G M_{*} r

. Therefore the disk angular momentum is

J = \int_{0}^{R} G M_{*} r Σ 2 π r d r = G M_{*} B,

where

B = \int_{0}^{R} r^{1/2} Σ 2 π r d r .

The absence of an external or inner-boundary torque makes

J

, and hence

B

, constant.

From

\overset{ν}{ˉ} = A r^{9/2} Σ^{2}

[A] = L^{3/2} M^{- 2} T^{- 1}, [B] = M L^{1/2} .

A B^{2} t

L^{5/2}

R \propto (A B^{2} t)^{2/5} \propto t^{2/5} .

For the supplied similarity solution, put

τ = t / t_{0}

and

u = r / R (t)

. Then

r = R u^{2}

and

d r = 2 R u d u

. The total mass is

M_{D} = 2 π \int_{0}^{R} Σ r d r = 4 π Σ_{0} R_{0}^{3/2} τ^{- 2/5} R^{1/2} \int_{0}^{1} (1 - u)^{1/2} d u = \frac{8 π}{3} Σ_{0} R_{0}^{3/2} τ^{- 2/5} R^{1/2} .

Since

R^{1/2} = R_{0}^{1/2} τ^{1/5}

M_{D} = \frac{8 π}{3} Σ_{0} R_{0}^{2} (\frac{t}{t _{0}})^{- 1/5} .

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