= Solution
At distance $r$, isotropic luminosity $L$ produces radiative acceleration $a_{\rm rad}=\kappa L/(4\pi r^2c)$ in material of <opacity> $\kappa$. In fully ionized hydrogen, one <electron> supplies a <Thomson scattering> cross-section $\sigma_T$ for approximately one <proton> mass of material, so $\kappa=\sigma_T/m_p$. Balancing radiation against the <Newtonian gravitational field> of the black hole gives the <Eddington luminosity>
$$
\boxed{L_{\rm Edd,H}=\frac{4\pi GM_{\rm BH}m_pc}{\sigma_T}}.
$$
Below this luminosity gravity can confine optically thin ionized hydrogen; above it, electron scattering alone accelerates the gas outward.
An electron--positron pair has mass $2m_e$ and total scattering cross-section $2\sigma_T$, so its opacity is $\sigma_T/m_e$. Therefore
$$
\boxed{L_{\rm Edd,pair}
=\frac{4\pi GM_{\rm BH}m_ec}{\sigma_T}
=\frac{m_e}{m_p}L_{\rm Edd,H}}.
$$
The limit is lower because the radiative force per unit inertial mass is larger in a pair plasma.
For a cloud with arbitrary constant opacity per unit mass,
$$
\boxed{L_{\rm Edd}=\frac{4\pi GM_{\rm BH}c}{\kappa}}.
$$
Suppose the luminosity jumps to a constant $L>L_{\rm Edd}$ while the cloud is at rest at $r_0$. With $\Gamma=L/L_{\rm Edd}$, its radial acceleration is
$$
\ddot r=\frac{GM_{\rm BH}}{r^2}(\Gamma-1).
$$
Multiplication by $\dot r$ and <integral>[integration] from $r_0$ to <infinity> gives <conservation of energy>
$$
\frac12v_\infty^2
=\frac{GM_{\rm BH}}{r_0}(\Gamma-1),
\qquad
\boxed{v_\infty=\left[\frac{2GM_{\rm BH}}{r_0}
\left(\frac L{L_{\rm Edd}}-1\right)\right]^{1/2}}.
$$
This result assumes that $L$ and $\kappa$ remain constant and that the cloud stays optically thin.
Back to article page
