r = x 2 + y 2 and h = z y = r sin sin ϕ

Σχετικά έγγραφα
e t e r Cylindrical and Spherical Coordinate Representation of grad, div, curl and 2

Curvilinear Systems of Coordinates

Solutions Ph 236a Week 2

Laplace s Equation in Spherical Polar Coördinates

Strain and stress tensors in spherical coordinates

Fundamental Equations of Fluid Mechanics

Example 1: THE ELECTRIC DIPOLE

Analytical Expression for Hessian

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

ANTENNAS and WAVE PROPAGATION. Solution Manual

Space Physics (I) [AP-3044] Lecture 1 by Ling-Hsiao Lyu Oct Lecture 1. Dipole Magnetic Field and Equations of Magnetic Field Lines

Tutorial Note - Week 09 - Solution

(a,b) Let s review the general definitions of trig functions first. (See back cover of your book) sin θ = b/r cos θ = a/r tan θ = b/a, a 0

21. Stresses Around a Hole (I) 21. Stresses Around a Hole (I) I Main Topics

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

General Relativity (225A) Fall 2013 Assignment 5 Solutions

The Laplacian in Spherical Polar Coordinates

Section 8.3 Trigonometric Equations

Example Sheet 3 Solutions

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

Oscillating dipole system Suppose we have two small spheres separated by a distance s. The charge on one sphere changes with time and is described by

2 Composition. Invertible Mappings

Finite Field Problems: Solutions

Parametrized Surfaces

Uniform Convergence of Fourier Series Michael Taylor

Problems in curvilinear coordinates

EE512: Error Control Coding

Math221: HW# 1 solutions

Homework 8 Model Solution Section

4.2 Differential Equations in Polar Coordinates

Inverse trigonometric functions & General Solution of Trigonometric Equations

b. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!

Exercises to Statistics of Material Fatigue No. 5

1 String with massive end-points

Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

Srednicki Chapter 55

VEKTORANALYS. CURVILINEAR COORDINATES (kroklinjiga koordinatsytem) Kursvecka 4. Kapitel 10 Sidor

Approximation of distance between locations on earth given by latitude and longitude

Section 7.6 Double and Half Angle Formulas

4.6 Autoregressive Moving Average Model ARMA(1,1)

ST5224: Advanced Statistical Theory II

Solutions to Exercise Sheet 5

6.3 Forecasting ARMA processes

Slide 1 of 18 Tensors in Mathematica 9: Built-In Capabilities. George E. Hrabovsky MAST

dx x ψ, we should find a similar expression for rθφ L ψ. From L = R P and our knowledge of momentum operators, it follows that + e y z d

Second Order Partial Differential Equations

Orbital angular momentum and the spherical harmonics

Integrals in cylindrical, spherical coordinates (Sect. 15.7)

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

Tutorial problem set 6,

Problem Set 3: Solutions

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

Statistical Inference I Locally most powerful tests

Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1

Matrices and Determinants

C.S. 430 Assignment 6, Sample Solutions

Partial Differential Equations in Biology The boundary element method. March 26, 2013

Other Test Constructions: Likelihood Ratio & Bayes Tests

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Homework 3 Solutions

derivation of the Laplacian from rectangular to spherical coordinates

Areas and Lengths in Polar Coordinates

Section 9.2 Polar Equations and Graphs

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)

( y) Partial Differential Equations

Reminders: linear functions

Every set of first-order formulas is equivalent to an independent set

Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequality for metrics: Let (X, d) be a metric space and let x, y, z X.

Geometry of the 2-sphere

Parallel transport and geodesics

1 Full derivation of the Schwarzschild solution

Areas and Lengths in Polar Coordinates

9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr

Numerical Analysis FMN011

Congruence Classes of Invertible Matrices of Order 3 over F 2

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

D Alembert s Solution to the Wave Equation

Matrix Hartree-Fock Equations for a Closed Shell System

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

Differentiation exercise show differential equation

Exercise 1.1. Verify that if we apply GS to the coordinate basis Gauss form ds 2 = E(u, v)du 2 + 2F (u, v)dudv + G(u, v)dv 2

CRASH COURSE IN PRECALCULUS

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

MATH 38061/MATH48061/MATH68061: MULTIVARIATE STATISTICS Solutions to Problems on Matrix Algebra

CHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD

Written Examination. Antennas and Propagation (AA ) April 26, 2017.

Lecture 34 Bootstrap confidence intervals

1 3D Helmholtz Equation

Geodesic Equations for the Wormhole Metric

Answer sheet: Third Midterm for Math 2339

Trigonometry 1.TRIGONOMETRIC RATIOS

PARTIAL NOTES for 6.1 Trigonometric Identities

The challenges of non-stable predicates

On a four-dimensional hyperbolic manifold with finite volume

Transcript:

Homewok 4. Solutions Calculate the Chistoffel symbols of the canonical flat connection in E 3 in a cylindical coodinates x cos ϕ, y sin ϕ, z h, b spheical coodinates. Fo the case of sphee ty to make calculations at least fo components Γ, Γ θ, Γ ϕ, Γ θθ,..., Γ ϕϕ Remak One can calculate Chistoffel symbols using Levi-Civita Theoem. Thee is a thid way to calculate Chistoffel symbols: It using appoach of Lagangian. This is the easiest way. see the Homewok 6 In cylindical coodinates, ϕ, h we have { x cos ϕ y sin ϕ z h x 2 + y 2 and ϕ actna y x h z We know that in Catesian coodinates all Chistoffel symbols vanish. Hence in cylindical coodinates see in detail lectue notes: Γ 2 x 2 x + 2 y 2 y + 2 z 2 z 0, since Γ ϕ Γ ϕ 2 x ϕ x + 2 y ϕ y + 2 z sin ϕ cos ϕ + sin ϕ cos ϕ 0. ϕ z Γ ϕϕ 2 x 2 ϕ x + 2 y 2 ϕ y + 2 z 2 ϕ z xx y y. Γ ϕ 2 x ϕ 2 x + 2 y ϕ 2 y + 2 z ϕ 2 z 0. Γ ϕ ϕ Γ ϕ ϕ 2 x ϕ ϕ x + 2 y ϕ ϕ y + 2 z ϕ sin ϕ y ϕ z 2 + cos ϕ x 2 All symbols Γ h, Γ h vanish 2 x h... 2 y h... Γ ϕ ϕϕ 2 x ϕ 2 ϕ x + 2 y ϕ 2 ϕ y + 2 z ϕ 2 ϕ z x x 2 y y 2 0. 2 z h... 0 Fo all symbols Γ h Γ h 2 z all symbols Γ h vanish. Γ h Γ h Γ hh Γ ϕh Γ hϕ Γ ϕ h dots 00 h since x h h y 0 and y. On the othe hand all 2 z vanish. Hence b spheical coodinates { x sin cos ϕ y sin sin ϕ z cos θ x 2 + y 2 + z 2 z θ accos ϕ actan y x x 2 +y 2 +z 2 We aleady know the fast way to calculate Chistoffel symbol using Lagangian of fee paticle and this method wok fo a flat connection since flat connection is a Levi-Civita connection fo Euclidean metic So pefom now bute foce calculations only fo some components. Then late in homewok 6 we will calculate using vey quickly Lagangian of fee paticle.

Γ 0 since 2 x i 2 0. Γ θ Γ θ 2 x θ x + 2 y θ y + 2 z θ z cos θ cos ϕx + cos θ sin ϕy sin θ z 0, Γ θθ 2 x 2 θ x + 2 y 2 θ y + 2 z 2 θ z sin θ cos ϕx sin θ sin ϕy cos θ z Γ ϕ Γ ϕ 2 x ϕ x + 2 y ϕ y + 2 z ϕ z sin θ sin ϕx + sin θ cos ϕy 0 and so on... 2 a Conside a connection such that its Chistoffel symbols ae symmetic in a given coodinate system: Γ i km Γi mk. Show that they ae symmetic in an abitay coodinate system. b Show that the Chistoffel symbols of connection ae symmetic in any coodinate system if and only if X Y Y X [X, Y] 0, fo abitay vecto fields X, Y. c Conside fo an abitay connection the following opeation on the vecto fields: and find its popeties. Hence Solution a Let Γ i km Γi mk But Γ i km Γi mk and SX, Y X Y Y X [X, Y]. to pove that Γi k m Γi m k Γ i k m Γ i m k x x x m xk x k x xi x i xi x i x k x k x m x m m. Hence x m x m Γ i km + x k x k Γ i mk + x x i x k x m x. x x i x m x k x Γ i m k xi x i x m x m x k x k Γ i mk + x x i x m x k x xi x i x m x m x k x k Γ i km + x x i x k x m x Γi k m. b The elation X Y Y X [X, Y] 0 holds fo all fields if and only if it holds fo all basic fields. One can easy check it using axioms of connection see the next pat. Conside X x, Y i x then since [ j i, j ] 0 we have that X Y Y X [X, Y] i j j i Γ k ij k Γ k ji k Γ k ij Γ k ji k 0 We see that commutato fo basic fields X Y Y X [X, Y] 0 if and only if Γ k ij Γk ji 0. c One can easy check it by staightfowad calculations o using axioms fo connection that SX, Y is a vecto-valued bilinea fom on vectos. In paticulaly SfX, Y fsx, Y fo an abitay smooth function. Show this just using axioms defining connection: SfX, Y fx Y Y fx [fx, Y] f X Y f Y X Y fx + [Y, fx] 2

f X Y f Y X Y fx + Y fx + f [Y, X] f X Y Y X [X, Y] fsx, Y 3 Let, 2 be two diffeent connections. Let Γ i km and 2 Γ i km be the Chistoffel symbols of connections and 2 espectively. a Find the tansfomation law fo the object : Tkm i Γ i km 2 Γ i km unde a change of coodinates. Show that it is tenso. 2 b? Conside an opeation 2 on vecto fields and find its popeties. Chistoffel symbols of both connections tansfom accoding the law. The second tem is the same. Hence it vanishes fo thei diffeence: T i k m Γ i k m 2 Γ i k m xi x i x k x k We see that T i km tansfoms as a tenso of the type 2 x m Γ i x m km 2 Γ i km xi x i. x k x k x m x m T i km b One can do it in invaiant way. Using axioms of connection study T 2 is a vecto field. Conside T X, Y X Y 2X Y Show that T fx, Y ft X, Y fo an abitay smooth function, i.e. it does not possesses deivatives: T fx, Y fx Y 2fX Y X fy + f X Y X fy f 2X Y ft X, Y. 4 a Conside t m Γ i im. Show that the tansfomation law fo t m is t m b Show that this law can be witten as t m xm x m t m + xm x m t m + x m 2 x x k x m x k x. x log det x. Solution. Using tansfomation law we have that xi x i x k x i t m Γ i i m δk i. Hence xi x k x i x i x m x m Γ i km + x x i x i x m x t m Γ i i m xi x k x i x i x m Γ i x km+ x x i m x i x m x δk i x m Γ i x km+ x x i m x i x m x xm t m + x x i x m x i x m x. b When calculating x log det x m x use vey impotant fomula: δ det A det AT A δa δ log det A T A δa. Hence x m x log det x xi 2 x x x i x m 3

and we come to tansfomation law fo. To deduce the fomula fo δ det A notice that deta + δa det A det + A δa and use the elation: det + δa + T δa + Oδ 2 A 5 Calculate Chistoffel symbols of the connection induced on the suface M in E n equipped with canonical flat connection. a M S in E 2 b M paabola y x 2 in E 2 c M cylinde,cone,sphee in E 3. d saddle z xy Solution. a Conside pola coodinate on S, x R cos ϕ, y R sin ϕ. to define the connection on S induced by the canonical flat connection on E 2. It suffices to define ϕ ϕ Γϕ ϕϕ ϕ. Recall the geneal ule. Let u α : x i x i u α is embedded suface in Euclidean space E n. The basic u α vectos. To take the induced covaiant deivative X Y fo two vectos X, Y we take a usual deivative of vecto Y along vecto X the deivative with espect to canonical flat connection: in Catesian coodiantes is just usual deivatives of components then we take the component of the answe, since in geneal deivative of vecto Y along vecto X is not to suface: u u α u α u β Γγ αβ u γ canonical α u β 2 u u α u β canonical α u β is just usual deivative in Euclidean space since fo canonical connection all Chistoffel symbols vanish. In the case of -dimensional manifold, cuve it is just ial acceleation!: u u Γu uu u canonical u u Fo the cicle S, x R cos ϕ, y R sin ϕ, in E 2. ϕ ϕ x ϕ x + y ϕ y ϕ d 2 u du 2 R sin ϕ x + R cos ϕ y, ϕ Γϕ ϕϕ ϕ canonic. ϕ ϕ ϕ ϕ R sin ϕ ϕ x + ϕ R cos ϕ R cos ϕ y x R sin ϕ y a since the vecto R cos ϕ x R sin ϕ y is othogonal to the vecto ϕ. In othe wods it means that acceleation is centipetal: ial acceleation equals to zeo. We see that in coodinate ϕ, Γ ϕ ϕϕ 0. Additional wok: Pefom calculation of Chistoffel symbol in steeogaphic coodinate t: x 2tR2 R 2 + t 2, y Rt2 R 2 t 2 + R 2. 4 0,

In this case t t t x t t Γt tt t x + y t y 2R 2 R 2 + t 2 2 canonic. t t R 2 t 2 t t 4t t 2 + R 2 2R 2 t + R 2 + t 2 2 2t x + 2R y x + 2tR, x tt In this case tt is not othogonal to velocity: to calculate tt we need to extact its othogonal component: tt tt tt, n t n n t R 2 + t 2 2tRx + t 2 R 2 y, whee t, n 0. Hence tt, n t 4t t 2 + R 2 t + We come to the answe: 4R3 t 2 +R 2 and 2 2R 2 R 2 + t 2 2 2t x + 2R + y t t tt tt tt, n t n 4R 3 t 2 + R 2 2 2t t 2 + R 2 t, i.e.γ t tt 2t t 2 + R 2 R 2 + t 2 2tRx + t 2 R 2 y 2t t 2 + R 2 t Of couse we could calculate the Chistoffel symbol in steeogaphic coodinates just using the fact that we aleady know the Chistoffel symbol in pola coodinates: Γ ϕ ϕϕ 0, hence Γ t tt dt dϕ dϕ dϕ dx dx Γϕ ϕϕ + d2 ϕ dt 2 dt dϕ d2 ϕ dt dt 2 dϕ It is easy to see that t R tan π 4 + ϕ 2, i.e. ϕ 2 actan t R π 2 and Γ t tt d2 ϕ dt 2 d dt dϕ 2 ϕ dt 2 dϕ dt 2t t 2 + R 2. b Fo paabola x t, y t 2 t t t x t x + y t t Γt tt t canonic. t t t t y x + 2t y, tt 2 y To calculate tt we need to extact its othogonal component: tt tt tt, n t n, whee n is an othogonal unit vecto: n, t 0, n, n : n t tt tt tt, n t n 2 y + 4t 2 2t x + y. 2 y, 2t x + y + 4t 2 + 4t 2 2t x + y 5

We come to the answe: 4t + 4t 2 x + 8t2 + 4t 2 y 4t + 4t 2 x + 2t y 4t + 4t 2 t t t 4t + 4t 2 t, i.e.γ t tt 4t + 4t 2 Remak Do not be supised by esemblance of the answe to the answe fo cicle in steeogaphic coodinates. c cylinde, cone and sphee a Cylinde{ x a cos ϕ h, ϕ: y a sin ϕ. z h h h Calculate Hence Γ h hh Γϕ hh 0 0 0, ϕ ϕ Hence Γ h hϕ Γh ϕh Γϕ hϕ Γϕ ϕh 0. a sin ϕ a cos ϕ 0 h h Γ h hh h + Γ ϕ hh ϕ h ϕ ϕ h Γ h hϕ h + Γ ϕ hϕ ϕ ϕ ϕ Γ h ϕϕ h + Γ ϕ ϕϕ ϕ 2 h 2 0 since hh 0. 2 0 since hϕ 0 hϕ 2 a cos ϕ a sin ϕ ϕϕ 0 a cos ϕ since the vecto ϕϕ a sin ϕ is othogonal to the suface of cylinde. Hence Γ h hϕ Γh ϕh Γϕ hϕ 0 Γ ϕ ϕh 0 We see that fo cylinde all Chistoffel symbols in cylindical coodinates vanish. This is not big supise: in cylindical coodinates metic equals dh 2 a 2 dϕ 2. This due to Levi-Ciovita theoem one can see that Levi-Civita which equals to induced connection vanishes since allcoefficients ae constants. Fo cone: see Cousewok poblem 3. Fo the sphee θ, ϕ: θ θ { x R sin θ cos ϕ y R sin θ sin ϕ z R cos θ R cos θ cos ϕ R cos θ sin ϕ, R sin θ, we have ϕ ϕ 0 R sin θ sin ϕ R sin θ cos ϕ R sin θ cos ϕ, n R sin θ sin ϕ 0 R cos θ Calculate θ θ Γ θ θθ θ + Γ ϕ θθ ϕ 2 θ 2 0 6

since 2 θ Rn is othogonal to the sphee. Hence Γ θ 2 θθ Γϕ θθ 0. Now calculate θ ϕ Γ θ θϕ θ + Γ ϕ 2 θϕ ϕ. θϕ hence Γ θ θϕ 0, Γϕ θϕ cotan θ hence Now calculate θ ϕ Γ θ θϕ θ + Γ ϕ θϕ ϕ 2 θϕ cotan θ ϕ, ϕ θ Γ θ ϕθ θ + Γ ϕ ϕθ ϕ θ ϕ Γ θ θϕ θ + Γ ϕ θϕ ϕ 2 cotan θ ϕ, i.e. θϕ 2 ϕθ cotan θ ϕ, 2. ϕθ 2 cotan θ ϕ, i.e. θϕ Γ θ ϕθ 0, Γϕ ϕθ cotan θ. Of couse we did not need to pefom these calculations: since is symmetic connection and ϕ θ θ ϕ, i.e. and finally Γ θ ϕθ Γ θ θϕ 0 Γ ϕ ϕθ Γϕ θϕ cotan θ. ϕ ϕ Γ θ ϕϕ θ + Γ ϕ ϕϕ ϕ 2 ϕ 2. 2 R sin θ cos ϕ R cos θ cos ϕ sin θ cos ϕ ϕ 2 sin θ cos θ R sin 2 θ sin θ cos θ θ R sin 2 θn, hence R sin θ sin ϕ 0 Γ θ ϕϕ sin θ cos θ, Γ ϕ ϕϕ 0. ϕ ϕ Γ θ ϕϕ θ + Γ ϕ ϕϕ ϕ R cos θ sin ϕ R sin θ sin θ sin ϕ cos θ 2 sin θ cos θ θ, i.e. ϕϕ { x u Fo saddle z xy: u, v: y v, u u 0, v v 0 It will be useful also z uv v u to use the nomal unit vecto n v +u u. 2 +v 2 Calculate: u u Γ u uu u + Γ v 2 uu v u 2 uu 0 since uu 0. 7

Hence Γ u uu Γ v uu 0. Analogously Γ u vv Γ v vv 0 since vv 0. Now calculate Γ u uv, Γ v uv, Γ u vu, Γ v vu: u v v u Γ u uv u + Γ v uv v uv 0 0 Using nomal unit vecto n we have: uv uv uv, n n Γ u uv u + Γ v uv v 0 0 v + u 2 + v 2 u u 2 + v 2 0 0 0 0, v u v u + u2 + v 2 + u2 + v 2 v + u 2 + v 2 0 v + u + u 2 + v 2 0 u u v u + u v + u 2 + v 2. Hence Γ u uv Γ u v vu +u 2 +v and Γ v 2 uv Γ v u vu +u 2 +v. 2 Sue one may calculate this connection as Levi-Civita connction of the induced Riemannian metic using explicit Levi-Civita fomula o using method of Lagangian of fee paticle. 8