# 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.
PAllen said:
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 curves in spacetime.
A.T. said:
- 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
But they are two different curves. The latter is not even well defined as such - you have to specify that you mean radial path in an SC coordinate constant time slice. Then, the spatial the spatial path is a geodesic of both the slice and the spacetime. For other coordinates, a radial spatial path need not be a geodesic of the spacetime, though it would normally be a geodesic of the slice.

The claim under discussion was a specific path being a geodesic of the spacetime and the submanifold. Your example is simply not relevant in that you are talking about two different paths.

PeterDonis
PAllen said:
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.

I'd agree that dr/ds is not zero. But I wouldn't expect it to be zero.

When I talk about the Scwharzschild space-time, formally I'm talking about coordinate [t,r,theta,phi], geometric units, and a metric with the line element ##-(1-r_s/r) dt^2 + dr^2/(1-r_s/r) + r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2##

When I talk about the 3d surface of the Schwarzschild metric, I'm talking about projecting the Schwarzschild metric to 3d space. There are of course many possible projections, so this is vague. Specifically, then, it's a metric with coordinates (##r, \theta, \phi##)) and a line element of ##dr^2/(1-r_s/r) + r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2##. Then the main goal is to discuss it's curvature.

One can think of this as the 3d hypersurface of constant t in the Schwarzschild metric. This is the easiest to graps, but it's a coordinate-dependent definition. I had previously presented a coordinate independent definition, but it was admittadely a bit abstract. It was based on the idea of considering a subspace of the tangent vector space at some point on the manifold, then mapping this subspace to a submanifold of the Schwarzschild manifold itself, using the exponential map between the tangent space and the manifold. Probably for my purposes it would be better to stick with the coordinate dependent defintion.

When I talk about the 2d equatorial "r-theta" plane, and it's curvature, I'm talking about the plane with coordinates ##(r, \phi)## and the line element ##dr^2/(1-r_s/r) + r^2 d\phi^2##, which can be thought of as the surface of Flamm's paraboloid, which is presented in Wiki and other places as a way to visualize (part of) the Schwarzschild curvature in the r-phi plane. So I expect dr/ds to be nonzero, because the space in question is curved.

Ah, ok @pervect, I thought you were saying that a curve constant r, t, and theta with varying phi was a geodesic of spacetime. Now I see you meant that a spacetime geodesic tangent to this curve (which will have r varying) is also a geodesic of the indicated 3d and 2d submanifolds (metric induced from the 4d metric). That seems right. I did not understand what you were proposing.

pervect said:
I had previously presented a coordinate independent definition
The usual coordinate-independent definition of the spacelike 3-surfaces in question is that they are the 3-surfaces that are orthogonal to the integral curves of the timelike Killing vector field of the spacetime.

PAllen said:
Ah, ok @pervect, I thought you were saying that a curve constant r, t, and theta with varying phi was a geodesic of spacetime. Now I see you meant that a spacetime geodesic tangent to this curve (which will have r varying) is also a geodesic of the indicated 3d and 2d submanifolds (metric induced from the 4d metric). That seems right. I did not understand what you were proposing.
I verified explicitly from the general geodesic equations that any spacelike geodesic with initial tangent having no timelike component, and no theta component, and a starting position in the equatorial plane (theta = pi/2), will be fully contained in the equatorial plane. It will thus be a geodesic of the 4-manifold, a constant SC time spatial slice, as well as the equatorial (r,phi) 2-manifold. This includes a spatial radial geodesic as a subcase, as well any initial (r,phi) direction within the equatorial surface.

Then, by rotational symmetry arguments, one can say:

Any geodesic of a constant time SC slice is also a geodesic of the spacetime as well as of a particular 2-manifold subset of this slice containing this geodesic. In the case of a purely radial geodesic, there will be infinite such 2-manifolds, instead of just one.

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