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Take constant kinematic viscosity $\nu$ and the inner specific angular momentum found above. The Angular velocity of a circular Schwarzschild geodesic gives $-{\Omega }^{\prime }=3c\sqrt{m}/(2r^{5/2})$. Hence the inward radial component of four-velocity in the stated viscous model isTo convert proper time to Schwarzschild time, neglect ${\dot{r}}^2$ in the normalization as prescribed and insert the circular value of $h$. One obtainsThe change in observed Schwarzschild time equals this coordinate-time change when differential light travel time is neglected. Since $dr/d\tau =-u$,This time is nonnegative: $r^2-12m(r-3m)=(r-6m{)}^2\ge 0$ on the integration interval.

Put $x=r/(3m)$ and $X=r_1/(3m)$. The integral becomesPolynomial division gives $x^2/(x-1)=x+1+1/(x-1)$. Separately, writing $y=x-1$ gives $\int 2x/\sqrt{x-1}dx=(4/3)(x-1{)}^{3/2}+4\sqrt{x-1}$. Therefore the antiderivative inside the integral isIts value at $x=2$ is $-4/3$. Subtracting this lower endpoint yields the formal viscous inspiral time in Schwarzschild spacetimeThe logarithm has a positive coefficient. The minus sign in the printed final expression is inconsistent with its preceding integral. Differentiating the corrected bracket gives $6X[X-2\sqrt{X-1}]/(X-1)$, precisely the positive integrand after rescaling. The bracket vanishes at $X=2$; replacing $+6\log (X-1)$ by its negative would give derivative $-12$ there and a negative elapsed time for $X$ just above $2$. This is a sign error, not a choice of potential or time convention.

The result assumes constant kinematic viscosity; if $\nu$ varies with radius it remains inside the time integral. There is also a physical limitation to the stipulated approximation: $h-h_0$ vanishes quadratically at $6m$, so its formal $u$ diverges and eventually violates the neglected-radial-motion condition. The finite integral is the formal extrapolation of the nearly circular viscous model. A real accretion disk must instead match to its plunging region of a black-hole accretion disk; this approximation cannot describe that transition arbitrarily closely.

The radial equation for a circular geodesic, or equivalently $V^{\prime }(r)=0$, gives the Angular velocity of a circular Schwarzschild geodesicOne orbit therefore takes Schwarzschild coordinate time

Observer $O_3$ is static at infinity, where coordinate time equals proper time, so

For the static observer $O_2$ at radius $r$, $dr=d\theta =d\varphi =0$, and the Schwarzschild spacetime metric givesThus

Along $O_1$,where $d\varphi /dt=\Omega$ was used. Hence the Proper period of a circular Schwarzschild geodesic is

Interpret the orbit in the Schwarzschild metric as an unforced circular timelike geodesic and choose its plane as $\theta =\pi /2$. Put $f=1-2\mu /r$. The radial geodesic equation at constant radius uses the Christoffel symbolsand gives the Angular velocity of a circular Schwarzschild geodesicThe proper-time normalization then yieldsOver one revolution, the Proper period of a circular Schwarzschild geodesic is thereforeBoth the gravitational and orbital kinematic contributions to time dilation are included. Circular timelike geodesics exist for $r>3\mu$; they are stable for $r>6\mu$ and unstable for $3\mu <r<6\mu$, with marginal stability at $6\mu$.

If “circular orbit” instead means a forced trajectory with an externally chosen angular velocity, the radius alone does not fix the answer. Its elapsed time is $\Delta \tau =\Delta t\sqrt{1-2\mu /r-r^2{\Omega }^2/c^2}$, or $2\pi \sqrt{1-2\mu /r-r^2{\Omega }^2/c^2}/|\Omega |$ per revolution. The geodesic interpretation supplies the missing angular velocity.