Notes on Differential Geometry

Similar documents
First order differential equation Linear equation; Method of integrating factors

Einstein Equations for Tetrad Fields

Addition of angular momentum

MSLC Math 151 WI09 Exam 2 Review Solutions

The second condition says that a node α of the tree has exactly n children if the arity of its label is n.

Addition of angular momentum

Mathematics 1110H Calculus I: Limits, derivatives, and Integrals Trent University, Summer 2018 Solutions to the Actual Final Examination

Thomas Whitham Sixth Form

The Matrix Exponential

A RELATIVISTIC LAGRANGIAN FOR MULTIPLE CHARGED POINT-MASSES

The Matrix Exponential

The van der Waals interaction 1 D. E. Soper 2 University of Oregon 20 April 2012

Y 0. Standing Wave Interference between the incident & reflected waves Standing wave. A string with one end fixed on a wall

That is, we start with a general matrix: And end with a simpler matrix:

PHYS ,Fall 05, Term Exam #1, Oct., 12, 2005

SPH4U Electric Charges and Electric Fields Mr. LoRusso

Thomas Whitham Sixth Form

Quasi-Classical States of the Simple Harmonic Oscillator

Finite Element Analysis

u x v x dx u x v x v x u x dx d u x v x u x v x dx u x v x dx Integration by Parts Formula

perm4 A cnt 0 for for if A i 1 A i cnt cnt 1 cnt i j. j k. k l. i k. j l. i l

Division of Mechanics Lund University MULTIBODY DYNAMICS. Examination Name (write in block letters):.

PROBLEM SET Problem 1.

u 3 = u 3 (x 1, x 2, x 3 )

On the Hamiltonian of a Multi-Electron Atom

y cos x = cos xdx = sin x + c y = tan x + c sec x But, y = 1 when x = 0 giving c = 1. y = tan x + sec x (A1) (C4) OR y cos x = sin x + 1 [8]

Higher order derivatives

General Notes About 2007 AP Physics Scoring Guidelines

1.2 Faraday s law A changing magnetic field induces an electric field. Their relation is given by:

PHYSICS 489/1489 LECTURE 7: QUANTUM ELECTRODYNAMICS

Schrodinger Equation in 3-d

Pipe flow friction, small vs. big pipes

Sundials and Linear Algebra

UNTYPED LAMBDA CALCULUS (II)

There is an arbitrary overall complex phase that could be added to A, but since this makes no difference we set it to zero and choose A real.

Multiple Short Term Infusion Homework # 5 PHA 5127

Coupled Pendulums. Two normal modes.

Lie Groups HW7. Wang Shuai. November 2015

Self-Adjointness and Its Relationship to Quantum Mechanics. Ronald I. Frank 2016

Finite element discretization of Laplace and Poisson equations

Derivation of Electron-Electron Interaction Terms in the Multi-Electron Hamiltonian

a 1and x is any real number.

2. Background Material

INTEGRATION BY PARTS

ELECTRON-MUON SCATTERING

Lorentz force rotor formulation.

A Propagating Wave Packet Group Velocity Dispersion

BINOMIAL COEFFICIENTS INVOLVING INFINITE POWERS OF PRIMES. 1. Statement of results

EEO 401 Digital Signal Processing Prof. Mark Fowler

Additional Math (4047) Paper 2 (100 marks) y x. 2 d. d d

Chemistry 342 Spring, The Hydrogen Atom.

Indeterminate Forms and L Hôpital s Rule. Indeterminate Forms

High Energy Physics. Lecture 5 The Passage of Particles through Matter

San José State University Aerospace Engineering AE 138 Vector-Based Dynamics for Aerospace Applications, Fall 2016

COHORT MBA. Exponential function. MATH review (part2) by Lucian Mitroiu. The LOG and EXP functions. Properties: e e. lim.

The Transfer Function. The Transfer Function. The Transfer Function. The Transfer Function. The Transfer Function. The Transfer Function

BINOMIAL COEFFICIENTS INVOLVING INFINITE POWERS OF PRIMES

Alpha and beta decay equation practice

Grade 12 (MCV4UE) AP Calculus Page 1 of 5 Derivative of a Function & Differentiability

Introduction to the quantum theory of matter and Schrödinger s equation

SLAC KLYSTRON LECTURES

MA1506 Tutorial 2 Solutions. Question 1. (1a) 1 ) y x. e x. 1 exp (in general, Integrating factor is. ye dx. So ) (1b) e e. e c.

Analysis of Algorithms - Elementary graphs algorithms -

Appendix 2.3 General Solutions for the Step Response of Third- and Fourth-Order Systems (with some unpleasant surprises!)

Classical Magnetic Dipole

Elements of Statistical Thermodynamics

5.80 Small-Molecule Spectroscopy and Dynamics

Deepak Rajput

y = 2xe x + x 2 e x at (0, 3). solution: Since y is implicitly related to x we have to use implicit differentiation: 3 6y = 0 y = 1 2 x ln(b) ln(b)

Schematic of a mixed flow reactor (both advection and dispersion must be accounted for)

GÖDEL SPACETIME A THESIS SUBMITTED TO THE GRADUATE SCHOOL OF NATURAL AND APPLIED SCIENCES OF MIDDLE EAST TECHNICAL UNIVERSITY MEHMET KAVUK

1973 AP Calculus AB: Section I

u r du = ur+1 r + 1 du = ln u + C u sin u du = cos u + C cos u du = sin u + C sec u tan u du = sec u + C e u du = e u + C

First derivative analysis

Differential Equations

Part 7: Capacitance And Capacitors

Introduction to Arithmetic Geometry Fall 2013 Lecture #20 11/14/2013

SECTION where P (cos θ, sin θ) and Q(cos θ, sin θ) are polynomials in cos θ and sin θ, provided Q is never equal to zero.

cycle that does not cross any edges (including its own), then it has at least

1 Quaternion Analysis

Brief Introduction to Statistical Mechanics

von Neumann-Wigner theorem: level s repulsion and degenerate eigenvalues.

Computing and Communications -- Network Coding

ECE 407 Computer Aided Design for Electronic Systems. Instructor: Maria K. Michael. Overview. CAD tools for multi-level logic synthesis:

EXST Regression Techniques Page 1

Fourier Transforms and the Wave Equation. Key Mathematics: More Fourier transform theory, especially as applied to solving the wave equation.

Analysis of Algorithms - Elementary graphs algorithms -

Calculus II (MAC )

2008 AP Calculus BC Multiple Choice Exam

Lecture Outline. Skin Depth Power Flow 8/7/2018. EE 4347 Applied Electromagnetics. Topic 3e

COUNTING TAMELY RAMIFIED EXTENSIONS OF LOCAL FIELDS UP TO ISOMORPHISM

ME 321 Kinematics and Dynamics of Machines S. Lambert Winter 2002

GEOMETRICAL PHENOMENA IN THE PHYSICS OF SUBATOMIC PARTICLES. Eduard N. Klenov* Rostov-on-Don, Russia

A. Limits and Horizontal Asymptotes ( ) f x f x. f x. x "±# ( ).

As the matrix of operator B is Hermitian so its eigenvalues must be real. It only remains to diagonalize the minor M 11 of matrix B.

1997 AP Calculus AB: Section I, Part A

Middle East Technical University Department of Mechanical Engineering ME 413 Introduction to Finite Element Analysis

Exercise 1. Sketch the graph of the following function. (x 2

4. Money cannot be neutral in the short-run the neutrality of money is exclusively a medium run phenomenon.

Voltage, Current, Power, Series Resistance, Parallel Resistance, and Diodes

Transcription:

Nots from phz 6607, Spcial an Gnral Rlativity Univrsity of Floria, Fall 2004, Dtwilr Nots on Diffrntial Gomtry Ths nots ar not a substitut in any mannr for class lcturs. Plas lt m know if you fin rrors. I. CHRISTOFFEL SYMBOLS A covariant rivativ oprator which is compatibl with a mtric ncssarily satisfis so that a g bc = 0 = a g bc abg c acg b, 1 a g bc = abg c + acg b, b g ca = bcg a + bag c, c g ab = cag b + cbg a. 2 Th scon two of ths follow from th first with cyclic prmutations of th inics. A th first two an subtract th thir to obtain a g bc + b g ca c g ab = 2 abg c, 3 whr th symmtry of th Christoffl symbols is us. Now, raising th c inx rsults in ab = 1 2 gc a g bc + b g ca c g ab. 4 A. Intitis involving th Christoffl symbols an covariant rivativs a ab = 1 g x b g 5 a ξ a = 1 g x a gξ a 6 [a ξ b] = x ξ [a b] 7 a a ψ = 1 gg ab ψ 8 g x a x b a F [ab] = 1 g x a gf [ab] 9 for A ab = A ba : a A ab = gba c gaa gbc g a g x c 2 x c Aa 10 1

In a gnric coorinat systm 2 φ = a a φ = g ab a b φ g ab c ab b φ = 1 g a gg ab b φ. 11 It is oftn convnint to fin c g ab c ab = 1 g a gg ab, 12 so that 2 φ = g ab a b φ c c φ. 13 II. THE RICCI IDENTITY AND THE RIEMANN TENSOR Lt ξ a b an arbitrary vctor fil, an consir a b ξ c b a ξ c = a b ξ c bcξ ab ξ c cξ ac b ξ bξ b a ξ c acξ + ba ξ c cξ + bc a ξ aξ = a bc ξ ab ξ c cξ ac b ξ bξ + b ac ξ + ba ξ c cξ + bc a ξ aξ = a bc ξ ac b ξ bξ + b ac ξ + bc a ξ aξ = a bc ξ + ac bξ + b ac ξ bc aξ = a bc ξ + b ac ξ bc aξ + ac bξ, 14 whr th first quality follows from th scription of th covariant rivativ in trms of th Christoffl symbols, th scon from th commutation of partial rivativs, th thir from th symmtry of th Christoffl symbol, th fourth from th Libnitz rul for iffrntiation, an th fifth by rarranging trms. From th Ricci intity w also hav With ξ a bing arbitrary, it is ncssary that aftr a rarrangmnt of trms an inics. a b ξ c b a ξ c = R abc ξ. 15 R abc = a bc + b ac a bc + b ac, 16 A. Algbraic intitis of th Rimann tnsor Consir [a b c] ψ for an arbitrary scalar fil ψ. From th Ricci intity [a b] c ψ = 1 2 R abc ψ 17 2

Antisymmtrizing ovr a, b, an c givs [a b c] ψ = 1 2 R [abc] ψ 18 but [b c] ψ = 0 bcaus th connction bc is symmtric. Bcaus ψ is arbitrary, it follows that R [abc] = 0, 19 for any Rimann tnsor. A scon way of stating this sam intity rlis upon th antisymmtry of th first two inics of th Rimann tnsor an is R abc + R cab + R bca = 0. 20 Consir a b ξ c λ c, for arbitrary vctor fils ξ c an λ c. It follows that a b ξ c λ c = a λ c b ξ c + ξ c b λ c = λ c a b ξ c + a λ c b ξ c + ξ c a b λ c + a ξ c b λ c. 21 Antisymmtriz ovr a an b: th lft han si is zro bcaus th connction ab is symmtric; th scon an fourth trms on th right han si cancl ach othr. Us th Ricci intity on th thir trm on th right to obtain Th vctor λ c is arbitrary, an w conclu that 0 = λ c [a b] ξ c + ξ c R abc λ. 22 [a b] ξ c = R abc ξ c. 23 Officially, this is th Ricci intity for contravariant vctors. Our intrst is usually on a rivativ oprator which is compatibl with th spactim mtric g ab, in which cas w can rais an lowr inics on ithr si of a rivativ oprator. Thus, it also follows that From a straight application of th Ricci intity W conclu that At this point w hav shown that an that W may now conclu that [a b] ξ c = R ab c ξ c. 24 [a b] ξ c = R ab c ξ c. 25 R abc = R ab[c]. 26 R abc = R [ab][c] 27 R abc + R cab + R bca = 0. 28 R abc = R cab R bca = R cab + R bca = R cab R acb R bca R cba = R cba + R abc + R bac + R cab = R cba R bac R bac + R bac + R cab 29 3

whr w us Eq. 28 in th first quality, intrchang th first two inics of vry trm to obtain th scon quality, us Eq. 28 on both trms to obtain th thir quality, intrchang th last two inics of vry trm to obtain th fourth quality, an us Eq. 28 on only th scon trm to obtain th fifth quality. In th fifth lin th thir an fourth trms cancl, an w ar lft with R abc = R cba R bac + R cab = R cab R abc + R cab ; 30 th scon lin follows from intrchanging ach pair of inics on th right han si. This asily simplifis, an w finally hav R [ab][c] = R abc = R cab = R [c][ab]. 31 To simplify th unrstaning of th inpnnc of ths algbraic intitis, assum that Eq. 31 hols for a 4-inx covariant tnsor, R abc, an s how much aitional information is larn if Eq. 28 also hols. Eq. 31 immiatly implis that Eq. 28 hols if any two of th inics a, b or c ar qual. So nothing nw is larn from Eq. 28 unlss a, b an c ar all iffrnt. But, Eq. 31 also immiatly implis that Eq. 28 hols if is qual to any of a, b or c. W may conclu that Eq. 28 givs nw information, byon that contain in Eq. 31, only if a, b, c an ar all istinct. Now w ar in position to count up th numbr of algbraically inpnnt componnts of th Rimann tnsor of a four imnsional manifol. An antisymmtric pair of inics [ab] may b chosn in 6 iffrnt ways, so thr ar 6 ways to chos th [ab] in R abc an 6 ways to pick th [c]. Also, R abc = R [c][ab], so R abc looks lik a 6 6 symmtric matrix, which has 21 algbraically inpnnt componnts. Eq. 28 provis prcisly on mor algbraically inpnnt rlationship whn all four inics ar istinct, an w conclu that th Rimann tnsor has twnty algbraically inpnnt componnts on a four imnsional manifol. an B. Diffrntial intitis of th Rimann tnsor Th Bianchi intity is a iffrntial intity of th Rimann tnsor. Start with a [b c] µ = 1 2 a R bc m µ m = 1 2 µ m a R bc m + 1 2 R bc m a µ m, 32 [a b] c µ = 1 2 R abc m m µ + 1 2 R ab m c µ m. 33 Now, antisymmtriz ach of ths ovr [abc]. Th righthan sis of th antisymmtriz vrsions ach of ths quations ar qual, µ m [a R bc] m + R [bc m a] µ m = R [abc] m m µ + R [ab m c] µ m 34 4

Th algbraic intitis of th Rimann tnsor implis that th first trm on th right han si is zro an that two of th othr trms cancl with th rsult that Bcaus µ m is arbitrary, w conclu that µ m [a R bc] m = R [bc m a] µ m + R [ab m c] µ m = 0. 35 [a R bc] m = 0, 36 which is th Bianchi intity. Th contract form of th Bianchi intity follows from first contracting ovr a an m, Now contract ovr b an, 0 = 3 [a R bc] a = a R bc a + c R ab a + b R ca a = a R bc a c R b + b R c. 37 0 = a R bc ba c R b + b R c = a R c a c R + a R c a, 38 which may b writtn as a R c a 1 2 g c a R = 0. 39 This last rsult is also rfrr to as th Bianchi intity, or somtims th contract Bianchi intity. III. THE ALTERNATING TENSOR In four imnsional spac-tim ɛ abc = ɛ [abc] = ±1, 0 g 40 pning upon whthr a, b, c, is an vn, o or no prmutation of t, x, y, z. Also ɛ abc = ɛ [abc] 1 = ±1, 0. 41 g Th contractions of th prouct of th two ɛ s hav simpl xprssions. First, with no contractions ɛ abc ɛ fgh = 4!δ a [ δ f = 4! δ 4 aδ [f δf aδ [ δg aδ [f b δ cδ h] δh aδ [f b δg c δ ]. 42 For th scon quality, not that th right han si is xplicitly antisymmtric in [, f, g, h]; th factor of 1 is th rquir normalization. A similar stp is prform at th n of 4 th following quations. 5

an For on contraction, ɛ abc ɛ afgh = 3! For two contractions, ɛ abc ɛ abgh = 2! 1! δaδ a [f δf aδ [a δg aδ [f b δa c δ h] δh aδ [f b δg c δ a] = 3! 4 3 δ [f = 2! 1! δ f b δ[g c δ h] = 3! 1! δ[f δg b δ[f c δ h] δh b δ c [g δ f] δbδ b c [g δ h] δg b δ[b c δ h] δh b δ c [g δ b] 43 = 2! 1! 4 2δ c [g δ h] = 2! 2! δ[g c δ h] = 1! 2! δc g δ h δc h δ g.44 For thr contractions, ɛ abc ɛ abch = 1! 2! δ c cδ h δ h c δ c = 1! 2! 4 1δ h = 1! 3! δ h. 45 Finally for all four pairs of inics contract, ɛ abc ɛ abc = 3! δ = 4! 46 Summarizing ths formula, with iffrnt labling of th inics, w hav ɛ abc ɛ fgh = 4!δ [ a δ f 47 ɛ abc ɛ fg = 3! 1! δ [ a δ f b δg] c 48 ɛ abc ɛ fc = 2! 2! δ a [ δ f] b 49 ɛ abc ɛ bc = 1! 3! δa. 50 ɛ abc ɛ abc = 4!. 51 IV. THE PROJECTION OPERATOR An obsrvr with four-vlocity u a may construct a spcial spatial altrnating tnsor ɛ abc = ɛ [abc] ɛ abc u = ±1, 0u t or u t g 52 An an obsrvr may us a projction oprator h a b to projct tnsor inics prpnicular to his four-vlocity, h a b g a b + u a u b. 53 Not that h a bu b = g a b + u a u b u b = u b u b = 0, 54 whr th scon quality follows from th normalization of th four-vlocity, u b u b = 1. It asily follows that h a bh b c = h a c, 55 as woul b xpct for a projction oprator. Any tnsor inx which is prpnicular to u a may b rais or lowr by ithr g ab an g ab or by h ab an h ab. If a normaliz, timlik vctor fil u a is hyprsurfac orthogonal, thn th projction oprator h ab also plays th rol of th mtric of th thr imnsional, spatial hypr-surfac which is prpnicular to u a. 6

V. ELECTRICITY AND MAGNETISM an Maxwll s quations for th lctromagntic fil F ab = F [ab] with a sourc J a : [a F bc] = 0 56 b F ab = 4πJ a. 57 For a givn obsrvr with four-vlocity u a, th lctromagntic fil may b compos into its lctric E a an magntic B a parts by projcting F ab paralll an prpnicular to u a, E a = F ab u b 58 an B a = 1 2 ɛ abcf bc u. 59 It is asy to show that E a u a = 0 an that B a u a = 0 so th lctric an magntic fils ar spatial vctors to th obsrvr, u a. W may irctly writ F ab in trms of its componnts as F ab = 2u [a E b] + ɛ abc B c u ; 60 this may b vrifi by substituting this xprssion into th abov quations for E a an B a. Th forc on a charg particl of charg q an mass m moving with four-vlocity v a is th right han si of mv b b v a = qf ab v b, 61 which is th quation of motion of a charg particl in fr-fall through an lctromagntic fil. VI. MAXWELL S EQUATIONS an Maxwll s quations for th lctromagntic fil F ab = F [ab] with a sourc J a : [a F bc] = 0 62 b F ab = 4πJ a. 63 Imagin a clou of charg ust moving through spactim with four-vlocity v a an co-moving numbr nsity n. Each bit of ust has a mass m an charg q. Th consrvation of ust implis that a nv a = 0. 64 An obsrvr with four-vlocity u a ss a charg nsity qnu a v a an currnt nsity qnh a bv b. To s that Maxwll s quations rquir th consrvation of charg, valuat a b F c b a F c = R abc F + R ab F c 65 7

from th Ricci intity. Now contract Eq. 65 with g ac an g b an us th antisymmtry of F ab to s that th lft han si of th contract Eq. 65 is Th right han si of th contract Eq. 65 is a b F ab b a F ab = 2 a b F ab. 66 g ac g b R abc F + g ac g b R ab F c = R b F b R a F a = 0 67 whr th first quality follows from th finition of th Ricci tnsor, R b R a ba, an th scon follows from th symmtry of R ab an th anti-symmtry of F ab. Thus, from Eq. 63 4π a J a = a b F ab = 0, 68 an th four-currnt nsity must b consrv for th consistncy of Maxwll s quations. Eq. 62, [a F bc] = 0, is th intgrability conition for th local xistnc of a vctor potntial A a, such that a A b b A a = F ab. 69 This is vry similar to th Euclian gomtry thorm that if F = 0 thn thr xists a vctor A such that F = A. Lt us fin th quation govrning th vctor potntial. From Eq. 69 c F ab = c a A b b A a = R cab A + a c A b c b A a, 70 from th Ricci intity. Aftr contraction with g bc, this bcoms b F ab = 4πJ a = R a A + a b A b b b A a. 71 This is quivalnt to b b A a a b A b R a A = 4πJ a. 72 In th Lorntz gaug, whr b A b = 0, this simplifis to b b A a R a A = 4πJ a. 73 A. Altrnativ E&M Gnrally, whn w consir an ara of physics which is wll unrstoo in spcial rlativity, such as lctricity an magntism, an try to fin th gnralization of th rlvant quations to curv spactim, w o as littl as possibl to th quations an typically just rplac orinary rivativs with rspct to Minkowski coorinats by covariant rivativs with rspct to an arbitrary coorinat systm. Howvr, this procss is not unambiguous. With this in min consir th vctor potntial A a fin in trms of th lctromagntic fil by a A b b A a = F ab. 74 In flat spactim th vctor potntial satisfis b b A a a b A b = 4πJ a. 75 8

It might appar rasonabl to vlop curv-spactim lctricity an magntism, by starting irctly with ths quations for A a an F ab whil consiring th rivativs to b covariant rivativs of curv spactim. Show that this vrsion of curv-spactim lctricity an magntism has an unplasant fatur. Hint: Look at th nots to s th accpt vrsion of curv-spactim lctricity an magntism. From Eq. 75, w can valuat th ivrgnc of J a. Spcifically, Focus on th first trm on th right han si. 4π a J a = a b b A a a a b A b 76 a b b A a = b a b A a + R a b b A a + R a ba b A = b a b A a R a A a + R b b A = b a b A a = b b a A a + R ab a A = b b a A a + b R b A 77 whr th first quality follows from th Ricci intity aftr intrchanging th orr of th covariant rivativs, th scon follows from th finition of th Ricci tnsor in trms of th Rimann tnsor, th thir follows from th symmtry of th Ricci tnsor, th fourth follows from again intrchanging rivativs an using th Ricci intity, an th fifth from again using th finition of th Ricci tnsor in trms of th Rimann tnsor. Now substitut this final rsult back into Eq. 76 to obtain 4π a J a = b b a A a + b R b A a a b A b = b R b A. 78 Gnrally, th right han si of this last xprssion is not zro, an w s that in this othr vrsion of lctricity an magntism in curv spactim charg is not consrv. This is consir an unplasant fatur. VII. STRESS-ENERGY TENSOR Th strss-nrgy tnsor for th lctromagntic fil is Th consrvation of strss-nrgy thn implis 4πT ab = F a c F bc 1 4 g abf c F c. 79 4π a T ab = a F a c F bc + F ac a F bc 1 2 F c b F c = a F a c F bc + 1 2 F ac a F bc F ac c F ba F ac b F ac = a F a c F bc 1 2 f ac a F cb + c F ba + b F ac = a F a c F bc, = 4πJ a F ba = 4πJ a F ab 80 9

whr th pnultimat lin follows from Eq. 62 an th last lin from Eq. 63. Exprss T ab in trms of th vctor potntial as 4πT ab = a A c c A a b A c c A b 1 4 g ab c A A c c A A c = a A c c A a b A c c A b 1 2 g ab c A c A A c. 81 10