A Short Introduction to Tensor Analysis

Σχετικά έγγραφα
A Short Introduction to Tensors

A Short Introduction to Tensor Analysis 2

Parallel transport and geodesics

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3

Cosmological Space-Times

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

Homework 8 Model Solution Section

derivation of the Laplacian from rectangular to spherical coordinates

Tutorial problem set 6,

Geodesic Equations for the Wormhole Metric

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

Space-Time Symmetries

Riemannian Curvature

Symmetry. March 31, 2013

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:

Areas and Lengths in Polar Coordinates

Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.

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

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

Second Order Partial Differential Equations

EE512: Error Control Coding

2 Composition. Invertible Mappings

D Alembert s Solution to the Wave Equation

Example Sheet 3 Solutions

Parametrized Surfaces

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

Reminders: linear functions

Areas and Lengths in Polar Coordinates

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

Geometry of the 2-sphere

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

Higher Derivative Gravity Theories

Spherical Coordinates

4.6 Autoregressive Moving Average Model ARMA(1,1)

Appendix A. Curvilinear coordinates. A.1 Lamé coefficients. Consider set of equations. ξ i = ξ i (x 1,x 2,x 3 ), i = 1,2,3

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

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

Concrete Mathematics Exercises from 30 September 2016

Sixth lecture September 21, 2006

1 String with massive end-points

Section 9.2 Polar Equations and Graphs

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

Section 8.3 Trigonometric Equations

Finite Field Problems: Solutions

PARTIAL NOTES for 6.1 Trigonometric Identities

If we restrict the domain of y = sin x to [ π, π ], the restrict function. y = sin x, π 2 x π 2

You may not start to read the questions printed on the subsequent pages until instructed to do so by the Invigilator.

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

Symmetric Stress-Energy Tensor

If we restrict the domain of y = sin x to [ π 2, π 2

Matrices and Determinants

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)

Integrals in cylindrical, spherical coordinates (Sect. 15.7)

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

1. (a) (5 points) Find the unit tangent and unit normal vectors T and N to the curve. r(t) = 3cost, 4t, 3sint

Inverse trigonometric functions & General Solution of Trigonometric Equations

THE ENERGY-MOMENTUM TENSOR IN CLASSICAL FIELD THEORY

Math221: HW# 1 solutions

Homework 3 Solutions

Statistical Inference I Locally most powerful tests

Solutions to Exercise Sheet 5

Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007

Variational Wavefunction for the Helium Atom

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.

C.S. 430 Assignment 6, Sample Solutions

The kinetic and potential energies as T = 1 2. (m i η2 i k(η i+1 η i ) 2 ). (3) The Hooke s law F = Y ξ, (6) with a discrete analog

( y) Partial Differential Equations

5. Choice under Uncertainty

The Simply Typed Lambda Calculus

ECE Spring Prof. David R. Jackson ECE Dept. Notes 2

Solution Series 9. i=1 x i and i=1 x i.

1 Lorentz transformation of the Maxwell equations

= {{D α, D α }, D α }. = [D α, 4iσ µ α α D α µ ] = 4iσ µ α α [Dα, D α ] µ.

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

Problem 3.1 Vector A starts at point (1, 1, 3) and ends at point (2, 1,0). Find a unit vector in the direction of A. Solution: A = 1+9 = 3.

Orbital angular momentum and the spherical harmonics

6.3 Forecasting ARMA processes

w o = R 1 p. (1) R = p =. = 1

Mock Exam 7. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q2 (a) (1 + kx) n 1M + 1A = (1) =

Second Order RLC Filters

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

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

CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS

Pg The perimeter is P = 3x The area of a triangle is. where b is the base, h is the height. In our case b = x, then the area is

Γενική Θεωρία της Σχετικότητας

( ) 2 and compare to M.

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

Problem Set 3: Solutions

Answer sheet: Third Midterm for Math 2339

CRASH COURSE IN PRECALCULUS

On a four-dimensional hyperbolic manifold with finite volume

Curved spacetime and General Relativity

A Note on Intuitionistic Fuzzy. Equivalence Relation

Physics 582, Problem Set 2 Solutions

Strain gauge and rosettes

CHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD

Transcript:

June 20, 2008

Scalars and Vectors An n-dim manifold is a space M on every point of which we can assign n numbers (x 1,x 2,...,x n ) the coordinates, in such a way that there will be a one to one correspondence between the points and the n numbers. The manifold cannot be always covered by a single system of coordinates and there is not a preferable one either. The coordinates ( of the point P are connected by relations of the form: x µ = x µ x 1, x 2,..., x n) for µ = 1,..., n and their inverse x µ = x µ (x 1, x 2,..., x n ) for µ = 1,..., n. If there exist µ x A µ ν = x ν and A ν µ = x ν det A µ ν (1) x µ then the manifold is called differential.

Any physical quantity, e.g. the velocity of a particle, is determined by a set of numerical values - its components - which depend on the coordinate system. Studying the way in which these values change with the coordinate system leads to the concept of tensor. With the help of this concept we can express the physical laws by tensor equations, which have the same form in every coordinate system. Scalar field : is any physical quantity determined by a single numerical value i.e. just one component which is independent of the coordinate system (mass, charge,...)

Any physical quantity, e.g. the velocity of a particle, is determined by a set of numerical values - its components - which depend on the coordinate system. Studying the way in which these values change with the coordinate system leads to the concept of tensor. With the help of this concept we can express the physical laws by tensor equations, which have the same form in every coordinate system. Scalar field : is any physical quantity determined by a single numerical value i.e. just one component which is independent of the coordinate system (mass, charge,...) Vector field (contravariant): an example is the infinitesimal displacement vector, leading from a point A with coordinates x µ to a neighbouring point A with coordinates x µ + dx µ. The components of such a vector are the differentials dx µ.

Vector Transformations From the infinitesimal vector AA with components dx µ we can construct a finite vector v µ defined at A. This will be the tangent vector of the curve x µ = f µ (λ) where the points A and A correspond to the values λ and λ + dλ of the parameter. Then v µ = dx µ dλ (2) Any transformation from x µ to x µ (x µ x µ ) will be determined by n equations of the form: x µ = f µ (x ν ) where µ, ν = 1, 2,..., n. This means that : d x µ = ν x µ x ν dx ν = ν f µ x ν dx ν for ν = 1,..., n (3) ṽ µ = d x µ dλ = ν x µ dx ν x ν dλ = ν x µ x ν v ν (4)

Contravariant and Covariant Vectors Contravariant Vector: is a quantity with n components depending on the coordinate system in such a way that the components a µ in the coordinate system x µ are related to the components ã µ in x µ by a relation of the form ã µ = ν x µ x ν aν (5) Covariant Vector: eg. b µ, is an object with n components which depend on the coordinate system on such a way that if a µ is any contravariant vector, the following sums are scalars b µ a µ = b µ ã µ for any x µ x µ [Scalar Product] (6) µ µ The covariant vector will transform as: b µ = x ν x µ b ν (7) ν

Tensors: at last A conravariant tensor of order 2 is a quantity having n 2 components T µν which transforms (x µ x µ ) in such a way that, if a µ and b µ are arbitrary covariant vectors the following sums are scalars: T λµ a µ b λ = T λµ ã λ bµ φ (8) Then the transformation formulae for the components of the tensors of order 2 are (why?): T αβ = x α x µ x β x ν T µν, T α β = x α x µ x ν x β T µ ν & T αβ = x µ x α x ν x β T µν The Kronecker symbol δ λ µ is a mixed tensor having frame independent values for its components.

Tensor algebra Tensor multiplication : The product of two vectors is a tensor of order 2, because ã α bβ = x α x β x µ x ν aµ b ν (9) in general: T µν = A µ B ν or T µ ν = A µ B ν or T µν = A µ B ν (10) Contraction: for any mixed tensor of order (p, q) leads to a tensor of order (p 1, q 1) T λµν λα = T µν α (11) Symmetric Tensor : T λµ = T µλ ort (λµ), T νλµ = T νµλ or T ν(λµ) Antisymmetric : T λµ = T µλ or T [λµ], T νλµ = T νµλ or T ν[λµ] Number of independent components : Symmetric : n(n + 1)/2, Antisymmetric : n(n 1)/2

Tensors: Differentiation The simplest tensor field is a scalar field φ = φ(x α ) and its derivatives are the components of a covariant tensor! φ x λ = x α φ x λ x α we will use: φ x α = φ,α (12) i.e. φ,α is the gradient of the scalar field φ. The derivative of a contravariant vector field A µ is : A µ,α Aµ x α = ( ) x µ x α x ν Ãν = x ρ ( x µ x α x ρ = x ν Ãν 2 x µ x ρ x ν x ρ x α Ãν + x µ x ρ Ãν x ν x α x ρ (13) Without thefirst term in the right hand side this equation would be the transformation formula for a covariant tensor of order 2. )

Tensors: Connections The transformation (x µ x µ ) of the derivative of a vector is: A µ,α = x µ x ρ x ν x α (Ãν,ρ + 2 x κ x ν } x σ x {{ ρ x κ Ã σ ) (14) } γ σρ ν in another coordinate (x µ x µ ) we get again A µ,α = x µ x ν x ρ x α ( A ν,ρ + γ ν σρa σ). (15) Suggesting that the transformation ( x µ x µ ) will be: Ã µ,α + Γ µ αλãλ = x µ x ν x ρ x α (A ν,ρ + Γ ν σρa σ ) (16) The necessary and sufficient condition for A µ,α to be a tensor is: Γ λ ρν = 2 x µ x ν x ρ x λ x µ + x κ x ρ x σ x ν x λ x µ Γµ κσ. (17) Γ λ ρν is the called the connection of the space.

Covariant Derivative According to the previous assumptions, the following quantity transforms as a tensor of order 2 A µ ;α = A µ,α + Γ µ αλa λ (18) and is called covariant derivative of the contravariant vector A µ. In similar way we get (how?) : φ ;λ = φ,λ (19) A λ;µ = A λ,µ Γ ρ µλ a ρ (20) T λµ ;ν = T λµ,ν + Γ λ ανt αµ + Γ µ ανt λα (21) T λ µ;ν = T λ µ,ν + Γ λ ανt α µ Γ α µνt λ α (22) T λµ;ν = T λµ,ν Γ α λνt µα Γ α µνt λα (23) T λµ νρ ;σ = T λµ νρ,σ + Γ λ ασt αµ νρ + Γ µ ασt λα νρ + Γ α νσt λµ αρ Γ α ρσt λµ να (24)

Parallel Transport Γ λ µν helps to determine a vector A λ, at a point P, which has to be considered as equivalent to the vector a λ given at P. δ (1) a µ = a µ (P ) a µ (P) = a µ (P) + a µ,ν dx ν a µ (P) = a µ,ν dx ν a µ (P ) A µ (P ) }{{} vector = a µ + δ (1) a µ }{{} at point P (a µ + δa µ ) }{{} at point P = a µ,ν dx ν δa µ = ( a µ,ν C λ ) }{{} µνa λ dx ν vector i.e. = δ (1) a µ δa µ }{{} vector δa µ = C λ µνa λ dx ν

δa µ = Γ λ µνa λ dx ν for covariant vectors (25) δa µ = Γ µ λν aλ dx ν for contravariant vectors (26) The connection Γ λ µν allows to define the transport of a vector a λ from a point P to a neighbouring point P (Parallel Transport). The parallel transport of a scalar field is zero! δφ = 0 (why?)

Curvature Tensor The trip of a parallel transported vector along a closed path The parallel transport of the vector a λ from the point P to A leads to a change of the vector by a quantity Γ λ µν(p)a µ dx ν and the new vector is: a λ (A) = a λ (P) Γ λ µν(p)a µ dx ν (27) A further parallel transport to the point B will lead the vector a λ (B) = a λ (A) Γ λ ρσ(a)a ρ (A)δx σ = a λ (P) Γ λ µν(p)a µ dx ν Γ λ ρσ(a) [ a ρ (P) Γ ρ βν (P)aβ dx ν] δx σ

Since the quantities dx ν and δx ν are small we can use the following expression Γ λ ρσ(a) Γ λ ρσ(p) + Γ λ ρσ,µ(p)dx µ. (28) Thus we have estimated the total change of the vector a λ from the point P to B via A (all terms are defined at the point P). a λ (B) = a λ Γ λ µνa µ dx ν Γ λ ρσa ρ δx σ + Γ λ ρσγ ρ βν aβ dx ν δx β + Γ λ ρσ,τ a ρ dx τ δx σ Γ λ ρσ,τ Γ ρ βν aβ dx τ dx ν δx σ If we follow the path P C B we get: a λ (B) = a λ Γ λ µνa µ δx ν Γ λ ρσa ρ dx σ + Γ λ ρσγ ρ βν aβ δx ν dx β + Γ λ ρσ,τ a ρ δx τ dx σ and the effect on the vector will be δa λ a λ (B) a λ (B) = a β (dx ν δx σ dx σ δx ν ) ( ) Γ λ ρσγ ρ βν + Γλ βσ,ν (29)

By exchanging the indices ν σ and σ ν we construct a similar relations ( ) δa λ = a β (dx σ δx ν dx ν δx σ ) Γ λ ρνγ ρ βσ + Γλ βν,σ (30) and the total change will be given by the following relation: δa λ = 1 2 aβ Rβνσ λ (dx σ δx ν dx ν δx σ ) (31) where Rβνσ λ = Γ λ βν,σ + Γ λ βσ,ν Γ µ βν Γλ µσ + Γ µ βσ Γλ µν (32) is the curvature tensor.

Geodesics For a vector u λ at point P we apply the parallel transport along a curve on an n-dimensional space which will be given by n equations of the form: x µ = f µ (λ); µ = 1, 2,..., n If u µ = dx µ dλ is the tangent vector at P the parallel transport of this vector will determine at another point of the curve a vector which will not be in general tangent to the curve. If the transported vector is tangent to any point of the curve then this curve is a geodesic curve of this space and is given by the equation : du ρ dλ + Γρ µνu µ u ν = 0. (33) Geodesic curves are the shortest curves connecting two points on a curved space.

Metric Tensor The distance ds of two points P(x µ ) and P (x µ + dx µ ) is given by ds 2 = ( dx 1) 2 + ( dx 2 ) 2 + ( dx 3 ) 2 (34) In another coordinate system, x µ, we will get dx ν = α x ν x α d x α (35) which leads to: ds 2 = g µν d x µ d x ν = g αβ dx α dx β. This gives the following transformation relation: g µν = x α x µ x β x ν g αβ (36) suggesting that the quantity g µν is a symmetric tensor, the so called metric tensor. Properties: g µν A µ = A ν, g µν T µα = T α ν, g µν T µ α = T αν, g µν g ασ T µα = T νσ

Metric element for Minkowski spacetime ds 2 = dt 2 + dx 2 + dy 2 + dz 2 (37) ds 2 = dt 2 + dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2 (38) For a sphere with radius R : ds 2 = R 2 ( dθ 2 + sin 2 θdφ 2) The metric element of a torus with radii a and b ds 2 = a 2 dφ 2 + (b + a sin φ) 2 dθ 2 (39) The contravariant form of the metric tensor: g µα g αβ = δµ β where g αβ 1 = det g µν G αβ minor determinant (40) The angle, ψ, between two infinitesimal vectors d (1) x α and d (2) x α is: cos(ψ) = g αβ d (1) x α d (2) x β gρσ d (1) x ρ d (1) x σ g µν d (2) x µ d (2) x ν. (41)

The Determinant of g µν The quantity g det g µν is the determinant of the metric tensor. The determinant transforms as x g = det g µν = det (g α x β ) ( ) ( ) x α x β αβ x µ x ν = det g αβ det x µ det x ν = (det x α ) 2 x µ g = J 2 g (42) where J is the Jacobian of the transformation. This relation can be written also as: g = J g (43) i.e. the quantity g is a scalar density of weight 1.

Christoffel Symbols In Riemannian space there is a special connection derived directly from the metric tensor. This is based on a suggestion originating from Euclidean geometry that is : If a vector a λ is given at some point P, its length must remain unchanged under parallel transport to neighboring points P. a 2 P = a 2 P or g µν(p)a µ (P)a ν (P) = g µν (P )a µ (P )a ν (P ) (44) Since the distance between P and P is dx ρ we can get g µν (P ) g µν (P) + g µν,ρ (P)dx ρ (45) a µ (P ) a µ (P) Γ µ σρ(p)a σ (P)dx ρ (46)

By substituting these two relation into equation (44) we get ( gµν,ρ g µσ Γ σ νρ g σν Γ σ ) µρ a µ a ν dx ρ = 0 (47) This relations must by valid for any vector a ν and any displacement dx ν which lead to the conclusion the the relation in the parenthesis is zero. Closer observation shows that this is the covariant derivative of the metric tensor! g µν;ρ = g µν,ρ g µσ Γ σ νρ g σν Γ σ µρ = 0. (48) i.e. g µν is covariantly constant. This leads to a unique determination of the connections of the space (Riemannian space) Γ α µρ = 1 2 g αν (g µν,ρ + g νρ,µ g ρµ,ν ). (49) which are called Christoffel Symbols. It is obvious that Γ α µρ = Γ α ρµ.

Geodesic in a Riemann Space The geodesics of a Riemannian space have the following important property. If a geodesic is connecting two points A and B is distinguished from the neighboring lines connecting these points as the line of minimum or maximum length. The length of a curve connecting A and B is: S = B A ds = B A [g µν (x α ) dx µ ds dx ν ] 1/2 ds ds A neighboring curve x µ (s) connecting the same points will be described by the equation: x µ = x µ + ɛξ µ (50) where ξ µ (A) = ξ µ (B) = 0. The length of the new curve will be: S = B A [g µν ( x α ) d x µ ds d x ν ] 1/2 ds (51) ds

For simplicity we set: f (x α, u α ) = [g µν (x α )u µ u ν ] 1/2 and f ( x α, ũ α ) = [g µν ( x α )ũ µ ũ ν ] 1/2 u µ = ẋ µ = dx µ /ds and u µ = x µ = d x µ /ds = ẋ + ɛ ξ µ (52) Then we can create the difference δs = S S 1 δs = = ɛ B A B A δf ds = B A [ f x α d ds ( f f ) ds ( )] f B u α ξ α ds + ɛ A ( ) d f ds u α ξα ds The last term does not contribute and the condition for the length of S to be an extremum will be expressed by the relation: 1 We make use of the following relations: f ( x α, ũ α ) = f (x α + ɛξ α, u α + ɛ ξ α ) = f (x α, u α ) + ɛ ξ α f x α + ξ α f «u α + O(ɛ 2 ) d ds «f u α ξα = f ξ α u α + d «f ds u α ξ α

B [ f δs = ɛ A x α d ( )] f ds u α ξ α ds = 0 (53) Since ξ α is arbitrary, we must have for each point of S: ( ) d f ds u µ f x µ = 0 (54) The Langrangian of a freely moving particle with mass m = 2, is: L = g µν u µ u ν f 2 this leads to the following relations f u α = 1 L 2 u α L 1/2 and f x α = 1 L 2 x α L 1/2 (55) and by substitution in (54) we come to the condition ( ) d L ds u µ L x µ = 0. (56)

Since L = g µν u µ u ν we get: thus µ L u u α = g µν u α uν + g µν u µ u ν u α = gµν δµ α uν + g µν u µ δ ν α = 2gµαuµ (57) L x α = g µν,αu µ u ν (58) d ds (2g µαu µ ) = 2 dg µα du µ ds uµ + 2g µα = 2g µα,ν u ν u µ du µ + 2g µα ds ds = g µα,ν u ν u µ + g να,µ u µ u ν du µ + 2g µα (59) ds and by substitution in (56) we get g µα du µ ds + 1 2 [g µα,ν + g αµ,ν g µν,α ] u µ u ν = 0 if we multiply with g ρα we the geodesic equations because du ρ /ds = u ρ,µu µ. du ρ ds + Γρ µνu µ u ν = 0, or u ρ ;νu ν = 0 (60)

Euler-Lagrange Eqns vs Geodesic Eqns The Lagrangian for a freely moving particle is: L = g µν u µ u ν and the Euler-Lagrange equations: ( ) d L ds u µ L x µ = 0 are equivalent to the geodesic equation du ρ ds + Γρ µνu µ u ν d 2 x ρ = 0 or ds 2 + dx µ dx ν Γρ µν = 0 ds ds Notice that if the metric tensor does not depend from a specific coordinate e.g. x k then ( ) d L ds ẋ κ = 0 which means that the quantity L/ ẋ κ is constant along the geodesic. Then eq (57) implies that L ẋ = g κ µκ u µ that is the κ component of the generalized momentum p κ = g µκ u µ remains constant along the geodesic.

Tensors : Geodesics If we know the tangent vector u ρ at a given point of a known space we can determine the geodesic curve. Which will be called: if u 2 < 0 timelike if u 2 = 0 null if u 2 > 0 spacelike where u 2 = g µν u µ u ν If g µν η µν then the light cone is affected by the curvature of the spacetime. For example, in a space with metric ds 2 = f (t, x)dt 2 + g(t, x)dx 2 the light cone will be drawn from the relation dt/dx = ± g/f which leads to STR results for f, g 1.

Null Geodesics For null geodesics ds = 0 and the proper length s cannot be used to parametrize the geodesic curves. Instead we will use another parameter λ and the equations will be written as: and obiously: d 2 x κ dλ 2 + dx µ dx ν Γκ µν dλ dλ = 0 (61) dx µ dx ν g µν dλ dλ = 0. (62)

Geodesic Eqns & Affine Parameter In deriving the geodesic equations we have chosen to parametrize the curve via the proper length s. This choice simplifies the form of the equation but it is not a unique choice. If we chose a new parameter, σ then the geodesic equations will be written: d 2 x µ dσ 2 where we have used dx µ ds = dx µ dσ dσ ds + dx α dx β Γµ αβ dσ dσ = d 2 σ/ds 2 dx µ (dσ/ds) 2 dσ and d 2 x µ ds 2 = d 2 x µ dσ 2 (63) ( ) 2 dσ + dx µ d 2 σ ds dσ ds 2 (64) The new geodesic equation (63), reduces to the original equation(61) when the right hand side is zero. This is possible if d 2 σ ds 2 = 0 (65) which leads to a linear relation between s and σ i.e. σ = αs + β where α and β are arbitrary constants. σ is called affine parameter.

Riemann Tensor When in a space we define a metric then is called metric space or Riemann space. For such a space the curvature tensor R λ βνµ = Γ λ βν,µ + Γ λ βµ,ν Γ σ βνγ λ σµ + Γ σ βµγ λ σν (66) is called Riemann Tensor. R κβνµ = g κλ R λ βνµ = 1 2 (g κµ,βν + g βν,κµ g κν,βµ g βµ,κν ) ) + g αρ (Γ α κµγ ρ βν Γα κνγ ρ βµ Properties of the Riemann Tensor: R κβνµ = R κβµν, R κβνµ = R βκνµ, R κβνµ = R νµκβ, R κ[βµν] = 0 Thus in an n-dim space the number of independent components is: n 2 (n 2 1)/12 (67) For a 4-dimensional space only 20 independent components

Ricci and Einstein Tensors The contraction of the Riemann tensor leads to Ricci Tensor R αβ = R λ αλβ = g λµ R λαµβ = Γ µ αβ,µ Γµ αµ,β + Γµ αβ Γν νµ Γ µ ανγ ν βµ (68) which is symmetric R αβ = R βα. Further contraction leads to the Ricci Scalar R = R α α = g αβ R αβ = g αβ g µν R µανβ. (69) The following combination of Riemann and Ricci tensors is called Einstein Tensor G µν = R µν 1 2 g µνr (70) with the very important property: ( G µ ν;µ = R µ ν 1 ) 2 δµ νr ;µ = 0. (71)

Flat and Empty Spacetimes R αβµν = 0 flat spacetime R µν = 0 empty spacetime Prove that : a λ ;µ;ν a λ ;ν;µ = R λ κµνa κ

Weyl Tensor Important relations can be obtained when we try to express the Riemann or Ricci tensor in terms of trace-free quantities. S µν = R µν 1 4 g µνr S = S µ µ = g µν S µν = 0. (72) or the Weyl tensor C λµνρ : R λµνρ = C λµνρ + 1 2 (g λρs µν + g µν S λρ g λν S µρ g µρ S λν ) + 1 12 R (g λρg µν g λν g µρ ) (73) C λµνρ = R λµνρ 1 2 (g λρr µν + g µν R λρ g λν R µρ g µρ R λν ) + 1 6 R (g λρg µν g λν g µρ ). (74) and we can prove that : g λρ C λµνρ = 0 (75)

The Weyl tensor is called also conformal curvature tensor because it has the following property. Consider besides the Riemannian space M with metric g µν a second Riemannian space M with metric g µν = e 2A g µν, where A is a function of the coordinates. The space M is said to be conformal to M. One can prove that R α βγδ R α βγδ while C α βγδ = C α βγδ. (76) I.e. a conformal transformation does not change the Weyl tensor. It can be verified at once from equation (74) that the Weyl tensor has the sam symmetries as the Riemann tensor. That is it has 20 independent components, but because it is traceless [condition (75)] there are 10 more conditions for the components therefore the Weyl tensor has 10 independent components.

Tensors : An example for parallel transport A vector A = A 1 e θ + A 2 e φ be parallel transported along a closed line on the surface of a sphere with metric ds 2 = dθ 2 + sin 2 dφ 2 and Christoffel symbols Γ 1 22 = sin θ cos θ and Γ2 12 = cot θ. The eqns δa α = Γ α µνa µ dx ν for parallel transport will be written as: A 1 x 2 = Γ 1 22A 2 A1 = sin θ cos θa2 φ A 2 x 2 = Γ 2 12A 1 A2 = cot θa1 φ

The solutions will be: 2 A 1 φ 2 = cos2 θa 1 A 1 = α cos(φ cos θ) + β sin(φ cos θ) A 2 = [α sin(φ cos θ) β cos(φ cos θ)] sin 1 θ and for an initial unit vector (A 1, A 2 ) = (1, 0) at (θ, φ) = (θ 0, 0) the integration constants will be α = 1 and β = 0. The solution is: A = A 1 e θ + A 2 e φ = cos(2π cos θ) e θ sin(2π cos θ) e φ sin θ i.e. different components but the measure is still the same A 2 = g µν A µ A ν = ( A 1) 2 + sin 2 θ ( A 2) 2 = cos 2 (2π cos θ) + sin 2 θ sin2 (2π cos θ) sin 2 θ = 1