= Solution
Take constant <kinematic viscosity> $\nu$ and the inner <specific angular momentum> found above. The <Angular velocity of a circular Schwarzschild geodesic> gives $-\Omega'=3c\sqrt m/(2r^{5/2})$. Hence the inward radial component of <four-velocity> in the stated viscous model is
$$
u=\frac{3\nu/2}{\sqrt r\,[r/\sqrt{r-3m}-\sqrt{12m}]}.
$$
To convert <proper time> to <Schwarzschild time>, neglect $\dot r^2$ in the normalization as prescribed and insert the circular value of $h$. One obtains
$$
\frac{dt}{d\tau}\simeq\sqrt{\frac{1+h^2/(c^2r^2)}{1-2m/r}}
=\sqrt{\frac r{r-3m}}.
$$
The change in observed <Schwarzschild time> equals this coordinate-time change when differential light travel time is neglected. Since $dr/d\tau=-u$,
$$
\Delta t=\int_{6m}^{r_1}\frac{dt/d\tau}{u}\,dr
=\boxed{\frac2{3\nu}\int_{6m}^{r_1}\left[r-\sqrt{12m(r-3m)}\right]\frac r{r-3m}\,dr.}
$$
This time is nonnegative: $r^2-12m(r-3m)=(r-6m)^2\geq0$ on the integration interval.
Put $x=r/(3m)$ and $X=r_1/(3m)$. The integral becomes
$$
\Delta t=\frac{6m^2}{\nu}\int_2^X\left[\frac{x^2}{x-1}-\frac{2x}{\sqrt{x-1}}\right]dx.
$$
Polynomial division gives $x^2/(x-1)=x+1+1/(x-1)$. Separately, writing $y=x-1$ gives $\int2x/\sqrt{x-1}\,dx=(4/3)(x-1)^{3/2}+4\sqrt{x-1}$. Therefore the antiderivative inside the integral is
$$
\frac{x^2}{2}+x+\log(x-1)-\frac43(x-1)^{3/2}-4\sqrt{x-1}.
$$
Its value at $x=2$ is $-4/3$. Subtracting this lower endpoint yields the <formal viscous inspiral time in Schwarzschild spacetime>
$$
\boxed{\Delta t=\frac{m^2}{\nu}\left[3X^2+6X+6\log(X-1)-8(X-1)^{3/2}-24\sqrt{X-1}+8\right].}
$$
\b[The 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.
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