Orthonormal tetrad frame for circular motion in Schwarschild spacetime

Difficulty level:   ★ ★ ★

Suppose a test particle undergoes circular motion in Schwarzschild spacetime, and not necessarily at the freefall rate. We assume its worldline is “centred on r=0” 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” \theta=\pi/2, and also with increasing \phi-coordinate. Note the r-coordinate is constant. Define the angular velocity as the coordinate ratio

    \[\Omega := \frac{d\phi}{dt} > 0\]

assumed to be constant. We can solve for the 4-velocity of the particle:

    \[u^\mu = \frac{1}{\sqrt{1-2M/r-r^2\Omega^2}}\big(1, 0, 0, \Omega\big)\]

which by definition is also the “time” vector \mathbf{e_{\hat 0}} of the particle’s orthonormal frame. We must have

    \[\Omega<\frac{1}{r}\sqrt{1-2M/r}\]

for the motion to remain timelike. Note that while freefall circular motion requires r>3M to be outside the photon sphere, for accelerated circular motion any r>2M 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:

    \[(e_{\hat 1})^\mu := \big(0,\sqrt{1-2M/r},0,0\big)\]

    \[(e_{\hat 2})^\mu := \big(0,0,1/r,0\big)\]

This fixes the final vector \mathbf{e_{\hat 3}} in the tetrad up to orientation (overall +/- factor), so we choose the orientation with positive \phi-component:

    \[\frac{1}{\sqrt{1-2M/r-r^2\Omega^2}} \big(r\Omega/\sqrt{1-2M/r},0,0,\sqrt{1-2M/r}/r\big)\]

Then \mathbf{e_{\hat\alpha}}\cdot\mathbf{e_{\hat\beta}}=\pm\delta_{\alpha,\beta} as required, where \delta is the Kronecker delta function. The particle’s 4-acceleration is given by the covariant derivative \nabla_{\mathbf u}\mathbf u. I omit the result, but only the r-component is nonzero, which is not unexpected. The length of the 4-acceleration vector is the magnitude of proper acceleration:

    \[\frac{\lvert M-r^3\Omega^2\rvert \sqrt{1-2M/r}}{r^2(1-2M/r-r^2\Omega^2)}\]

For the special case of geodesic motion the acceleration must vanish, so \Omega=\sqrt{M/r^3}, and the factor 1-2M/r-r^2\Omega^2 reduces to 1-3M/r. See Hartle ( §9.3) for a very different derivation of this. For \Omega=0 the expression reduces to the familiar acceleration of a static observer.