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.