# Spatial curvature around a spherically symmetric mass

• B
• J O Linton
In summary, the circumference of a circular orbit round a massive object will be greater than expected if space is flat, since the proper distance traveled to increase the area of the sphere is greater than the distance traveled in flat space. If space around a spherically symmetric mass is curved, the circumference of the orbit will be greater than expected even if the radial coordinate is measured locally.
J O Linton
TL;DR Summary
Spatial curvature around a spherically symmetric mass
Suppose I measure the circumference of a circular orbit round a massive object and find it to be c. Suppose I then move to a slightly higher orbit an extra radial distance δr as measured locally. If space was flat I would expect the new circumference to be c + 2πδr. Will the actual measurement (taking the spatial curvature into account) be greater or less than this? and what would this measurement tell me about the local curvature of space in the region? Is there a simple formula relating the change in c with change in r and how might this formula be related to the Schwarzschild metric?

J O Linton said:
Summary:: Spatial curvature around a spherically symmetric mass

Suppose I measure the circumference of a circular orbit round a massive object and find it to be c. Suppose I then move to a slightly higher orbit an extra radial distance δr as measured locally. If space was flat I would expect the new circumference to be c + 2πδr. Will the actual measurement (taking the spatial curvature into account) be greater or less than this? and what would this measurement tell me about the local curvature of space in the region? Is there a simple formula relating the change in c with change in r and how might this formula be related to the Schwarzschild metric?
The Schwarzschild coordinate ##r## is defined as the radial coordinate at which a spherical surface has area ##\pi r^2##. In particular, it is not a radial distance. This is for ##r## outside the event horizon. If we want to go from coordinate ##r_0## to coordinate ##r_1##, then the proper distance traveled is greater than ##r_1 - r_0## (see below). This means that we have to travel further in order to increase the area of the sphere than we would in flat spacetime.

To see this, we look at the spatial part of the Schwartzschild metric: $$\big ( 1 - \frac{2M}{r} \big)^{-1} dr^2 + r^2(d\theta^2 + \sin^2 \theta d \phi^2)$$
If we fix ##r = r_0##, say, then we see that the area for fixed ##r_0## is simply ##4\pi r_0^2##.

And, if we fix ##\theta## and ##\phi## and take the line integral from ##r_0## to ##r_1##, we see that the integrand is always ##> 1##, so the measured distance will be ##> r_1 - r_0##. For a small change, where ##r_1 = r_0 + \delta r##, we would measure the radial distance as approximately ##(\frac{r_0}{r_0-2M})\delta r##, which we see is greater than ##\delta r##.

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vanhees71, cianfa72 and Dale
J O Linton said:
Suppose I measure the circumference of a circular orbit round a massive object and find it to be c. Suppose I then move to a slightly higher orbit an extra radial distance δr as measured locally. If space was flat I would expect the new circumference to be c + 2πδr. Will the actual measurement (taking the spatial curvature into account) be greater or less than this?
By ##r##, do you mean the radial coordinate, or the actual radial distance? The answer @PeroK gave you assumes that ##r## is the radial coordinate and ##\delta r## is the change in the radial coordinate, which means that the new circumference is exactly ##c + 2 \pi \delta r##, since the radial coordinate is defined so the circumference of a circle with radial coordinate ##r## is ##2 \pi r##. @PeroK then shows, using the Schwarzschild metric, that the actual radial distance is larger than would be expected from Euclidean geometry, i.e., it is larger than ##\delta r##.

However, the way your question is phrased makes it seem to me that you are using ##r## to mean the actual radial distance, in which case it would probably be better to pick a different symbol for it to avoid confusion, since ##r## is standardly used for the radial coordinate. If we use ##d## instead for the radial distance, then for two circles separated by a physical distance ##\delta d## radially, if the circumference of the inner one is ##c##, the circumference of the outer one will be less than ##c + 2 \pi \delta d##.

J O Linton said:
what would this measurement tell me about the local curvature of space in the region?
It tells you that the "space" defined by observers who are at rest with respect to the central body is not Euclidean; in fact it has the shape of the Flamm paraboloid, which is curved.

Note, however, that I carefully said "space" as defined by a particular family of observers (more technically, it is the "space" defined by spacelike hypersurfaces that are orthogonal to the worldlines of those observers). A different family of observers will define a different "space"; for example, the family of observers freely falling radially inward from rest at infinity will define a "space" that is flat, i.e., Euclidean.

vanhees71 and Orodruin
For visual reference, this is the 2-dimensional Flamm paraboloid:

Many thanks for these replies. Two further questions.
The curvature of a 2D surface is completely specified by a sungle parameter which can be determined locally by measuring the circumference of a small circle of known radius. How many parameters are required to completely specifiy the local curvature of 3-space at a point? and what local measurements would be required to determine them?
It occurs to me that the above questions are entirely mathematical and are independent of any particular physical situation so perhaps I should also ask - how many parameters are required to specify the curvature of 3-space at a point near a massive object from the point of view of an observer at rest with respect to that body, and what measurements should he make to determine those parameters?

The curvature of a manifold is specified by the rank-4 curvature tensor. Its (anti-)symmetries and the Bianchi identity results in it not having the full ##n^4## independent parameters. Instead, it has ##n^2(n^2-1)/12## independent parameters. For ##n=3## this means ##9(9-1)/12 = 6##. You can measure individual components of the curvature tensor by parallel transport of vectors around different infinitesimal loops.

PeroK
Interesting. But I am surprised that the answer is 6 not 3. I can imagine carrying out parallel transport round loops whose normals are oriented along the three different axes but where are the other three loops?

J O Linton said:
Interesting. But I am surprised that the answer is 6 not 3. I can imagine carrying out parallel transport round loops whose normals are oriented along the three different axes but where are the other three loops?
This is not sufficient. Vectors have three components and you need to parallel transport each around each loop.

I see. In the special case of a spherically symmetric 3-space around a massive object, I imagine that some of the 6 parameters are zero. How many are non-zero and which loops would be required to determine them?

I am sorry to see this thread die because I am trying very hard to understand the spatial curvature around a gravitating mass. I strongly suspect that the answer to my query in #9 is that the spatial curvature is completely described by a nsingle parameter and that the curvature is analagous to the curvature of a sphere (albeit in 3 dimensions rather than two).
When viewed from a distant point above the north pole in polar coordinates, the arc length ##\delta s## of a sphere of radius ##R## is related to the increment in radial coordinate ##\delta r## by the expression $$\delta s^2 = (1 - r^2/R^2)^{-1} \delta r^2$$ This is very similar to the spatial component of the Schwartzchild metric which I believe can be expressed as $$\delta s^2 = (1 - r_s/r)^{-1} \delta r^2$$ If we were to view an ant crawling from the north pole of the sphere towards the equator we would see it foreshortened, becoming a single point when it reaches the equator and then disappearing from sight. In the same way, if we were to view a spaceship descending towards the event horizon of a black hole we would see it foreshorten like the ant, becoming infinitely compressed at the event horizon before disappearing from view. Of coures, neither the ant, nor the occupants of the spaceship would see anything strange as they passed this point. Both the equator of the sphere and the event horizon are what is known as coordinate singularities. Things only blow up to infinity from the stationary observer's point of view.
What I would dearly like to know is - are there simple formulae for a) the proper distance between a radial coordinate ##r_0## and the Schwartzchild radius ##r_s##; b) the proper time for a particle in free fall from ##r_0## to ## r_s## and c) the time for this to happen from the point of view of the external stationary observer?

In addition to Bianchi and Schwartzchild, let's not forget Webster: "Farther" (distance); "Further" (additional).

J O Linton said:
I am sorry to see this thread die because I am trying very hard to understand the spatial curvature around a gravitating mass. I strongly suspect that the answer to my query in #9 is that the spatial curvature is completely described by a nsingle parameter and that the curvature is analagous to the curvature of a sphere (albeit in 3 dimensions rather than two).
The non-zero components of the Riemann curvature tensor are given here:

https://en.wikipedia.org/wiki/Schwarzschild_metric#Curvatures

J O Linton said:
In the same way, if we were to view a spaceship descending towards the event horizon of a black hole we would see it foreshorten like the ant, becoming infinitely compressed at the event horizon before disappearing from view.
Physics does not need to be decribed or understood by what one "observer" "sees". We study any physics using a mathematical model - and, among other things, the model predicts what local measurements are obtained by observers.

How, for example, do you encapsulate Newton's theory of gravity by what one or other observers see?

The way SR is sometimes taught with an extreme over-reliance on observers and light signals is not a useful preparation for GR. You are one of several posters on here who tackle GR from this "what does an observer see" approach. It's preventing you from understanding the nature of curved spacetime, which more than anything requires an overall geometric model.

J O Linton said:
What I would dearly like to know is - are there simple formulae for a) the proper distance between a radial coordinate ##r_0## and the Schwartzchild radius ##r_s##; b) the proper time for a particle in free fall from ##r_0## to ## r_s## and c) the time for this to happen from the point of view of the external stationary observer?
There is no definitive, absolute radial coordinate with which to compare the Schwarzschild radial coordinate. Calculations for free fall proper times for infalling objects are found in most texts on GR. Again, what an external stationary observer sees may be important in one sense, but it adds little understanding to the concept of a curved spacetime manifold.

vanhees71
J O Linton said:
This is very similar to the spatial component of the Schwartzchild metric
No, it isn't. The first has (fixing your expressions so the ##-1## exponent doesn't appear, which makes their actual behavior more apparent):

$$\frac{R^2}{R^2 - r^2}$$

The second has:

$$\frac{r}{r - r_s}$$

These are not at all "similar" functions of ##r##.

I strongly suggest not trying to base your understanding on this kind of thinking.

vanhees71
J O Linton said:
What I would dearly like to know is - are there simple formulae for a) the proper distance between a radial coordinate ##r_0## and the Schwartzchild radius ##r_s##;
This depends on your choice of coordinates, or more precisely your choice of how to foliate the region outside the horizon with spacelike hypersurfaces. If you are using Schwarzschild coordinates, i.e., you have picked the set of spacelike hypersurfaces of constant Schwarzschild coordinate time, just integrate your formula for ##ds## in terms of ##dr## from ##r_0## to ##r_s##.

Note, however, that in different coordinate charts you can get different answers: for example, in Painleve coordinates the spacelike hypersurfaces of constant coordinate time are flat Euclidean 3-spaces, so the distance you want is just ##r_0 - r_s##.

J O Linton said:
b) the proper time for a particle in free fall from ##r_0## to ## r_s##
Yes, although the answer will depend on what velocity the particle has at ##r_0##. The simplest general method of solution is to use the effective potential, which can be found by using constants of motion to obtain an equation for ##\left( dr / d\tau \right)^2## (Wikipedia [1] gives a decent discussion of this, and it can also be found in most GR textbooks and, IIRC, in Carroll's online lecture notes):

$$\left( \frac{dr}{d\tau} \right)^2 = \frac{E^2}{m^2} - 1 + \frac{r_s}{r}$$

where I have used units in which ##c = 1## and I have set the angular momentum ##L## to zero since we are talking about radial free fall. ##E## is the energy at infinity, which is a constant of the motion, and ##m## is the particle's rest mass; we can find the ratio ##E^2 / m^2## by using our initial condition, which, if we assume the particle is dropped from rest at ##r_0## will be ##dr / d\tau = 0## at ##r = r_0##. This gives

$$\frac{E^2}{m^2} = 1 - \frac{r_s}{r_0}$$

Plugging that back into the above equation gives:

$$\frac{dr}{d\tau} = \sqrt{ r_s \left( \frac{1}{r} - \frac{1}{r_0} \right) }$$

This can be converted into an integral for ##\tau##, which, after some algebra, becomes:

$$\tau = \sqrt{ \frac{r_0}{r_s} } \int_{r_0}^{r_s} \sqrt{ \frac{r}{r_0 - r} } dr$$

This looks messy, but it can be solved with the aid of a table of integrals.

[1] https://en.wikipedia.org/wiki/Schwarzschild_geodesics

J O Linton said:
and c) the time for this to happen from the point of view of the external stationary observer?
This depends on your choice of coordinates; there is no invariant answer. You can make the answer "infinity" by choosing Schwarzschild coordinates, or you can make it various finite numbers by choosing other coordinates such as Painleve or Eddington-Finkelstein.

Or you can look for some invariant criterion, such as when the external observer's worldline leaves the past light cone of the event where the infalling particle crosses the horizon (which will give a finite answer).

PAllen, vanhees71 and PeroK
Many thanks for your replies. Even if you haven't quite satisfied all my queries, you have pointed me in the right direction.

PeterDonis said:
This depends on your choice of coordinates, or more precisely your choice of how to foliate the region outside the horizon with spacelike hypersurfaces. If you are using Schwarzschild coordinates, i.e., you have picked the set of spacelike hypersurfaces of constant Schwarzschild coordinate time, just integrate your formula for ##ds## in terms of ##dr## from ##r_0## to ##r_s##.

Note, however, that in different coordinate charts you can get different answers: for example, in Painleve coordinates the spacelike hypersurfaces of constant coordinate time are flat Euclidean 3-spaces, so the distance you want is just ##r_0 - r_s##.
Note that a key distinction between the Schwarzschild foliation versus those that are regular at the horizon (e.g. Panlieve or Kruskal) is that the former slices all intersect the horizon at the same event. In the case of an eternal black hole, this event is the boundary between the white hole and black hole horizons. It is physically peculiar, to say the least, to measure distance from an external body with continued existence, to a BH horizon with continued existence and a causal history on the horizon, by always measuring to the beginning of the BH horizon. Thus, I would claim any of the other foliations, without this anomaly, are far more physically meaningful.

vanhees71 and PeterDonis
PAllen said:
I would claim any of the other foliations, without this anomaly, are far more physically meaningful.
As a counterpoint to this, though, the Schwarzschild foliation has the nice property that the radial spacelike curves along which the distance to the horizon is being measured are geodesics of the spacetime, whereas in the others the radial spacelike curves are not. So there is not a single foliation that has all of the nice properties we would like. This kind of thing often happens in curved spacetime.

vanhees71 and PAllen
PeterDonis said:
As a counterpoint to this, though, the Schwarzschild foliation has the nice property that the radial spacelike curves along which the distance to the horizon is being measured are geodesics of the spacetime, whereas in the others the radial spacelike curves are not. So there is not a single foliation that has all of the nice properties we would like. This kind of thing often happens in curved spacetime.
Just a thought on this - I might calculate this later, but from any exterior event, there are infinite radial spacelike geodesics, the Schwarzschild one being just one. It is unique only by being orthogonal to a stationary world line. I suspect that geodesics orthogonal to the radial infaller from infinity passing through some event have completely different properties.

vanhees71
PAllen said:
from any exterior event, there are infinite radial spacelike geodesics, the Schwarzschild one being just one. It is unique only by being orthogonal to a stationary world line.
Yes.

PAllen said:
I suspect that geodesics orthogonal to the radial infaller from infinity passing through some event have completely different properties.
Yes, they would. Unfortunately, though, such geodesics do not lie within a spacelike hypersurface of constant coordinate time in any of the known charts (Painleve would be the most obvious one to consider since its coordinate time is the same as the proper time of the radial infallers from infinity).

vanhees71
J O Linton said:
I am sorry to see this thread die because I am trying very hard to understand the spatial curvature around a gravitating mass. I strongly suspect that the answer to my query in #9 is that the spatial curvature is completely described by a nsingle parameter and that the curvature is analagous to the curvature of a sphere (albeit in 3 dimensions rather than two).

I'm not sure why you think that. But I can give you a brief summary of what goes on.

In order to decompose the space-time curature tensor into various pieces, one piece of which can be thought of as representing the spatial curvature, the tool to use is the Bel decomposition. There's a short stub at https://en.wikipedia.org/w/index.php?title=Bel_decomposition&oldid=1016066388

In general, using this decomposition, there are 21 nonzero components which satisfy one constraint, for a total of 20 independent components. These are distributed as

6 for the electrogravitic components, which can be thought of as representing tidal forces
9 for the magnetogravitic components, which represent "frame dragging" effects which are formally similar to the magnetic field in electromagnetism
6 for the topogravitic tensor, which represents what you are interested in, spatial curvature.

In Schwazschild coordinates, it turns out that there are 3 nonzero components for the electrogravitic tensor, i.e. the tidal forces, no magnetogravitic components, and 3 nonzero components for the topogravitic tensor, which represents the spatial curvature. The Schwarzschild metric is a vacuum metric, and it turns out that for any vacuum metric, the topogravitic tensor is the same as the electrogravitic tensor.

There's a reference to such a decomposition (which doesn't use the same terminology) in MTW's textbook Gravitation, which I could dig out if there is interest. MTW's approach is just to use an orthonormal basis, and pick out the relevant components, much as one picks out electric and magnetic fields from the rank 2 Faraday tensor for electromagnetism.

Note that this is all from memory - I'm going to take a chance and go by my memory rather than look it up, which is a bit risky - last time I did that I got burnt, but I guess I haven't learned from that :).

With three nonzero components and one constraint, there are two independent degrees of freedom for the Schwarzschild metric. The electrogravitic part is easy, it's just the stretching tidal forces, one radial, two traverse. The topogravitic tensor has the same values for the components, but I'd have to think more how to interpret it physically. I'm thinking along the lines that if one took the coordinates seriously (and one probably shouldn't), one could regard the topogravitic tensor as sort of a velocity-squared "force". I'll mull this over and see if I can come up with something better thought out.

Anyway, considering all the symmetries, you should have 3 nonzero components and one constraint, both for the entire space-time curvature tensor, and for the spatial part. Unless I've made an error, which is possible as I am rushing this.

vanhees71
Alright, here is a calculation, which is unfortunately at the A level, to at least confirm some of what I said. Trying to explain it more simply (in terms of geodesic deviation) is something I may or may not have luck with.

I start with the isotropic form of the Schwarzschild metric, using https://core.ac.uk/download/pdf/81932666.pdf. However, I'll take the liberty of using "r" where the paper uses ##\rho##.

The 4-d metric is
$$ds^2 = - \left( \frac{1 - \frac{M}{2r} }{ 1 + \frac{M}{2r} } \right) ^2 dt^2 + \left( 1 + \frac{M}{2r} \right)^4 \left( dr^2 + r^2\,d\theta^2 + r^2\,\sin^2 \theta \, d\phi^2 \right)$$

Using the obvious and simple projection of the 4d spacetime into a 3d space by ignoring t, we get the spatial part of the metric

$$\left( 1 + \frac{M}{2r} \right)^4 \left( dr^2 + r^2\,d\theta^2 + r^2\,\sin^2 \theta \, d\phi^2 \right)$$

We recognize this as polar coordinates, with a scale factor of (1 + M/2r)^2. Note that r here is the isotropic r coordinate, which is different than the Schwarzschild r coordinate, as per the reference.

We can create an orthonormal cobasis (dx^1, dx^2, dx^3) as follows:

$$dx^1 = \left(1+\frac{M}{2r} \right)^2 \,dr \quad dx^2 = r\,\left(1+\frac{M}{2r} \right)^2\,d\theta \quad dx^3 = r\,\left( 1+\frac{M}{2r} \right)^2 \, \sin \theta$$

We can calculate the Riemann in this cobasis. And grinding it through the automated GR tensor, we get three nonzero components. If we let
$$q = \frac{M}{r^3 \left(1+\frac{M}{2r} \right)^6}$$

we have ##R_{1212}## = -q, ##R_{1313}## = -q, ##R_{2323}## = 2q (along with the usual Bianci identies, of course).

We could probably have done the same thing in non-isotropic coordinates, but the isotropic coordinates are more "physical", as seen by the fact that the only difference is a r-dependent scaling of the distance, with infinitesimal changes in the coordinates representing larger changes in the distance for "low" r.

Dale and vanhees71
PeterDonis said:
Yes, they would. Unfortunately, though, such geodesics do not lie within a spacelike hypersurface of constant coordinate time in any of the known charts (Painleve would be the most obvious one to consider since its coordinate time is the same as the proper time of the radial infallers from infinity).
Yes, I already figured out that such geodesics are not contained within Panlieve spatial slices. However, they are still geodesics with a nice definition, and constructing them along each point of an exterior stationary world line (which they would not be orthogonal to), would map each world line point to a different horizon event.

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vanhees71
PAllen said:
they are still geodesics with a nice definition, and constructing them along each point of an exterior stationary world line (which they would not be orthogonal to), would map each world line point to a different horizon event.
Yes, this would provide an interesting alternative foliation that I do not think I have seen in the literature.

vanhees71
I was thinking about how to explain curvature in 3d space, and I decided a good approach was to first describe curvature in 2d space, then generalize a bit. Curvature in 2d is simple, since there is only one non-zero component of the curvature tensor. We can thus think of the curvature of a 2d surface as being a single number.

Let's start with manifolds. Manifolds are a mathematical entity, which we can visualize most easily via an embedding. In particular, we can regard the surface of a 3d object as a 2d manifold. As a mathematical entity, manifolds exist without such an embedding, which is basically a visual aid. But this visual aid makes things easier.

Let's continue with the curvature of a sphere, one of the simplest 2d cases. In GR, when we talk about curvature, we talk about intrinsic curvature. One method of describing curvature is via geodesic deviation. Another method is to consider the angular excess, or angular deficits, of geodesic triangles, but I won't be getting into that here in detail. I will refer the interested reader to the wikipedia on spherical trignometry and "angular excess", https://en.wikipedia.org/w/index.ph...ry&oldid=1051975245#Area_and_spherical_excess.

I also won't discuss how to draw geodesics in general, I will simply utilize the fact that geodesics on the sphere are great circles. Motivationally, it is sufficient to regard spatial geodesics as curves which are the shortest distance between two nearby points on some manifold. Since we are using embeddings as a visual aid, this is the shortest distance of a curve lying entirely in some surface. This is not the best way, and it actually only works when the points are sufficiently close together. "Antipodal points" are an example of where the idea breaks down when the points are too far apart. There can be occasions where there is more than one straight line between two antipodal points on a surface, and they are not necessarily of the same length. We actually need to add a constraint on the size of the region to make this idea work. And it turns out there are better ways - I will mention that the better way involves a concept called a "connection", but I won't get into more detail than that mention here.

On the sphere, geodesics are simple - they are great circles. If a space is flat, geodesics that start out parallel remain the same distance apart. On the sphere, we can see that space is not flat. If we imagine the set of great circles running through the north and south poles, we can see that these geodesics are all parallel to each other at the equator, as the geodesics go north-south, and they are everywhere orthogonal to the equator, which runs east -west. However, the initially parallel geodesics at the equator do not remain at a constant distance apart, in fact they meet at the north and south poles. This is how we know the surface of a sphere is curved - if it were flat, parallel geodesics would stay at a constant separation.

We want to parameterize the geodesics by some affine parameter. Because we are considering only spatial curvature, we can get away with regarded this affine parameter as time, though it is more rigorous to regard it as an affine parameter. If we regard the affine parameter as time, then the tangent vector at any point p of the geodesic is the velocity u of an object following the purely spatial geodesic. Note we are not considering space-time geodesics, just spatial geodesics, and we are re-purposing time as an affine parameter for ease of exposition. Thus we are not really treating time in the sense that it is used in special relativity, it's simply a parameter that identifies points on our curves. The notation I am using is not really standard, it's been heavily modified with the intent to be easier to grasp. Hopefully this will actually help, rather than cause more confusion, but - time will tell.

With this idea of time as an affine parameter in mind, we can say that the spatial geodesics "accelerate" towards each other. This gives rise to the next section, the geodesic deviation equation. The geodesic equation says that

$$\frac{D^2 x}{dt^2} = -R u^2 x$$

This says that the relative acceleration between objects following these spatial geodesics of the geometry in question (in this case, great circles on the sphere) is proportional to the square of the velocity, u, and the initial separation of the geodesics.

If we had more than two dimensions, we'd need tensor notation and it's associated indices, but with only two dimensions, we can oversimplify things by omitting all the indices, and still have a useful equation.

By dimensional analysis, we can see that R must have units of 1 / distance^2. R is equivalent to the Gaussian curvature, see for instance https://en.wikipedia.org/w/index.php?title=Curvature&oldid=1047300795#Gaussian_curvature.

We can also use some Newtonian physics to get the same conclusion. The 3-d acceleration vector would be v^2/r, pointed out of the surface, towards the center of the sphere. The component that we are interested in is only the component that lies in the plane of the sphere (or more precisely, it's in the plane tangent to the sphere, the tangent plane). The compnent in the tangent plane is v^2 sin(theta) / r_sphere. We can approximate sin(theta) as (x/r_sphere), so we get that the acceleration of nearby geodesics towards each other is v^2 x / r_sphere^2. We can see this matches the geodesic equation I have previously presented, with R = 1/r_sphere^2.

This procedure illustrates how we can use the geodesic deviation equation to find the intrinsic curvature. We can construct geodesics (great circles) on the sphere, and perform all measurements on the sphere, without considering the geometry in which the sphere is embedded. The sphere has its own intrinsic geometry, the embedding allows us to leverage our knowledge of Euclidean geometry of the globe to understand the non-Euclidean geometry of the surface of the globe, the sphere.

Now that we have describe curvature in 2 dimensions via geodesic deviation, let's talk a little bit about three dimensions. If we have a set of unit length and orthogonal basis vectors at some point p, ##\hat{x}, \hat{y}, \hat{z}##, we can consider the xy plane, the xz plane, and the yz plane as each being a 2 dimensional surface. To be a bit more precise, each vector generates a geodesic, and the set of vectors in the xy plane generate a set of geodesics which lie in some surface. After generating this surface, we can measure it's curvature. We can do the same for the xz and yz surfaces as well.

Each of these generated surfaces can have its own curvature. This gives us three components of curvature for 3d space. This is actually not quite enough in general - but it IS enough if we pick our axes correctly. We can reduce the 6 components in the general case to three, by a proper alignment of our coordinate axes with the underlying geometry, though I won't attempt to prove that here.

The mathematical details underlying this are interesting, but a bit advanced. We take the 4x4x4x4 Riemann tensor, and with a bit of special pleading that works for 3 dimensions, reduce it to a 3x3 symmetric matrix. For instance, we can map R_(12)(12) to R'_33. After this mapping of the Riemann to a 3x3 real matrix, we note that the eigenvectors of any real symmetric matrix are orthogonal. Using these orthogonal eigenvectors, we can diagonalilze the matrix, leaving the eigenvalues as the diagonal elements. In this application , these diagonal eigenvalues are the three curvature parameters that I mentioned, the curavatures of the xy, xy, and yz planes.

J O Linton said:
I strongly suspect that the answer to my query in #9 is that the spatial curvature is completely described by a nsingle parameter and that the curvature is analagous to the curvature of a sphere (albeit in 3 dimensions rather than two).

Are you talking about the spatial curvature inside a body of uniform density?

https://en.wikipedia.org/wiki/Interior_Schwarzschild_metric#Visualization

I had a few remarks to add to my previous comments. Curvature as geodesic deviation always involves a deviation transverse to the direction of motion. In 2 dimensions, there is only one direction transverse to the motion, and a single component of curvature, but in 3 dimensions, there is more freedom of choice.

It turns out to be possible to choose a set of three orthogonal "principal axis" in three dimensions such that the separation vector and the acceleration vector point in the same direction, which simplifies the mental picture considerably. The proper choice of axes makes the geodesic deviation, and thus the curvature (as we are using geodesic deviation to define the curvature) much easier to get a handle on. In general one cannot count on an arbitrary choice of axes to have this useful property, but one can always find such a set of principle axes.

Given the correct choice of principal axes, we can reduce the description of curvature in three dimensions into a set of three numbers, which as stated previously can be thought of as the intrinsic, or Gaussian, curvature of the projection on the manifold of the xy, xz, and yz planes.

A bit of an aside here. Vectors must commute, which is why we must talk about the tangent space to define vectors, as displacements on a manifold do not add in a commutative manner. As Misner remarks, going 500 miles east and 500 miles north does not lead to the same destination on the curved sphere as going 500 miles north and 500 miles east - displacements do not commute on the sphere, because it is curved. Thus vectors cannot represent displacements of finite magnitude. Taking the appropriate limit of small displacements is the work-around to make vectors possible on a manifold.

To go from the tangent space, where these infinitesimal vectors are defined, to the manifold, where the curvature exists, we propagate the infinitesimal vectors along geodesics which generate a set of points that represent a curved surface in the manifold. Thus, we can measure the intrinsic curvature of this curved subspace via geodesic deviation. As it is a planar (2 dimensional) submanifold, we can represent this curvature via a single number.

The curvature of these three surfaces determines the geodesic deviation for a geodesics lying in any of the associated planes. We can handle the case where the separation vector is not along a principle axis by decompose the separation vector into components which lie along each of the principle axes. Then we can compute the net contribution of each of these components to the total deviation (acceleration) vector and add them together, as the curvature tensor is linear in each part, i.e. multi-linear. The quadratic part of the deviation generated by the multi-linear tensor is generated by duplicating a linear relationship twice. The end result in the general case is a deviation vector (acceleration vector) that is not in the same direction as the displacement vector between geodesics.

If we consider specifically the 3d spatial surface of the 4d Schwarzschild metric, the r-theta and r-phi planes both share a negative curvature. The magnitude of the curvature proportional to the mass and inversely proportional to the cube of the Schwarzschild r coordinate, so the curvature becomes small at large r. The curvature is a property that we measure at each point, not a generic property of the whole manifold.

Because the curvature in the r-theta plane is negative, spatial geodesics in this plane do not remain parallel, but diverge. It's the opposite to what happens on a sphere, where the geodescis (great circles) converge. On the r-theta plane, the geodesics diverge, they "accleerate" away from each other. Spherical symmetry also means that any plane which incorporates a vector in the radial direction has the same properties and is described by the same negative curvature. In essence, the spherical symmetry means we can make any plane the r-theta plane and/or the r-phi plane by a convenient coordinate choice.

The theta-phi plane, unlike the two radial planes, has a positive curvature, so geodesics in that plane converge. This curvature in this plane is also proportional to the mass of the central body and inversely proportional to r^3, but the magnitude of the deviation (accleration) is double that of the r-theta and/or r-phi planes.

pervect said:
The theta-phi plane, unlike the two radial planes, has a positive curvature, so geodesics in that plane converge.
The curves you refer to here are not geodesics of either the spacelike 3-surface or of the spacetime. They are only geodesics of a 2-sphere with constant ##r## (and constant ##t##, if we are looking at the spacetime as a whole). So it's somewhat misleading to look at these curves when you are trying to understand the curvature of the spacelike 3-surface or the spacetime as a whole. All you are really showing here is that 2-spheres of constant ##t## and ##r##, considered as manifolds in their own right, have positive curvature.

cianfa72
PeterDonis said:
The curves you refer to here are not geodesics of either the spacelike 3-surface or of the spacetime. They are only geodesics of a 2-sphere with constant ##r## (and constant ##t##, if we are looking at the spacetime as a whole). So it's somewhat misleading to look at these curves when you are trying to understand the curvature of the spacelike 3-surface or the spacetime as a whole. All you are really showing here is that 2-spheres of constant ##t## and ##r##, considered as manifolds in their own right, have positive curvature.

I use spheres a lot because they're simple to describe. Any 2 dimensional surface has a curvature that can be described by a single number. On a sphere, this number for curvature is a global constant, on a non-spherical shape each point on the manifold is associated with its own curvature at that specific point. The method I chose to describe the curvature, geodesic deviation, is not unique to the sphere, though I used the sphere as an example. I did oversimplify the method of geodesic deviation by omitting the tensor indices. This may cause some confusion, but including the indices in my judgment makes the response A-level, which I tried to avoid as much as I could (which was probably not entirely).

Also lacking is a fully technical description of how to draw a geodesic, though I do allude to the fact that geodesics are curves of "shortest distance". More accurately would be to say that geodesics using the Levi-Civita connection are curves of "shortest distance", but the Levi-Civita connection is always implied in General Relativity. It'd also be more accurate to talk about curves of "extremal" distance, but I believe this also significantly raises the complexity level. I suppose I could have used Baez's approach about talking about geodesics as curves that move "in the same direction", but I've always found this unsatisfying, though it does have its merits.

A geodesic can be a geodesic both manifolds and sub-manifolds. We can describe a geodesic by giving a starting point, and a direction at that starting point. Then for an example of what I mean, consider, a geodesic in the 4d Schwarzschild space-time, starting at the point r=2, theta=0, phi=0, t=0, with an inital velocity/direction given by d/dt = d/dr = d/d theta=0 and d/ d phi nonzero. This curve is a geodesic in the 4d Schwarzschild space-time, the 3d spatial surface of the 4d spacetime (a submanifold of the Schwarzschild spacetime with t=0), and the 2d manifold of the r-theta equatorial plane (a 2d submanifold of the Schwarzschild metric with t=0 and theta=0).

The geodesics I've discussed are of this type, they are geodesics of the 3-manifold and the 2-manifold both.

Because it is easier to describe geodesics in only 2 dimensions, I took the approach to attempt to leverage the knowledge of 2d to address the issue of geodesics in spaces of three dimensions.

A proof that this is sufficient is A-level, but I discussed briefly (at A-level) the reasoning that led me to the specific choices of submanifolds I made to describe the 3d curvature. This is the fact that the Riemann tensor in 3 dimensions can be identified with a 3x3 symmetric matrix. The trick unfortunately doesn't generate to spaces of higher dimensions. A 3x3 symmetric matrix has eigenvectors that may be familiar from other applications of such symmetric 3x3 matrices, including but not limited to the moment of inertia tensor. If we diagonalize this matrix, we can use the diagonals to describe the entire matrix. The process of diagonalizing the matrix is done via a principle axis transformation using these eigenvectors. The principle axes are just those axes where the geodesic deviation acceleration vector , d^2 x/ ds^2, points in the same direction as the displacement vector x.

Once we diagonalize the matrix by choosing the right coordinates, we can specify the entire matrix by specifying it's diagonals, which can be physically interpreted as those 2d curvatures I discussed.

PeterDonis said:
The curves you refer to here are not geodesics of either the spacelike 3-surface or of the spacetime.
IIUC the point is that the curves we are considering belong to the spacelike 2-surfaces of constant coordinate time ##t## and reduced radius ##r## (Schwartzchild spacetime in Schwartzchild coordinate chart). These spacelike 2-surfaces are actually 2-spheres and those curves are actually (spacelike) geodesics of.

PeterDonis said:
So it's somewhat misleading to look at these curves when you are trying to understand the curvature of the spacelike 3-surface or the spacetime as a whole. All you are really showing here is that 2-spheres of constant ##t## and ##r##, considered as manifolds in their own right, have positive curvature.
Ok, this is misleading since those geodesic curves are restricted to the ##r## slices of the ##t=const## spacelike 3-hypersurface.

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cianfa72 said:
IIUC the point is that the curves we are considering belong to the spacelike 2-surfaces of constant coordinate time ##t## and reduced radius ##r## (Schwartzchild spacetime in Schwartzchild coordinate chart). These spacelike 2-surfaces are actually 2-spheres and those curves are actually (spacelike) geodesics of.
Yes. The point is, so what? Why would you even consider those curves in the first place if you're interested in the spatial curvature of a 3-surface of constant ##t##?

PeterDonis said:
Yes. The point is, so what? Why would you even consider those curves in the first place if you're interested in the spatial curvature of a 3-surface of constant ##t##?
Ok yes, got it.

pervect said:
A geodesic can be a geodesic both manifolds and sub-manifolds. We can describe a geodesic by giving a starting point, and a direction at that starting point. Then for an example of what I mean, consider, a geodesic in the 4d Schwarzschild space-time, starting at the point r=2, theta=0, phi=0, t=0, with an inital velocity/direction given by d/dt = d/dr = d/d theta=0 and d/ d phi nonzero. This curve is a geodesic in the 4d Schwarzschild space-time, the 3d spatial surface of the 4d spacetime (a submanifold of the Schwarzschild spacetime with t=0), and the 2d manifold of the r-theta equatorial plane (a 2d submanifold of the Schwarzschild metric with t=0 and theta=0).
While your first statement above is correct as a possibility, in the particular case it is not. To make matters easy,
reference:

www.physics.usu.edu/Wheeler/GenRel2015/Notes/GRSchwarzschildGeodesics.pdf

and putting in an initial tangent of (0,0,0,1) , one finds the geodesic equations do not vanish for dr’/ds. Thus even though the starting tangent is only in the phi direction, the radial coordinate will change along the geodesic of the spacetime. Thus your proposed curve is definitely not a geodesic of the spacetime. Whether it is a geodesic of the 3-d spatial slice requires additional analysis, which I will make if I have time and interest.

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pervect said:
A geodesic can be a geodesic both manifolds and sub-manifolds. We can describe a geodesic by giving a starting point, and a direction at that starting point. Then for an example of what I mean, consider, a geodesic in the 4d Schwarzschild space-time, starting at the point r=2, theta=0, phi=0, t=0, with an inital velocity/direction given by d/dt = d/dr = d/d theta=0 and d/ d phi nonzero. This curve is a geodesic in the 4d Schwarzschild space-time, the 3d spatial surface of the 4d spacetime (a submanifold of the Schwarzschild spacetime with t=0), and the 2d manifold of the r-theta equatorial plane (a 2d submanifold of the Schwarzschild metric with t=0 and theta=0).
PAllen said:
While your first statement above is correct as a possibility, in the particular case it is not.
A trivial example of a geodesic in spacetime and space is radial free fall.

A.T. said:
A trivial example of a geodesic in spacetime and space is radial free fall.
Not really. For radial free fall geodesic, both t coordinate and r coordinate change as a function of an affine parameter. For a radial spatial geodesic, the t coordinate does’t change at all. Thus, they are completely different curves in spacetime. What is actually true is that a radial spatial geodesic in Schwarzschild geometry is both a geodesic of spacetime and a geodesic of a constant SC time slice.

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PeterDonis
PAllen said:
A.T. said:
A trivial example of a geodesic in spacetime and space is radial free fall.
Not really.
- The world-line of a radial free fall is a geodesic in space-time
- The spatial-path of a radial free fall is a geodesic in space

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