Potential due to thin disk

Transcription

1 Stellar Dynamics & Structure of Galaxies hanout #9 Potential ue to thin isk H << Coul use above an write own potential as sum over rings. But metho oes not work at r = a. Instea use cylinrical polar,,φ,z), coorinates Expect Φ Φ, z) an Φ, z) = Φ, z) by symmetry. Outsie isk 2 Φ =. 1 Φ ) + 2 Φ z 2 = We solve this by separation of variables, letting Φ,z) = J)Zz) Zz) 1 J) ) 1 J ) = J }{{} 1 Z function of so 2 Z + J) 2 z 2Zz) = 2 Z z 2 }{{} function of z = k 2, say z 2 k2 Z = 9-1) Z = A expkz) + B exp kz) an 1 J ) + k 2 J) = 9-2) 47

2 We woul quite like Φ, ) an Φ, ) to be zero, so is the appropriate solution for Zz). Zz) = A exp k z ) The equation 9-2) is the efining equation for a Bessel function. These are the analogues of sines an cosines now for cylinrical as oppose to linear problems e.g. rum beats). So while similarly 2 y z y = has solutions sinkz), coskz)9-3) 1 s y ) + k 2 y = s s s has solutions J ks),y ks) 9-4) which you can look up in e.g. Abramowitz & Stegun Hanbook of Mathematical Functions. Examples are given on the next page. 48

3 Note that as x J x) 1 an Y x). More generally the equation 1 s s s y s ) + k 2 ν s 2 ) y = has solutions J ν ks),y ν ks), so we get while a whole family of Bessel functions characterize by the inex ν. Also there are moifie Bessel functions where k ik 2 y z 2 k2 y = has solutions sinikz), cosikz) or expkz) Similarly 1 s s s y ) k 2 y = s I ks) K ks) an 1 s s y ) ) k 2 + ν2 y = s s s 2 I ν ks),k ν ks) see Abramowitz + Stegun Hanbook of Mathematical functions An we can take this even further. By analogy with Fourier transforms where sin, cos form the basis, we have J,Y Hankel transforms. Given a function gr), then the Hankel transform of g is gk) = gr)j ν kr)rr 49

4 an the inverse transform is: gr) = gk)j ν r)kk [ look these up in books of Hankel transforms! ] eturning to the axisymmetric plane istribution, we have 9-1) Zz) = exp k z ) 9-2) J) = J kr) choose J to get Φ finite at = Let k > then Φ k,z) = Ce kz J k) z > Ce kz J k) z < This is true k >, but a specific k for each Φ k. General potential k Φ k Φ,z) = fk)e k z J k)k 9-5) Here fk) is a weighting function, corresponing to the C values in the sum. So what we nee to o for a particular mass istribution is fin fk). If we are going to relate it to a mass istribution, the next thing we shoul o is look at the z = plane, i.e. the region we have neglecte so far since we have taken 2 Φ = an so consiere regions outsie the plane. Note that Φ k is continuous across z = but Φ k is not ue to z epenence.that is where the mass is, so that is not a surprise. 2 Φ k = except at z = an Φ k as z, satisfies conitions for potential from an isolate mass istribution. Still nee to link with ρ or Σ)) in the plane. 5

5 Use Gauss Theorem Poisson s equation plus ivergence theorem) to etermine Σ in the z = plane. Over the cyliner 4πGρV = 2 ΦV =. Φ)V = Φ. ˆn 2 S Consier the limit in which the cyliner height. Then if A is the area of an en of the cyliner LHS = 4πGΣA HS = [ ] Φ z z=+ [ ] ) Φ A z z= Equating these [ ] + Φ = 4πGΣ) z LHS = kfk)e k+ J k)k kfk) z= e k J k)k = kfk)j k)k kfk)j k)k = 2 kfk)j k)k Σ) = 1 fk)j k)kk 2πG 51

6 Hence etermine fk) [an hence Φ] from inverse Hankel transform fk) = 2πG Σ)J k)r Thus the process for termining Φ from ρ in this case is Σ f Φ. Note: For etermining the circular velocity nee Φ, whch becomes J x) an for Bessel function J have JJ x x) = J 1 x) [Example]. x, This has been a bit longwine, but the steps are clear. They are: Summary of erivation of Φ for thin axisymmetric isk 1. 2 Φ = outsie isk. Solve by separation of variables. 2. Solutions of form Φ k,z) = Ce k z J k) k > 3. Φ k as,z an satisfies 2 Φ = is potential of an isolate ensity istribution 4. General Φ can be written as Φ,z) = Φ k,z)fk)k where fk) is an appropriate weight function. 5. Use Gauss theorem to etermine Σ) = 1 fk)j k)kk 2πG 6. Hence fk) = 2πG Σ)J k) So given Σ, use item 6) to etermine fk), an then 5) to obtain Φ. The circular velocity in the plane of a plane istribution of matter is given by vc) 2 = Φ z= 52

7 an we have x = k J k) = k x J x = kj 1 k) Then since we have equation 9-5) [Φ,z) = fk)e k z J k)k] then Examples a) Mestel isk v 2 C) = fk)j 1 k)kk A Mestel isk has the surface ensity istribution Σ) = Σ Thus M< ) = 2πΣ ) = 2π Σ = 2πΣ as fk) = 2πGΣ J k) = 2πGΣ k From Grashteyn an yzkik J ν bx)x = 1 b 53 eν) > 1 b >

8 Φ,z) = 2πGΣ e k z J k) k k an v 2 c) = 2πGΣ J 1 k)k v 2 c) = 2πGΣ = const Note that vc 2 GM) ) = exactly in this case even though istribution is a isk, not spherical. More generally, fin vc 2 GM) to within 1% [reasonable accuracy] for most smooth Σ istributions. see figure on next page) Conclue that measurement of v c ) is a goo measure of mass insie. 54

9 Exponential Disk Here Σ) = Σ exp [ /] 9-6) This has finite mass M = = 2πΣ 2 = 2πΣ 2 2πΣ exp [ /] e x xx } {{ } =1 Then fk) = 2πGΣ e / J k) [ Grashteyn + yzhik : e αx J βx) xx = α [β 2 + α 2 ] 3/2 ] [ Actually they have something ) which requires a little work: J ν βx)x ν+1 x = 2α2β)ν Γν ) π α2 + β 2 ) ν+3 2 an you nee to put in ν =, Γ3/2) = π/2. Then put α = 1/ an β = k.] Hence Use fk) = 2πGΣ 2 [1 + k ) 2 ] 3/2 Φ,z) = 2πGΣ 2 J k)e k z 1 + k ) 2) 3/2 k ) ) Jνxy)x 1 1 x 2 + a 2 ) = I 1/2 ν/2 2 ay K ν/2 2 ay 55

10 You can o this with help from Grashteyn + yzhik again, using ) for ea) >, y >, eν) > 1, an I z) = I 1z), K z) = K 1z) J ν xy) x x2 + a 2 = I ν/2 1 2 ay)k ν/2 1 2 ay) a of this gives J ν xy) x a x 2 + a = y 2 3/2 2 I ν/2 1 2 ay)k ν/2 1 2 ay) + y 2 I ν/2 1 2 ay)k ν/2 1 2 ay) so for ν = or J xy) x a x 2 + a = y 2 3/2 2 I 1 2 ay)k ay) + y 2 I ay)k 1 2 ay) J ν xy) x x 2 + a = y [I 2 3/2 1 2a 2 ay)k ay) I ay)k 1 ] 2 ay) Then with x = k, y = an a = 1/ this becomes J ν k) k 1 + k) = [ I 23/2 2 2 )K 1 ) I 1 )K ] ) Also you fin for the circular velocity with y = /2 ) v 2 C = Φ = 4πΣ y 2 [I K I 1 K 1 ] which is helpfully left as an example... 56