Exercises in field theory

Transcription

1 Exercises in field theory Wolfgang Kastaun June 12, 2008

2 Vectors and Tensors Contravariant vector Coordinate independent definition, e.g. as tangent of a curve. A contravariant vector A is expressed with respect to given coordinates as an ordinary vector with components A µ. It has to satisfy a coordinate transformation rule à µ = x µ x ν Aν Covariant vector Expressed with respect to coordinates, again an ordinary vector A µ. Demand that A µ B µ transforms like a scalar for any contravariant vector B. Coordinate transformation rule à µ = x ν x µ A ν

3 Vectors and Tensors Example: Transformation from Cartesian coordinates x i = (x, y, z) to spherical coordinates x i = (r, θ, φ). x = r sin(θ) cos(φ) y = r sin(θ) sin(φ) z = r cos(θ) A covariant vector transforms like A r A θ = A φ x r x θ x φ y r y θ y φ z r z θ z φ A x A y A z Note the transformation matrix is the inverse of the transformation matrix for a contravariant vector.

4 Vectors and Tensors Example: Transformation from Cartesian coordinates x i = (x, y, z) to spherical coordinates x i = (r, θ, φ). x = r sin(θ) cos(φ) y = r sin(θ) sin(φ) z = r cos(θ) A covariant vector transforms like A r sin θ cos φ sin θ sin φ cos θ A x A θ = r cos θ cos φ r cos θ sin φ r sin θ A y A φ r sin θ sin φ r sin θ cos φ 0 A z Note the transformation matrix is the inverse of the transformation matrix for a contravariant vector.

5 Vectors and Tensors Tensors are defined as linear mappings from a number of co- and contravariant vectors to scalars. E.g. rank-2 tensors map A µ, B ν C µν A µ B ν A µ, B ν C µ νa µ B ν A µ, B ν C µν A µ B ν Coordinate transformation rule for tensors C µν = x µ x α x ν x β Cαβ, Cµ ν = x µ x α x β C µν = x α x µ x β x ν C αβ x ν Cα β

6 Vectors and Tensors We introduce a metric tensor g µν satisfying g µν = g νµ, g µα g αν = δ µ ν..to define the length of a vector A 2 = A µ A ν g µν..to map between co- and contravariant vectors A µ = g µν A ν, A µ = g µν A ν

7 Vectors and Tensors Example: Transformation of the Euclidean metric g µν = δ µν from Cartesian coordinates x i = (x, y, z) to cylindrical coordinates x i = (ρ, z, φ). We obtain x = ρ cos(φ), y = ρ sin(φ), z = z g ρρ = x x ρ ρ g xx + y y ρ ρ g yy + z z ρ ρ g zz = cos 2 (φ) + sin 2 (φ) + 0 = 1 g φφ = x x φ φ g xx + y y φ φ g yy + z z φ φ g zz = ρ 2 sin 2 (φ) + ρ 2 cos 2 (φ) = ρ 2 g zz = x x z z g xx + y y z z g yy + z z z z g zz = 1

8 The covariant derivative Φ ;µ of a scalar Φ is identical to the ordinary partial derivative Φ,µ.

9 The covariant derivative Φ ;µ of a scalar Φ is identical to the ordinary partial derivative Φ,µ. The covariant derivative of a scalar is a covariant vector.

10 The covariant derivative Φ ;µ of a scalar Φ is identical to the ordinary partial derivative Φ,µ. The covariant derivative of a scalar is a covariant vector. This can be shown using Φ( x) = Φ(x( x)) and the chain rule of differentiation Φ ;ν = Φ,ν = x Φ( x) ν = = x µ x ν x ν Φ(x( x)) x µ Φ(x) = x µ x ν Φ,µ = x µ x ν Φ ;µ

11 The partial derivatives A µ,ν of a contravariant vector do not transform like a tensor.

12 The partial derivatives A µ,ν of a contravariant vector do not transform like a tensor. The covariant derivative A µ ;ν of a contravariant vector is a tensor A µ ;ν = A µ,ν + Γ µ νσa σ

13 The partial derivatives A µ,ν of a contravariant vector do not transform like a tensor. The covariant derivative A µ ;ν of a contravariant vector is a tensor A µ ;ν = A µ,ν + Γ µ νσa σ For this, Γ µ νσ has to obey a certain coordinate transformation rule.

14 The partial derivatives A µ,ν of a contravariant vector do not transform like a tensor. The covariant derivative A µ ;ν of a contravariant vector is a tensor A µ ;ν = A µ,ν + Γ µ νσa σ For this, Γ µ νσ has to obey a certain coordinate transformation rule. The connection Γ µ νσ is not a tensor.

15 The partial derivatives A µ,ν of a contravariant vector do not transform like a tensor. The covariant derivative A µ ;ν of a contravariant vector is a tensor A µ ;ν = A µ,ν + Γ µ νσa σ For this, Γ µ νσ has to obey a certain coordinate transformation rule. The connection Γ µ νσ is not a tensor. If f is a scalar, we obviously have the product rule (fa µ ) ;ν = f ;ν A µ + fa µ ;ν

16 The covariant derivative of a covariant vector can be defined by demanding a product rule (A µ B µ ) ;σ = A µ ;σb µ + A µ B µ;σ

17 The covariant derivative of a covariant vector can be defined by demanding a product rule From that, it follows (A µ B µ ) ;σ = A µ ;σb µ + A µ B µ;σ A µ B µ;σ = (A µ B µ ) ;σ A µ ;σb µ = (A µ B µ ),σ A µ ;σb µ = A µ,σb µ + A µ B µ,σ ( A µ,σ + Γ µ σαa α) B µ = A µ ( B µ,σ Γ α σµb α )

18 The covariant derivative of a covariant vector can be defined by demanding a product rule From that, it follows (A µ B µ ) ;σ = A µ ;σb µ + A µ B µ;σ B µ;σ = B µ,σ Γ α σµb α

19 The covariant derivative of a covariant vector can be defined by demanding a product rule From that, it follows (A µ B µ ) ;σ = A µ ;σb µ + A µ B µ;σ B µ;σ = B µ,σ Γ α σµb α For any covariant vectors A µ, C µ, A µ C σ B µ;σ = C σ (A µ B µ ) ;σ C σ B µ A µ ;σ is a scalar B µ;σ is a tensor.

20 The covariant derivative of a covariant vector can be defined by demanding a product rule From that, it follows (A µ B µ ) ;σ = A µ ;σb µ + A µ B µ;σ B µ;σ is a tensor. If f is a scalar, we have B µ;σ = B µ,σ Γ α σµb α (fb µ ) ;ν = f ;ν B µ + fb µ;ν

21 The covariant derivative of a tensor can be defined by demanding a product rule (C µν A µ B ν ) ;σ = C µν ;σ A µ B ν + C µν A µ;σ B ν + C µν A µ B ν;σ

22 The covariant derivative of a tensor can be defined by demanding a product rule Thus, (C µν A µ B ν ) ;σ = C µν ;σ A µ B ν + C µν A µ;σ B ν + C µν A µ B ν;σ C µν ;σ A µ B ν = (C µν A µ B ν ),σ C µν A µ;σ B ν C µν A µ B ν;σ = C µν,σ A µ B ν + C µν A µ,σ B ν + C µν A µ B ν,σ C µν ( A µ,σ Γ α σµa α ) Bν C µν A µ (B ν,σ Γ α σνb α ) = A µ B ν ( C µν,σ + Γ µ σαc αν + Γ ν σαc µα)

23 The covariant derivative of a tensor can be defined by demanding a product rule Thus, (C µν A µ B ν ) ;σ = C µν ;σ A µ B ν + C µν A µ;σ B ν + C µν A µ B ν;σ C µν ;σ = C µν,σ + Γ µ σαc αν + Γ ν σαc µα

24 The covariant derivative of a tensor can be defined by demanding a product rule Thus, (C µν A µ B ν ) ;σ = C µν ;σ A µ B ν + C µν A µ;σ B ν + C µν A µ B ν;σ C µν ;σ = C µν,σ + Γ µ σαc αν + Γ ν σαc µα By multiplying the product rule with another vector D σ, we easily show that C µν ;σ is a tensor (of rank 3).

25 Similarly, we demand (C µ νa µ B ν ) ;σ = C µ ν;σa µ B ν + C µ νa µ;σ B ν + C µ νa µ B ν ;σ (C µν A µ B ν ) ;σ = C µν;σ A µ B ν + C µν A µ ;σb ν + C µν A µ B ν ;σ It follows C µ ν;σ = C µ ν,σ + Γ µ σαc α ν Γ α σνc µ α C µν;σ = C µν,σ Γ α σµc αν Γ α σνc µα Again, these covariant derivatives are tensors.

26 We will typically use a special connection, the Levi-Civita-connection, given by Γ µ νσ = 1 2 gµα (g αν,σ + g ασ,ν g νσ,α ) The Levi-Civita-connection components Γ µ νσ are called Christoffel symbols. It has the property g µν;α = 0 Thus, (g µν A µ B ν ( ) ;σ = g µν A µ ;σ B ν + A µ B;σ ν )

27 Example: the Christoffel symbols for the metric g vv = 1, g uu = u 2 v 2, g uv = 0 From Γ µ νσ = 1 2 gµα (g αν,σ + g ασ,ν g νσ,α ) We find Γ u uu = u u 2 v 2, Γv uu = v, Γ u vu = Γ u uv = v u 2 v 2