Introduction to GR: Newtonian Gravity

Introduction to GR: Newtonian Gravity General Relativity @99 DPG Physics School Physikzentrum Bad Honnef, 14 19 Sept 2014 Clifford Will Distinguished Professor of Physics University of Florida Chercheur Associé Institut d Astrophysique de Paris http://phys.ufl.edu/~cmw/ cmw@physics.ufl.edu

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)

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

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. the difference in time is a very small one 1553 Giambattista Benedetti proposed equality 1586 Simon Stevin experiments 1589-92 Galileo Galilei Leaning Tower of Pisa? 1670-87 Newton pendulum experiments 1889, 1908 Baron R. von Eötvös torsion balance experiments (10-9 ) 1990 2010 UW (Eöt-Wash) 10-13 2010 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

Tests of the Weak Equivalence Principle 10-8 Eötvös Matter waves 10-9 Renner Free-fall 10-10 10-11 Princeton Boulder Fifth-force searches Eöt-Wash APOLLO (LLR) 10-13 Microscope 10-15 (2015) Future: STEP, GG, STE-QUEST 10-12 Moscow LLR Eöt-Wash 10-13 10-14 a 1 -a 2 (a 1 +a 2 )/2 1940 1920 1900 1980 1970 1960 1990 2000 2010 YEAR OF EXPERIMENT

Newtonian equations of Hydrodynamics Writing ma = mru, U = GM/r, Equation of motion Field equation Generalize to multiple sources (sum over M s) and continuous matter @ @t dv dt = ru + r ( v) =0, r 2 U = 4 G, d dt := @ @t + 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

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

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

Spherical and nearly spherical bodies Spherical symmetry 1 r 2 @ @r r 2 @U @r = 4 G (t, r) @U @r = 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

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 0j @ j + 1 1 r 2 x0j x 0k @ j @ k r ( 1)` 1 x 0L @ L `! r Then the Newtonian potential outside the body becomes 1X ( 1)` 1 U ext (t, x) =G I hli @ hli, `! r `=0 I hli (t) := (t, x 0 )x 0hLi d 3 x 0 x L := x i x j...(l times) @ L := @ i @ j...(l times) h...i := symmetric tracefree product

Symmetric tracefree (STF) tensors A hijk...i Symmetric on all indices, and ija hijk...i =0 Example: gradients of 1/r @ j r 1 = n j r 2, @ jk r 1 = 3n j n k jk r 3, i @ jkn r 1 = h15n j n k n n 3 n j kn + n k jn + n n jk r 4 @ 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)!!]

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,

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

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

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 0 2 3 4 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

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 A @ jl A m A r AB I hli i 1 A IhL0 B @A 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 2 @ jl 1 r

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 effective one-body problem a = Gm r 2 n Energy and angular momentum conserved: E = 1 2 m 1v 2 1 + 1 2 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

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 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)

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

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

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

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

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

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

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

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

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 = 3 2 3 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 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)

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 277.8 Earth 90.0 Mars 2.5 Jupiter 153.6 Saturn 7.3 Total 531.2 Discrepancy 42.9 Modern measured value 42.98 ± 0.04 General relativity prediction 42.98

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 = 3 2 3 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!

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

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

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)