# Schwarzschild metric and spherical symmetry

1. Aug 12, 2012

### grav-universe

In deriving the Schwarzschild metric, the first assumption is that the transformation of r^2 (dθ^2 + sin^2 θ dψ) remains unchanged due to the spherical symmetry. What does that mean exactly? What is the logic behind it? Please apply any math involved in algebraic form. Thanks.

2. Aug 12, 2012

### bcrowell

Staff Emeritus
http://www.lightandmatter.com/html_books/genrel/ch07/ch07.html#Section7.2 [Broken]

Last edited by a moderator: May 6, 2017
3. Aug 12, 2012

### grav-universe

Sorry, I still don't understand.

Last edited by a moderator: May 6, 2017
4. Aug 12, 2012

### grav-universe

Let's say that according to a distant observer, there is an observer on a planet. Now I can see that since the distant observer is in nearly flat space, the planet will be spherical since it has spherical symmetry, the gravity acts symmetrically in all directions and so forth and shapes the planet that way, say. The distant observer measures the radius of the planet to be r.

Okay, so let's say that the surface observer measures the radius to be r' = r / a, inferred by a ruler in the radial direction that is contracted by a factor of 'a' according to the distant observer. Let's also say that the surface observer measures the planet to be spherical. That would require that measurements along the surface are also 1 / a greater than the distant observer measures. dθ r is just the measurement along the circumference, so that would be factored by 1 / a in the metric also. Basically then, according to the distant observer, the surface observer's ruler is contracted by a factor of 'a' in any direction. I don't see anything about spherical symmetry according to the distant observer that would rule this out. So by what logic isn't it permitted, requiring that the distant observer and surface observer measure the same distances along the surface?

5. Aug 12, 2012

### Muphrid

Under what transformation are the angular parts of the line element supposed to be invariant? I don't understand the "transformation" that you refer to in the first post.

6. Aug 12, 2012

### Mentz114

Spherical symmetry means that the field can only depend only on r and no other spatial coordinates.

7. Aug 12, 2012

### grav-universe

Well, the local surface observer measures

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

where v_r' is the locally measured radial speed of a photon and v_t' is the locally measured tangent speed of a photon, whereby

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

dr'^2 + dθ'^2 r'^2 = c^2 dt'^2

We assume that due to spherical symmetry, the quantity dθ' r' remains unchanged, so dθ' r' = dθ r. That is the distance measured in the tangent direction, which this transformation says will be measured the same by both the local observer and the distant observer. This is the part I'm asking about. From there, dr / dr' and dt' / dt are each derived to be sqrt(1 - 2m/r), giving

[dr / sqrt(1 - 2m/r)]^2 + dθ^2 r^2 = c^2 [dt sqrt(1 - 2m/r)]^2

dr^2 / (1 - 2m/r) + dθ^2 r^2 = c^2 dt^2 (1 - 2m/r)

which is the metric.

8. Aug 12, 2012

### grav-universe

Right, that I see easily see. Due to the spherical symmetry, all measurements can only vary by some function of r. But how does that translate to dθ' r' = dθ r?

9. Aug 12, 2012

### bcrowell

Staff Emeritus
That doesn't work, because the definition needs to be coordinate-independent. See the link in #2 for a coordinate-independent definition.

10. Aug 12, 2012

### pervect

Staff Emeritus
Very loosely speaking, spherical symmetry means that the metric isn't changed by rotations. If you'd prefer a formal definition, try Wald, who says

Note that the only assumption being made here is the existence of the spherical symmetry. It's not an "assumption" to say that the line element can be put into the form you quote (which is the same line element from flat space-time), it's a consequence of the initial assumption of spherical symmetry.

The basic idea in very loose and informal terms is that if you take the set of points of constant distance from the black hole they must form a set which is isomorphic to a sphere. Thus we can use the same line element that we use for a sphere in flat space-time for the subset of the manifold of "points of constant r" without loss of generality.

Note that we are attempting to do our best to take advantage of the underlying symmetry to find the simplest possible solution. So we are deliberately restricting our coordinate choices to correspond to the underlying symmetry in an effort to find the set of equations that is easiest to solve. If you want to make a different coordinate choice after you've solved the equations, all you need to do is to provide a transformation of coordinates - for example, consider isotropic coordinates.

11. Aug 12, 2012

### grav-universe

I still don't see how it would be a consequence of the spherical symmetry, but I think I see something else in your reply. Are you really saying that it is not an assumption at all, but a given? That the rest of the metric is derived according to what would be required to keep the transformation of dθ r constant, in order to keep the derivation as simple as possible? And then, after deriving it this way, we can then find for other coordinate systems in which dθ r will vary?

12. Aug 13, 2012

### pervect

Staff Emeritus
Metrics exist in which the coefficient of d$\theta$ is different, for instance the isotropic solutions.

$$ds^2 = -\left(\frac{1-M/2R}{1+M/2R}\right)^2\,dt^2 + \left(1+M/2R\right)^4 \, \left[dR^2 + R^2 \,d\theta^2 + R^2 sin^2 \theta d \phi^2\right]$$

The point is that we know the solution must have spherically symmetric shells. Furthermore, once we have a metric, the shells must have an area. It's a coordinate choice to say that we can define a radial coordinate such that area = 4 pi r^2.

It's not the only coordinate choice possible, but it's a convenient one for getting a solution.

13. Aug 13, 2012

### DrGreg

I think maybe the question you are asking is this. Given a static spherically symmetric spacetime, why do we assume the metric is of the form$$ds^2 = A(r)^2 \, dt^2 - B(r)^2 \, dr^2 - r^2 \, d\Omega^2$$where$$d\Omega^2 = d\theta^2 + \sin^2 \theta \, d\phi^2$$rather than the more general form$$ds^2 = C(R)^2 \, dt^2 - D(R)^2 \, dR^2 - E(R)^2 \, d\Omega^2 \mbox{ ?}$$The answer is that if we have a metric of the second form, we can perform a change of variables to get the first form, viz.\begin{align} r &= E(R) \\ R &= E^{-1}(r) \\ A(r) &= C(R) \\ B(r) &= \frac{D}{\left(\frac{dE}{dR}\right)}(R) \end{align}Also, as pervect indicates, its also possible to perform a different change of variables to get isotropic coordinates of the form$$ds^2 = F(\rho)^2 \, dt^2 - G(\rho)^2 \left( d\rho^2 + \rho^2 \, d\Omega^2 \right)$$

14. Aug 13, 2012

### grav-universe

Looks interesting. Could you explain the mathematical manipulations involved with what you posted here in more detail please?

15. Aug 13, 2012

### grav-universe

I had always thought of the Schwarzschild metric as the "true" solution. We could switch to other coordinate systems to make things easier in some way to find an answer to a problem, a "psuedo" coordinate system, but then we must switch that answer back to Schwarzschild to get the actual result. In SR, we can switch between coordinate systems because synchronization is not absolute, but we do not generally change ruler lengths and tick rates of clocks between two frames because it would not be "natural", not representing actual length contraction and time dilation, even though with different synchronizations, different length contractions and time dilations would be measured, but we would only transport identical clocks and rulers from the first frame to the second and then leave them "as is" and measure accordingly after the clocks have been synchronized some particular way while leaving their tick rates alone.

In GR, all hovering observers are in static space, and while there is time dilation between radiuses, all hovering observers agree that all clocks at a particular radius are synchronized to read the same, so there are no synchronization issues between radiuses. Well, if all local clocks are synchronized using the Einstein synchronization method, granted, but all GR coordinate systems apply that I think. So what we are really changing is distance measurements as far as I can tell. But as far as tangent distances being the same in Schwarzschild as a coordinate choice, if local observers at r were to measure the circumference around r by physically laying very small rulers end to end around the circumference, then transport the very same set of rulers to the radius s, the local observers there would physically measure the same distance 2 pi in relation to s around the circumference, so having the same ruler length in the tangent direction, or they will not. By comparing the local measurements at all radiuses this way, there would be only one function that relates them, and that would give us the length contraction of a ruler in the tangent direction at a particular radius. So we could not set that value to any coordinate choice we want in order to derive the metric if we want to find the "natural" length contraction of rulers in the tangent direction, could we?

16. Aug 13, 2012

### PAllen

This is the opposite of what GR is about. There is no preferred coordinate system, and all physical observables are coordinate independent. The modern preference is coordinate independent operations whenever feasible.

17. Aug 13, 2012

### grav-universe

Oh wait, here's a surprisingly simple mathematical manipulation that would easily get rid of the function tied to the angle.

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

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

dr'^2 + dθ'^2 r'^2 = c^2 dt'^2

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

where all three functions are time independent and angle independent, leaving them only as functions of r. From there we can gain

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

A(r) = f(r) / h(r) and B(r) = g(r) / h(r)

A(r) c^2 dt^2 - B(r) dr^2 - dθ^2 r^2 = 0

I don't know how I didn't see that before. But then, even after solving for A(r) and B(r) and gaining the actual metric by plugging in those values, we still have the problem of not knowing what h(r) really is to begin with. The overall metric must still work out using A(r) and B(r), and if we leave the metric as is, we have Schwarzschild, but how do we know if the tangent length contraction is really unity or something else?

18. Aug 13, 2012

### grav-universe

Okay, well, this is what I want to know. Since the distant observer is in very nearly flat space-time, I want to apply his coordinate system the same way as we would in Euclidean space. According to his Euclidean coordinate system, there is a gravitating body at a point in space and hovering observers at radiuses r and s from that point. Observers at r physically lay rulers around the circumference at r and measure C_r. The same set of rulers are then transported to s and observers there do the same thing, measuring C_s. What I want to know is what will the ratio (C_r / C_s) (s / r) be? And how do you determine that?

19. Aug 13, 2012

### PAllen

There is no global inertial coordinates in GR. Depending on which feature of an inertial coordinates system you want to try to extend globally, you will get different global coordinate systems. How do you pick which to prefer? In SR, this problem does not exist - all 'reasonable' ways of building a global frame produce the same result. In GR, each 'feature' you want to emphasize produces different global coordinates.

For example, why favor hovering observers? They have proper acceleration, which is not at all like a global inertial frame. As for rulers, if one end is under different proper acceleration than the other, no material consistent with relativity will be unafffected, so the result depends on the material.

IMO, SC coordinates are one of less informative - they have a coordinate singularity that has no physical meaning; they are anisotropic, while no observer actually sees local anisotropy.

20. Aug 13, 2012

### PAllen

As for your question, in SC coordinates, r and s are defined by the result of such a measurement. You cannot measure to the center, as that is a singularity. So r and s are defined as C_r/2π and C_s/2π. If you then want to measure the distance between r and s, it depends on who measures it and how. Some ways of measuring it will produce the proper distance integrated using SC coordinates along r for dt=0. For any given way of measuring distance between shells, any coordinates will produce the same answer for the distance - because this is a function of the measurement method, not the coordinates used.

21. Aug 13, 2012

### DrGreg

It all follows almost directly from equating the first two equations for ds2 in post #13, term by term.

$r \, d\Omega = E(R) \, d\Omega$ gives you $r = E(R)$.

$R = E^{-1}(r)$ is the inverse (assuming, of course, that E is invertible).

$A(r) \, dt = C(R) \, dt$ gives you $A(r) = C(R) = C(E^{-1}(r))$.

$B(r) \, dr = D(R) \, dR$ gives you
$$B(r) = D(R) \frac{dR}{dr} = \frac{D(R)}{\left( \frac{dr(R)}{dR} \right)} = \left. \frac{D}{\left(\frac{dE}{dR}\right)}\right|_{R = E^{-1}(r)}$$

22. Aug 13, 2012

### pervect

Staff Emeritus
As others have remarked, you can't do this literally. And it's not really clear what features of the coordinate system you want to preserve, i.e. if we try to come "as close as we can", it's not clear what features you consider important, and what features you don't.

None of what you ask for - measuring distances - requires you to use a particular coordinate system. But, there is one thing you do have to specify.

Do you recall from special relativity that space and time intermix for different observers? I'm afraid I don't know your background.

In order to split space-time into space + time, you need to specify some notion of simultaneity, because of the fact that, speakig loosely, distances depend on one's state of motion in relativity - i.e. length contraction.

The most common notion is the shared notion of simultaneity of static observers, one who hover at constant distances and are all stationary relative to one another.

With this much set, distances are defined (loosely speaking again) as the shortest distance between two points in a surface of constant time, said surface being defined by the agreed-upon notion of simultaneity of the hoovering observers, for example.

In general you have to solve the geodesic equation, but for the simple examples the path you need to integrate over is obvious.

All you need to do then is to construct the proper curve, and measure its length using the metric.

I'll go into the math more later on, at the moment I don't have the time.

Meanwhile, as far as coordinate systems go, consider the equivalent problem on the Earth. You can't define a coordinate system or make a map of the Earth on a flat sheet of paper that preserves all distances. Distances on the surface of the Earth are defined, and you can make a flat map match up with any particular local neighborhood you want it to - but you can't cover the whole globe with a flat map that preserves distances.

23. Aug 13, 2012

### pervect

Staff Emeritus
OK - the way you integrate the length of a curve in general coordinates isn't much different from the way you'd do it, in for example, polar or spherical coordinates.

[1] It's easiest if you parametirize the curve, say for example r(s), theta(s), phi(s). THe derivatives of the curve will be the tangents to it, i.e. dr/ds, dthet/ds, dphi/ds.

You also have some line element, let say it's A(r)dr^2 + B(r)dtheta^2 + C(ri)D(theta)*dphi^2

For spherical coordinates A(r) would be 1, B(r)=r^2, C(r)=r^2, D(theta)=sin^2(theta) for example

Then you can find the length of the curve segment ds by taking

length^2 = A(r)*(dr/ds)^2 ds^2 + B(r)*(dtheta/ds)^2*ds^2 + C(r)D(theta) (dphi/ds)^2 ds^2

where dr/ds, dtheta/ds and dphi/ds are the tangents to the parameterized curve in [1].

so you just integrate sqrt(length^2) over ds.

And that's all there is to it.

Making sure that the curve is the shortest one mathematically is a bit trickier - you have to know about the geodesic equation. But you can often guess the geodesic correctly by inspection, for instance, you know to measure the circumference around great circles (r=phi=constant), or (theta=0,r=constant), and you know to measure radial distances radially (theta=phi=constant).

24. Aug 14, 2012

### jarod765

Another way to think about it is the space-time can be foliated by submanifolds with a metric equivalent to a 2-sphere metric.

25. Aug 14, 2012

### grav-universe

Okay, let me ask a different question then. I understand that distances can't actually be physically and directly measured in curved space using only the distant observer's own ruler, but only inferred. So I suppose one could say that the inferred distance depends upon the coordinate system and how distances relate according to the metric associated with that coordinate system. I am looking for a particular coordinate system in that case.

Let's say that in flat space-time, there are three distant stationary observers that are equidistant from each other. They can triangulate to find the location of a point at the center of their positions, the center of the triangle. They also locate that position relative to the fixed stars. They chart their own positions in relation to each other and the fixed stars by drawing a two dimensional map with their triangle lying upon the plane of the map.

Okay, so one of the distant observers, call her observer A, sends a probe to a distance r from that point at the center of the triangle, directly along the radial line between the point and observer A. In flat space-time, the distance r can be directly measured using observer A's own ruler. The position of the probe is also charted on the map.

But now, at some point and in some manner, a small spherical planet is moved so that its center lies directly upon the point at the center of the triangle. The distant observers can still triangulate the position of the point and the center of the planet and chart this position on another map, along with their own positions in relation to each other and the fixed stars. They lay this map on top of the first map and all of the positions almost perfectly coincide. The distant observers are not infinitely remote but at such a great distance that the difference between the maps is infinitesimal, so insignificant.

Okay, so now, the distant observers will employ the Schwarzschild coordinate system in relation to the gravity of the planet. Observer A sends a probe as she did before to a distance r from the center of the planet, the distance r that will correspond to her current coordinate system. She infers what that distance r will be in relation to herself and the other two distant observers, and to the fixed stars, and charts that position upon the second map as well, according to her current coordinate system, where her coordinate system says r will lie. She then places the second map over the first map again and compares positions. Will the charted position of the first probe at r in the flat space-time of the first map match the charted position of the second probe at r in the curved space-time of the second map? If not, what coordinate system would be required so that the positions correspond?