lecture notes8

In order to go further we will approximate the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: f hydrogen (13.6 eV or 3.29 PHz). This is of order 1048−1050 photons s−1 for an O star. € In a spherically symmetric nebula with density nH, one may then write the equation for the ionization rate. The electron and proton densities are both nHξ. Thus: ∞ 4 πJν ( r) 2 ˙ n Hξ ( r) = −α (H 0 , T ) n Hξ 2 ( r) + n H[1 − ξ ( r)] ∫ν ( H 0 ) aν dν , i hν where the first term accounts for recombinations and the second for ionizations. In order to go further, we will approximate the radiation field Jν to be entirely from € the central star. It is attenuated by the inverse ­square law and by absorption as one moves outward, r L 4 πJν ( r) = ν 2 exp − ∫ 0 n H[1 − ξ ( r' )]aν dr' , 4 πr [ € where the integral in the exponential represents the optical depth for absorption of extreme UV photons.1 In this formula we have neglected changes in the structure of the nebula on the timescale for the photons to propagate through it (almost always valid) and radiation emitted within the nebula (generally wrong at the tens of percents level – more on this later). A common further approximation is to neglect the frequency dependence of the cross section. This is not good in general – the photoionization cross section decreases with frequency – but since for stellar sources most of the radiation is emitted near 13.6 eV one can get an approximate solution this way and gain conceptual insight. This leads to r Q(H 0 ) a ˙ ξ ( r) = −α (H 0 , T ) n Hξ 2 ( r) + [1 − ξ ( r)] exp − ∫ 0 n H [1 − ξ ( r' )]a dr' , 2 4 πr which is the basic equation that we will examine. € B. RECOMBINATION COEFFICIENTS AND CROSS SECTIONS We next need to know the actual values of the recombination coefficient and the photoionization cross section. Recombination: Recombination is simply due to the radiative decay of an electron from an unbound to a bound state of hydrogen. Therefore, we might imagine that the rate of the reaction (in units of recombinations cm−3 s−1) is simply the probability to find a proton and electron “near” each other times the atomic decay rates. Here “near” means within a few Bohr radii (i.e. a volume of V~10−23 cm3) and so the probability is P~nenpV. The typical atomic decay rate is A~109 s−1, so we would expect a recombination coefficient of α~10−14 cm3 s−1. In fact, at T=104 K, the recombination coefficient for hydrogen is: α A (H 0 ,10 4 K) = 4.2 × 10−13 cm3 s−1 . The larger value than our naïve estimate is due to Coulomb focusing of the incoming electron. This focusing becom...
View Full Document

This document was uploaded on 03/08/2014 for the course AY 102 at Caltech.

Ask a homework question - tutors are online