= Dust-burst accretion pulse
{title2=$\dot M_p=\frac{m_d(4-\alpha)}{T[1-\eta^{4-\alpha}]}(t/T)^{3-\alpha}$}
Let the initial dust mass be $m_d$, let its differential number distribution be $KD^{-\alpha}$ with $\alpha<4$ between $D_{\min}$ and $D_{\max}$, and put $\eta=D_{\min}/D_{\max}$. Under <burst transport under inverse-size radial drag>, complete absorption at a fixed radius occurs at $t=TD/D_{\max}$, with $T=t_r(D_{\max})$. The arriving <mass> is proportional to $D^{3-\alpha}dD$. Substitute $D=D_{\max}t/T$ and $dD/dt=D_{\max}/T$ and use <mass normalization of a power-law size distribution> to prove the displayed rate for $\eta T\le t\le T$, with zero rate elsewhere. Its integral is exactly $m_d$. Resonant delays, scattering losses, a finite release duration or grain destruction change this ideal pulse.
Back to article page
