Nebular heating and thermal equilibrium

• Nebulae are heated by photoionizations only. • Every photoionization creates an electron with energy 1 mυ 2 = h(ν −ν ) 2 p 0 1 • Every recombination removes mυ 2 . 2 r • Difference€ must be lost by radiation if nebula is in thermal equilibrium. • Average energy of photoionized€ electron is

∞ Jν ∫ h(ν −ν 0 )dν 3 ν 0 hν kTi = 2 ∞ J ∫ ν dν ν 0 hν

€ Ionization temperature • Suppose ionizing spectrum is on Wien part (UV tail) of blackbody curve: 3 3 2hν 1 2hν −hν / kT* Jν ∝ Bν (T* ) = ≈ e . c2 ehν / kT* −1 c2 • After some math, can show that 3 2 kTi ≈ kT* ⇒ Ti ≈ T*. € 2 3 • So Ti depends on shape not strength of ionizing spectrum – Energy of ionized electron depends on energy of photon that hit it, € not on how many photons there are. –3 • Photoionization x-section decreases as n so Ti may increase with distance from star. Heating rate • Energy gained by nebula through photoionization is

∞ 0 4πJν 0 G(H) = N(H ) ∫ h(ν −ν 0 )aν (H )dν ν 0 hν – Photoionization x-sect [m2] for H0 • Which through ionization equilibrium is ∞ J € ν 0 ∫ h(ν −ν 0 )aν (H )dν 0 ν 0 hν G(H) = N e N pαA (H ,T) ∞ Jν 0 ∫ aν (H )dν ν 0 hν 3 = N N α (H0 ,T) kT ≈ N N α (H0 ,T)kT . e p A 2 i e p A *

• Similarly for He; hence total G = G(H) + G(He0 ) + G(He1 ).

€ € Cooling mechanisms • Free-free (bremsstrahlung) – Least important -- usually negligible • Recombination emission. – Electrons recombine with ions to form excited atoms. – Atom cascades to lower levels. – Emitted photons that escape serve to cool the nebula. – Calculated from kinetic energy-weighted recombination coefficients to each level of atom. • Collisional cooling. – Most important mode of nebular cooling. – Metals have many low-lying energy states. – These states can be populated by collisions with electrons having kT ~ 1 eV. (NB kT = 0.86 eV at T = 10000 K) – If spontaneous de-excitation takes place before another collision de- excites the atom, photon escapes and cools nebula.

G = L ff + LR (H) + LC .

€ Radiative loss (Cox & Daltabuit, 1971) function

• We need to sum radiative losses from several processes to yield the radiative loss function = power per unit volume radiated by a plasma with given element abundances in equilibrium. –3 • Plot shows results for ne=nH=1 cm . Important features: – Forbidden transitions dominate for T £ 104 K. – Permitted transitions dominate for 104 K £ T £ 107 K. – Free-free emission (Bremsstrahlung) is most important at high T ³ 107 K. – Gas at T ~ 105 K cools quickly -- lots of atomic losses at these temperatures. Cooling by recombination emission

• Kinetic energy lost by electron gas per unit volume per unit time:

0 LR (H) = N e N pkTβA (H ,T) ∞ n−1 Maxwell-Boltzmann 0 0 where βA (H ,T) = ∑ ∑βnL (H ,T) n=1 L=0 1 ∞ 1 and β (H 0 ,T) = υσ (H 0 ,T) mυ 2 f (υ)dυ. nL kT ∫0 nL 2 0 • bnL(H ,T) is effectively a kinetic energy-averaged recombination coefficient. • n,L are the shell number and total angular momentum € quantum number of the state concerned. Collisional cooling and detailed balance - I

• Common ions of O, N, C, Ne, Ar all have levels that are ~ 1 eV above ground state (~ kT at T~104 K) – Efficient for cooling nebulae. • How many collisional excitations m–3 s–1? • Answer involves:

– Electron, ion densities Ne, Ni – Volume swept out by an electron each second which depends on velocity –1/2 so multiply NeNi by a factor q ~ Te . 2 – The Boltzmann factor. If meu /2< DE, collision won’t have enough energy to excite atom/ion from ground state. Larger DE implies fewer electrons with enough energy at fixed T. No energy barrier for downward collisional transitions. – Statistical weights of levels involved. Collision rate out of a level with statistical weight g is g times smaller than out of a similar level with just 1 state. – A dimensionless quantum mechanical collision strength: Ω(i, j)

€ Collisional cooling and detailed balance - II

• Putting it all together, collision rate from level i to j is Ne Ni qij

• If DE > 0 then 1/ 2 # 2π & ! Ω(i, j) −12 Ω(i, j) 3 -1 qij = % ( 3/ 2 = 8.269 ×10 1/ 2 m s $ kTe ' me gi giTe • But if DE < 0 then

−12 Ω(i, j) -ΔE/kT 3 -1 qij = 8.269 ×10 1/ 2 e m s . € giTe • Collision strength is symmetrical: Ω(i, j) = Ω( j,i) g q j e-ΔE/kT q . € ⇒ ij = ji gi €

€ The low-density limit

• If particle density is low, then it’s unlikely that a collision will occur before a spontaneous decay occurs. • So every collisional excitation creates a photon that escapes the nebula, and cooling rate is:

LC = ∑N e N 1q1 j hν1 j j

All excited levels Energy diff between state j (typically 4) and ground state € • But we need to consider collisional de-excitation too. Lecture 17 Revision quiz

• Why is the distribution of kinetic energies for electrons liberated by photoionization different from the distribution of kinetic energies removed by recombination? • Try the challenge of Slide 2 to show that 3 2 kT ≈ kT ⇒ T ≈ T . 2 i * i 3 *

• Why does the photoionization cross-section appear in the integrals€ on Slide 3, but not in slide 1?