Suppose a test particle undergoes circular motion in Schwarzschild spacetime, and not necessarily at the freefall rate. We assume its worldline is “centred on ” so to speak, or to be technical we could speak of the integral of a Killing vector field which is timelike at the given location: this also includes the case of no rotation at all but that’s fine. For simplicity, orient Schwarzschild-Droste coordinates so the motion coincides with the “equator” , and also with increasing -coordinate. Note the -coordinate is constant. Define the angular velocity as the coordinate ratio
assumed to be constant. We can solve for the 4-velocity of the particle:
which by definition is also the “time” vector of the particle’s orthonormal frame. We must have
for the motion to remain timelike. Note that while freefall circular motion requires to be outside the photon sphere, for accelerated circular motion any is permissible.
Now for the “space” vectors, it is natural, given our orientation of coordinates, to define two of them as simply normalised coordinate vectors:
This fixes the final vector in the tetrad up to orientation (overall factor), so we choose the orientation with positive -component:
Then as required, where is the Kronecker delta function. The particle’s 4-acceleration is given by the covariant derivative . I omit the result, but only the -component is nonzero, which is not unexpected. The length of the 4-acceleration vector is the magnitude of proper acceleration:
For the special case of geodesic motion the acceleration must vanish, so , and the factor reduces to . See Hartle ( §9.3) for a very different derivation of this. For the expression reduces to the familiar acceleration of a static observer.