AST4320 - UiO [PDF]

AST4320: LECTURE 12 M. DIJKSTRA

1. Support Material Lecture 12: Photoionization Heating & Supernova Feedback 1.1. Photoionization Heating. The goal of this section is to show that the heating rate per unit volume due to photoionisation is independent of the intensity of the background. The heating rate per unit volume is (1)

H = nHI H

, where H is the heating rate per neutral hydrogen atom. This heating rate is given by Z 1 J(⌫) (⌫) (2) H= d⌫ [h⌫ h⌫L ], h⌫ ⌫L where h⌫L = 13.6 eV denotes the hydrogen ionisation potential (from the ground state), J(⌫) denotes the angle averaged intensity (which is simply the intensity in the radiation field if the radiation field is ‘isotropic’, i.e. the same in all directions). The units of J(⌫) are erg s 1 cm 2 Hz 1 . Finally, (⌫) denotes the photoionisation cross-section. This cross-section is given by (⌫) = 0 (⌫/⌫L ) 3 , with 0 = 6.3 ⇥ 10 18 cm2 . If we assume that J(⌫) = A(⌫/⌫L ) ↵ , then Z 1 ⇣ ⌘ ↵ 3 ⇣ h⌫ ⌫ h⌫L ⌘ (3) H=A 0 d⌫ , ⌫L h⌫ ⌫L ⌘g(↵)

where we have replaced the integral (which is just a number that depends on ↵) with g(↵). You can check that the dimensions of the right hand side of the equation are ‘erg s 1 ’, as it should be. This equation states that the heating rate per neutral hydrogen atom is proportional to the intensity in the ionising background To get the heating rate per unit volume we need to compute nHI which depends on the ionising background intensity. The total number density of hydrogen nuclei (i.e. in neutral and ionised state) is given by nH = nHI + nHII = nHI + np (where np = ne is the number density of protons, and ne denotes the number density of electrons). We further introduce the neutral fraction xHI ⌘ nHI /nH . The neutral fraction xHI is set by an equilibrium between photoionisation and recombination. 1

2

M. DIJKSTRA

(4)

(5)

• The total photoionisation rate per unit volume is nHI , where photoionisation rate per neutral atom, and is given by Z 1 J(⌫) (⌫) = d⌫ . h⌫ ⌫L

denotes the

If we again substitute the same expressions for J(⌫) and (⌫) then Z 1 ⇣ ⌘ ↵ 3 1 ⌫ =A 0 d⌫ . ⌫L h⌫ ⌫L ⌘h(↵)

This expression shows that the photoionisation rate per atom is proportional to the intensity in the ionising background. • The total recombination rate per unit volume is equal to ne np R(T ), where R(T ) denotes the recombination coefficient. This coefficient R(T ) is a quantity that can be obtained from quantum physics. If we balance photoionisation and recombination (6) ne np R(T ) = nHI ) (1

xHI )2 n2H R(T ) = xHI nH ) (1

xHI )2 nH R(T ) = xHI

. Under the assumption that the medium is mostly ionised (xHI ⌧ 1) - which is a reasonable approximation in typical conditions in the intergalactic medium - we can approximate (1 xHI )2 ⇠ 1, and (7)

xHI =

nH R(T )

=

nH R(T ) . A 0 h(↵)

This expression states that the neutral fraction increases linearly with density (recombination more efficient), and inversely with the ionising background intensity. Thus we can finally write for the volume heating rate (8)

H = nHI H = xHI nH H = n2H

A

g(↵)R(T ) 0 g(↵)R(T ) ) H = n2H . A 0 h(↵) h(↵)

That is, the heating rate per unit volume does not depend on the intensity of the ionising background! 1.2. A (very simple) model for Supernova Feedback. The aim of the section is to show with a simple model how supernova feedback may preferentially suppress star formation in low mass dark matter halos. The idea is that a dark matter halo of mass M has a total baryonic content Mb = M ⌦⌦mb . Suppose that a fraction f⇤ of all these baryons is turned into stars, while fraction (1 f⇤ ) remains in gaseous phase. The total gravitational binding energy of the baryonic gas is

AST4320: LECTURE 12

3

GM Mb (1 f⇤ ) ⌦b GM 2 (1 f⇤ ) = , R ⌦m R where we R denotes the virial radius (this radius is not really correct. In reality the gas will have assembled deeper in the potential well at a smaller radius. For simplicity, we ignore this). The (maximum) total energy that can be deposited into this gas by supernova is equal to ⌦b (10) Esn = ✏sn ⌘M⇤ = ✏sn ⌘f⇤ M, ⌦m where ✏sn denotes the total energy released per supernova explosion, ⌘ = 0.01 is the number of supernovae per unit solar mass of gas that forms stars. The total energy deposited by supernovae equals the binding energy of the remaining gas when (9)

(11)

Ub =

⌦b GM 2 (1 ⌦m R

f⇤ )

= ✏sn ⌘f⇤

v2 ⌦b f⇤ GM M) = = circ . ⌦m 1 f⇤ R✏sn ⌘ ✏sn ⌘

2 For f⇤ ⌧ 1, we therefore have f⇤ / vcirc / M 2/3 . This means that the fraction of baryons that is needed to unbind/blow out all gas from the dark matter potential halo increases with mass. In other words, it is easier to evacuate all gas from the lower mass halos with supernova explosions.

1.3. Reading Material. Reading material: suggested review article by Weinberg et al, as mentioned in the slides provides a good, brief summary of both missing satellite and too-big-to-fail problems (discussed next lecture).

AST4320: LECTURE 13

1. Warm Dark Matter & Mass Power Spectrum A particle becomes non-realistic (roughly) when its kinetic energy is less than its restmass energy, i.e. 3 (1) kB T < mX c2 , 2 where mX denotes the mass of the particle, c the speed of light, and kB denotes the Boltzmann constant. The kinetic energy of the particle is of course also given by Ekin = 12 mX v 2 . Particle that are relativistic therefore have v ⇠ c. Particles become non-relativisitic when (2)

mX (in eV) = TUniverse (in eV).

Relativistic particles propagate at the speed of light, and cannot be confined to gravitational potential wells. Therefore the ‘free-streaming’ of relativistic particles washes out any structure on scales smaller than the particle horizon of the Universe when they first become non-relativistic (this is the maximum distance these particles can travel). Denote this time with tnr , where the subscript ‘nr’ stands for ‘non-relativistic’. The scale factor of the Universe at time tnr corresponds to anr . If anr ⌧ aeq (where aeq denotes the scale factor of the Universe at matter-radiation equality), the the particle horizon rH / a. The particle horizon for particles at tnr therefore is Teq anr (3) rH (anr ) = rH (aeq ) = rH (aeq ) , aeq Tnr where in the last equality we used that T / a 1 during radiation domination. Now, we know that Teq ⇠ 1 eV. Therefore, the particle horizon size at tnr equals eV 1 1 (4) rH (anr ) ⇠ rH (aeq ) = 10 3 rH (aeq ) = 10 3 rH (aeq ) , Tnr (eV) Tnr (keV) mX (keV) For example, for a particle mass mX = 1 keV, the horizon size when the particle becomes non-relativistic is ⇠ 10 3 rH (aeq ). Structures smaller than this scale would have been washout out completely by the free streaming of the particles. We therefore expect structure formation to be suppressed at scales r < 10 3 rH (aeq ). Another way of phrasing this is that the structure is washed out at wave numbers larger than k > 103 k0 , where k0 ⇠ 0.01 cMpc, which corresponds to the wavenumber k0 ⌘ 1/rH (aeq ). We therefore expect dark matter particles with mass mX to suppress the power spectrum at wave numbers (5)

kfs ⇠ 10(mX /keV) cMpc

1

.

At these wave numbers the mass power spectrum is best constrained by Ly↵ forest data. 1

AST4320: LECTURE 14

1. Ly↵ Forest 1.1. The Mean Number Density of Hydrogen Nuclei/Atoms in the Universe. We know that the expansion of the Universe dilutes the mean density of baryons, ⇢¯b , as ⇢¯b,0 a 3 = ⇢¯b,0 (1 + z)3 . We also know that the mean baryon density is given by (1)

⌦b ⌘

⇢¯b ) ⇢¯b = ⌦b ⇢crit = ⌦b h2 ⇥ 1.88 ⇥ 10 ⇢crit

where I used that ⇢crit = 1.88 ⇥ 10 29 h2 g cm nuclei as a function of z is therefore

(2)

n ¯ H = fH

⇢¯b = (1 mp

YHe )

3.

29

g cm

3

,

The mean number density of hydrogen

⇢¯b,0 (1 + z)3 = (1 mp

YHe )⇢crit ⌦b (1 + z)3 /mp ,

where fH denotes the mass-fraction of baryons that is hydrogen. Here, fH = 1 YHe = 0.76, in which YHe = 0.24 denotes the max fraction in helium. Note that fH + YHe = 1.0, i.e. all baryons are either hydrogen or Helium. Substituting numbers we get n ¯ H ⇡ 2.0 ⇥ 10

(3)

7

(1 + z)3 cm

3

.

1.2. Expected Optical of the IGM to Photons with ⌫ > ⌫↵ . Caution: I use a slightly di↵erent notation than in the lecture.

ds HI zs

za absorption by H atoms 1

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AST4320: LECTURE 14

We address the question: what is the optical depth ⌧IGM (⌫) in the intergalactic medium (IGM) to photons which are emitted at a frequencies exceeding the Ly↵ resonance frequency. Look at a small patch of IGM with proper1 length ds at redshift za (shown in the Figure). As shown in the Figure, we denote the intensity in the radiation field - before it entered the patch - with I(⌫). Here ⌫ denotes the frequency of the photon at za . Absorption that occurs inside the patch modifies the intensity to I(⌫) exp[ d⌧ (⌫)] upon leaving the patch. Radiative transfer theory then states that d⌧ (⌫) = nHI (za ) ↵ (⌫)ds. Here, nHI (za ) denotes the number density of HI atoms in the patch, ↵ (⌫) denotes the Ly↵ absorption cross-section evaluated at ⌫. If this photon was emitted at frequency at redshift zs (which 1+zs denotes the redshift of the source), its frequency at the time of emission was ⌫e = ⌫ 1+z . a The total opacity of the IGM to photons emitted at this frequency ⌫e is obtained by summing the opacity over all patches of length ds. It is important to realise that the cosmological expansion of the Universe redshifts the photon during its flight, and the photon frequency is redshifted further in each successive step ds (each step ds corresponds to a lower redshift za (s)): (4)

⌧IGM (⌫e ) =

Z

observer

ds nHI (za )

↵ (⌫[za ]),

source

a where za = za (s), and ⌫[za ] = ⌫e 1+z 1+zs . The two ends of patch ds expand away from each other by an amount dv = H(za )ds. This redshifts the photons by an amount

d⌫ dv H(za )ds cd⌫ = = ) ds = . ⌫ c c H(za )⌫ If we substitute this into the integral we get Z ⌫obs Z ⌫ cd⌫ cd⌫ (6) ⌧IGM (⌫e ) = nHI (za ) ↵ (⌫[za ]) = nHI (za ) ↵ (⌫), H(z )⌫ H(z a a )⌫ ⌫ ⌫obs Rb Ra where in the last step we used that a = b , and replaced ↵ (⌫[za ]) = ↵ (⌫) (because we perform an integral over the variable ⌫, and do not care about its associated value of za in the evaluation of ↵ ). (5)

The integrand contains the Ly↵ absorption cross-section ↵ (⌫[za ]. We showed on one of the lecture slides that this function is sharply peaked at the Ly↵ resonance ⌫res . When we compute the integral, we only get noticeable contribution from a very narrow range a of frequencies ( ⌫/⌫ ⇠ 10 4 ) - which via ⌫[za ] = ⌫e 1+z 1+zs - corresponds to a very narrow range in redshift centered on zres . The redshift zres is the redshift at which the photon has res redshifted into the resonance (i.e. ⌫ = ⌫res ), which is therefore given by ⌫res = ⌫e 1+z 1+zs . 1Proper means ‘physical’, i.e. not comoving.

AST4320: LECTURE 14

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Because H(z) and nHI (z) barely evolve over the narrow range of z (around zres ) over which the integral is non-zero, we take them outside the integral: cnHI (zres ) ⌧IGM (⌫e ) = H(zres )

(7)

Z

In the last step we used that since

⌫ ⌫obs

d⌫ ⌫

↵ (⌫)

cnHI (zres ) ↵ (⌫) ⇡ H(zres )⌫res

⌫

d⌫

↵ (⌫).

⌫obs

is sharply peaked on ⌫res .

Using the radiative transfer theory of everything, (8)

Z

⌧IGM (⌫e ) =

R

2

⇡e (⌫)d⌫ = f m , we get. ec

cnHI (zres ) ⇡e2 f↵ , H(zres )⌫res me c

where f↵ = 0.416 is the Ly↵ ‘oscillator strength’, e = 4.80 ⇥ 10 10 esu (yes, esu), me = 9.1 ⇥ 10 28 g is the mass of the electron, c = 3 ⇥ 1010 cm/s, and ⌫res = 2.46 ⇥ 1015 Hz. Substituting the numerical values and the mean density of hydrogen in the Universe (Eq 3 above), we obtain ⇣ 1 + z ⌘1.5 res (9) ⌧IGM (⌫e ) = 1.7 ⇥ 105 , 4

where zres is the redshift at which the photon redshifts into the resonance. We discuss the implications of this number on the lecture slides (slide 20 and onwards). 1.3. Optical of the Photoionized IGM to Photons with ⌫ > ⌫↵ . . In ionising equilibrium, photoionisation balance recombination: (10)

nHI = ne np ↵rec (T ).

Let us define the neutral fraction x ⌘ nHI /[nHI + np ] ⌘ nHI /n, which gives the fraction of hydrogen that is neutral. The remaining fraction 1 x is ionised. We have nHI = xn, ne = np = (1 x)n. Substituting, we have (11)

xn = (1

x)2 n2 ↵rec (T ).

Assuming that x ⌧ 1 (as indicated by observations of the Ly↵) forest, 1 (12)

x=

n↵rec (T )

x ⇠ 1, and

.

To get the opacity of the ionised IGM, we use that we derived the opacity in the neutral IGM, and at mean density. For an arbitrary density2 n ⌘ n ¯ (1 + b ), the ionised fraction is x( b ) = The opacity of the IGM at some arbitrary density is then 2

b

denotes the overdensity in baryons.

n ¯ (1+

b )↵rec (T )

.

4

(13)

AST4320: LECTURE 14

⌧IGM (⌫e ) = (1 +

b [zres ])

¯ (zres )↵rec (T ) 2n (zres )

1.7 ⇥ 105

⇣1 + z 4

res

⌘1.5

.

If we assume that the baryons trace the dark matter (and hence that b = , in which denotes the overdensity in both baryons + dark matter), substitute our expression for n ¯ (zres ) (from Eq 3), and plug-in numerical values, we obtain ↵rec (T ) ⇣ 1 + zres ⌘4.5 (14) ⌧IGM (⌫e ) ⇡ 2(1 + [zres ])2 , (zres ) 4 which is shown on slide 30.

AST4320: LECTURE 15

1. Ly↵ Forest 1.1. Derivation of Opacity of the IGM in Ionized Gas. In the previous lecture we derived the opacity in the IGM in neutral hydrogen atoms. We found that in order to explain observations of the Ly↵ forest, we needed the IGM to be highly ionised. Here, we compute the opacity of the ionised component of the IGM (as discussed on slide 8 of the lecture slides). We continue the same geometry as in the previous lecture and compute the change in intensity I(⌫) (evaluated at frequency ⌫) of radiation passing through a patch of ionised gas of length ds.

ds ezs

za absorption by electrons

The change in intensity is I(⌫) ) I(⌫) exp[ d⌧ (⌫)], in which d⌧ (⌫) denotes the (di↵erential) optical depth across the patch. The optical depth d⌧ (⌫) is given by (1)

d⌧ (⌫) = ne (za )

T ds,

2 in which ne (za ) denotes the number density of free electrons in the patch, T = 8⇡ 3 re = 2 6.65 ⇥ 10 25 cm2 denotes the cross-section for electron scattering, in which re = mee c2 (see assignment 7.) denotes the classical electron radius. This cross-section is often referred to Thomson scattering cross-section. The total optical depth is obtained by summing over all patches ds (i.e. integrating over ds) Z observer (2) ⌧ (⌫) ⌘ ⌧e = ne (za ) T ds. source 1

2

AST4320: LECTURE 15

Note that we have dropped the frequency dependence. The Thomson scattering crosssection has no frequency dependence. The right-hand-side of the equation therefore has no frequency dependence. Therefore, the left-hand-side should also not depend on ⌫. We labelled this frequency-independent opacity with subscript ‘e’ ⌧e to emphasis that this is the opacity in free electrons. To get to the expression on slide 8 of the lecture notes, we need to convert the integral over ds into an integral over dz. A photon that propagates a distance ds does so over a finite time interval dt = ds/c. During this time interval dt, the scale factor of the Universe changed by an amount da = da dt dt. In other words (3)

ds = cdt = c

da da =c = a˙ H(a)a

where we used that H ⌘ aa˙ , and a ⌘ (4)

⌧e (z) =

c

Z

1 1+z .

zobserver zsource

c

dz , H(z)(1 + z)

Substituting this into Eq 2 we have

ne (z) T dz =c (1 + z)H(z)

Z

z 0

ne (z) T dz , (1 + z)H(z)

which is the expression listed in the lecture notes. Note that we derived in the previous lecture that n¯HI (z) ⇠ 2.0 ⇥ 10 7 cm s 3 . If for simplicity we ignore helium, then we can say electrons are supplied exclusively by hydrogen, and when hydrogen is highly ionised ne (z) = n¯HI (z). 1.2. The Photoionization Rate and Volume Emissivity of Ionizing Photons. This calculation accompanies slide 27 and beyond of lecture slides. In previous lectures we derived that the photoionisation rate (units s 1 ) was related to the angle-averaged intensity J(⌫) (units erg s 1 cm 2 Hz 1 ) of the ionising radiation field as Z

1

J(⌫) (⌫) ⌘ J(⌫L )G(↵), h⌫ 0 which assumes that⇣ the ⌘↵ionising radiation background has a power-law spectrum of the form J(⌫) = J(⌫L ) ⌫⌫L , where ⌫L denotes the frequency above which hydrogen can be photoionized, i.e. h⌫L = 13.6 eV (h denotes Planck’s constant). The function G(↵) contains all relevant numerical constant, and shows explicitly that the overall value of the integral depends on the assumed slope of the power-law, ↵ Our next goal is to connect J(⌫L ) to the total energy in ionising photons that are emitted per unit volume (the so-called ‘volume-emissivity’). (5)

=

d⌫

AST4320: LECTURE 15

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We evaluate the intensity J(⌫L ) (in erg s 1 cm 2 Hz 1 ) at a point x. We first compute the contribute dJ(⌫L ) from all sources of ionising photons separated from x by a distance [r, r + dr]. Note that dr is infinitesimally small. Each source ‘i’ within this shell has a luminosity Li (⌫) (units erg s 1 Hz 1 ), and to compute the total contribution to the ionising intensity at x we have to sum over all sources. We have X Li (⌫) Vshell ✏(⌫L ) (6) dJ(⌫L ) = ⌘ . 4⇡r2 4⇡r2 i

In the second step we used the definition of the volume emissivity, denoted with ✏(⌫L ): which is just the total energy in ionising photons emitted per unit volume. We know the volume of the shell to be Vshell = 4⇡r2 dr, and therefore (7)

dJ(⌫L ) = ✏(⌫L )dr.

Summing over all shells gives as the following integral Z 1 (8) J(⌫L ) = ✏(⌫L )dr. 0

This integral diverges. This is known as Olbers paradox (illustrated visually on slide 28). In the lecture we discussed two solutions to this paradox. • In the first, we noticed that in reality, we have to take into account that we observe the shell at r at some earlier redshift z. Exactly as we argued above, a physical step dr corresponds to a step dz = H(z)(1+z)ds towards higher redshift. We should c then take into account that the volume emissivity is changing with redshift (as we know that both the number density of galaxies, as well as their properties, change with redshift). • In the second, we used that ionising photons are not free to travel uninterrupted through the IGM. Ly↵ forest observations indicate that the IGM is highly - but not completely - ionised. We see absorption features in the Ly↵ forest arising because of tiny left-over amounts of hydrogen. This hydrogen also places a limit on how far ionising photons can travel through the IGM. The ionized IGM is not fully transparent to ionising photons. A way to include the fact that ionising photons ion have a finite ’mean free path’ ion mfp is simply by cutting o↵ the integral at r = mfp , i.e. Z ion mfp (9) J(⌫L ) = ✏(⌫L )dr = ion mfp ✏(⌫L ). 0

Substituting Eq 9 into Eq 5 we are left with

(10)

=

ion mfp ✏(⌫L )G(↵)

which is shown on slide 29 of the lecture notes.

,

AST4320: LECTURE 19

1. Black Holes Part I 1.1. Derivation of Chandrasekhar Mass. In a previous lecture in which we derived the density profile for the ‘isothermal’ sphere, we introduced the ‘distribution function’ f (x, v) which denotes the density of particles in 6-dimensional phase-space. We had to invoke this distribution function to describe a system of stars. The ordinary number density (number per unit volume), n(x) [we will drop the x from now on], relates to the distribution function as (1)

n=

Z

d3 p f (x, v).

We are going to use this phase-function to describe the electron degeneracy pressure in a gas. To do this we start with Heisenberg’s uncertainty principe that states that particles cannot be confined in a phase-space volume element smaller than (2)

( x p)3

~3 .

You can think of ~3 as the minimum volume in phase-space into which we can squeeze a particle. Electrons (and also neutrons & protons) are fermions, and for (spin 1/2) fermions we can only fit a maximum of 2 particles in such a volume element. The maximum phasespace density of fermions is therefore fmax (x, v) = 2~ 3 ⇠ ~ 3 (where we drop factors of order unity). Let us now compute the number density, n, under the assumption that the phase-space density is maximal. This number density is given by Z Z Z pF (3) n = d3 p fmax (x, v) ⇠ d3 p ~ 3 = dp p2 ~ 3 ⇠ p3F ~ 3 , 0

where we truncated the integral over momentum at the so-called ‘Fermi-momentum’ pF . Inverting this equation gives (4)

pF ⇠ ~n1/3

This equation is important: it shows that for a degenerate gas, as we increase the number density, we add electrons with increasingly large momentum. This is a strange statement since momentum is associated with motion of particles, which is associated with temperature. Strangely, if you have taken a course on statistical physics, you know that the 1

2

AST4320: LECTURE 19

assumption that the phase-space density is maximal corresponds to having zero temperature! That is, we assumed zero-temperature gas, yet as we increase the density of this gas, an increasing fraction of electrons obtains larger momenta. p2F The Fermi-energy of the particles relates to momentum via EF = 2m in the nonrelativistic regime. However, if we increase the density enough (via e.g. gravitational compression) there can be particles with large enough momenta that they are relavisitic1. In this case EF = pF c. Let us focus on the non-relatvistic regime first. Consider as star of radius R that contains N particles (say N electrons and N protons). The number density of particles is then n ⇠ N/R3 . We can then compute the Fermi-energy per electron (which is higher than that of the protons) from 1 ~2 N 2/3 ~2 N 2/3 (~n1/3 )2 = ⇠ . 2me 2me R2 me R2 We compare this to the gravitational binding energy per electron which is GM mp (6) Eg ⇠ , R where we used the mass of the proton instead of the mass per electron. If we were to ‘lift’ an electron out of the gravitational potential, then a small displacement of the electron would create an electric field, which would pull along the proton. The e↵ective mass of the electron is then set by that of the proton. The total energy of the electron is (5)

EF =

~2 N 2/3 GM mp ~2 N 2/3 GN m2p = . me R2 R me R2 R Nature wants to minimise the energy of this electron, which sets a radius Rmin (we will look into this in the assignment). One thing we can see is that as we increase N the total energy decreases (the assignment will make this clear as well): the objects becomes more compact and the density of electrons increases. This in turn increases the Fermi-momentum until the gas becomes relativistic. For relativistic gas, we have to write a di↵erent set of equations. (7)

Etot = EF + Eg =

For a relativistic gas we have EF = pF c ⇠ ~n1/3 c ⇠ ~N 1/3 cR 1 . The total energy of the electron is then ~N 1/3 c GN m2p ~N 1/3 c GM mp (8) Etot = = . R R R Minimizing the energy now has two possible implications: (i) if ~N 1/3 c GN m2p > 0 then minimising Etot corresponds to increasing R. Degeneracy pressure of the relativistic gas pushes the star apart (possibly back into the non-relativistic regime [?]); (ii) if ~N 1/3 c GN m2p < 0 then minimising Etot corresponds to decreasing R. As R decreases, the energy 1In other words, formally we can have a relativistic degenerate gas with zero temperature.

AST4320: LECTURE 19

3

of the system decreases, and run-away collapse occurs. The condition ~N 1/3 c GN m2p = 0 therefore marks the onset of the stability. Solving for N gives us the maximum number of particles which we can have in a degenerate gas

Nmax ⇠

(9)

⇣ ~c ⌘3/2 ⇠ 2 ⇥ 1057 ) Mmax ⇠ 1.5M . Gm2p

This Maximum mass is the so-called Chandrasekhar mass. Note that pF is identical for protons. However, protons have a higher mass, and at fixed momentum they have a lower velocity. The pressure exerted by particles generally scales as P / pv, where v denotes the velocity. The proton degeneracy pressure is therefore lower. Further note that runaway collapse of the electron-degenerate gas actually does not lead to a black hole directly. At sufficiently high densities the free electrons are forced into the protons (this is known as inverse beta-decay) to make neutrons. We can then do the same calculation as we did here for a neutron degenerate gas. At the level of the discussion in class, this calculation is identical. A more detailed calculation -taking into account the ‘equation of state’ of nuclear matter - gives rise to the so-called ‘Tolman-OppenheimerVolko↵ limit’, which is slightly di↵erent than what we derived. Objects above this limit would collapse into a black hole. 1.2. Geometrically Thin Accretion Disks. Accretion disks are often depicted as geometrically thin. The goal of this calculation is to check what physical conditions this requires. The thickness of the disk is determined by pressure inside the disk balancing the gravitational pull by the black hole. We therefore consider the equation of hydrostatic equilibrium (rp = ⇢gas r ), and use cylindrical coordinates. We are interested in the thickness of the disk, and hence focus on the z component (see slide 15.). Under the assumption that gravity is dominated by the black hole we have dP = dz

(10)

⇢gas

GM sin ✓ . r2

P (z=H) P (z=0) P We approximate the pressure gradient in the z direction with dP ⇠ H . dz = H We also know that sin ✓ = H/R, where H denotes the half-thickness of the disk (see slide 15.). The equation of hydrostatic equilibrium becomes

⇢gas H GM P GM H = ⇢gas 2 = . H r r r2 r

(11) We know that (12)

GM r

2 . Rearranging some terms gives us = vrot

⇣ H ⌘2 P 2 = vrot . ⇢gas r

4

AST4320: LECTURE 19

The term on the left-hand side therefore have

P ⇢gas

⇠ c2s (see our lectures on the Jeans mass2), and we c2s =

(13)

⇣ H ⌘2 r

2 vrot .

This equation is important. In accretion disks the rotational velocity close to the black holes usually greatly exceeds the sound-speed, i.e cs /vrot = H/r ⌧ 1. This suggests that at any r, the half-thickness of the disk is much smaller than r. However, this analysis also implies that the geometrically thin disk approximation breaks down when the disk becomes very hot, which is the case very close to the BH (see next analysis), and/or for very high accretion rates (also see next analysis). 1.3. Temperature Structure in Geometrically Thin Accretion Disks. The total energy per particle of mass m on a circular orbit around a BH is 1 GM m GM m 2 Etot = mvrot = 2 r 2r The change in the energy dEtot of the particle as it moves inward by dr (14)

dEtot GM mdr GM mdr dr = ⇠ , 2 dr 2r r2 where again I drop constants of order unity. The total luminosity dL from gas moving from r to r dr is (15)

dEtot =

GM mdr ˙ , 2 r where m ˙ denotes the total mass that is flowing through the shell at r per unit time. Now we assume that this luminosity comes out in black body radiation (which is the case if the gas is ‘optically thick’ to the radiation). Under this assumption dL = dA SB T 4 = 2⇡rdr SB T 4 . Setting this equal to the equation above, and dropping constants of order unit we find (16)

dL =

(17)

T =

⇣ GM m ˙ ⌘1/4 r3

SB

/

⇣Mm ˙ ⌘1/4 r3

.

This equation has several implications: (i) the temperature increases towards smaller r, and (ii) the temperature increases with mass accretion rate, and (iii) accretion disks become hotter around lower mass black holes M . To see this latter point, we evaluate the temperature at some fixed radius R in terms of the Schwarzschild Radius Rs of the black hole, i.e. R = xRs (where we keep x fixed). We then know that at fixed x r / Rs / M . We 2Formally c2 = s

dP d⇢

. For an ideal gas this introduces an extra factor of order unit which we can ignore in the present analysis.

AST4320: LECTURE 19

5

will see later that it is also reasonable to assume that the mass accretion rate is proportional to the BH mass M , i.e. m ˙ / M . Under these assumptions we have ⇣ M 2 ⌘1/4 (18) T (xRs ) / / M 1/4 . M3

AST4320: LECTURE 19

1. Black Holes Part II 1.1. Spectra Emerging from Accretion Disks. With the temperature profile T (r, M ) determined for an optically thick (but geometrically thin) accretion disk, we can compute the spectrum emerging from the accretion disk as a whole. We know that each shell emits as a black body, for which the spectrum is determined by its temperature, B(⌫, T [r]). Summing over all shells gives us the full spectrum. That is

(1)

L(⌫) = 2⇡

Z

Rout

dr rB(⌫, T [r]), Rin

where Rin denotes the inner radius of the accretion disk, and Rout denotes the outer radius. We know the blackbody spectrum to be ⇣ h h⌫ i B(⌫, T ) / ⌫ 3 exp kb T

(2)

1

⌘

1

Throughout the accretion disks there exists a one-to-one relation between radius r and temperature T as T / r 3/4 (see slide 11). When T / r 3/4 we have r / T 4/3 and dr / T 1/3 d[1/T ]. This yields (dropping all constants of proportionality): L(⌫) / ⌫

(3) Substituting x =

(4)

h⌫ kB T ,

3

Z

1/Tout

d[1/T ] 1/Tin

which means T / ⌫x L(⌫) / ⌫

1/3

Z

h⌫/kTout

1

T 4/3 T 1/3 . exp(h⌫/[kbT ]) 1

and d[1/T ] =

dx x5/3 [exp(x)

kdx h⌫ .

1]

We are then left with

1

.

h⌫/kTin

The gas has the highest temperature Tin at Rin . We expect the spectrum to be cuth i h⌫ o↵ as L(⌫) / ⌫ 3 exp at h⌫ k b Tin . On the other hand, the gas has the lowest kb T ⇣ h i ⌘ 1 temperature Tout at Rout . At h⌫ ⌧ kb Tout , we expect that L(⌫) / ⌫ 3 exp kbh⌫ 1 / Tout ⌫ 2 (because exp(x) ⇠ 1 + x for small x).

1

2

AST4320: LECTURE 19

1.2. Derivation of Eddington Luminosity. As we increase the gas accretion rate m, ˙ we increase the total luminosity if the accretion rate L. This radiation can exert a pressure on the in falling material. Consider gas at a distance r from the central black hole. The GM m gravitational force on an electron equals agrav = r2 p (where I deliberately used the proton mass, see the discussion above). On the other, the electron experience a force outwards because of the incoming radiation. This force is given by (5)

Frad =

dp , dt

dp

in which dt denotes the rate at which transfer momentum to the electron. Each time an electron ‘absorbs’ a photon it absorbs its momentum, which is given by h⌫/c. Electrons do of course not absorb photons, but instead scatter them. The precise amount of momentum that each photon gives to the electron depends on the angle at which the photon is scattered. However, it can be shown that on average each scattering event transfers an amount h⌫/c. The total force by the radiation on the electron is then dp hh⌫i = , dt c where denotes the total rate at which the electron scatters photons, and hh⌫i denotes the average energy of the photon. We have compute scattering rates before. The scattering rate (with units s 1 ) is related to the luminosity of the accretion disk as (6)

(7)

=

N˙ T L T = 2 4⇡r 4⇡r2 hh⌫i

where N˙ is the total number of emitted photons. Substituting this into the Equation 23 gives dp L T = , dt 4⇡r2 c This radiation pressure force balances gravity when (8)

(9)

Frad =

L=

4⇡GM mp c T

⌘ Ledd ,

which is what we used on slide 28. 1.3. Radiative Efficiency ⌘. The total energy of a proton on a circular orbit of radius R around a black hole of mass M is (see Eq 14 in Lecture notes 19) GM mp . 2R In general relativity there is a minimum radius at which particles can be on circular orbits. This radius is referred to as the ‘ISCO’ radius (Innermost Stable Circular Orbit), RISCO . (10)

Etot =

AST4320: LECTURE 19

3

This is a general realistic e↵ect. In Newtonian gravity there is no analogous radius1. As we showed on slide 10, the ISO radius depends on the black hole spin parameter a. For a stationary (non-spinning) black hole a = 0 and RISCO ⇠ 6GM . The total energy of c2 a proton orbiting at RISCO is therefore GM mp 1 (11) Etot = = mc2 ⇠ 0.08mp c2 . 2 12GM/c 12 That is, the total kinetic + binding energy of the proton is 8% of its rest-mass energy. As the plot on slide 10 shows, RISCO depends on black hole spin, and as a result, so does Etot . It is common to express this total energy as Etot = ⌘mc2 .

(12)

The standard assumption is that ⌘ ⇠ 0.1, which corresponds to a stationary black hole. The parameters ⌘ is the ‘radiative efficiency’, as it indicates how much binding energy is radiated away. Finally, mass-energy equivalence states that this negative kinetic + binding energy corresponds to a negative mass of ⌘mp . If a proton finally is ‘absorbed’ by the black hole, it contributes a total of MBH = (1 ⌘)mp to the mass of the black hole. 1.4. Black Hole Mass Growth in the Eddington Limit. Section 1.2 presents a derivation of the Eddington luminosity Ledd , which is the maximum luminosity that accretion is allowed to generate, without stopping further accretion onto the black hole. This luminosity increases proportionally to the mass accretion rate m, ˙ and the Eddington luminosity therefore has a corresponding maximum mass accretion rate. The goal of this section is to compute how the mass of a black hole grows, under the assumption that it accretes at the maximum rate possible. Our starting point is that each particle of mass m contributes (1 the black hole. In other words, we can write

⌘)mp to the mass of

dMBH = (1 ⌘)m, ˙ dt where m ˙ denotes the total mass accretion rate onto the black hole. We also know that each particle of mass m needs to radiate away ⌘mc2 to reach the ISCO, RISCO . After the particle reaches the ISCO it is dragged into the black hole without emitting any more energy. The total luminosity that is generated by mass accreting at some rate m ˙ is therefore L = ⌘ mc ˙ 2, 2 i.e. m ˙ = L/[⌘c ]. Substituting this into Eq 12 we have (13)

dMBH 1 ⌘L = . dt ⌘ c2

(14)

1You could say that the maximum circular velocity of a particle is c. Then gravity balance the centrifugal

acceleration when relevant.

GM r2

=

c2 , r

and r =

GM c2

which is half the Schwarzschild radius, and is therefore not

4

AST4320: LECTURE 19

We wanted to know the accretion rate for the maximum mass accretion rate, which had a corresponding Eddington luminosity. If we replace L with Ledd , we get a di↵erential expression for the mass growth rate for the maximum accretion rate: dMBH 1 ⌘ Ledd 1 ⌘ 4⇡GM mp 1 ⌘ MBH (15) = = ⌘ , 2 dt ⌘ c ⌘ c T ⌘ tE where we substituted the expression for Ledd , and where we defined the ‘Eddington time’ tE ⌘ c T /[4⇡Gmp ] ⇠ 0.45 Gyr. Eq 14 is a first order di↵erential equation with the solution ⇣1 ⌘ t ⌘ ⇣t⌘ (16) MBH (t) = MBH (t = 0) exp ⇠ MBH (t = 0) exp ⌘ tE ts where ts ⇠ 50 Myr. This is an important result. A black hole that is accreting at its maximum possible rate, increases its mass by a factor of e over ts ⇠ 50 Myr. This was used on slide 39 to argue that the existence of super massive black holes of mass MBH ⇠ 109 M at z ⇠ 7.1 [when the Universe was only 0.77 Gyr old!] is difficult to explain by having stellar mass black holes grow by ⇠ 7 8 orders of magnitude via the maximum allowed accretion rate over its entire life time.

AST4320: LECTURE 21

1. 21-cm Astrophysics The goal of these notes is to compute the change in the intensity of radiation passing through a ‘cloud’ of neutral hydrogen gas at redshift z, due to radiative transitions between the hyperfine levels (see below)of atomic hydrogen. The geometry of the problem is sketched on slide 24 of the lecture slides. 1.1. Background. Note that the ground state of atomic hydrogen is split into two hyperfine levels (see slide 7 of slides) with energy di↵erence corresponding to a wavelength of 10 ⇠ 21 cm. We denote the number density of hydrogen atoms in the higher energy state with n1 , and the number density of hydrogen atoms in the lower energy state with n0 . Throughout I use ‘hyperfine transition’ and ‘21-cm transition’ interchangeably. 1.2. Derivation of 21-cm Surface Brightness. In general the radiative transfer equation reads: dI⌫ (1) = ⌫ I ⌫ + j ⌫ ds where I⌫ denotes the intensity (with units erg s 1 Hz 1 cm 2 sr 1 ), ⌫ is the opacity (with units cm 1 ), and j⌫ denotes the volume emissivity (with units erg s 1 cm 3 Hz 1 sr 1 ). The general solution to this equation is given by (2)

I⌫ (s) = I⌫ (0) exp( ⌫ s) +

j⌫ [1 ⌫

exp( ⌫ s)].

The volume emissivity of 21-cm radiation is given by j⌫ = h⌫ 4⇡ n1 A10 (⌫), while the h⌫ opacity is given by ⌫ = 4⇡ [n0 B01 n1 B10 ] (⌫). Here, the A’s and B’s denote the Einstein coefficients of the 21-cm transition. For the 21-cm transition A10 ⇡ 2.87 ⇥ 10 15 s 1 . c2 Moreover, the Einstein B’s can be obtained from the Einstein A via B10 = A10 2h⌫ 3 , B01 =

g1 g0 B10

= 3B10 . These relations hold generally, except for

g1 g0

specifically to the 21-cm transition. We first use these expressions to rewrite the term

(3)

j⌫ = ⌫

h⌫ 4⇡ n1 A10 h⌫ 4⇡ [n0 B01

(⌫)

n1 B10 ] (⌫)

=

n1 A10 n1 A10 = . n0 B01 n1 B10 B10 [3n0 n1 ]

1

10

= 3 which applies j⌫ ⌫ :

2

AST4320: LECTURE 21

We next introduce the spin temperature Ts - also known as the excitation temperature, which is defined as ⇣ h⌫ ⌘ ⇣ h⌫ ⌘ n1 g1 10 10 (4) ⌘ exp = 3 exp . n0 g0 k B Ts k B Ts This spin temperature is an extremely important parameter in radio-astronomy. Using the spin temperature, we can recast the expression for (5)

j⌫ = ⌫

c2 3 2h⌫10

j⌫ ⌫

h⌫10 3 3n0 A10 exp exp 2h⌫10 kB T s = ⇥ ⇤ 2 h⌫10 c 1 exp A10 3n0 1 exp kB T s

as h⌫10 kB T s h⌫10 kB T s

.

Now, j⌫⌫ is expressed entirely as a function of this spin temperature. n1 10 Next we assume that h⌫ kTs ⌧ 1 ) n0 ⇠ 3. This is reasonable, during the lecture we saw that the spin temperature is a weighted mean of the gas and CMB temperature, which is much larger than h⌫k10 ⇠ 0.07 K. Under this assumption we can use that exp(x) ⇠ 1 + x + O(x2 ), and we can write (6)

3 3 k T 2 j⌫ 2h⌫10 1 2h⌫10 2⌫10 2kTs B s = = = k B Ts = 2 . 2 2 2 h⌫ 10 ⌫ c c h⌫10 c 10 k T B s

Substituting this back into Eq 2 we get (7) .

I⌫ (s) = I⌫ (0) exp( ⌫ s) +

2kTs 2 10

[1

exp( ⌫ s)]

Note that ⌫ s is the usual optical depth ⌧⌫ (which is dimensionless). Assuming that ⌫ s ⌧ 1 - which again is very reasonable - we can again use exp(x) ⇠ 1 + x + O(x2 ) and rewrite the radiative transfer equation as: 2kTs (8) I⌫ (s) = I⌫ (0)[1 ⌫ s] + ⌫ s 2 . 10

We now specify that the radiation irradiation the HI cloud is that Cosmic Micro wave background. The 21-cm transition of atomic hydrogen only a↵ects radiation near these frequencies. At these long-wavelength frequencies, the intensity of the CMB is accurately described I⌫ = 2kT 2 (see slide 28). Eq 8 then can be expressed as: (9)

T (s) = T (s = 0) + ⌫ s[Ts

T (s = 0)],

where T (s = 0) = TCMB . Note that T (s) is not really the changed temperature of the CMB. Rather, it represents the change in intensity at 10 as this radiation passes through

AST4320: LECTURE 21

3

the hydrogen gas. The changed temperature at s simply corresponds with the temperature associated with this changed intensity, using the relation I⌫ = 2kT 2 . The total change in intensity, or equivalently the temperature, is therefore (10)

Tb = ⌫ s[Ts

T (s = 0)].

The change in intensity is referred to as the ‘brightness’ temperature. It gives the brightness contrast with which we can see the hydrogen gas compared to the background CMB. In the assignment we shall compute the term ⌫ s, and that this will lead to the following brightness temperature: TCMB i , Ts where the constant C will be evaluated in the assignment. Note that this expression shows that Tb > 0 if Ts > TCMB , and vice versa. (11)

Tb = C

hT

s

AST4320: LECTURE 22

1. The Deflection Angle in the Newtonian Approximation We can derive the angle under which gravity bends the path of photons in Newtonian gravity. We assume that a particle of mass m is initially propagating at the speed of light c in some direction es . If gravity would not a↵ect the trajectory of the particle, it would pass the central object of mass M at a minimum distance b. This distance is called ’impact parameter’.

es v=c Fg

s b

r

v=c

M Figure 1. Adopted geometry for Newtonian derivation of ↵ ˆ (note that ✓ 6= ↵ ˆ . They just appear very similar, which is just coincidence). As the particle moves past the central object, gravity exerts a force along r. This cause the particle to accelerate in a direction perpendicular to es . After the interaction, the particle will have a velocity component v? perpendicular to es . The magnitude of this velocity component is Z 1 Z dv? 1 1 (1) v? = dt = a? ds, c 1 1 dt F grav

where we used that cdt = ds. The gravitational acceleration a? = ?m . The gravitational m mb force is given by F?grav = GM cos ✓ = GM . Substituting this into the Equation, and r2 r2 r 2 2 2 using that r = b + s we have Z is!1 1 1 GM b GM b h s 2GM p (2) v? = ds 2 = = , 3/2 2 c 1 (b + s ) c bc b2 b2 + s2 s! 1 1

2

AST4320: LECTURE 22

The deflection angle in the Newtonian approximation is now v? 2GM = c bc2 Einstein famously showed that in general relativity the deflection angle is twice this:

(3)

sin ↵ ˆ⇡↵ ˆ=

(4)

↵ ˆ=

4GM bc2

2. The Lens Equation The ‘Lens equation’ relates the deflection angle ↵ ˆ to an apparent shift of the location of a source, denoted with ↵, on the sky. The geometry of the lensing problem is shown on slide 14 of the lecture slides. Note that this slide has been modified from the original to avoid our confusion in the lecture about the angles. The current slide shows the correct geometry. The angle denotes the angular separation between lens and source if there were no gravitational lensing. The angle ✓ denotes their angular separation in the presence of lensing. The di↵erence - the angular shift - between these two angles is ↵ ⌘ ✓ . Note that all angles in this problem are extremely small. In this case, the distance SS 0 (denoted in green on slide 14) is given by SS 0 = ↵ ˆ DDS , where DDS denotes the (angular diameter) distance between the lens and source. The distance SS 0 is also given by SS 0 = ↵ ˆ DS , where DS denotes the distance to the source. Combining the two gives: (5)

↵=

Substituting this into the equation ↵ ⌘ ✓

DDS ↵ ˆ. DS , and reshu✏ing gives:

DDS DDS 4GM DDS 4GM ↵ ˆ=✓ =✓ . 2 DS DS bc DS DD ✓c2 Here, we used that the impact parameter b (indicated in red on slide 14) is in turn given by b = ✓DD . This equation can be written in a more compact form: (6)

=✓

✓E2 , ✓ where ✓E denotes the ‘Einstein radius’, which is defined as (7)

(8)

=✓

✓E2 ⌘

4GM DDS . c2 DS DD