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Supplemental Lecture 5

Thomas Precession and Fermi-Walker Transport in and Geodesic Precession in

Abstract

This lecture considers two topics in the of tops, spins and gyroscopes: 1. and Fermi-Walker Transport in Special Relativity, and 2. Geodesic Precession in General Relativity. A gyroscope (“spin”) attached to an accelerating particle traveling in Minkowski - is seen to precess even in a torque-free environment. The angular rate of the precession is given by the famous Thomas precession frequency. We introduce the notion of Fermi-Walker transport to discuss this phenomenon in a systematic fashion and apply it to a particle propagating on a circular orbit. In a separate discussion we consider the geodesic motion of a particle with spin in a -time. The spin necessarily precesses as a consequence of the space-time . We illustrate the phenomenon for a circular orbit in a . Accurate experimental measurements of this effect have been accomplished using earth satellites carrying gyroscopes.

This lecture supplements material in the textbook: Special Relativity, Electrodynamics and General Relativity: From Newton to Einstein (ISBN: 978-0-12-813720-8). The term “textbook” in these Supplemental Lectures will refer to that work.

Keywords: Fermi-Walker transport, Thomas precession, Geodesic precession, Schwarzschild metric, Probe B (GP-B).

Introduction

In this supplementary lecture we discuss two instances of precession: 1. Thomas precession in special relativity and 2. Geodesic precession in general relativity. To set the stage, let’s recall a few things about precession in classical mechanics, Newton’s , as we say in the textbook.

The precession of a “spin” vector at angular about a fixed axis is described by the

first order differential equation, 𝑠𝑠⃗ 𝜔𝜔

= × (1) 𝑑𝑑𝑠𝑠⃗ 𝑑𝑑𝑑𝑑 𝜔𝜔��⃗ 𝑠𝑠⃗ If points in the z direction, = , then rotates around it at angular velocity . The quantities𝜔𝜔��⃗ and are conserved.𝜔𝜔��⃗ 𝜔𝜔 To𝑘𝑘�⃗ see this,𝑠𝑠⃗ “dot” into Eq. 1, 𝜔𝜔 𝑧𝑧 𝑠𝑠⃗ ∙ 𝑠𝑠⃗ 𝑠𝑠 ( ) 𝑠𝑠⃗ = = ( × ) = 0 (2a) 𝑑𝑑𝑠𝑠⃗ 1 𝑑𝑑 𝑠𝑠⃗∙𝑠𝑠⃗ 𝑠𝑠⃗ ∙ 𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑 𝑠𝑠⃗ ∙ 𝜔𝜔��⃗ 𝑠𝑠⃗ and if we dot into Eq. 1,

𝑘𝑘�⃗ = = ( × ) = 0 (2b) 𝑑𝑑𝑠𝑠⃗ 𝑑𝑑𝑠𝑠𝑧𝑧 𝑘𝑘�⃗ ∙ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑘𝑘�⃗ ∙ 𝜔𝜔��⃗ 𝑠𝑠⃗ which proves the two points. Finally, to find ( ) and ( ) write out Eq.1 in Cartesian

coordinates and use matrix notation, 𝑠𝑠𝑥𝑥 𝑡𝑡 𝑠𝑠𝑦𝑦 𝑡𝑡 0 0 = 0 0 𝑥𝑥 𝑥𝑥 (3) 𝑑𝑑 𝑠𝑠 0 −0𝜔𝜔 0 𝑠𝑠 𝑦𝑦 𝑦𝑦 𝑑𝑑𝑑𝑑 �𝑠𝑠 � �𝜔𝜔 � �𝑠𝑠 � 𝑧𝑧 𝑧𝑧 Note that the matrix in Eq. 2 is antisymmetric𝑠𝑠 ( ) and𝑠𝑠 this guarantees that time 𝑖𝑖𝑖𝑖𝑖𝑖 𝑗𝑗 𝑘𝑘 evolution preserves the norm of the spin (this point∑𝑗𝑗𝑗𝑗 𝜖𝜖was𝜔𝜔 also𝑠𝑠 made in Eq. 2a). The two coupled equations in Eq. 3 read,

= = + 𝑑𝑑𝑠𝑠𝑥𝑥 𝑑𝑑𝑠𝑠𝑦𝑦 −𝜔𝜔𝑠𝑠𝑦𝑦 𝜔𝜔𝑠𝑠𝑥𝑥 To solve this coupled set of differential𝑑𝑑𝑑𝑑 equations, diff𝑑𝑑𝑑𝑑erentiate the first one, substitute into the second one and find a harmonic oscillator equation for . Similarly for . Then for the initial

conditions (0), (0), (0) = ( , 0,0), the solution𝑠𝑠 𝑥𝑥is ( ), ( ),𝑠𝑠𝑦𝑦 ( ) =

( cos , �𝑠𝑠sin𝑥𝑥 𝑠𝑠,𝑦𝑦0). The𝑠𝑠𝑧𝑧 spin� precesses𝑠𝑠 as shown in Fig.� 1.𝑠𝑠𝑥𝑥 𝑡𝑡 𝑠𝑠𝑦𝑦 𝑡𝑡 𝑠𝑠𝑧𝑧 𝑡𝑡 �

𝑠𝑠 𝜔𝜔𝜔𝜔 𝑠𝑠 𝜔𝜔𝜔𝜔

Fig. 1 Precessing spin in two

When we turn to Thomas precession [1] we will consider the spin of a particle on a curve in Minkowski space. The particle will experience a four velocity ( ) and four acceleration ( ) 𝜇𝜇 𝜇𝜇 on the curve and the spin ( ) = ( ), ( ) will be seen to𝑢𝑢 precess𝑡𝑡 in the lab frame, 𝑎𝑎 𝑡𝑡 𝜇𝜇 0 𝑆𝑆 𝑡𝑡 �𝑠𝑠 𝑡𝑡 𝑠𝑠⃗ 𝑡𝑡 � = × (4a) 𝑑𝑑𝑠𝑠⃗ 𝑑𝑑𝑑𝑑 𝜔𝜔��⃗𝑇𝑇 𝑠𝑠⃗ where is the Thomas angular velocity [1],

𝑇𝑇 𝜔𝜔��⃗ × = 2 (4b) 𝛾𝛾 𝑎𝑎�⃗ 𝑣𝑣�⃗ 2 𝜔𝜔��⃗𝑇𝑇 𝛾𝛾+1 𝑐𝑐 Here is the particle’s velocity in the lab frame and is its acceleration. Eq 4a and 4b are exact

in special𝑣𝑣⃗ relativity. We see that is different from𝑎𝑎 ⃗zero if the acceleration “curves” the particle’s trajectory. Recall from 𝜔𝜔the𝑇𝑇 textbook discussion of Thomas precession that a calculation of the precession of the spin over a full cycle of its closed simple path was possible just knowing Lorentz contraction [2,3]. The result in Eq. 4 goes much further and shows how it develops differentially. This equation has wider and more general applicability. Now consider precession in general relativity. There are many examples of precession in general relativity but we will consider only the simplest: the precession of the spin of a massive particle as it moves freely outside a fixed , described by the Schwarzschild metric derived in

Chapter 12 of the textbook. We already know to expect precession due to the curvature of 𝜇𝜇 the space: of a vector around an infinitesimal closed path produces a non𝑅𝑅 -𝛼𝛼𝛼𝛼zero𝛼𝛼 result when 0 inside the path. Here we will be interested in the precession of a 𝜇𝜇 gyroscope on𝑅𝑅 a𝛼𝛼𝛼𝛼 satellite𝛾𝛾 ≠ orbiting the earth as it undergoes many polar orbits over a long period of time [4]. The general relativistic predictions for such an experiment were well confirmed by the Gravity Probe B (GP-B) launched and operated by NASA from 2004-2011 [5]. In addition to confirming geodesic precession the probe also measured “frame-dragging” [6] a phenomenon that goes beyond the Schwarzschild metric and accounts for the spinning of the earth on its axis. The is described by the which contains gravito-magnetic effects that produce additional precession effects (Lense-Thirring precession) which is approximately one- hundredth the magnitude of the effect studied here. This phenomenon will be the subject of a later supplementary lecture.

Fermi-Walker Transport and Thomas Precession

We seek the special relativistic generalization of precession. We are particularly interested in the first order equation of motion of the spin of a massive particle as it travels on a curve and experiences acceleration. This is the situation in the study of relativistic spectroscopy of atoms where the electrons experience the electromagnetic field of the nucleus.

We begin with a more general situation. Consider a four vector attached to the particle 𝜇𝜇 which is traveling on a curve in the lab inertial frame. Its time evolution𝐴𝐴 is expected to be given by a first order differential equation of the covariant form [4],

= 𝜇𝜇 (5) 𝑑𝑑𝑑𝑑 𝜇𝜇 𝛼𝛼 𝑑𝑑𝑑𝑑 ∑𝛼𝛼 𝐾𝐾 𝛼𝛼 𝐴𝐴 where is the length variable. It will become the differential of , , in

the applications𝑑𝑑𝑑𝑑 that follow. Here is a second rank which must be constructed𝑐𝑐𝑐𝑐 𝑐𝑐out of 𝜇𝜇𝜇𝜇 variables that describe the particle:𝐾𝐾 the particle’s four velocity and acceleration . The 𝜇𝜇 𝜇𝜇 𝑢𝑢 𝑎𝑎 tensor must also be anti-symmetric, so that it preserves the length of , as illustrated for 𝛼𝛼 Newtonian mechanics in the Introduction. Let’s apply Eq. 5 to the time𝐴𝐴 evolution of a four vector we understand thoroughly, , 𝜇𝜇 𝑢𝑢 = = 𝜇𝜇 𝑑𝑑𝑢𝑢 𝜇𝜇 𝜇𝜇 𝛼𝛼 𝑎𝑎 𝑐𝑐 � 𝐾𝐾 𝛼𝛼𝑢𝑢 𝑑𝑑𝑑𝑑 𝛼𝛼 This equation works if we choose,

1 = ( ) 𝜇𝜇𝜇𝜇 𝜇𝜇 𝜌𝜌 𝜇𝜇 𝜌𝜌 𝑐𝑐𝑐𝑐 2 𝑎𝑎 𝑢𝑢 − 𝑢𝑢 𝑎𝑎 In this case, 𝑐𝑐

1 = ( ) = 𝜇𝜇 𝛼𝛼 𝜇𝜇 𝜇𝜇 𝛼𝛼 𝜇𝜇 𝑐𝑐 � 𝐾𝐾 𝛼𝛼𝑢𝑢 2 � 𝑎𝑎 𝑢𝑢𝛼𝛼 − 𝑢𝑢 𝑎𝑎𝛼𝛼 𝑢𝑢 𝑎𝑎 𝛼𝛼 𝑐𝑐 𝛼𝛼 where we used = 0 and = . (The result = 0 was discussed in the 𝛼𝛼 𝛼𝛼 2 𝛼𝛼 textbook and follows∑𝛼𝛼 𝑎𝑎𝛼𝛼 𝑢𝑢from the fact∑ that𝛼𝛼 𝑢𝑢𝛼𝛼 in𝑢𝑢 the particle’s𝑐𝑐 instantaneous∑𝛼𝛼 𝑎𝑎𝛼𝛼𝑢𝑢 rest frame, the spatial components of the four velocity vanish and the temporal component of the four acceleration 𝜇𝜇 vanishes.) 𝑢𝑢 𝜇𝜇 𝑎𝑎 We arrive at the Fermi-Walker transport equation [4],

= ( ) 𝜇𝜇 (6) 𝑑𝑑𝐴𝐴 1 𝜇𝜇 𝜇𝜇 𝛼𝛼 2 𝑑𝑑𝑑𝑑 𝑐𝑐 ∑𝛼𝛼 𝑎𝑎 𝑢𝑢𝛼𝛼 − 𝑢𝑢 𝑎𝑎𝛼𝛼 𝐴𝐴 Our task is to apply Eq. 6 to the time evolution of the spin of the traveling particle. Denote the spin four vector . 𝜇𝜇 In order to compare𝑆𝑆 to the results on Thomas precession obtained in the text, we place the particle on a circular trajectory [4],

= cos , = sin , = 0

The four velocity components,𝑥𝑥 𝑟𝑟 𝜔𝜔𝜔𝜔= 𝑦𝑦, are𝑟𝑟 then,𝜔𝜔 𝜔𝜔 𝑧𝑧 𝜇𝜇 𝜇𝜇 = 𝑑𝑑𝑥𝑥 sin⁄𝑑𝑑𝑑𝑑 𝑢𝑢 = cos = 0 1 2 3 𝑢𝑢 −𝜔𝜔𝜔𝜔 𝜔𝜔𝜔𝜔 𝑢𝑢 𝜔𝜔𝜔𝜔 𝜔𝜔𝜔𝜔 𝑢𝑢 The value of follows from the fact that the four vector has norm = = ( ) 0 𝜇𝜇 2 0 2 ( ) ( )𝑢𝑢 which implies that = 1 + ( ) = . ∑𝜇𝜇 𝑢𝑢 𝑢𝑢𝜇𝜇 𝑐𝑐 𝑢𝑢 − 1 2 2 2 0 2 𝑢𝑢 We− can𝑢𝑢 also write the phases𝑢𝑢 of the𝑐𝑐 �orbital 𝜔𝜔motion,𝜔𝜔⁄𝑐𝑐 𝑐𝑐𝑐𝑐, in term of , the time variable in the

inertial lab frame. Note that = , so = and 𝜔𝜔𝜏𝜏= , where𝑡𝑡 = .

The Fermi-Walker transport𝑑𝑑𝑑𝑑⁄𝑑𝑑𝑑𝑑 equation𝛾𝛾 𝑡𝑡for the𝛾𝛾𝛾𝛾 spin four𝜔𝜔𝜔𝜔 vectorΩ𝑡𝑡 reads,Ω 𝜔𝜔⁄𝛾𝛾

1 = ( ) 𝜇𝜇 𝑑𝑑𝑆𝑆 𝜇𝜇 𝜇𝜇 𝛼𝛼 2 � 𝑎𝑎 𝑢𝑢𝛼𝛼 − 𝑢𝑢 𝑎𝑎𝛼𝛼 𝑆𝑆 𝑑𝑑𝑑𝑑 𝑐𝑐 𝛼𝛼 which simplifies because = 0 ( In the particle’s rest frame = ( , 0,0,0) and = 𝜇𝜇 ′𝜇𝜇 ′𝜇𝜇 (0, ), so = 0. Since∑𝜇𝜇 𝑢𝑢𝜇𝜇 𝑆𝑆the inner product is Lorentz invariant𝑢𝑢, the result𝑐𝑐 must hold𝑆𝑆 in all ′ 𝜇𝜇 frames.)𝑠𝑠⃗ Now∑𝜇𝜇 𝑢𝑢 we𝜇𝜇𝑆𝑆 ′have,

= 𝜇𝜇 (7) 𝑑𝑑𝑆𝑆 1 𝜇𝜇 𝑑𝑑𝑢𝑢𝛼𝛼 𝛼𝛼 2 𝑑𝑑𝑑𝑑 − 𝑐𝑐 ∑𝛼𝛼 �𝑢𝑢 𝑑𝑑𝑑𝑑 � 𝑆𝑆 Since = 0, we learn that = 0 so = 0. So if we take = 0 initially then it 3 3 3 3 remains𝑢𝑢 zero for later t. The 𝑑𝑑value𝑆𝑆 ⁄𝑑𝑑 of𝑑𝑑 follows𝑑𝑑𝑆𝑆 ⁄from𝑑𝑑𝑑𝑑 the fact that =𝑆𝑆 0, 0 = ( ) 0 ( ) ( ) 0 0 1 1 , so = + . So,𝑆𝑆 if we can find and 𝑆𝑆 ∙ 𝑢𝑢, then we are𝑢𝑢 done.𝑆𝑆 − 𝑢𝑢 𝑆𝑆 − 2 2 0 1 1 2 2 0 1 2 𝑢𝑢 𝑆𝑆 Now𝑆𝑆 we can𝑢𝑢 write𝑆𝑆 𝑢𝑢Eq.𝑆𝑆 7 fo⁄𝑢𝑢r ( ) and ( ), 𝑆𝑆 𝑡𝑡 𝑆𝑆 𝑡𝑡 1 2 𝑆𝑆 𝑡𝑡 𝑆𝑆 𝑡𝑡 = = + 1 (8) 𝑑𝑑𝑆𝑆 1 1 𝑑𝑑𝑢𝑢𝛼𝛼 𝛼𝛼 𝜔𝜔𝜔𝜔 sin 𝜔𝜔𝜔𝜔 𝑑𝑑𝑢𝑢1 1 𝑑𝑑𝑢𝑢2 2 2 2 𝑑𝑑𝑑𝑑 − 𝑐𝑐 ∑𝛼𝛼 �𝑢𝑢 𝑑𝑑𝑑𝑑 � 𝑆𝑆 �− 𝑐𝑐 � � 𝑑𝑑𝑑𝑑 𝑆𝑆 𝑑𝑑𝑑𝑑 𝑆𝑆 � where the term = 0 because = 1 + ( ) is a constant. We can work on the 2 right hand side of𝑑𝑑𝑢𝑢 Eq.0⁄𝑑𝑑 8,𝑑𝑑 𝑢𝑢0 � 𝜔𝜔𝜔𝜔⁄𝑐𝑐

+ = cos + sin 𝑑𝑑𝑢𝑢1 1 𝑑𝑑𝑢𝑢2 2 2 1 2 2 𝑆𝑆 𝑆𝑆 𝜔𝜔 𝑟𝑟 𝜔𝜔𝜔𝜔 𝑆𝑆 𝜔𝜔 𝑟𝑟 𝜔𝜔𝜔𝜔 𝑆𝑆 Substituting into Eq. 8 and𝑑𝑑𝑑𝑑 doing 𝑑𝑑similar𝑑𝑑 manipulations for we arrive at a matrix 2 differential equation with sinusoidal coefficients, 𝑑𝑑𝑆𝑆 �𝑑𝑑𝑑𝑑 sin t cos t sin t = ( 1) (9) 𝑑𝑑 1 cos t sin 2t cos t 1 𝑆𝑆 2 Ω Ω Ω 𝑆𝑆 𝑑𝑑𝑑𝑑 � 2� 𝛾𝛾 − Ω � 2 � � 2� 𝑆𝑆 − Ω − Ω Ω 𝑆𝑆 where we used = 1, = and = t to simplify the prefactor and the 2 2 2 2 arguments of the𝜔𝜔 sinusoidal𝑟𝑟 ⁄𝑐𝑐 𝛾𝛾functions.− 𝑑𝑑𝑑𝑑 ⁄𝑑𝑑𝜏𝜏 𝛾𝛾 𝜔𝜔𝜔𝜔 Ω One way to solve a differential equation with sinusoidal coefficients is with Fourier transforms. In the appendix to this lecture we solve it by trial and error. For the initial conditions (0) = and (0) = 0, the solution is [4], 1 2 𝑆𝑆 𝑠𝑠 𝑆𝑆 ( ) = [(1 + ) cos(1 ) t + (1 ) cos(1 + ) t] 1 1 𝑆𝑆 (𝑡𝑡) = 2 𝑠𝑠[(1 + 𝛾𝛾) sin(1 − 𝛾𝛾)Ωt + (1 − 𝛾𝛾) sin(1 + 𝛾𝛾)Ωt] (10) 2 1 𝑆𝑆 𝑡𝑡 2 𝑠𝑠 𝛾𝛾 − 𝛾𝛾 Ω − 𝛾𝛾 𝛾𝛾 Ω Eq. 10 has some familiar features. We read off the Thomas angular frequency from the first terms in Eq. 10,

= ( 1) (11)

𝑇𝑇 This is the result for Thomas precssionΩ found𝛾𝛾 in− the Ωtextbook using Lorentz contraction [2,3]. For non-relativistic motion, 1 2 1 and we have the expression used in calculations 2 2 of relativistic corrections𝛾𝛾 of− atomic≈ 𝑣𝑣 spectroscopy.⁄ 𝑐𝑐 ≪ A plot of the “large” first terms in Eq. 10, those proportional in amplitude to (1 + ), shows that the spin precesses in the opposite

direction to the motion of the particle around𝛾𝛾 its orbit. This point was also made in the cryptic argument in the textbook. Another feature of the solution Eq. 10 is that it consists of a “large” amplitude “slow” term and a “small” amplitude “fast” term in the non-relativistic limit. It is not a simple single frequency rotation. This behavior is typical of gyroscopic precession [7]: as the spin rotates in the clockwise direction with the Thomas angular frequency, it simultaneously executes a small amplitude rapid rotation in the opposite direction.

The student is encouraged to master reference [8], which is “linked” below in the references, for a derivation of the Thomas precession equation, Eq. 4. This is the most useful and fundamental result. The derivation is at the level of the textbook and shows clearly that the Thomas precession occurs because the curvature in the particle’s flight, as measured in the lab inertial frame, makes the Lorentz boost from the lab frame to the rest frame of the particle time dependent. This is the subtle effect which produces a non-zero . The treatment in [8] obtains

the equation of motion of the spin in its instantaneous rest frame.Ω𝑇𝑇 There the motion is a pure precession with a single frequency, the Thomas value. The approach presented here considers the spin four vector in the lab frame. Its equation of motion is more complicated, Eq. 9 is not anti- symmetric so it is not a pure rotation, as we saw in the solution Eq. 10.

Geodesic Precession in General Relativity

We are interested in the motion of a particle of a non-zero mass around a large static mass M [4]. The metric describing space-time outside of M is the Schwarzschild metric,

= (1 ) (1 ) ( + sin ) (12) 2 2 2 −1 2 2 2 2 2 𝑠𝑠 𝑠𝑠 where is the 𝑑𝑑Schwarzschild𝑑𝑑 − 𝑟𝑟 ⁄ radius𝑟𝑟 𝑐𝑐 𝑑𝑑 𝑑𝑑 =− 2 − 𝑟𝑟 ⁄𝑟𝑟. In the𝑑𝑑𝑑𝑑 textbook− 𝑟𝑟 𝑑𝑑𝑑𝑑 we determined𝜃𝜃 𝑑𝑑𝑑𝑑 the conserved 2 quantities,𝑟𝑟𝑠𝑠 the energy e and the angular𝑟𝑟𝑠𝑠 𝐺𝐺𝐺𝐺⁄𝑐𝑐 h,

(1 ) = , = (13) 𝑑𝑑𝑑𝑑 2 − 𝑟𝑟𝑠𝑠⁄𝑟𝑟 𝑑𝑑𝑑𝑑 𝑒𝑒 𝑟𝑟 𝜑𝜑̇ ℎ Then in Chapter 12 we obtained the first integral of the equation of motion (the geodesic equation),

= ( ) , = (1 ) + 1 2 (14) 2 2 ℎ 2 𝑟𝑟̇ 𝑒𝑒 − 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 𝑟𝑟 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 − 𝑟𝑟𝑠𝑠⁄𝑟𝑟 �𝑟𝑟 � has the shape shown in Fig. 2 The “dot” in Eq. 14 denotes differentiation with respect to the𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 proper time . The shape of follows from the following effects: For large r, is normalized to unity,𝜏𝜏 as r decreases𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 Newton’s 1 potential pulls down, then as r𝑉𝑉 decreases𝑒𝑒𝑒𝑒𝑒𝑒 more the centrifugal barrier pushes up and⁄ 𝑟𝑟finally the attractive𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 non-linear relativistic 1 3 term pulls to minus infinity. 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 ⁄𝑟𝑟 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒

Fig. 2 The effective one dimensional potential for a massive particle in the Schwarzschild metric

We are particularly interested in circular orbits where = 0. The expression for 2 ( ) = 1 + 1 3 = 0 shows that 2 is a local𝑟𝑟 minimuṁ of , , ℎ 2 2 2 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 𝑟𝑟 > 0, as 𝑟𝑟shown𝑟𝑟𝑠𝑠 𝑟𝑟 𝑠𝑠in� the figure� −. (In𝑟𝑟𝑠𝑠 ⁄thisℎ �case the minimum, labeled𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 r in𝑑𝑑 the𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 figure⁄𝑑𝑑𝑑𝑑, 2 2 𝑑𝑑occurs𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 ⁄near𝑑𝑑𝑑𝑑 4.31. The minimum is quite flat in this example.) Then h and e can be solved in terms of r. Algebra yields,

( ) = , = ( ) ( ) (15) 𝑟𝑟𝑟𝑟𝑠𝑠⁄2 1−𝑟𝑟𝑠𝑠⁄𝑟𝑟 ℎ � 1−3𝑟𝑟𝑠𝑠⁄2𝑟𝑟 𝑒𝑒 1−3𝑟𝑟𝑠𝑠⁄2𝑟𝑟 Our last preliminary is to find the period of the circular orbit in terms of its radius r. The

= = = 2 , = conservation law implies that 2 so for a full revolution, 2 𝑟𝑟 2 . Eq. 15 then𝑟𝑟 𝜑𝜑 ̇ gives,ℎ 𝑑𝑑𝑑𝑑 𝑐𝑐ℎ 𝑑𝑑𝑑𝑑 ∆𝜑𝜑 𝜋𝜋 ∆𝜏𝜏 2 𝜋𝜋𝑟𝑟 ⁄𝑐𝑐ℎ = 2 = 1 (16) 2 2𝜋𝜋𝜋𝜋 2𝑟𝑟 3𝑟𝑟𝑠𝑠 ∆𝜏𝜏 𝜋𝜋𝑟𝑟 ⁄𝑐𝑐ℎ 𝑐𝑐 � 𝑟𝑟𝑠𝑠 � − 2𝑟𝑟 � The first conservation law in Eq. 13,(1 ) = , implies, using Eq. 16, 𝑑𝑑𝑑𝑑 − 𝑟𝑟𝑠𝑠⁄𝑟𝑟 𝑑𝑑𝑑𝑑 𝑒𝑒 = = ( ) (17) 𝑒𝑒 2𝜋𝜋𝜋𝜋 2𝑟𝑟 ∆𝑡𝑡 1−𝑟𝑟𝑠𝑠⁄𝑟𝑟 ∆𝜏𝜏 𝑐𝑐 �𝑟𝑟𝑠𝑠 With these preliminaries done, we are ready to consider geodesic precession. Suppose that the mass in orbit at radius r carries a gyroscope or classical spin. There are no torques on the spin, so it moves on a geodesic, and it parallel transports and satisfies,

+ = 0 𝑑𝑑 𝜇𝜇 𝜇𝜇 𝛼𝛼 𝛽𝛽 𝑠𝑠 𝑐𝑐 � Γ𝛼𝛼𝛼𝛼𝑢𝑢 𝑠𝑠 𝑑𝑑𝑑𝑑 𝛼𝛼𝛼𝛼 The student should review Chapter 12 of the textbook for the discussion of geodesic propagation in a curved space-time. In this application is the four velocity of the mass moving freely in a 𝜇𝜇 space-time described by the Schwarzschild𝑢𝑢 metric. We suppose that the particle moves on a circular orbit at fixed = 2, and using the notation where a dot indicates , for the

Scwarzschild mtric. 𝜃𝜃 𝜋𝜋⁄ 𝑑𝑑⁄𝑑𝑑𝑑𝑑

/ = , , , = ( , 0, 0, ) = 1 , 0,0, (18) −1 2 ( ) 𝜇𝜇 3𝑟𝑟𝑠𝑠 1 𝑟𝑟𝑟𝑟𝑠𝑠⁄2 2 𝑢𝑢 �𝑐𝑐𝑡𝑡̇ 𝑟𝑟̇ 𝜃𝜃̇ 𝜑𝜑̇ � 𝑐𝑐𝑡𝑡̇ 𝜑𝜑̇ �� − 2𝑟𝑟 � 𝑟𝑟 � 1−3𝑟𝑟𝑠𝑠⁄2𝑟𝑟 � where we used Eq. 15.

Finally we need the Christoffel symbols. These were computed in Chapter 12 of the textbook and read,

= (1 ), = (1 ), = 1 , = 1 (19) 1 𝑟𝑟𝑠𝑠 1 2 3 2 Γ00 2𝑟𝑟 − 𝑟𝑟𝑠𝑠⁄𝑟𝑟 Γ33 −𝑟𝑟 − 𝑟𝑟𝑠𝑠⁄𝑟𝑟 Γ12 ⁄𝑟𝑟 Γ13 ⁄𝑟𝑟 with other components of the Christoffel symbols vanishing.

We have additional constraints on the spin four vector. First, = 1, which reads, 𝜇𝜇 for = 2, ∑𝜇𝜇 𝑠𝑠 𝑠𝑠𝜇𝜇 −

𝜃𝜃 𝜋𝜋⁄ ( ) + ( ) + ( ) + ( ) = 1 0 2 1 2 2 2 3 2 00 11 22 33 so, 𝑔𝑔 𝑠𝑠 𝑔𝑔 𝑠𝑠 𝑔𝑔 𝑠𝑠 𝑔𝑔 𝑠𝑠 −

(1 )( ) (1 ) ( ) ( ) ( ) = 1 0 2 −1 1 2 2 2 2 2 3 2 𝑠𝑠 𝑠𝑠 In addition there is− the𝑟𝑟 ⁄invariant,𝑟𝑟 𝑠𝑠 − − 𝑟𝑟 ⁄=𝑟𝑟 0, which𝑠𝑠 −follow𝑟𝑟 𝑠𝑠s from− 𝑟𝑟the 𝑠𝑠fact that− u is purely time- 𝜇𝜇 like in the particle’s rest frame and∑ 𝜇𝜇s 𝑢𝑢is purely𝑠𝑠𝜇𝜇 space-like there. Writing out this expression = 0, 𝜇𝜇 ∑𝜇𝜇 𝑢𝑢 𝑠𝑠𝜇𝜇 1 2 (1 )(1 3 2 ) / = 0 (1 3 2 ) −1 2 0 2 𝑟𝑟𝑟𝑟𝑠𝑠⁄ 3 − 𝑟𝑟𝑠𝑠⁄𝑟𝑟 − 𝑟𝑟𝑠𝑠⁄ 𝑟𝑟 𝑠𝑠 − 𝑟𝑟 2 � 𝑠𝑠 𝑟𝑟 − 𝑟𝑟𝑠𝑠⁄ 𝑟𝑟 So, in the lab frame,

= ( ) (20) 0 �𝑟𝑟𝑟𝑟𝑠𝑠⁄2 3 𝑠𝑠 1−𝑟𝑟𝑠𝑠⁄𝑟𝑟 𝑠𝑠 With these preliminaries we can focus on the geodesic equations for the spatial components of the spin,

+ = + + = 0 𝑑𝑑 1 1 𝛼𝛼 𝛽𝛽 𝑑𝑑 1 1 0 0 1 3 3 𝑠𝑠 𝑐𝑐 � Γ𝛼𝛼𝛼𝛼𝑢𝑢 𝑠𝑠 𝑠𝑠 𝑐𝑐 Γ00𝑢𝑢 𝑠𝑠 𝑐𝑐 Γ33𝑢𝑢 𝑠𝑠 𝑑𝑑𝑑𝑑 𝛼𝛼𝛼𝛼 𝑑𝑑𝑑𝑑 Now substitute Eq. 18, 19 and 20 into this expression and do some algebra to find,

= 1 3 2 (21a) 𝑑𝑑 1 𝑐𝑐 𝑟𝑟𝑟𝑟𝑠𝑠 3 𝑑𝑑𝑑𝑑 𝑠𝑠 𝑟𝑟 � 2 � − 𝑟𝑟𝑠𝑠⁄ 𝑟𝑟 𝑠𝑠 Next the geodesic equation for is particularly simple because = = 0 eliminates the non- 2 1 2 trivial terms, 𝑠𝑠 𝑢𝑢 𝑢𝑢

= 0 (21b) 𝑑𝑑 2 𝑑𝑑𝑑𝑑 𝑠𝑠 Finally, the geodesic equation for is, 3 𝑠𝑠 + = + = 0 𝑑𝑑 3 3 𝛼𝛼 𝛽𝛽 𝑑𝑑 1 3 3 1 𝑠𝑠 𝑐𝑐 � Γ𝛼𝛼𝛼𝛼𝑢𝑢 𝑠𝑠 𝑠𝑠 𝑐𝑐 Γ31𝑢𝑢 𝑠𝑠 𝑑𝑑𝑑𝑑 𝛼𝛼𝛼𝛼 𝑑𝑑𝑑𝑑 Using Eq. 18 and 19 produces,

= (21c) 𝑑𝑑 3 𝑐𝑐 𝑟𝑟𝑟𝑟𝑠𝑠⁄2 1 3 𝑑𝑑𝑑𝑑 𝑠𝑠 𝑟𝑟 �1−3𝑟𝑟𝑠𝑠⁄2𝑟𝑟 𝑠𝑠 So, we end up with three first order differential equations with constant (time independent) coefficients. If we differentiate Eq. 21c and substitute Eq. 21a into the result, we find that 3 satisfies a harmonic oscillator equation, 𝑠𝑠

+ = 0 2 (22) 3 𝑐𝑐 𝑟𝑟𝑠𝑠 3 3 𝑠𝑠̈ 2𝑟𝑟 𝑠𝑠 where the double dot is short hand for . If initially (0) = 0, then 2 2 3 𝑑𝑑 ⁄(𝑑𝑑𝑑𝑑) = sin 𝑠𝑠 3 where 𝑠𝑠 𝜏𝜏 𝑠𝑠 𝜔𝜔𝜔𝜔

= = (23) 𝑟𝑟𝑠𝑠 𝐺𝐺𝐺𝐺 3 3 𝜔𝜔 𝑐𝑐�2𝑟𝑟 � 𝑟𝑟 and we used the value for the Schwarzschild radius, = 2 . 2 𝑠𝑠 We need to compare the frequency of the spin precession𝑟𝑟 𝐺𝐺𝐺𝐺⁄ to𝑐𝑐 the frequency of the orbital motion. The orbital frequency follows from Eq. 16 for the period of the orbital motion,

= = (1 3 2 ) / = (1 3 2 ) / (24) 2𝜋𝜋 𝑟𝑟𝑠𝑠 −1 2 −1 2 3 𝜔𝜔𝑜𝑜𝑜𝑜𝑜𝑜 ∆𝜏𝜏 𝑐𝑐�2𝑟𝑟 − 𝑟𝑟𝑠𝑠⁄ 𝑟𝑟 𝜔𝜔 − 𝑟𝑟𝑠𝑠⁄ 𝑟𝑟 We learn that the rotation of the spin of the particle lags slightly behind the orbital rotation. The precession angle per revolution is then,

2 = 2 1 1 3 /2 3 2 (25)

𝜋𝜋 − 𝜔𝜔∆𝜏𝜏 𝜋𝜋� − � − 𝑟𝑟𝑠𝑠 𝑟𝑟� ≈ 𝜋𝜋𝑟𝑟𝑠𝑠⁄ 𝑟𝑟 where we assumed 1 in the final inequality. 𝑟𝑟𝑠𝑠 Eq. 25 has been� tested𝑟𝑟 ≪ with exquisite accuracy by the satellite Gravity Probe B (GP-B) which was operated by NASA from 2004 to 2011 [5]. This space mission launched a satellite containing four gyroscopes in low earth orbit and accumulated data on their precession relative to a distant star, IM Pegas. The experiment measured geodesic precession to approximately 1%. It also measured another precession effect, the Lense-Thirring effect [6], due to the rotation of the earth to approximately 10%. This effect will be discussed in another Supplementary lecture in the future.

Geodesic precession has also been measured in strongly curved space-time through the observation of the spins of binary pulsars [9].

Appendix

Let’s solve the differential equation Eq. 9 which describes Thomas precession in special relativity,

sin t cos t sin t = ( 1) 1 cos t sin 2t cos t 1 𝑑𝑑 2 �𝑆𝑆2� 𝛾𝛾 − Ω � Ω 2 Ω Ω � �𝑆𝑆2� with the initial conditions𝑑𝑑𝑑𝑑 𝑆𝑆 (0) = and −(0) =Ω0. − Ω Ω 𝑆𝑆 1 2 Let’s proceed by “trial𝑆𝑆 and error”𝑎𝑎 using𝑆𝑆 some observations about precession made in the introductory comments to the lecture.

We try a solution that describes simple precession.

= cos , = sin (A.1) 1 2 where the parameters a and 𝑆𝑆should𝑎𝑎 be determined𝛼𝛼𝛼𝛼 𝑆𝑆 from𝑎𝑎 the𝛼𝛼 𝛼𝛼differential equation. The first component of Eq. 9 reads, 𝛼𝛼

= ( 1) (sin t cos t + sin t ) 𝑑𝑑 1 2 1 2 2 𝑆𝑆 𝛾𝛾 − Ω Ω Ω 𝑆𝑆 Ω 𝑆𝑆 Substituting the guess Eq.𝑑𝑑𝑑𝑑 A.1 into this expression,

sin = ( 1) a sin t (cos t cos + sin t sin ) 2 −𝑎𝑎𝑎𝑎 𝛼𝛼𝛼𝛼 sin𝛾𝛾 − = (Ω 1Ω) a sinΩ t cos𝛼𝛼(𝛼𝛼 ) Ω 𝛼𝛼𝛼𝛼 2 where we used the trig identity−𝑎𝑎𝑎𝑎 cos𝛼𝛼𝛼𝛼 cos𝛾𝛾 −+ sinΩ sin Ω= cosΩ( − α 𝑡𝑡). Next use the trig ( ) ( ) identity 2 sin cos = sin 𝐴𝐴+ sin𝐵𝐵 + 𝐴𝐴, so the𝐵𝐵 differential𝐴𝐴 − 𝐵𝐵equation becomes, 𝐴𝐴 𝐵𝐵 𝐴𝐴 − 𝐵𝐵 1 𝐴𝐴 𝐵𝐵 sin = (1 ) [sin + sin(2 ) ] 2 2 −𝛼𝛼 𝛼𝛼𝛼𝛼 − 𝛾𝛾 Ω 𝛼𝛼𝛼𝛼 Ω − α 𝑡𝑡 We learn that = in order to match the time dependences on both sides of the equation, but then the coefficients𝛼𝛼 Ω do not match and the trial solution fails! We learn that the trial solution was too simple to accommodate both the frequencies and amplitude of the differential equation Eq. 9. Let’s try a slightly more general trial function: the sum of two precession terms so we can accommodate those two aspects of the differential equation, = cos + cos , = sin + sin 1 2 Now the left hand side𝑆𝑆 of the𝑎𝑎 first𝛼𝛼 component𝛼𝛼 𝑏𝑏 𝛽𝛽 𝛽𝛽of the𝑆𝑆 differential𝑎𝑎 𝛼𝛼 equation𝛼𝛼 𝑏𝑏 reads,𝛽𝛽𝛽𝛽

( + ) ( ) sin + sin = (sin + sin ) + (sin sin ) 2 2 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎 𝛼𝛼𝛼𝛼 𝑏𝑏𝑏𝑏 𝛽𝛽𝛽𝛽 ( ) 𝛼𝛼𝛼𝛼( ) 𝛽𝛽𝛽𝛽 ( ) 𝛼𝛼𝛼𝛼 −( 𝛽𝛽) 𝛽𝛽 = ( + ) sin cos + ( ) sin cos (L) 𝛼𝛼+𝛽𝛽 𝑡𝑡 𝛼𝛼−𝛽𝛽 𝑡𝑡 𝛼𝛼−𝛽𝛽 𝑡𝑡 𝛼𝛼+𝛽𝛽 𝑡𝑡 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 2 2 𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏 2 2 where we used the trig identities, sin + sin = 2 sin cos and sin sin = 𝐴𝐴+𝐵𝐵 𝐴𝐴−𝐵𝐵 2sin cos . 𝐴𝐴 𝐵𝐵 2 2 𝐴𝐴 − 𝐵𝐵 𝐴𝐴−𝐵𝐵 𝐴𝐴+𝐵𝐵 2 2 On the right hand side,

(1 ) [a sin t cos t cos + sin t cos t cos + sin t sin + sin t sin ] 2 2 2 − 𝛾𝛾The ΩquantityΩ in backetsΩ can𝛼𝛼𝛼𝛼 be rewritten𝑏𝑏 Ω in theΩ same𝛽𝛽𝛽𝛽 form𝑎𝑎 as theΩ left side,𝛼𝛼𝛼𝛼 𝑏𝑏 Ω 𝛽𝛽𝛽𝛽

[a sin t (cos t cos + sin t sin ) + a sin t(cos t cos + sin t sin )]

Using Ωthe trig Ωidentity𝛼𝛼 𝛼𝛼 cos cosΩ +𝛼𝛼sin𝑡𝑡 sin Ω= cos(Ω )𝛽𝛽 twice,𝛽𝛽 thisΩ expression𝛽𝛽𝛽𝛽 becomes,

[a sin𝐴𝐴 t cos𝐵𝐵 ( 𝐴𝐴) +𝐵𝐵a sin cos𝐴𝐴(− 𝐵𝐵 ) ] (R)

Now compare the left (L)Ω and rightΩ − α(R)𝑡𝑡 hand sides.Ω TheirΩ − timeβ 𝑡𝑡 dependences match if

= , a = b (A.2) 𝛼𝛼+𝛽𝛽 2 Ω α β The magnitudes of the two sides match if,

2 = (1 ) ( + ) 2 The right hand side of this expression𝑎𝑎 can𝑎𝑎 be simplified− 𝛾𝛾 Ω 𝑎𝑎 using𝑏𝑏 Eq. A.2,

2 = (1 ) ( + ) = (1 ) a(1 + ) = (1 ) a(1 + ) 2 2 2 Finally, canceling𝑎𝑎𝛼𝛼 a −on𝛾𝛾 bothΩ sides,𝑎𝑎 𝑏𝑏 multiplying− 𝛾𝛾 throughΩ by𝑏𝑏⁄ 𝑎𝑎 and using− 𝛾𝛾 Eq.Ω A.2 again,𝛼𝛼⁄𝛽𝛽

2 = (1 ) ( + ) = 2(1 𝛽𝛽 ) (A.3) 2 2 2 which reduces to = 𝛼𝛼(1𝛼𝛼 )(1−+𝛾𝛾 )Ω 𝛼𝛼, 𝛽𝛽 − 𝛾𝛾 Ω 2 𝛼𝛼𝛼𝛼 − 𝛾𝛾 𝛾𝛾 Ω So, our final task is to solve Eq. A.2 and A.3,

= (1 )(1 + ) , + = 2 (A,4) 2 The resulting quadratic𝛼𝛼 equation𝛼𝛼 − gives,𝛾𝛾 𝛾𝛾 Ω 𝛼𝛼 𝛽𝛽 Ω

= (1 ) , = (1 + ) (A.5)

Finally, the parameters a and b𝛼𝛼 follow− from𝛾𝛾 Ω Eq. A.2,𝛽𝛽 𝛾𝛾 Ω

( ) = = ( ) (A.6) 𝑏𝑏 𝛼𝛼 1−𝛾𝛾 𝑎𝑎 𝛽𝛽 1+𝛾𝛾 Since the overall scale of the spin is not determined by the precession equation, we can choose,

= (1 + ) , = (1 ) (A.7)

Collecting everything, the solution𝑎𝑎 for the𝛾𝛾 spin𝑏𝑏 reads, − 𝛾𝛾

1 = [(1 + ) cos(1 ) + (1 ) cos(1 + ) ] 2 1 𝑆𝑆 1 𝑠𝑠 𝛾𝛾 − 𝛾𝛾 Ω 𝑡𝑡 − 𝛾𝛾 𝛾𝛾 Ω 𝑡𝑡 = [(1 + ) sin(1 ) + (1 ) sin(1 + ) ] 2 2 𝑆𝑆 𝑠𝑠 𝛾𝛾 − 𝛾𝛾 Ω𝑡𝑡 − 𝛾𝛾 𝛾𝛾 Ω 𝑡𝑡 as stated in the text.

One can check that the second differential equation for ( ), Eq. 9, is also satisfied by this 2 result. 𝑆𝑆 𝑡𝑡

References

[1] J. D. Jackson, Classical Electrodynamics, John Wiley & Sons, New York, 1962. [2] W. Rindler, Introduction to Special Relativity, Oxford University Press, Oxford, 1991.

[3] W. Rindler, Essential Relativity, Springer-Verlag, Berlin, 1971.

[4] P. Hoyng, Relativistic Astrophysics and Cosmology, A Primer, Springer Publishers, The Netherlands, 2006. [5] J. D. Fairbanks et. al, (eds.), Chapter 6.1-6.3, Freeman and Co, 1988. See the GP-B website:

http://einstein.stanford.edu/ [6] I. Ciufolini and E. C. Pavlis, Nature, 431, 950 (2004).

[7] H. Goldstein, Classical Mechanics, Addison-Wesley Publication Co., Inc., London, 1950.

[8] J. P. Lambare, Eur. J. Phys. 38, 045610 (2017). https://relativitydoctor.com/wp-

content/uploads/2018/04/Lambare_2017_Eur._J._Phys._38_045602.pdf

[9] J.M. Weisberg and J. H. Taylor, Ap. J. 576, 942 (2002).