Cluster X-ray Gas Galaxies only contain about % of the baryons present in rich clusters. The other 8% resides in the intracluster medium. These baryons can find their way into the intracluster medium in 3 ways: They can fall into the cluster as it forms They can be blown out of small galaxies by winds They can be swept out of larger galaxies by ram pressure They can be ejected from galaxies by the jets and winds of AGN Let s consider the first three of these mechanisms one at a time. The first is simple. A rich cluster contains M 1 15 M of material within a 1 Mpc radius. If an atom of mass m H falls into this cluster from infinity, then its thermalization temperature is given by GMm H 1 m Hv 3 kt = T 18 K 1.1) In the absense of cooling, this gas will emit x-rays. 1
Galactic Winds [Mathews & Baker 1971, Ap.J., 17, 41] Consider an elliptical or spheroidal) galaxy of luminosity, L, and mass-to-light ratio, Υ. The stars in the galaxy are losing mass at the normal rate of Ṁ ISM 1 11 M per year per unit solar luminosity of the galaxy. Because the stars are moving isotropically within this galaxy, the gas will thermalize at a temperature given by GΥLm H 1 m Hv 3 kt ISM 1.) In other words, T ISM 1 3 m H k GΥL 1.3) Now assume that supernovae occasionally go off in the galaxy, and eject Ṁ SN per year per unit solar luminosity into the ISM/null. This matter will likely be thrown off at v SN 1, km s 1 and thermalize with an efficiency, ǫ. In other words, 1 ǫ of the bulk motion energy of the ejecta will be radiated away.) The temperature of this ejecta, if thermalized, would be or ǫ 1 m Hv SN 3 kt SN 1.4) T SN = ǫ 1 m H 3 k v SN 1.5) When this much hotter) material mixes with the ambient ISM, it heats it. The final temperature of the system will become T new = ṀISMT ISM + ṀSNT SN M ISM + ṀSN 1.6)
Explicitly, this is T new = 1 3 m H k Ṁ ISM GΥL + ṀSN ǫvsn Ṁ ISM + M 1.7) SN Next, we consider the escape velocity for the system, v esc = GM 1.8) which means that the escape temperature is given by 1 m Hvesc = 3 kt esc = T esc = m H 3 k GΥL 1.9) Now we ask the question: when is the ISM temperature greater than the escape temperature? 1 m H 3 k Ṁ ISM GΥL + ṀSN ǫv SN Ṁ ISM + ṀSN > 3 m H k GΥL 1.1) which, with a little math, becomes L wind < ǫv SN GΥ Ṁ SN M ISM + ṀSN 1.11) For reasonable numbers, such as ǫ.1, Ṁ ISM 1 11 M, Ṁ SN 1 13 M both per year per solar luminosity), Υ 1, and 1 kpc, L wind 1 1 L. Small galaxies therefore cannot hold onto their interstellar medium. Once a supernova goes off, it will heat the ISM, and create a galactic wind sometimes called a Mathews-Baker wind). 3
Galaxy Sweeping [Gunn & Gott 197, Ap.J., 176, 1] Interstellar matter may be swept out of a galaxy by ram pressure, as the classic calculation of Gunn & Gott shows. To repeat their analysis, consider a spiral galaxy moving through a cluster that already has a low-density x-ray gas. Also, following their prescription, assume that most of the mass of the spiral resides in its disk. This is not true, of course, but since the paper was written before the discovery of dark matter halos, we ll go along with it.) Now let s approximate a spiral galaxy as an infinite plane, with surface mass density σ. From freshman physics, the force required to lift a particle of mass m from the surface of the galaxy to infinity is r θ h F m = = G πrσ r + h ) πgσ h cosθ dr = πrσ G r + h ) h dr r + h ) 1/ r r + h ) 3/dr = πgσ 1.1) Now let the surface density of gas in the plane of the spiral galaxy be σ gas. Since the mass of gas contained within a surface area da is related to the surface density by dm = σ gas da, the pressure required to strip gas from the galaxy is P = F dm da = πgσ σ gas 1.13) 4
Now when a galaxy moves through a cluster, the gas in the disk will feel a ram pressure due to x-ray gas. If the density of the x-ray gas is ρ x, then, from fluid mechanics, the ram pressure is P = ρ x v 1.14) Consequently, the gas inside the galaxy will get stripped out by the ram pressure when ρ x v > πgσ σ gas 1.15) This is the Gunn-Gott criterion for sweeping of gas from spirals. When does this occur? Well, there are 1 11 stars in the Milky Way, and the size of the Milky Ways disk is kpc. If a typical Milky Way star has one solar mass, then the approximate stellar surface density is σ 1 11 /π.3 gm cm. Also, if the density of particles in the local interstellar medium is 1 atom cm 3 and the thickness of the disk is pc, then the surface density of gas is σ gas 6 1 atoms cm. Plugging these numbers into 1.15), and picking a typical velocity for a galaxy moving through a cluster v 13 km s 1 ), yields a requisite x-ray gas density of 5 1 4 atoms cm 3. This calculation underestimates by up to an order of magnitude), the density needed strip a galaxy, since it ignores the presence of dark matter. However, the qualitative result is the correct: it does not take much of an intercluster medium to strip spiral galaxies of their gas. When this occurs, star formation in the spirals will rapidly come to a halt since they ll be no gas to form stars), and the galaxy will presumably) change into a lenticular. At the same time, the increased amount of intracluster material will cause the next spiral galaxy to be stripped that much easier. As time goes on, the amount and x-ray luminosity) of the intracluster medium can be increased substantially. 5
The X-ray Gas What will be the state of the x-ray gas? Given that the galaxy velocity dispersion in a cluster is km s 1, a typical virial temperature of the gas will be T 1 8 K. Also, for reference, a typical density for intracluster gas is n e 1 3 cm 3. TIMESCALE FO EQUIPATITION Let us first ask how long it takes an electron or proton) to share its energy with its surrounding and come to thermal equilbrium. It is fairly straightforward to show that the mean free path of a charged particle in a plasma is λ = 33/ kt) 4π 1/ ne 4 lnλ 1.16) where e is the charge of the electron or proton), n is the number density of particles and Λ is related to the Debeye length, i.e., the ratio of the largest to smallest impact parameters. The derivation of this equation is straightforward; you will undoubtably see it, or the equivalent for a gravitational plasma of stars, in another class). Numerically, Λ is { ) T n ) } 1/ e Λ = 37.8 + ln 1 8 1 3 cm 3 1.17) Note that this is a log quantity, hence it is virtually independent of n e and T. When we plug in the numbers, the mean free path of an electron or ion) is λ e λ i 3 kpc ) T n ) 1 e 1 8 1 3 cm 3 1.18) This is almost the length of a galaxy! 6
Now, a reasonable estimate of the equipartition time is t eq = λ/ v rms = λ/ 3kT m 1.19) You can do the computation in a lot more detail, but the result will stay the same, to a factor of a couple.) Thus, from 1.16) t eq = 3m1/ kt) 3/ 4π 1/ ne 4 lnλ 1.) For proton-proton interactions m = m p ), this works out to be t eq = 1.4 1 7 yr Tp 1 8 ) 3/ n p 1 3 cm 3 ) 1 1.1) while for electron-electron interactions m = m e ) t eq = 3.3 1 5 yr Te 1 8 ) 3/ n e 1 3 cm 3 ) 1 1.) Note that these timescales are very short cosmologically speaking). Consequently, any new matter introduced into the intracluster medium will shortly come into equipartition with that medium. 7
TIMESCALE FO COOLING The next question to ask concerns the cooling timescale for the x-ray gas. In other words, how long with it take the temperature of the gas to decrease by 1/e. / ) 1 dt d lnt t cool = T dt = 1.3) dt Because most of the ions in an x-ray gas are either hydrogenor helium-like, there are no low-lying levels to enable collisional cooling to work. Thus, the principal cooling mechanism is thermal Bremstrahlung. ecall that energy emitted by free-free emission is Thus, ǫ ff = 3πe6 3m e c 3 = 3πe6 3hm c c 3 ) 1/ π Z n i n e g ff e hν/kt dν 3m e kt ) 1/ πk ḡ T 1/ n e 3m e i Z i n i 3 1 7 n p T 1/ ergs cm 3 s 1 1.4) ǫ ff = de dt = 3 nkdt dt = 3 1 7 n p T 1/ ergs cm 3 s 1 1.5) which means that t cool = kt 1/ 1 7 n =. 11 yr ) 1/ T 1 8 n e 1 3 cm 3 ) 1 1.6) This is more than a Hubble time. So, unless we re talking about gas in the very center of the cluster where the density may be high), the x-ray gas will not be cooling significantly. 8
TIMESCALE FO PESSUE WAVE POPAGATION Finally, let s ask how long it takes a pressure wave to cross the cluster and distribution any inhomogeneity over the entire system). This number is simply the size of the cluster divided by the sound speed, i.e., / γkt t p = D/c s = D 1.7) µm H where γ is the ratio of specific heats γ = 5/3) and µ H is the mean molecular weight of the order of one). Plugging in the numbers gives ) 1/ ) T D t p = 6.5 1 8 yr 1 8 1.8) Mpc Once again, this is short compared to a Hubble time. Thus, the gas will be in pressure equilibrium with its surroundings. In other words, hydrostatic equilibrium will hold. 9
X-ray Gas and Hydrostatic Equilibrium [Fabricant et al. 198, Ap.J., 41, 55] Given that an x-ray gas is in thermal and pressure equilibrium, it is fairly straightforward to use the distribution of gas to derive the mass distribution of the cluster. Begin with the equation of hydrostatic equilibrium and the ideal gas equation dp dr = GMr) r ρ 1.9) P = ρ µm H kt 1.3) If you take the derivative of 1.3) with respect to r, then dp dr = k { T dρ } m H dr + ρdt = GMr) dr r ρ 1.31) Now if we multiply through by r/ρ, and take T outside the bracket, then dp dr = kt { r dρ µm H ρ dr + r } dt = GMr) T dr r = ktr) { d lnρ µm H G d lnr + d lnt } = Mr) 1.3) d lnr r So Mr) = ktr) µm H G { d lnρ d lnr + d lnt } r 1.33) d lnr Note that if the gas is truly isothermal due to the short equipartition timescale), the last term in the bracket disappears, making the mass a function of only one derivative. 1
Converting from Projected adius to True adius Unfortunately, equation 1.33) is of limited use. In particular, the density, ρ, is not an observable quantity; what is measured is the projected x-ray emissivity, I ν ), on the plane of the sky. However, it is possible go from projected intensity to three-dimensional emissivity, ǫ ν r) via an Abel integral. If the emitting x-ray plasma is spherically symmetric, then the flux one observes at any projected position in the cluster is z r I ν ) = ǫ ν r)dz = ǫ ν r)r r ) 1/dr 1.34) Now let s introduce a dummy variable ξ which takes the place of r), and multiple each side of the equation so that d I ν ) ξ ) = 1/ ǫ ν r)rdr r ) 1/ d ξ ) 1/ 1.35) Now let s integrate both sides over d from ξ to infinity ξ I ν )d ξ ) = d 1/ ξ ǫ ν r)rdr r ) 1/ ξ ) 1/ 1.36) Now comes the tricky part. Let s switch the order of the integration. As the figure below shows, in order to do this, we must 11
also change the limits of the integration. In the first case, r is integrated from to infinity parallel to the x-axis) and is integrated from ξ to infinity. In the second case, to cover the same area, goes from ξ to r, and is taken from ξ to infinity. Thus ξ I ν )d ξ ) 1/ = ξ ǫ ν rdr r ξ d r ) 1/ ξ ) 1/ 1.37) r= =r ξ ξ r ξ r Now the second integral on the right hand side can be evaluated. If we substitute η = ξ r ξ 1.38) For the integral is over, r is held constant, so d r ) 1/ ξ ) 1/ = 1 1 r r ξ )ηdη r ξ ) 1/ 1 η ) 1/ r ξ ) 1/ η dη = 1 η ) = π 1.39) 1/ So I ν )d ξ ξ ) = π ǫ 1/ ν r)rdr 1.4) ξ or, if we differentiate, ǫ ν ξ) = 1 πξ d dξ ξ d I ν ) 1.41) ξ ) 1/ 1
Converting from Density to Flux The projected x-ray flux distribution can be converted to the true x-ray flux distribution via the Abel integral. To go from flux to density, consider that the x-ray emissivity from thermal bremsstrahlung is ǫ ff = 3πe6 3m e c 3 Consequently, ) 1/ π Z n e n i g ff e hν/kt dν = n 3m e kt eλt) 1.4) ǫ ff = n e ΛT) = ρ ΛT) 1.43) where the angle brackets represent an integral over a energy bandpass from ν 1 to ν ). Thus d ln ǫ ff d lnr = d lnρ d lnr + d ln ΛT) d lnr 1.44) Now we solve for ρ and re-write the derivative of ΛT), so d lnρ d lnr = 1 d ǫ ff d lnr 1 d ln ΛT) d lnt d lnt d lnr 1.45) If we then substitute this in 1.33) we get Mr) = ktr µm H G or µm HGMr) ktr { d lnt d lnr + 1 d ln ǫ ff d lnr 1 d ln ǫ ff d lnr = d lnt d lnr 1 d ln ΛT) d lnt { 1 1 d lnt } d lnr 1.46) } d ln ΛT) d ln T 1.47) 13
Note the terms in the above equation. The derivative of ΛT) details how the free-free emission in an x-ray band changes with temperature; it is well known from basic physics via 1.4). Also known is the derivative of the x-ray emission with radius; this comes from the Abel inversion of the observed surface brightness. Therefore, given a function for Tr) and a boundary condition usually the central temperature or pressure at infinity), this ordinary differential equation can be solved for mass. 14
Isothermal X-ray Gas If an x-ray gas is in hydrostatic equilibrium, then dp dr = GMr) r ρ = dω dr ρ 1.48) where Ω is the local potential. In addition, if we take the derivative of the ideal gas law, dp dr = kt dρ 1.49) µm H dr Putting these two equations together yield kt d lnρ µm H dr = dω dr 1.5) Now consider: the x-ray gas isn t the only thing effected by the cluster potential. The galaxies are also moving about. Let s define a characteristic temperature of the galaxies, T gal, via 1 µm H v = 3 kt gal 1.51) where v is the mean square velocity of the galaxies. Note, however, that this is not the observed velocity dispersion: since we only measure radial velocities, all we can see is one component of the motion. The observed velocity dispersion, σ gal, is related to the true dispersion by v = 3σ gal. Of course, this is only true if the orbits are isotropic v x = v y = v z ) and the cluster is virialized.) So, from 1.51) 1 µm H3σgal = 3 kt gal = T gal = µm Hσgal k 1.5) 15
If the galaxies are in virial equilibrium, then this temperature is related to the potential by K + Ω = = σ gal = Ω 3 = T gal = µm H 3k Ω 1.53) But how does this temperature compare to that of the x-ray gas. One can imagine that the x-ray gas may be cooler than the galaxies, due to its Bremstralung radiation, or the inclusion of nonvirialized, possibly foreground or background galaxies in the velocity dispersion measurement. Or, one can hypothesize that the gas is hotter, due to infall. ecall that the kinematic energy an object has from the virial theorum is half that of a particle dropped from infinity.) To parameterize the problem, let s define β as the ratio of the virial temperature of the galaxies to the x-ray temperature of the gas, i.e., β = µm H kt σ gal 1.54) Since both the gas and the galaxies must feel the same potential, then, from 1.5) and 1.5) kt d lnρ x µm H dr = σ gal d lnρ gal dr = dω dr 1.55) where ρ x is the density of the x-ray gas, and ρ gal the density of the galaxies. e-writing this, we get d lnρ x dr µm H d lnρ gal kt σ gal dr If we trivially integrate over r, then = β d lnρ gal dr 1.56) ρ x = ρ β gal 1.57) 16
The Total Mass of an X-ray Cluster If the gas is isothermal, then its radial distribution should be that of an isothermal sphere. In that case, a reasonable approximation for the density distribution over the inner core radii is ρ ρr) = 1.58) 1 + r ) 3/ where ρ is the central density, r is r/r, and r is the core radius. This is sometimes called a modified Hubble Law. The modified Hubble Law departs from an isothermal sphere at large radii. Nevertheless, it is common to extrapolate this law to infinite radius, and derive the total x-ray gas mass of the cluster in terms of the core radius and central density. But note: this only works if β > 1; otherwise, the mass is infinite, even with this approximation.) The calculation is fairly straightforward. One can measure r by counting galaxies and fitting the radial distribution to an isothermal profile or a Hubble law). One can also measure the central density from the observed amount of x-ray emission. So M x = If we let x = r, then = 4πρ r 3 M x = 4πρ r 3 = πρ r 3 4πr ρ dr 1 + r/r ) 3β/ ) Γ 3 r 1 + r ) 3β/ d r 1.59) 1 x1/ 1 + x) 3β/ dx ) ) Γ 3β 1) Γ 3β ) 17
or = π 3/ ρ r 3 M x = 3.15 1 1 M where Γ is the gamma function. Γ 3β 1) Γ 3β ) n ) 1 3 cm 3 ) 1.6) r.5 Mpc ) ) Γ 3β 1) Γ 3β ) 1.61) 18
β and the Flux Distribution of an X-ray Cluster One can determine β by directly measuring the temperature of the intracluster medium and comparing it to the velocity dispersion measured for the galaxies. However, an alternative method, which is often used is to integrate 1.34) over the line-of-sight, and compare the result to the observed x-ray surface brightness. From 1.34) this surface brightness should be ne ) n i I) = n e n i ΛT)dz = ΛT) r)r n i r ) 1/dr 1.6) If we substitute for n i using 1.57) and 1.58), and let x = r )/r + ), then I) = n ΛT) ne n i ) rdr r ) 1/ 1 + r /r ))3β = n r 6β ne n i ) rdr r ) 1/ r + r ) 3β = n r 6β = n r6β = π ne ne ne n i n i n i ) 1 r + ) 3β+ 1 ) r + ) 3β+ 1 ) { n r r 1 + r Γ 1 ) } 3β+ 1 x 1/ 1 + x) 3β dx ) Γ 3β 1 Γ 3β) ) Γ 3β 1 Γ 3β) ) 1.63) 19
When one fits the x-ray surface brightness for many clusters in this manner, one finds β.65. On the other hand, when one measures the galactic velocity dispersion and compares that number to the gas temperature as determined directly by fitting x-ray spectra, then β 1.. Either a) The galaxies and/or the hot gas are not isothermal. b) The galaxies are not in isotropic orbits. c) The galactic velocity dispersion is overestimated by the inclusion of contaminating foreground and background galaxies. d) The x-ray gas and galaxies do not see the same potential. [Castillo-Morales & Schindler 3] [Leccardi & Molendi 8]