**Difficulty level:**★ ★ ★

Suppose a test particle undergoes circular motion in Schwarzschild spacetime, centred on (loosely speaking) but not necessarily freely-falling. 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:

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.