Advanced Newtonian gravity

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1 Foundations of Newtonian gravity Solutions Motion of extended bodies, University of Guelph h treatment of Newtonian gravity, the book develops approximation methods to obtain weak-field solutions es the motion of self-gravitating bodies, the physics of Poisson and Will eral relativity and their consequences, this textbook offers ory by developing powerful methods that can be applied radiative losses on gravitating systems. The book concludes ories of gravity. Gravity Newtonian, Post-Newtonian, Relativistic al relativity and relativistic astrophysics, the book planetary motion around the Sun, the timing of binary d by binary black holes. Text boxes explore related topics r 100 exercises present interesting and challenging tests of ngton University in St. Louis. He is a member of the US or-in-chief of Classical and Quantum Gravity. He is well broad audiences. a SuperStock/Alamy. Gravity iversity of Guelph. He is a Fellow of the American Physical ds of Physical Review Letters and Classical and Quantum or of Physics at the University of Florida and J. S. and Clifford M. Will Celestial mechanics

2 Outline Foundations of Newtonian gravity Solutions for isolated bodies Motion of extended bodies Celestial mechanics These lectures rely heavily on Poisson and Will (PW).

3 Foundations of Newtonian gravity (PW 1.2, 1.3, 1.4) Defining equations Global conservation laws

4 Defining equations The matter sourcing gravity is taken to consist of a perfect fluid. Main variables: Newtonian potential U, mass density ρ, pressure p, velocity field v, density of internal energy ɛ, specific entropy s, temperature T, etc. ρ(t, x 2 ) U = 4πGρ U(t, x) = G x x d3 x ρ dv dt ρ t + (ρv) = 0 = ρ U p, d dɛ ɛ + p ρ dt = t + v dρ = ρt ds = 0 p = p(ρ, s, )

5 Global conservation laws Integral identity: d ρ(t, x)f(t, x) d 3 x = dt ( ρ f ( ρ f = ( = = t + f ρ ) d 3 x t ) f (ρv) d 3 x t ρ f t ρ df dt d3 x + ρv f ) d 3 x fρv ds

6 Total mass and momentum M := ρ d 3 x = constant, P := ρv d 3 x = constant dm dt = ρ d1 dt d3 x = 0 dp dt = ρ dv dt d3 x = ρ U d 3 x p d 3 x = G ρρ x x x x 3 d3 x d 3 x = 0 p ds

7 Total energy and angular momentum E := T + Ω + E int = constant, J := ρx v d 3 x = constant T (t) := 1 ρv 2 d 3 x 2 Ω(t) := 1 ρu d 3 x = G E int (t) := ɛ d 3 x ρρ x x d3 x d 3 x

8 Virial theorem I jk (t) := ρx j x k d 3 x T jk (t) := 1 ρv j v k d 3 x 2 Ω jk (t) := 1 2 G ρρ (x x ) j (x x ) k x x 3 d 3 x d 3 x P (t) := p d 3 x 1 d 2 I jk 2 dt 2 This is usually stated in scalar form, = 2T jk + Ω jk + P δ jk 1 d 2 I = 2T + Ω + 3P 2 dt2

9 Solutions for isolated bodies (PW 1.5) Spherical bodies Nearly spherical bodies Symmetric-tracefree tensors

10 Spherical bodies In spherical symmetry, Poisson s equation reduces to 1 r 2 r r2 U(t, r) = 4πGρ(t, r) r Solution inside (r < R) U(t, r) = Solution outside (r > R) Gm(t, r) r m(t, r) = 4π + 4πG r 0 R r ρ(r )r 2 dr ρ(t, r )r dr U = GM r, M := m(t, r = R) = constant

11 Nearly spherical bodies Express the external potential (r > R) as a multipole expansion, 1 x x = 1 r 1 x j j r x j x k 1 jk r = with r := x and L := j 1 j 2 j l. So U(t, x) = G or U(t, x) = G ρ(t, x ) x x d3 x = G ( 1) l l=0 l! l=0 1 L r ( 1) l I L 1 (t) L l! r, IL (t) := l=0 ( 1) l l! x L L 1 r ρ(t, x )x L d 3 x ρ(t, x )x L d 3 x

12 Symmetric-tracefree (STF) tensors Let n := x/r. Then j r 1 = n j r 2 ( jk r 1 = 3 n j n k 1 ) 3 δ jk r 3 = 3n jk r 3 [ jkn r 1 = 15 n j n k n n 1 ( ) ] nj δ kn + n k δ jn + n n δ jk r 4 5 = 15n jkn r 4 L r 1 = ( 1) l (2l 1)!! n L r l+1 U(t, x) = G l=0 (2l 1)!! I L (t) n L l! r l+1

13 STF tensors and spherical harmonics Only the STF piece of the multipole moments I L is required in the potentials. For example, I jk n jk = ( I jk δjk I nn) n jk = I jk n jk An STF tensor of rank l has 2l + 1 independent components. The expansion of U in STF tensors is equivalent to an expansion in spherical harmonics, U(t, x) = G I lm (t) = l l=0 m= l 4π 2l + 1 I lm(t) Y lm(θ, φ) r l+1 ρ(t, x)r l Y lm (θ, φ) d3 x

14 Motion of extended bodies (PW 1.6) Assumptions Center-of-mass variables Equations of motion

15 Assumptions Each body is made up of a perfect fluid. Bodies are well separated: r R. Each body is in equilibrium; the time scale for internal dynamics is much shorter than the external time scale. Under these conditions we have a near decoupling of the internal and external dynamics. External problem: Determine the motion of each body as a whole. Sensitive to internal dynamics only through a limited number of multipole moments. Internal problem: Determine the internal motions of each body. Sensitive to external dynanics only through a weak tidal interaction.

16 Center-of-mass variables The external problem involves replacing the original fluid variables with coarse-grained, center-of-mass variables. Body label: A = 1, 2, N m A := A r A := 1 m A v A := 1 m A a A := 1 m A := I L A A ρ d 3 x ρx d 3 x A A A ρv d 3 x = dr A dt ρ dv dt d3 x = dv A dt ρ(x r A ) L d 3 x

17 Equations of motion (1/3) The motion of body A follows from Euler s equation, m A a A = ρ dv dt d3 x = ρ U d 3 x p ds A The Newtonian potential is decomposed as U = U A + U A with ρ(t, x ) U A = G A x x d3 x, U A = ρ(t, x ) G B A B x x d3 x Because A ρ U A d 3 x = 0, only the external potential enters the equations of motion, m A a A = ρ U A d 3 x A A A

18 Equations of motion (2/3) Because r R, the external potential can be Taylor expanded about r A : This yields U A (t, x) = U A (t, r A ) + (x r A ) j j U A (t, r A ) = (x r A) j (x r A ) k jk U A (t, r A ) + l=0 m A a j A = 1 l! (x r A) L L U A (t, r A ) l=0 1 l! I L A (t) jlu A (t, r A )

19 Equations of motion (3/3) After evaluation of the external potential, a j A = G B A + l=2 + 1 m A { m B r 2 AB n j AB 1 [ ( 1) l I L B l! + m ] B I L 1 A jl m A r AB l=2 l =2 ( 1) l l!l I L A! I L B 1 jll r AB } r AB := r A r B, r AB := r AB, n AB = r AB /r AB

20 Two-body system For a two-body system, the position of each body can be measured relative to the barycenter Then R := m 1 m r 1 + m 2 m r 2 = 0, m := m 1 + m 2 r 1 = m 2 m r, r 2 = m 1 m r, r := r 1 r 2 When the first body has vanishing multipole moments, a j = d2 r j dt 2 = Gm r 2 nj + Gm ( 1) l l=2 l! I L 2 1 jl m 2 r For an axisymmetric body (e = symmetry axis) I jk 2 = m 2 R 2 2J 2 (e j e k 1 3 δjk), J 2 dimensionless

21 Celestial mechanics (PW 3.2, 3.3, 3.4) Kepler s problem Perturbed Kepler problem Perturbation by a third body Effects of oblateness

22 Kepler s problem (1/2) Two bodies, vanishing multipole moments, a = Gmn/r 2 Constants of the motion h = r v = r 2 φ ez ε = 1 2ṙ2 + h2 2r 2 Gm r (fixed orbital plane) = 1 2ṙ2 + V eff (r) Veff/(Gm/p) ε < 0 ε = 0 ε > 0 r/p

23 Kepler s problem (2/2) Conic sections p r =, f = φ ω = true anomaly 1 + e cos f h = Gmp, ε = Gm 1 e2, ω = constant (Runge-Lenz) 2p Motion in time: elliptical orbits a 3 r = a(1 e cos u), t = T + (u e sin u) Gm a = p, u = eccentric anomaly, T = constant 1 e2

24 Orbit in space Z m 1 pericenter f m 2 Ω ω ι Y ascending node X Orbital elements: p, e, ι, Ω, ω, T

25 Perturbed Kepler problem Add a perturbing force f to the equations of motion. How is the orbital motion affected? Secular changes: accumulate over many orbits Oscillatory changes: average out over each orbit Osculating orbits The perturbed orbit is at all times described by the Keplerian relations r = r n, v = ṙ n + h r 2 λ, h = h e z r = p 1 + e cos f ṙ = e Gm p sin f, h = Gmp but with varying orbital elements: p(t), e(t), ι(t), Ω(t), ω(t).

26 Osculating equations f = R n + S λ + W e z dp p dt = Gm 1 + e cos f S [ de p dt = sin f R + 2 cos f + e(1 + ] cos2 f) S Gm 1 + e cos f dι p dt = Gm dω p dt = 1 Gm sin ι dω dt = 1 p e Gm ( ) df df dt = dt cos(ω + f) 1 + e cos f W sin(ω + f) 1 + e cos f W [ cos f R e cos f 1 + e cos f ) Kepler ( dω + cos ιdω dt dt ] + f) sin f S e cot ιsin(ω 1 + e cos f W

27 Perturbation by a third body f = Gm 3r R 3 [ n 3(n N)N + O(r/R) ] r := r 12, R := r 23 For simplicity we place the third body in the same orbital plane (ι = 0) and on a circular orbit with N = cos F e X + sin F e Y. The osculating equations return the secular changes (per orbit) p = 15π m 3p 4 mr 3 e2 (1 e 2 ) 7/2 sin 2(ω F ) e = 15π m 3 p 3 2 mr 3 e(1 e2 ) 5/2 sin 2(ω F ) ω = 3π m 3 p 3 2 mr 3 (1 e2 ) 5/2[ cos 2(ω F ) ]

28 Perihelion advance of Mercury Average pericenter advance ω = 3π 2 m 3 m ( ) a 3 (1 e 2 ) 1/2 R Planet Advance (arcsec/century) Venus Earth 90.0 Mars 2.5 Jupiter Saturn 7.3 Total Discrepancy 42.9 Modern measured value ± 0.04 General relativity prediction 42.98

29 Effects of oblateness Take one member of the two-body system to have a significant oblateness measured by the dimensionless quadrupole moment J 2. Then f = 3 2 J GmR 2 { [5(e 2 n) 2 1 ] } n 2(e n)e r 4 e = symmetry axis The only secular change is in ω + cos ι Ω = 3πJ 2 ( R p ) 2 ( 1 3 ) 2 sin2 ι For the Sun, J and the effect on Mercury is 0.03 arcsec/century, below the observational uncertainties.

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