Introduction to GR: Newtonian Gravity

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1 Introduction to GR: Newtonian Gravity General DPG Physics School Physikzentrum Bad Honnef, Sept 2014 Clifford Will Distinguished Professor of Physics University of Florida Chercheur Associé Institut d Astrophysique de Paris cmw@physics.ufl.edu

2 Outline of the Lecture Part 1: Newtonian Gravity Foundations Equations of hydrodynamics Spherical and nearly spherical bodies Motion of extended fluid bodies Part 2: Newtonian Celestial Mechanics Two-body Kepler problem Perturbed Kepler problem *Based on Gravity: Newtonian, post-newtonian and General Relativistic, by Eric Poisson and Clifford Will (Cambridge U Press, 2014)

3 *Based on Gravity: Newtonian, post-newtonian and General Relativistic, by Eric Poisson and Clifford Will (Cambridge U Press, 2014)

4 Foundations of Newtonian Gravity Newton s 2nd law and the law of gravitation: The principle of equivalence: a = m G GMr m I r 3 If m I a = F F = Gm G Mr/r 3 m G = m I (1 + ) Then, comparing the acceleration of two different bodies or materials a = a 1 a 2 = ( 1 2 ) GMr r 3

The Weak Equivalence Principle (WEP) 400 CE Ioannes Philiponus: let fall from the same height two weights of which one is many times as heavy as the other.

5 The Weak Equivalence Principle (WEP) 400 CE Ioannes Philiponus: let fall from the same height two weights of which one is many times as heavy as the other. the difference in time is a very small one 1553 Giambattista Benedetti proposed equality 1586 Simon Stevin experiments Galileo Galilei Leaning Tower of Pisa? Newton pendulum experiments 1889, 1908 Baron R. von Eötvös torsion balance experiments (10-9 ) UW (Eöt-Wash) Atom inteferometers matter waves vs macroscopic object Bodies fall in a gravitational field with an acceleration that is independent of mass, composition or internal structure

6 Tests of the Weak Equivalence Principle 10-8 Eötvös Matter waves 10-9 Renner Free-fall Princeton Boulder Fifth-force searches Eöt-Wash APOLLO (LLR) Microscope (2015) Future: STEP, GG, STE-QUEST Moscow LLR Eöt-Wash a 1 -a 2 (a 1 +a 2 )/ YEAR OF EXPERIMENT

7 Newtonian equations of Hydrodynamics Writing ma = mru, U = GM/r, Equation of motion Field equation Generalize to multiple sources (sum over M s) and dv dt = ru + r ( v) =0, r 2 U = 4 G, d + v r, p = p(, T,... ) rp, Formal solution of Poisson s field equation: Write U(t, x) =G G(x, x 0 ) (t, x 0 )d 3 x 0, Euler equation of motion Continuity equation Poisson field equation Total or Lagrangian derivative Equation of state Green function r 2 G(x, x 0 )= 4 (x x 0 ) ) G(x, x 0 )=1/ x x 0 U(t, x) =G (t, x 0 ) x x 0 d3 x 0

8 Rules of the road Consequences of the continuity equation: for any f(x,t): d (t, x)f(t, x) d 3 x + d 3 fr ( v) d 3 I + v rf d 3 x f v = df dt d3 x. (t, x 0 )f(t, x, x 0 ) d 3 x 0 + v0 r 0 f d 3 x 0, (t, x 0 )f(t, x, x 0 ) d 3 x 0 = 0 + v rf + v0 r 0 f d 3 x 0 = 0 df dt d3 x 0

9 Global conservation laws M := (t, x) d 3 x = constant P := (t, x)v d 3 x = constant E := T (t)+ (t)+e int (t) = constant T (t) := 1 v 2 d 3 x 2 (t) := 1 2 G 0 x x 0 d3 x 0 d 3 x, E int (t) := d 3 x J := x v d 3 x = constant d dt vd 3 x = ( ru = G =0 rp) d 3 x d( V)+pdV =0 r v = V 1 dv/dt 0 x x 0 x x 0 3 d3 xd 3 x 0 I pnd 2 S R(t) := 1 M (t, x)x d 3 x = P M (t t 0)+R 0

10 Spherical and nearly spherical bodies Spherical symmetry 1 = 4 G = Gm(t, r) r 2 m(t, r) := r 0 4 (t, r 0 )r 02 dr 0 U(t, r) = Gm(t, r) r +4 G R r (t, r 0 )r 0 dr 0. Outside the body U = GM/r

11 Spherical and nearly spherical bodies Non-spherical bodies: the external field x 0 < x Taylor expansion: 1 x x 0 = 1 r 1X = `=0 1 x j r 2 x0j x k r ( 1)` 1 x L `! r Then the Newtonian potential outside the body becomes 1X ( 1)` 1 U ext (t, x) =G I hli, `! r `=0 I hli (t) := (t, x 0 )x 0hLi d 3 x 0 x L := x i x j...(l L j...(l times) h...i := symmetric tracefree product

12 Symmetric tracefree (STF) tensors A hijk...i Symmetric on all indices, and ija hijk...i =0 Example: gradients of j r 1 = n j r jk r 1 = 3n j n k jk r 3, jkn r 1 = h15n j n k n n 3 n j kn + n k jn + n n jk r L r 1 hli r 1 =( 1)`(2` 1)!! n hli r`+1 General formula for n <L> : n hli = [`/2] X p (2` 2p 1)!! h ( 1) 2P n +sym(q)i L 2P (2` 1)!! p=0 q := `!/[(` 2p)!(2p)!!]

13 Symmetric tracefree (STF) tensors Link between n <L> and spherical harmonics e hli n hli = n hli := `! (2` 1)!! P`(e n) 4 `! (2` + 1)!! Y hzi 10 = r 3 Y hxxi 20 = `X m= ` 4, r 5 16, Average of n <L> over a sphere: 8 I < hhn L ii := 1 4 n L d = : 1 (2`+1)!! Y hli `m Y`m(, ) Yhxi r 3 11 = 8, Yhyyi 20 = r 5 16, r 3 Yhyi 11 = i 8, Yhzzi 20 =2 L/2 +sym[(` 1)!!] ` =even 0 ` =odd r 5 16,

14 Spherical and nearly spherical bodies Example: axially symmetric body I hli A = m ARÀ(J`) A e hli J` := r 4 2` +1 1 MR` (t, x)r`y `0(, e ) d 3 x U ext (t, x) = GM r " 1 1X `=2 J` R r ` P`(cos ) # Note that: J 2 = C A MR 2 C A

15 Motion of extended fluid bodies Main assumptions: Bodies small compared to typical separation (R << r) isolated -- no mass flow T int ~ (R 3 /Gm) 1/2 << T orb ~ (r 3 /Gm) 1/2 -- quasi equilibrium adiabatic response to tidal deformations -- nearly spherical External problem: determine motions of bodies as functions (or functionals) of internal parameters Internal problem: given motions, determine evolution of internal parameters Solve the two problems self-consistently or iteratively Example: Earth-Moon system -- orbital motion raises tides, tidally deformed fields affect motions

16 Motion of extended fluid bodies Basic definitions m A := (t, x) d 3 x A r A (t) := 1 (t, x)x d 3 x m A A dm A /dt =0 v A (t) := dr A dt a A (t) := dv A dt = 1 m A = 1 m A Is the center of mass unique? pure convenience, should not wander outside the body not physically measurable almost impossible to define in GR m A a A = G G A A 0 x x 0 A x x 0 3 d3 xd 3 x X 0 x x 0 B6=A B x x 0 3 d3 x 0 5 d 3 x A A v d 3 x dv dt d3 x Define: x := r A (t)+ x x 0 := r B (t)+ x 0 r AB := r A r B

17 N-body point mass terms Motion of extended fluid bodies a j A = G X B6=A + 1X `=2 + 1 m A ( m B r 2 AB n j AB 1 h ( 1)`I hli B `! 1X 1X `=2 `0=2 ( 1)`0 `!`0! Moments of other bodies + m i B I hli 1 jl A m A r AB I hli i 1 A IhL0 jll 0 r AB ) Effect of body s own moments Two-body system with only body 2 having non-zero I <L> r := r 1 r 2, r := r R := (m 1 r 1 + m 2 r 2 )/m Moment-moment interaction terms m := m 1 + m 2 µ := m 1 m 2 /m a j = Gm r 2 nj + Gm 1X `=2 ( 1)` `! I hli 2 m jl 1 r

18 Outline of the Lecture Part 1: Newtonian Gravity Foundations Equations of hydrodynamics Spherical and nearly spherical bodies Motion of extended fluid bodies Part 2: Newtonian Celestial Mechanics Two-body Kepler problem Perturbed Kepler problem *Based on Gravity: Newtonian, post-newtonian and General Relativistic, by Eric Poisson and Clifford Will (Cambridge U Press, 2014)

The two-body Kepler problem set center of mass at the origin (X = 0) ignore all multipole moments (spherical bodies or point masses) define r := r 1 r 2,r:= r,m:= m 1 + m 2,µ:= m 1 m 2 /m reduces to

19 The two-body Kepler problem set center of mass at the origin (X = 0) ignore all multipole moments (spherical bodies or point masses) define r := r 1 r 2,r:= r,m:= m 1 + m 2,µ:= m 1 m 2 /m reduces to effective one-body problem a = Gm r 2 n Energy and angular momentum conserved: E = 1 2 m 1v m 2v 2 2 G m 1m 2 r 1 r 2 = 1 2 µv2 G µm r L = m 1 r 1 v 1 + m 2 r 2 v 2 = µr v orbital plane is fixed

20 Effective one-body problem Make orbital plane the x-y plane y λ=dn/dφ r v = r 2 d dt := he z v = dr dt =ṙn + r r φ n From energy conservation: x ṙ 2 =2[" V e (r)] V e (r) = h2 r 2 Gm r Reduce to quadratures (integrals) t t i = ± i = h r dr 0 p r i 2[" Ve (r 0 )] t t i dt 0 r(t 0 ) 2

Keplerian orbit solutions Radial acceleration, or d/dt of energy equation: r h 2 r 3 = Gm r 2 Find the orbit in space: convert from t to φ: d/dt = d/d d 2 d 2 1 r + 1 r = Gm h 2 =(h/r 2 )d/d 1 r = 1

21 Keplerian orbit solutions Radial acceleration, or d/dt of energy equation: r h 2 r 3 = Gm r 2 Find the orbit in space: convert from t to φ: d/dt = d/d d 2 d 2 1 r + 1 r = Gm h 2 =(h/r 2 )d/d 1 r = 1 p (1 + e cos f) f :=! true anomaly p := h 2 /Gm semilatus rectum Elliptical orbits (e < 1, a > 0) r peri = p 1+e, =! r apo = p 1 e, =! + a := 1 2 (r peri + r apo )= p 1 e 2 Hyperbolic orbits (e > 1, a < 0) in out = 2 arcsin(1/e)

22 Useful relationships ṙ = he p sin f Keplerian orbit solutions v 2 = Gm p (1 + 2e cos f + e2 )=Gm E = Gµm 2a e 2 =1+ 2h2 E µ(gm) 2 a 3 P =2 Gm 1/2 2 r for closed orbits 1 a Alternative solution r = a(1 e cos u) n(t T )=u e sin u tan f 2 = r 1+e 1 e tan u 2 n =2 /P u = eccentric anomaly f = true anomaly n = mean motion

23 Dynamical symmetry in the Kepler problem e and p are constant (related to E and h) orbital plane is constant (related to direction of h) ω is constant -- a hidden, dynamical symmetry Runge-Lenz vector A := v h n Gm = e cos! e x +sin! e y ) = constant Comments: responsible for the degeneracy of hydrogen energy levels added symmetry occurs only for 1/r and r 2 potentials deviation from 1/r potential generically causes dω/dt

24 Keplerian orbit in space Six orbit elements z i = inclination relative to reference plane: cos = ĥ e Ω = angle of ascending node cos = ĥ e Y sin ω = angle of pericenter x λ h n f ω Ω i y e = A sin! = A e z e sin a = h^2/gm(1-e 2 ) T = time of pericenter passage T = t f 0 r 2 h df Comment: equivalent to the initial conditions x 0 and v 0

25 Osculating orbit elements and the perturbed Kepler problem a = Gmr r 3 + f(r, v,t) Define: r := rn, r := p/(1 + e cos f), p = a(1 e 2 ) v := he sin f p n + h r n := cos cos(! + f), h := p Gmp cos sin sin(! + f) e X osculating orbit + sin cos(! + f) + cos cos sin(! + f) e Y +sin sin(! + f) e Same x & v actual orbit := cos sin(! + f) cos sin cos(! + f) e X + sin sin(! + f) + cos cos cos(! + f) e Y +sin cos(! + f) e new osculating orbit ĥ := n =sin sin e X sin cos e Y + cos e e, a, ω, Ω, i, T may be functions of time

26 Osculating orbit elements and the perturbed Kepler problem Decompose: Example: a = A = v h Gm dh dt Gm da dt ḣ cos Gmr r 3 + f(r, v,t) h = r v =) dh dt = r f n =) Gm da dt f = Rn + S = rw + rs ĥ = f h + v (r f) + Wĥ =2hS n hr + rṙs rṙw ĥ. ḣ = rs d dt (h cos ) =ḣ e h d dt sin = rw cos(! + f)sin + rs cos

27 Osculating orbit elements and the perturbed Kepler problem Lagrange planetary equations r dp p dt =2 3 1 Gm 1+ecos f S, r apple de p dt = sin f R + 2 cos f + e(1 + cos2 f) Gm 1+ecos f d dt = sin d dt = d! dt = 1 e r p cos(! + f) Gm 1+ecos f W, r p sin(! + f) Gm 1+ecos f W, r apple p cos f R + Gm An alternative pericenter angle: 2+e cos f 1+e cos f S, sin f S e cot sin(! + f) 1+e cos f W $ :=! + cos d$ dt = 1 r apple p e Gm cos f R + 2+e cos f 1+e cos f sin f S

28 Osculating orbit elements and the perturbed Kepler problem Comments: these six 1st order ODEs are exactly equivalent to the original three 2nd order ODEs if f = 0, the orbit elements are constants if f << Gm/r 2, use perturbation theory yields both periodic and secular changes in orbit elements can convert from d/dt to d/df using df dt = df dt Kepler d! dt + cos d dt Drop if working to 1st order

29 Osculating orbit elements and the perturbed Kepler problem Worked example: perturbations by a third body a = Gmr r 3 r 12 a 1 = Gm 2 r12 3 r 12 a 2 =+Gm 1 Gm 3 r R 3 r 3 12 h n r 13 Gm 3 r13 3, r 23 Gm 3 r23 3 i 3(n N)N + O(Gm 3 r 2 /R 4 ) R := r 23,N:= r 23 / r 23,m:= m 1 + m 2 r 23 3 r 13 R := f n = S := f Gm 3r R 3 1 3(n N) 2, =3 Gm 3r R 3 (n N)( N), W := f ĥ =3 Gm 3r R 3 (n N)(ĥ N) Put third body on a circular orbit r df G(m + N = e X cos F + e Y sin F, dt = m3 ) R 3 2 df dt r=r 12 1

30 Osculating orbit elements and the perturbed Kepler problem Worked example: perturbations by a third body Integrate over f from 0 to 2π holding F fixed, then average over F from 0 to 2π: h ai =0 h ei = 15 2 h!i = a e(1 e 2 ) 1/2 sin 2 sin! cos! m R 3 a (1 e 2 ) 1/2h 5 cos 2 sin 2! +(1 e 2 )(5 cos 2! 3)i m 3 m 3 m h i = 15 2 h i = 3 2 R 3 a e 2 (1 e 2 ) 1/2 sin cos sin! cos! m R 3 a (1 e 2 ) 1/2 (1 5e 2 cos 2! +4e 2 ) cos R m 3 m 3 m Also: h $i = 3 2 m 3 m 3 a (1 e 2 ) 1/2h 1+sin 2 (1 5sin!)i 2 R

Osculating orbit elements and the perturbed Kepler problem Worked example: perturbations by a third body Case 1: coplanar 3 rd body and Mercury

31 Osculating orbit elements and the perturbed Kepler problem Worked example: perturbations by a third body Case 1: coplanar 3 rd body and Mercury s perihelion (i = 0) h $i = 3 2 m 3 m a R 3 (1 e 2 ) 1/2 For Jupiter: d$/dt = 154 as per century (153.6) For Earth d$/dt = 62 as per century (90)

32 Mercury s Perihelion: Trouble to Triumph 1687 Newtonian triumph 1859 Leverrier s conundrum 1900 A turn-of-the century crisis 575 per century Perturbing Planet Advance 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

33 Osculating orbit elements and the perturbed Kepler problem Worked example: perturbations by a third body Case 2: the Kozai-Lidov mechanism h ai =0 h ei = 15 2 h!i = a e(1 e 2 ) 1/2 sin 2 sin! cos! m R 3 a (1 e 2 ) 1/2h 5 cos 2 sin 2! +(1 e 2 )(5 cos 2! 3)i m 3 m 3 m h i = 15 2 m 3 m R a R 3 e 2 (1 e 2 ) 1/2 sin cos sin! cos! A conserved quantity: e cos h ei +sin h i =0 1 e2 =) p 1 e 2 cos =constant L!

34 Osculating orbit elements and the perturbed Kepler problem Worked example: perturbations by a third body Case 2: the Kozai-Lidov mechanism Eccentricity Inclination Pericenter P Kozai P ' m m 3 R a 3 1 e c sin c

=6 J 2 1 p 4 sin2 = 3 J 2 R p 2 cos For Mercury (J 2 = 2.2 X 10-7 ) d$ dt =0.

35 Osculating orbit elements and the perturbed Kepler problem Worked example: body with a quadrupole moment a = Gmr 3 r 3 2 J GmR 2 n 5(e 2 n) 2 r 4 1 o n 2(e n)e, a =0, e =0, =0 2 R 5! =6 J 2 1 p 4 sin2 = 3 J 2 R p 2 cos For Mercury (J 2 = 2.2 X 10-7 ) d$ dt =0.03 as/century For Earth satellites (J 2 = 1.08 X 10-3 ) 7/2 d R dt = 3639 cos deg/yr a LAGEOS (a=1.93 R, i = 109 o.8): 120 deg/yr! Sun synchronous: a= 1.5 R, i = 65.9

36 Outline of the Lecture Part 1: Newtonian Gravity Foundations Equations of hydrodynamics Spherical and nearly spherical bodies Motion of extended fluid bodies Part 2: Newtonian Celestial Mechanics Two-body Kepler problem Perturbed Kepler problem *Based on Gravity: Newtonian, post-newtonian and General Relativistic, by Eric Poisson and Clifford Will (Cambridge U Press, 2014)

The two-body Kepler problem

The two-body Kepler problem set center of mass at the origin (X = 0) ignore all multipole moments (spherical bodies or point masses) define r := r 1 r 2,r:= r,m:= m 1 + m 2,µ:= m 1 m 2 /m reduces to effective