 - 2m/r)

dθ' = dθ a = dθ sqrt(1 - 2m/r)

r' = r / a = r / sqrt(1 - 2m/r)

Last edited: Sep 3, 2012
13. Sep 3, 2012

### grav-universe

Actually, it looks like I did perform a sort of "reverse" proof in post #27.

14. Sep 3, 2012

### Staff: Mentor

grav-universe, there's a lot to digest in your latest posts here and in the other thread, so I've only had time to skim. But the following items in your latest post jumped out at me on a quick skim:

I see how you're deriving this, based on the distant observer seeing the rod as "bent"; the distant observer, to make a right triangle with the radial line from the center of the body and the tangential line with length dx, has to use a slightly shorter radial line.

However, this part is *not* correct. The correct transform is dθ' = dθ (and $d \phi ' = d \phi$ as well). Only with that transform will the physical area of the 2-sphere cover 4 pi units of solid angle for both the local and the distant observer's coordinates; and it's the area in units of solid angle covered by the 2-sphere that defines the angular coordinates, *not* the area in length units squared.

The definition of the r coordinate then matches up the labeling of a given 2-sphere with the metric coefficients, as a function of that r coordinates in such a way as to convert 4 pi units of solid angle into the physical area of the right 2-sphere (the one labeled with the same r coordinate as we plugged into the metric coefficient). But that's *all* the r coordinate does. I walked through how this works for standard Schwarzschild coordinates vs. isotropic coordinates in post #9; note that in both cases there, the 2-sphere is covered by 4 pi units of solid angle.

15. Sep 5, 2012

### grav-universe

Sorry I haven't posted. I'm trying to work through it more definitively, but it's taking some time.