Following the mathematics of Hernquist 1990

Transcription

1 Following the mathematics of Henquist 199 J. Babe Octobe 12, 21 1 Pupose of this document This guide is intended to be ead alongside the pape Henquist 199 to help explain some of the athe lengthy mathematical esults. It will not go into any depth on the physics content except whee an undestanding of the mathematics equies it. The only mateial used in addition to the pape itself is Galactic Dynamics (Binney and Temaine). This guide is meant to help you though some of moe awkwad and unfamilia pieces of maths. To stat with, though, most steps ae explained as the basic concepts ae dealt with. As the pape pogesses I concentate moe on the new maths than descibing, fo a 5 th time how to do integation by pats. 2 The model, its chaacteistics and the goal of the execise In ode to undestand the pocesses involved it is good to think fo a moment about what the goal of the upcoming calculations ae going to be. In this instance, we ae inteested in the use of this new model to find the distibution function of the spheical systems the Henquist model descibes. The distibution function is a vey impotant descipto of the system. It epesents coodinates in a six-dimensional position-velocity space (phase-space) such that at a given time t it can descibe a sub-population of stas with position x and velocity v. It will become vey impotant late. Howeve, fist we need to wok out a few basic chaacteistics of the model befoe we can stat making any pogess towads the distibution function (DF). The Henquist model in tems of a density distibution is: 1

2 ρ() = M a 1 (1) 2π ( + a) 3 whee a epesents a chaacteistic length scale of the system and M is the total mass of the system. The fist step is to tun the density distibution into a cumulative mass distibution. This involves integating the mass element dm = ρ().dv (2) ove all available volumes. This is a poblem you will have almost cetainly encounteed a numbe of times befoe and it cetainly doesn t look vey challenging. I m going to wok though the poblem in some detail to demonstate that even appaently tivial poblems, such as this, can thow up some unexpected poblems in thei solving. In this case, since the volume is spheical, M() = 2π π 2 π 2 M a 1 2π ( + a) 3.2 sin(θ)d.dθ.dφ = 2Ma.d (3) ( + a) 3 whee the standad volume element in spheical coodinates has been applied. Integation of this poceeds easily enough using the by-pats method. M 2a. ( + a).d = [2a ( + a) 2 ] M 2a. 1 2 ( + a) 2.d (4) Tidying up and integating the last tem leaves us with a a M() = [ ( + a) 2 ] [ ( + a) ] (5) This leaves us with a slight poblem that the equations poduce infinities at =. The way to combat this is to ecognise fist that = and M tot = V ρ().dv = 4π ρ() 2.d (6) This means we can e-wite equation 6 in tems of the total mass and the new limits. 2

3 a a a a M() = [ ( + a) 2 ] [ ( + a) ] = M t [ ( + a) 2 ] [ ( + a) ] (7) So, now we don t have poblems with infinities. a + a( + a) M() = M M( ) = M( ( + a)2 2 ) = M 2 (8) ( + a) 2 ( + a) 2 ( + a) 2 The pape solves fo the potential by integating Poisson s equation fo the system. A simple, equivalent method is to use the standad classical esult that F = Φ. In this instance F = GM() ˆ = Φ 2 (9) The answe hee is clealy to integate F with espect to. Howeve, the foce-field ceated by a shell of potential at adius affects space fom out to infinity so the integal wants to be between the limits of and infinity. We have ou expession fo M() so we can expess the potential thusly Φ = GM ( + a) = [ GM 2 ( + a) ] = GM ( + a) (1) The next few equations, poducing the dimensionless vaiables ae genuinely staightfowad so we shall now move on to equation [9] on the pape and concen ouselves with the Jeans equation. 3 Velocity dispesion and anisotopy Fo the next section we ae inteested in the velocity dispesion of the system. In ode to study it we will be using something called a Jeans equation which is equation [9] in the pape. The deivation of a Jeans equation is long, detailed and not actually that elevant to the undestanding of the pape. Because of this, the explanation of whee the Jeans equations come fom will be elegated to an Appendix. Howeve, it is a good idea to look in moe detail at what pat velocity dispesion plays in the system as well as the new tem, β. The anisotopy paamete, β, descibes the elative diffeences between the othogonal components of motion in the system and it does this by examining the elations between the 3

4 velocity dispesions along diffeent axes. Howeve, thee is one thing that needs examining in moe detail at this moment and that is why the symbol, v2 x, is used to denote velocity dispesion athe than mean squaed velocity. As it tuns out the answe to this nit-picking question actually ties in to ou oiginal statement of the pupose and use of the DF. Lets stat by defining a useful function ν(x) which descibes the pobability pe unit volume of finding a sta at x but without caing what velocity it has i.e. a numbe density. ν(x) = f(x, v).d 3 v (11) whee f(x,v) is ou DF. The velocity dispesion is chaacteised by both the changes in the mean velocity fom point to point and also the local spead of velocities aound the mean at those points. The mean velocity can be found by the familia fomula fo expectation values v(x) = v P x (v).d 3 v (12) whee P x (v) is the pobability distibution of velocities at the point x and can be expessed as the pobability of finding a sta at x with specific velocity v ove the pobability of finding a sta at x with any velocity. We aleady have functions fo both of those and so ou mean velocity becomes v(x) = 1 ν(x) v f(x, v).d 3 v (13) Ou velocity dispesion measue now has a desciption of the changes in mean velocity fom point to point so now we just modify it so that the spead of velocities (the vaiance of v) is being used athe than specific values. This leads us to the velocity dispesion tenso σ 2 1 ν(x) (v v) 2.f(x, v).d 3 v (14) Notice that this essentially govened entiely by the fom of the DF. Now we have a velocity dispesion measue and some idea of why being able to descibe it is impotant in ou attempt to find the DF thee just emains the question of the confusion with the m.s velocity. This is due to two things, the fact that the system is isotopic and govened by a single potential which is only a function of position and the fact that the desciption in wods we 4

5 gave fo the velocity dispesion can be witten diectly as σ 2 = v 2 v 2 (15) The fact that the system is govened by a single function Φ(x), the Hamiltonian= K + V descibes the motion. Replacing f(x, v) with v2 + Φ in equation 13 will descibe the mean 2 velocity. The integal is ove all velocities but since v is odd while ou new f(h) is even, the oveall integal is odd and will evaluate to. This means that v. v = so fom equation 15, σ 2 = v 2. In othe wods, in this scenaio the velocity dispesion and the mean squaed velocity ae equivalent. This finally bings us to undestanding the anisotopy paamete: β 1 σ2 θ + σ2 φ 2σ 2 = 1 v 2 θ + v 2 φ 2 v 2 (16) If obits ae pefectly cicula then v 2 = and β =. If the obits ae adial then v θ 2 = v φ 2 = and β = 1. Howeve, the case we ae examining with the Henquist is fo obits which ae egodic i.e. all the mean squae velocity components ae equal and β =. Thee is one last discepancy to clea up. We have so fa been using ν(x). In actual fact this is equivalent to ρ that you find in Henquist 199. The eason is that Henquist states that he chose to nomalise his density function to the total mass, not unity. 4 Velocity dispesion Now we have a good idea of what s going on with anisotopy and the concens of the pevious section we can stat making pogess with the mathematics again. We wee last consideing the Jeans equation. Now, that we have an undestanding of β we can, ionically, set it to and emove middle tem to obtain the equation we will be integating 1 d ρ d (ρ v ) 2 = dφ d theefoe v 2 = 1 ρ ρ() dφ d (17) whee the integal is taken ove those limits fo simila easons to the ealie evaluation of the potential. The poblem with this integal, and many of those to come is not that the integals ae paticulaly had, it is just that they ae vey unwieldy. This will, in fact, be the last one solved by hand as equation solves such as Maple will be used fo subsequent integals. Since 5

6 a lot of the following poof is aithmetic I will skip ove some sections to save eveyone s time. So, using equations 1 and 1 we can begin v 2 = GM 2 a 1.d (18) 2πρ() ( + a) 5 This is best solved by patial factions as the powe of 5 makes it had to find substitutions that impove mattes. So, expanding the integand gives a.( + a) = A 5 + B ( + a) + C ( + a) + + F (19) 2 ( + a) 5 Find the common factos, solve the 6 simultaneous equations (which ae mecifully simple) and the esult should be a.( + a) 5 = a 4 a 4 ( + a) a 3 ( + a) 2 1 ( + a) 5 (2) This can be integated, a tem at a time, to yield anothe seies which can be substituted into equation 18 to give us the final esult v 2 = GM 2 2πρ() {a 4. ln( + a) a 4 1. ln() [ a 3 ( + a) + 1 2a 2 ( + a) a( + a) ( + a) ]} 4 Expanding out the 1 ρ() in the font of the equation leads to (21) v 2 = GM 2 12a {12( + a)3 a 4. ln( + a ( + a)3 ) ( + a) [12 a a 2 ( + a) + 4 a( + a) + 3 ]} (22) 2 ( + a) 3 The last stetch is to take the squae backeted tem and change it into the fom equied by the esult in Henquist. This involves witing it as a single faction ove a common denominato of a 3 ( + a) 3. Subsequent expansion of the numeato into a polynomial in a and lets you emove the facto of a 3 fom the denominato to leave 12( a )3 + 42( a )2 + 52( a ) + 25 ( + a) 3 (23) 6

7 which we can substitute into equation 22 to give the final esult v 2 = GM 2 12a {12( + a)3 a 4. ln( + a ) ( + a) [12( a )3 + 42( a )2 + 52( ) + 25]} (24) a The last thing to do is to apply the limits of integation of infinity. Evaluating at infinity gives us which effectively eveses the sign on the function subtacting it fom. 5 Enegies Fom hee on out the functions become much too tiesome to do by hand. Fo example, look at the definition of T() given in the pape. To evaluate even the fist tem, A. 2. ln( + a) equies multiple instances of integation by pats. As such, Maple s inteactive function solve will be used which should speed things up consideably. The definition of T() seen hee follows fom simple enough logic. T (v) = 1 2 m.v2 T () = 1 2 M() v 2 (25) The only tick is to notice that, since ou system is egodic, all components of the velocity dispesion ae equal and so v 2 = v 2 + v 2 θ + v 2 φ = 3 v 2 (26) Then the fomula can be expessed in the usual spheical polas to poduce T () = 6π ρ v 2 2.d (27) Substituting in fo ρ and v 2 gives us the function we need to solve. To find a solution using Maple what we fist do is define the functions Maple will be woking with: > ho := M*a/((2*pi)**(+a)^3) and so on fo the othe given functions. Finding 7

8 > Tintegand := 6*pi*ho*vsquaed*^2 will poduce a lage, ungainly function that Maple will be unable to integate. The easiest shot-tem solution is to beak up the function into smalle pieces and let Maple integate each pat sepaately befoe adding them all up at the end. The downside to this appoach is how tedious and inflexible it is: any change in an ealie pat of the function will need to be copied manually down evey time it appeas. The final step is to ty and expess the esult of this pocess into something that esembles the desied functional fom. The esult in Maple fo the above is a polynomial with 12 vey simila tems in it. By way of example, the simplification of the tems in ln(f()) poceeded as follows [ GM 2 ln( a ) a + GM 2 3 ln(1 + a) + GM 2 ln(1 + a) a 4 a GM 2 ln( + a) ] (28) a = GM 2 a [( a )3 ln(1 + a/) ln(1 + a ) ln(a) + ln(1 + a )] (29) The constant tem will, by definition, be emoved when the limits ae applied and thus this section of the esult is the same as that found by Henquist. The same applies fo the othe tems. Potential enegy is much easie and can be solved without Maple. The total potential enegy is given by the usual integation in spheical coodinates Ω tot = 1 2 V ρ()φ().dv = 2π ρ()φ(). 2 d = GM 2 a.d (3) ( + a) 4 whee the pefacto of 1 is to pevent counting of the same paticles (o, moe exactly, the 2 same mass-containing volume element) twice. Integation by pats will emove the facto of and, afte a little wok 1 [ 3( + a) 3 ] [ 6( + a) 2 ] = [ ] [ 1 6a ] = GM 2 2 6a (31) The final piece of this section is to find the escape and cicula velocities. The equations ae given in the but it s woth a second o two to emind ouselves of whee they come fom. The cicula velocity comes fom equating the centipetal foce with the gavitational attaction thus 8

9 mv 2 c = GM()m GM() v 2 c = (32) and the escape velocity is the velocity equied to have exchanged all kinetic enegy fo potential at = 1 2 mv2 e = mφ() v e = 2φ() (33) Simple substitution and eaangement yields the final esults. 6 Back to the distibution function We have aleady dealt with the distibution function in ou discussion of velocity dispesion and anisotopy and it is hee that Henquist s choice to nomalise the function to the mass, athe than unity, is found. In those discussions we also dealt with the fact that since the system was spheical and isotopic (i.e. descibed by a single-valued adial potential) the adial coodinate could be mapped diectly to enegy. In ode to find this new function, f(e), we need to biefly go back to ealie in the pape whee I skipped ove the dimensionless vaiables. Fist, we need to undestand how, mathematically, this new function fits in with what we aleady know and the little paagaph befoe equation [17] in the pape leaves out a lot of impotant easoning. To begin with, ecall the Hamiltonian we used in the anisotopy section and how we said that the distibution of paticles was entiely dependent on that Hamiltonian. Well, that Hamiltonian was H = K + V = 1 2 v2 + Φ(x) (34) Now, lets deal in less absolute tems by using elative enegies instead, so now we define the potential elative to some constant Φ. This means that the enegy of a given sta will still be H, but now it will be H below the backgound potential Φ. So, now we have Ψ Φ Φ and ɛ H + Φ = Ψ 1 2 v2 (35) In ode to be consistent, we also apply the limit that as, Ψ Φ i.e. the potential at an infinite distance fom the system is the same as the natual backgound potential. 9

10 So, in ou new tems we can stat by e-witing equation 11 in tems of ou elative enegy tems. Ou f(x, v) is given by ou new enegy function f(ɛ) and when we substitute it in we now have ν() = f(ψ 1 2 v2 ).d 3 v = 4π f(ψ 1 2 v2 ).v 2 dv (36) v Fom equation 35 we can find dɛ dv left with = v. When we collect tems and substitute fo v we ae ν() = 4π Ψ f(ɛ) 2(Ψ ɛ).dɛ (37) whee the 2(Ψ ɛ) tem is equivalent to v and, as a fun check fo the physicality of what we ae doing, if you substitute H = K + V back into Ψ and ɛ then v = 2K which is what we would expect. The shap eyed among you may also have noticed that a minus sign went missing somewhee in equation 37 when we substituted in fo v. This is because kinetic enegy, which, as we have seen, is what we ae dealing in, only caes about the magnitude of velocity, not the diection. Since we have aleady discussed the fact that the potential completely descibes the adial component of the distibution in a spheical system we can call equation 37 a function of Ψ. At this point we ae still tying to get to the equation cited by Henquist. The next step is to diffeentiate both sides w..t Ψ. This gives us 1 d Ψ f(ɛ).dɛ (38) 8π dɛ Ψ ɛ This just so happens to be a kind of equation known as an Abel integal. f(x) = x g(t) (x t) sin(πα).dt g(t) = α π t [ dx (t x) 1 α df dx + f() ] (39) t1 α I am not going to go into the physical intepetation of these tansfoms hee as it suffices to teat it as a puely abstact and convenient esult that allows us to invet elations of the same fom as equation 38 so that we can say This esult is known as Eddington s fomula. f(ɛ) = 1 ɛ [ dψ d 2 ν 8π 2 ɛ Ψ dψ + 1 dν ] (4) 2 ɛ dψ Ψ= 1

11 The second-ode integals ae due to ou athe awkwad choice of vaiables to match the fom of the Abel fomula 39. Apat fom that, the above is obtained by a diect substitution fom equation 38 into equation 39. We now have a geneal fomula elating the DF to functions of potential. In ode to specify a esult fo the Henquist model we need to use the equations fo potential that wee woked out at the stat of the pape. If you look at equation [8] in the pape you ll see that it gives a elationship between the density and dimensionless potential that follows simply enough fom the definitions of the dimensionless quantities. Remembe that, due to Henquist s choice of nomalising to a mass value, ρ() M.ν(). We can use the dimensionless vaiables defined in equations [6-8] in the pape to state that ν() = Mρ() = ρ 2πa 3 = 1 2πa 3 Ψ4 1 Ψ (41) We need to diffeentiate this function twice in ode to be able to substitute into equation 41. The fist ode of diffeentiation is easy enough and yields dν dψ = 1 2πa 2 GM Ψ 3 (4 3 Ψ) (1 Ψ) 2 (42) Evaluating the above at Ψ = gives zeo. The pefactos come fom the fact that Ψ aψ GM. Now we diffeentiate again d 2 ν dψ 2 = d d Ψ ( Ψ 3 (4 3 Ψ)(1 Ψ) 2 ) = a 2 Ψ 2 (6 8 Ψ + 3 Ψ 2 ) GM (1 Ψ) (43) 3 Now we can finally substitute back into equation 4 to bing us to the final step. We need to integate f(ɛ) = 2 ɛ (2π) 3 (GM) 2 a 2 Ψ 2 (6 8 Ψ + 3 Ψ 2 ) (1 Ψ) 3 dψ ɛ Ψ (44) Using Maple to solve this equies, again, beaking it down into pieces as we have done befoe. We can easily expess Ψ in tems pf Ψ and so the integation can poceed. Eventually, we aive at the fomula given in the pape 11

12 1 2(2π)3 (GMa) 1.5 ɛ (1 ɛ) [(1 2 2 ɛ)(8 ɛ2 8 ɛ 3) + 3 ɛ ɛ(1 sin 1 ] (45) ɛ) whee ɛ q 2. In this espect the pevious equation can be thought of as a function of E instead of ɛ. The limiting cases discussed in the pape ae found by vaious Taylo expansions. Fo example, sin 1 (x), x ; sin 1 (x) (46) This now allows fo the (1 q 2 ) tems to be collected and a binomial expansion to be applied (1 q 2 ) 2 1 2q 2 + 3q 4 4q (47) Expanding the othe tems in the equation and collecting tems yields the function fo q. A simila pocess fo the othe limit povides that answe but the maths is equivalent i.e. using limiting cases fo sin 1 (x) and seies expansions. The only significant change is the addition of a π 2 tem fom the sin 1 (1). Remembe that we initially only cae about the fist few tems in the seies since inceasing powes of any numbe n 1 will tend to. So, just isolate the tems that poduce factos of up to q 4 and discad the est. Do the same fo all the seies and the coefficients should match up fo those tems. 7 Density of States We ae eaching the end and most of the mathematical tools that we will need to complete the pape have aleady been used at some pio point, such as Abel fomulae which will appea once moe. As such, I shall not into quite as much detail since the aim of this document is to tell what mathematical tools wee used, not to suffocate unde evey single detail. The next function we ae inteested in is the density of states, g(e). In this context the density of states descibes the volume of phase-space occupying a cetain enegy state. So, the mass of stas of a given binding enegy is the poduct of the volume of phase-space occupying a given enegy state and the mass occupying a given phase-space volume given as a function of enegy, o equation [22] in the pape. The density of states, as just descibed, is given mathematically by 12

13 g(e) = δ(h E)d 3 xd 3 v (48) This integal is then evaluated ove both anges up to some value E, o the adius and velocity that coespond to E. This means that the integal looks like g(e) = (4π) 2 (Φ=E) 2.d v 2 δ( 1 2 v2 + Φ E).dv (49) The tick hee is to notice that, if you make a change of vaiable to ɛ = v2 then the second 2 integal, being a delta function, can only be evaluated at one point, whee ɛ = (Φ E). The substitution and this fact mean that the integal is now adius(φ=e) g(e) = (4π) 2 2 2(E Φ).d (5) To solve this the way suggested in Binney and Temaine, we need to make anothe change of vaiable. Cuently, we ae integating between the cente of the system and a adius at which Φ = E. We can expess this instead in tems of anothe coodinate we shall call X. Stat fom the definition of potential in equation 1 and eaange to make the subject = GM Φ GM a; Let X be aφ GM (Φ = E) = E Substituting in = ax a gives us the integal we will be evaluating a = Aa a (51) A g(e) = (4π) 2 (ax a) 2 2E( GM Aa 1 GM Xa ).a.dx = (4π)2 a 3 2E A 1 (X 1) 2 ( A X 1).dX (52) This solves similaly to equation 44. Using Maple and putting in the limits will poduce (4π) 2 a 3 2E[ A 1( A2 8 5A ) + A 8 (A2 4A + 8) cos 1 (A 1 2 )] (53) The fomula in the pape is equivalent but equies substitutions fo A = q 2 and ν g. The est of this section is teading familia gound. g(e) has a vey simila fom to f(e) so the consideation of limiting cases is pactically identical also. Consequently, thee isn t much 13

14 point to futhe detail. Fom pape equation [22] it should be clea what we now need to do to find dm. In ode to find the limiting cases of dm you simply find the poduct of the dɛ dɛ same limited cases of g(e) and f(e). This pocess is long and amounts to much collection of tems and simplification. You may find it easie to multiply the two geneal equations togethe and expand the tems befoe applying the appoximations so make it a little easie to wok out what happens to the seies expansions. The steps in between wee caied out using equation solves and so detailing them at length hee would not actually help you undestanding of how to pocess these functions. If you eally must know how one would solve this with pen and pape then consult Wolfam Alpha (see Acknowledgments). 8 Reintoducing anisotopy The final stage in ou calculation is to eintoduce the possibility that the system has vaying isotopy. We ae going to state that the DF depends now on anothe function, Q, which is Q = ɛ L2 2 2 a (54) 2 a is some scale ove which the anisotopy becomes appaent i.e. if 2 a = then Q ɛ and we get the model we have been woking with up to this point as the anisotopy does not occu within. We shall also define a set of pola coodinates fo dealing with velocity as follows v = v cos η; v θ = v sin η cos ψ; v φ = v sin η sin ψ (55) The geomety of η and ψ becomes appaent fom the consideation that fo η =, v = v. η is the angula diffeence between the adial component of velocity and the adial. Similaly, ψ is the angle made by the velocity vecto and a cicumfeential ing. So, using these coodinates we can now wite this as Q = (Ψ 1 2 v2 ) ( (v. sin η)2 ) = Ψ 1 2a 2 2 v2 (1 + 2 sin η) (56) a 2 We ae going to stat by woking out a familia function, the numbe density ν() using ou new vaiable, Q, and the pola coodinates 14

15 ν() = π f(ɛ, L).d 3 v = 2π sin η.dη Ψ 2(Ψ Q) f(q).dq (57) [1 + ( a ) 2 sin 2 η] 3 2 We can chose to eaange this equation to poduce an integal in η which happens to solve as a standad integal by substitution So, we have int π sin η [1 + ( a ) 2 sin 2 η] 3 2.dη = ( a ) 2 (58) (1 + ( a ) 2 )ν() = 4π Ψ f(q) 2(Ψ Q).dQ (59) Notice that the ight hand side of this equation coesponds exactly to equation 37. We have aleady solved this befoe though the use of Abel integals. The only diffeence is that now we have a modified density function ν Q () (1 + ( a ) 2 )ν() (6) We can now substitute ou new functions Q and ν Q () f(ɛ) = 1 Q [ dψ d 2 ν Q 8π 2 Q Ψ dψ + 1 dν Q ] (61) 2 Q dψ Ψ= This can be solved in pecisely the same way as equation 45 using the new vaiables. Remembe that the pape uses ν() ρ() and that q is a constant tem. Fo the steps to M get fom the Abel fomula above to the esult in the pape just look back at how that was accomplished ealie as the methodology is the same. The final esult woth quoting hee is that the model we employed hee by the assumption in equation 54 is known as am Osipkov-Meit model and they have an anisotopy paamete given by β() = ( a ) 2 (62) This poduces egodic behaviou fo a and adial behaviou fo a. 15

16 This bings us to the end of the section of the pape I am going to cove. Hopefully this has allowed you to undestand the mathematics that undelies the deivations in the pape as well as the concepts and physics in it too. 9 Acknowledgments The majoity of the above is taken fom the detailed desciptions of the same by Binney and Temaine in Galactic Dynamics. This document mainly seves to gathe the infomation in Galactic Dynamics and e-examine it in an ode that is most applicable to the study of the Henquist model. Seveal equations wee solved using a combination of Maple 13 and Wolfam Alpha.I would ecommend using Maple 13 to this kind of wok due to its much moe eadable inteface but equally, if you ae inteested, Wolfam Alpha can tell you how it went about solving the integal. So if you need convincing that it is actually physically possible then I would ecommend giving a look. If you have any questions about this document o you spot any mistakes feel fee to me at jab22@st-andews.ac.uk. 16