GRAvitAtionAl potentiAls *definitions and Gauss Theorem *density-potential pairs *spherical potentials *axisymmetric potentials *triaxial potentials Gas: hydrostatic equilibrium G M(r) The downward gravitational force 4 πr 2 ρ(r)dr r 2 dP Outward pressure force 4 πr 2 dr dr r --- radius vector dP G M(r) M(r) --- mass within r = − 2 ρ(r) ρ(r) --- mass density dr r P(r) --- gas pressure at r Definitions: find force or potential field of a stellar distribution Describe mass distribution as a continuous function In a 1-D system: always possible to define potential energy U(x) corresponding to any given force f(x): x U(x) = − ∫dx' f (x') integral over closed path x0 vanishes where x0 is arbitrary position at which U=0. The choice of x0 does not affect the dynamics Gravitational potential is the gravitational energy per unit mass Hence, gravitational energy of mass m is U(x) = mΦ(x) Note, that because U depends on the endpoints only: f(x) = − ∇U conservative field In multi-dimensional space: gravitational force: vector field r M - mass 0 r’ For an arbitrary density distribution: dΦ(r) = −G dM(r') / r'− r (2-1a) (2-1b) Gauss Theorem (for gravity) Remember: divergence of a vector Divergence of A = div A = ∇ ⋅ A ∂ ∂ ∂ = (i + j +k )⋅(A i + A j+ A k) ∂x ∂y ∂z 1` 2 3 ∂A ∂A ∂A = 1 + 2 + 3 ∂x ∂y ∂z 5 Taking divergence of eq.(2-1b): Poisson eq. (inside M) Laplace eq. (outside M) Note, in 1-D this is trivial (spherical): dF = - G dM(r)/r2 = - 4 π G ρ (r)dr ∇F = − 4πGρ(r) = − ∇2 Φ ≡ − ∆ Φ 1 r'−r But in 3-D, you should remember that ∇ = (gradient) r 3 r'− r r'−r and (divergence): cancels product r'−r 3 3(r'−r)⋅(r'−r) ∇ ⋅ = − + = rule: r 3 3 5 0 r'− r r'−r r'−r when r' ≠ r So, to take the divergence of F(r): ∇r • F ∇r ⋅ ∇r ⋅ So, the only contribution to ∇ r • F comes from the point r’ = r Take a small sphere with radius |r’-r|=h centered on this point: r r’ Divergence theorem: to replace volume integral by integral over enclosed surface Poisson eq. where (on the surface): |r’-r|=h and Application of Gauss Theorem Note, relationship between Φ and ρ are linear For volume V with a surface A enclosing mass M (2-2) application of Gauss theorem The potential energy W of self-gravitating system can be defined by setting Φ = 0 at infinity, and So, W < 0 always Density-potential pairs Consider potential for an arbitrary spherical mass distribution: r r M (r') Φ(r) = − dr' a(r) = G dr' , with r =∞ Φ (r) < 0 ∫ ∫ r'2 0 r0 r0 everywhere with the enclosed mass r M(r) = 4π ∫ dr' r'2 ρ(r') Point mass (Keplerian potential) r0 Φ(r) = - GM/r F(r) = − ∇Φ = dΦ / dr = − GM / r2 2 vc (r) = GM/r = - Φ (r) circular velocity 2 vesc (r) = 2GM/r = - 2Φ (r) escape velocity Uniform spherical shell Outside: Φ (r) = - GM/r (Keplerian) Inside: Φ (r) = const.; F(r) = 0 10 Homogeneous (uniform) sphere const for r < a ρ(r) = 0 for r > a outside: Φ (r) = -GM/r (Keplerian) inside: Φ (r) = -2πGρ (a2-r2/3) 2 π Fr = - GM(r)/r = - (4/3) Gρr harmonic oscillator 2 So, k/m = (4/3) πGρ = ω and Pr = 2π/ω 1/2 radial period of oscillations Pr = (3π/Gρ) −1/2 and free-fall time tff ~ (1/4) Pr ~ (Gρ) 2 1/2 Because Fr = v /r vc(r) = ω r = [(4/3) πGρ] r We define Ω(r) = ω (= const in this case) solid body rotation Note that Pc = Pr Logarithmic potential We know that many rotation curves are flat at large radii, vc ~ v0, so r M (r') Φ(r) = G dr' , with r =∞ ∫ r'2 0 r0 meaning that potential behaves as logarithmic… Spherical systems -α For power law: ρ = ρ0 (r/a) we have: 2-α 2 Φ(r) = -[(4πGaρ0)/(3−α)] (r/a) = vc /(α-2) •for α > 3, M(<r) infinity for r 0 infinite mass at the origin •for α = 2, we have singular isothermal sphere with circular velocity 2 1/2 vc(r) = (4π G a ρ0) = const. at all radii, 2 yielding Φ(r) = 4πGa ρ0 ln(r/a) More specific spherical models •Hernquist •Jaffe •Plummer sphere ρ (R) for various spherical models Axisymmetric thin disks (cylindrical r, z) •Vertical (z) potential near the plane z = 0: within the disk: ρ0 the volume density at the z = 0 plane above the disk: surface density Σ(z) Using the Gauss theorem, eq.(2-2): ∂Φ − = g = 4πGρ z = 2πGΣ(z) (inside) ∂z z 0 = 2πGΣ (above) ∞ where Σ = ∫ dz Σ(z) 0 Note, unlike spherical potentials, disk potential depends on the mass outside r 15 Examples: 2 •Mestel disk: Σ(r) = Σ0 r/r0 has vc (r) = 2πGΣ0r0 = GM(<r)/r unusual case when vc is independent of M(>r) ! −r / rd •Exponential disk: Σ(r) = Σ0 e fits the light profile in a much more realistic way than Mestel disk, and has circular velocity (see analytical approximation we used!): where y = r/2rd, and st nd In, Kn are Bessel functions of the 1 and 2 kind •Kuzmin-Toomre disk: Note, because Poisson equation is linear in ρ, Φ: differences between density-potential pairs and differentials of density-potential pairs are also ρ−Φ pairs •Toomre disk sequence: of order n can be obtained from the above Kuzmin disks by differentiation with respect to a2: Here n=1 Kuzmin disk; n=infinity is Gaussian disk •Bessel disk: k Σ(r) = J (kr); Φ(r, z) = exp(−k z )J (kr) 2πG 0 0 Axisymmetric flattened systems Realistic bulge + disk, etc. systems are neither spherical nor thin disks Combining both we get flattened potentials •Miyamoto-Nagai flattened system: If a = 0, we get Plummer sphere, and if b = 0, we get Kuzmin disk the viRiAl theoRem *illustrations *general case *mass determination *binding energy *specific heat: gravothermal catastrophe Illustrations Circular orbits v Consider the mass m in a circular orbit around M (>> m) mv 2 GmM = r r 2 Multiply by r: GmM mv 2 = ===> 2K = − W or 2K + W = 0 r 2K -W E = -K, where Define the ratio η = K/ W = 1/2 E = K+W the total energy Note, that in this case instantaneous value is also time-averaged value 20 Time-averaged Keplerian orbit (elliptical orbits) rp ra In general, η changes along the Keplerian orbit Example: compare the pericentric, ηp, and apocentric, ηa, values: 2 η v r r p = p p = a ≠ 2 1 ηa vara rp using rpvp = rava (angular momentum conservation) Taking time averages over an orbit: η = 0.5 <-W> = <GM/r> = GM /<r> E = -K and <K> = <0.5v2> = GM/2<r> (as before) Note, that time averages for a single non-Keplerian orbit do not usually have η=0.5. But this always holds when averaged over all the particles. Above, m and M form the whole system, with K=0 for M. General case Consider a cluster of N stars with time-dependent potential Φ(r, t). Individual energies are not conserved but the total E is. To show this, we write the 2nd law of Newton: d Gm m = − i j − (2-3) (mivi ) ∑ 3 (ri rj ) dt j r − r i≠ j i j mi cancels out Next, take the scalar product of this equation with vi: d d Gm m ⋅ = = − i j − ⋅ ∑vi (mivi ) K ∑ 3 (ri rj ) vi (2-4) i dt dt i,j r − r i≠ j i j Repeating the same procedure with a star vj: 1 d Gm m ⋅ = − i j − ⋅ ∑v j (m jv j ) ∑ 3 (rj ri ) v j (2-5) 2 j dt i,j r − r i≠ j i j Adding the right-hand sides of eqs.(2-4) and (2-5) Adding the right-hand sides of eqs.(2-4) and (2-5) Gmim j d Gmim j − (ri − r j )⋅(vi + v j ) = − . ∑ 3 ∑ dt r − r i, j ri − rj i, j i j i≠ j i≠ j This is equal to 2W: 1 Gm m 1 1 = − i j = Φ ρ Φ W ∑ ∑mi (ri ) or ∫ (r) (r)dV. 2 i,j ri − rj 2 i 2 i≠ j Note: division by 2 means that each pair will contribute one term only to the sum Adding eqs.(2-4) and (2-5): d 1 Gmim j (2-6) 2 K − ∑ = 0. dt 2 i,j ri − rj i≠ j E = K + W = const According to eq.(2-6): the stars in an isolated cluster can change their kinetic and potential energies, as long as their sum remains constant The Virial Theorem: on average, the kinetic and potential energies are in a specific balance d Gm m = − i j − (mivi ) ∑ 3 (ri rj ) dt j r − r Proof: i≠ j i j Start again with eq.(2-3), with an addition of an external force Fext. Next, take scalar product with ri and sum over all stars: d Gmim j i (mi vi )⋅ri = − (ri − rj )⋅ri + Fext ⋅ri . (2-7) ∑ dt ∑ 3 ∑ i i,j ri − rj i i≠j A similar equation would result if we started with the j-force: d Gmim j (m v )⋅r = − (r − r )⋅r + F j ⋅r . ∑ dt j j j ∑ 3 j i j ∑ ext j (2-8) j i,j ri − rj j i≠j The left sides of these two equations are the same; each equal to 1 d2 1 d2I (rr)"= 2(rr"+ r'r') ⋅ − ⋅ = − ∑ 2 (miri ri ) ∑mivi vi 2 2K, 2 i dt i 2 dt where I is the moment of inertia of the system: I ≡ ∑miri ⋅ri i Averaging eqs.(2-7) and (2-8): the first term on the right-hand side is the potential energy W, so 1 d2I − = + i ⋅ (2-9) 2 2K W ∑Fext ri 2 dt i 25 Taking long-term average of eq.(2-9) over time interval 0 < t < τ: 1 dI dI i (2-10) (τ) − (0) = 2 < K > + < W > +∑< Fext ⋅ri > .