SEVERAL TOPICS FROM RELATIVITY

Transcription

1 Topics from Relativity 1 SEVERAL TOPICS FROM RELATIVITY FRANZ ROTHE 2010 Mathematics subject classification: 51Fxx; 51Pxx; 70H40. Keywords and phrases: Instructional exposition, General theory, Geometry and physics, special relativity. Contents 1 Riemannian geometry Curved coordinate systems About differentiable manifolds Tensors Riemannian manifold Lie derivative Special Relativity Relativity of time and length Discovery of Aberration and Parallax Aberration and the Doppler effect The one-dimensional Doppler effect Four-vectors and Minkowski metric The relativistic Doppler effect Four-velocity The energy-momentum vector The Compton effect Collision of particles The motion of particles The Lorentz Group Different aging of twins The Lorentz transformations Infinitesimal generators

2 2 F. Rothe 4 The Poincaré Half-Plane Model Poincaré half-plane and Poincaré disk The Euler-Lagrange equation The curve of minimal hyperbolic length The minimum of hyperbolic length Some useful reflections in the half-plane Equation of motion Affine geodesic Metric geodesic The quadratic Lagrangian Null geodesics The method Killing vector Geodesics in the Schwarzschild metric The equation for the shape of relativistic orbits Kepler s classical nonrelativistic orbits Scattering in Newtonian dynamics Perturbation expansion for relativistic bounded orbits The mercury perihelion rotation Perburtation expansion for the angle of deflection The bending of light Gauss Differential Geometry and the Pseudo-Sphere Introduction About Gauss differential geometry Riemann metric of the Poincaré disk Riemann metric of Klein s model A second proof of Gauss remarkable theorem Principal and Gaussian curvature of rotation surfaces The pseudo-sphere Poincaré half-plane and Poincaré disk Embedding the pseudo-sphere into Poincaré s half-plane Embedding the pseudo-sphere into Poincaré s disk About circle-like curves Mapping the boundaries References [1] Max Born, Einstein s Theory of Relativity, revised edition, Dover publications, 1924, [2], The Born-Einstein Letters , Friendship, Politics and Physics in Uncertain Times, Macmillan, [3] Ta-Pei Cheng, Relativity, Gravitation and Cosmology, second edtion, Oxford University Press, [4] Leo Corry, David Hilbert and the Axiomatization of Physics ( ), from Grundlagen der Geometrie to Grundlagen der Physik, Kluwer Academic Publishers, ISBN X(HB), [5] David Griffiths, Introduction to Elementary Particles, second edition, Wiley-VCH, [6] Stephen G. Gasiorowicz Jeremy Bernstein, Paul M. Fishbane, Modern Physics, Prentice-Hall, [7] G.Efstathiou M.P.Hobson and A.N.Lasenby, General Relativity, Cambridge University Press, [8] Abraham Pais, Subtle is the Lord the science and life of Albert Einstein, Oxford University Press, [9] David Park, The Grand Contraption the world in myth, number, and chance, second printing, Princeton University Press, 2005.

3 Topics from Relativity 3 [10] Matthew Sands Richard P. Feynman, Robert B. Leighton, The Feynman Lectures on Physics, Addison-Wesley, [11] Gerald t Hooft, Introduction to General Relativity, Rinton Press Princeton, [12] Hermann Weyl, Raum-Zeit-Materie, 5th revised edition, Springer, Franz Rothe, Department of Mathematics and Statistics University of North Carolina at Charlotte Charlotte, NC frothe@uncc.edu

4 4 F. Rothe 1. Riemannian geometry 1.1. Curved coordinate systems The conversion of spherical coordinates (r, θ, φ) to Cartesian coordinates (x, y, z) is x = r sin θ cos φ (1.1) y = r sin θ sin φ (1.2) z = r cos θ (1.3) These are the formulas normally used to define spherical coordinates, taking as their standard domain the values (r, θ, φ) with r 0, 0 θ π and π < φ π. According to the multivariable chain rule, the conversion of the differentials becomes dx dy dz = sin θ cos φ r cos θ cos φ r sin θ sin φ sin θ sin φ r cos θ sin φ r sin θ cos φ cos θ r sin φ 0 For example we consider a point particle moving along a path x = x(t), y = y(t), z = z(t), respectively r = r(t), θ = θ(t), φ = φ(t). The components of the velocity respectively v x = dx dt, vy = dy dt, vz = dz dt w r = dr dt, wθ = dθ dt, wz = dφ dt have the same transformation law given by formula (1.4). Indeed, one gets v x sin θ cos φ r cos θ cos φ r sin θ sin φ w r v y = sin θ sin φ r cos θ sin φ r sin θ cos φ w θ v z cos θ r sin φ 0 w φ The following is a totally different situation. A potential V is called scalar if the invariance W(r, θ, φ) = V(x, y, z) holds. Usually, we use the same letter, and indicate by a prime that the functions V and W are different. But they describe the same geometric or physical situation. According to the multivariable chain rule, the conversion of the gradient [ r W, θ W, φ W] to [ x V, y V, z V] becomes sin θ cos φ r cos θ cos φ r sin θ sin φ [ r W, θ W, φ W] = [ x V, y V, z V] sin θ sin φ r cos θ sin φ r sin θ cos φ (1.6) cos θ r sin φ 0 One sees that the same matrix as in formula (1.5) occurs, but now on the other side of the equation. Too, note that this coincidence is only made possible by using column vectors for the differentials, whereas we have used row vectors for the gradient. I use the symbol T meaning "transpose", to convert row vectors to column vectors, for typesetting convenience, and for transposition of matrices. Definition 1.1 (Components of a contravariant vector). The components of a contravariant vector have the same conversion (1.4) as the differentials. Definition 1.2 (Components of a covariant vector). The components of a covariant vector have the same conversion (1.6) as the components of the gradient of a scalar. dr dθ dφ (1.4) (1.5)

5 Topics from Relativity 5 For example, the electric field [E x, E y, E z ] in Cartesian coordinates, gets in spherical coordinates the components [F r, F θ, F φ ] such that sin θ cos φ r cos θ cos φ r sin θ sin φ [F r, F θ, F φ ] = [E x, E y, E z ] sin θ sin φ r cos θ sin φ r sin θ cos φ (1.7) cos θ r sin φ 0 Remark. The same conversion (1.6) as for a gradient, holds for partial derivatives in any general linear transformation, but only for the covariant derivatives in all nonlinear transformations About differentiable manifolds These laws for the conversion of coordinates systems apply in a more general context, indeed on any differential manifold. Definition 1.3 (Differential manifold). A differential manifold M of dimension n is a topological locally compact Hausdorff space, with the following additional structure. For every point there exists a neighborhood with a coordinate system (x 1,..., x n ). The coordinate systems for intersecting neighborhoods are compatible. There exists bijective differentiable transformations between any two different coordinate systems say (x 1,..., x n ) and (x 1,..., x n), which are valid in the intersection of such neighborhoods and make them compatible. Remark. Thus coordinate transformation are considered to be passive transformations, naming the same points with different coordinate labels, at least at first hand. We name the coordinates, respectively base vectors, of two system S and S by the same letters and use primes for distinguishing between them. Remark. The extension of such transformations to the entire manifold are called point transformations. Existence of such point transformations, indeed locally of any prescribed form, can be proved. The proof uses a tool called the partition of unity. Definition 1.4 (Tangent plane and tangent manifold). For every point P of a differential manifold of dimension n, there exists a tangent plane T P. This is an n-dimensional linear space, with some basis [e 1,..., e n ]. The contravariant differential [dx a ] corresponds to the vector n dx = e a dx a a=1 These vector are invariant under coordinate transformations, just like scalar quantities. The union T = P M of all tangent planes is called the tangent manifold. This is a differential manifold of dimension 2n. Lemma 1.1. Under any point transformation x a = x a(x 1,..., x n ) T P for a = 1,... n the contravariant components [v b ] have the transformation law v a = x a x b vb and the base vectors e b have the transformation law Hence the vector itself is invariant: e b = x a x b e a These are the vectors in the tangent plane T P. for a = 1,... n for b = 1,... n v = v b e b = v a e a

6 6 F. Rothe Remark. We have already used the Einstein sum convention: over any index which appears upper and lower-hand in an expression, is to be taken the sum, unless otherwise stated. For example, the differential is simply written dx = e a dx a Remark. The existence of the tangent manifold can be proved in general, by means of a some rather abstract (and farfetched) construction. The existence proof is easy under the assumption the manifold is, at least locally, embedded into any higher dimension flat space R N, and no restrictions on the dimension N are imposed. 1 This situation occurs for general relativity. Indeed, suppose an embedding of a manifold M R N is given by formulas X i = X i (x 1,... x n ) for i = 1,..., N (1.8) The tangent space basis at any point P = (x 1,..., x n ) is simply [ ] X 1 T e a = x a,..., XN x a for a = 1,..., n (1.9) Definition 1.5 (Dual linear space). The dual X of a real linear space X consists of all linear functionals X R. We get the natural bilinear form, which maps X X to R and assigns to any ordered pair of x X and x X the real number x, x. The cotangent plane TP is identified with the dual of T P, which is linear space of linear functionals T p R. One gets a convenient basis [ω a ] for TP by requiring for a, b = 1... n and extending by linearity. Lemma 1.2. Under any point transformation ω a, e b = δ a b = 1 if a = b; 0 if a b x a = x a(x 1,..., x n ) the covariant components [ f b ] have the transformation law f b = x a x b f a and the base vectors ω b have the transformation law Hence the vector ω a = x a x b ωb for a = 1,... n for b = 1,... n for a = 1,... n v = f b ω b = f aω a itself is invariant. These are the vectors in the cotangent plane T P. Lemma 1.3. For a particle moving in a scalar potential field S with velocity v, the rate of change of the potential felt by the particle is ds dt = ( a S )v a = S, v 1 Allowing any large number N, this not a too strong assumption. The situation is different, once restrictions on the number N are imposed.

7 Topics from Relativity 7 Proof. For the rate of change of the composite function S (t) = S (x 1 (t),..., x n (t)), the multivariable chain rule gives ds = S dx a dt x a dt but this is just the functional S = ( a S )ω a T P applied to the vector v = vb e b since Lemma 1.4. By the rule S, v = ( a S )ω a, v b e b = ( a S )v b δ a b = ( as )v a b, a = id(a), b (1.10) for all a T p and b TP, the natural bijection id : T P TP is given. Via this mapping id, one may identify the double-dual TP with the tangent plane T P, Hence one may view T P as the space of linear functionals TP R. Proof. The double-dual TP consists by definition of the linear functionals T P R. A basis [f a] of TP is given by requiring f a, ω b = δa b 1 if a = b; = (1.11) 0 if a b and extending by linearity. An injective linear mapping id : T P TP is defined by setting id(e a ) = f a for a = 1... n and extending by linearity. Since the linear spaces T P and TP both have the same finite dimension n, one obtains even a bijection. This bijection gives the identification TP = T P. Since f a, ω b = δ a b = ωb, e a linearity gives the rule (1.10) for all c T P = T P and b T P Tensors Definition 1.6 (Tensor). Let q 0 and r 0 be integers. A tensor T of type (q, r) is a multilinear mapping 1 q factors r factors { }} { T : TP T P T P { }} { T P T P T P R The type of such a tensor is q-fold contravariant and r-fold covariant. We see that a tangent vector v T P has become a tensor of type (1, 0). It has been "disguised" as the linear mapping., v : TP R. A generic tensor T of type (q, r) is a linear combination T = T a 1...a q b 1...b r e a1 e a2 e aq ω b 1 ω b 2 ω b r (1.12) Note the convenience of the Einstein sum convention. In shorthand, we often write [T a 1...a q b 1...b r ] for such a tensor, since all information is already contained in the set of components. The basis of the linear space of tensors of type (q, r) is the set of exterior products e a1 e a2 e aq ω b 1 ω b 2 ω b r = e b 1...b r a 1...a q 1 Some authors used to talk about a "machine", but mathematicians still need coffee machines to convert their thoughts into theorems, nevertheless.

8 8 F. Rothe with any a 1... a q and b 1... b r in 1... n. The basis tensors are defined by the requirements b 1...b r ( ea 1...a q ω c 1,..., ω c q ), e d1,..., e dr = δ c 1 a 1 δ c 1 a 1 δ d 1 b 1 δ d 1 b 1 1 if a 1 = c 1,..., a q = c q and b 1 = d 1,..., b r = d r ; = 0 otherwise with any a 1... a q, c 1... c q and b 1... b r, d 1... d q in 1... n and extending by linearity. dimension of this linear space is n q+r. The Lemma 1.5. The rule a 1 a 2 a q b 1 b 2 b r ( c 1,..., c q, d 1,..., d r ) = c 1, a 1 c q, a q b 1, d 1 b r, d r holds for any a 1... a q T p, c 1... c q T P and b1... b q T P, d 1... d q T P. Problem 1.1. In general, when transforming the components of a tensor of arbitrary type (q, r), the components for the S -system are obtained from those of the S -system putting for each superscript a Jacobian transformation matrix x a / x c, and for each subscript an inverse Jacobian x c / x a. Both Jacobians appear on the right-hand side together with the S -system tensor. Apply these rules to get the components t c ab from the components t f de. Answer. t c ab = xd x e x c x a t f x b x f de Problem 1.2. Suppose we may only use the Jacobian x a / x c but not its inverse. Under that restriction, the superscripts are handled in the same way as in problem 1.1, but for each subscript the Jacobian x a / x c appear on the left-hand side together with the S -system tensor. Apply these rules to relate the components t c ab and t f de. Answer. x a x b x d x e t c ab = x c t f x f de Remark. It helps to remember that upper and lower indices on the same side of the equation are always paired, when one counts the upper index in the denominator of a differential quotient like a lower index. Definition 1.7 (Affine connection). An affine connection defines the tangential part of the rate of change of the tangent-plane and its base vectors. The connection coefficients are defined by Lemma 1.6. An affine connection yields Γ a bc = ωa, c e b (1.13) c ω a, e b = Γ a bc (1.14) and hence determines, too, the cotangential part of the rate of change of the cotangent-plane and its base vectors.

9 Topics from Relativity 9 Lemma 1.7. The rate of change of the tangent base vectors are c e b = e a Γ a bc + n bc where the normal parts satisfy for all a, b, c = 1 n. ω a, n bc = 0 Lemma 1.8. The rate of change of the cotangent base vectors are c ω a = ω b Γ a bc + ma c where the normal parts satisfy for all a, b, c = 1 n. m a c, e b = 0 Definition 1.8 (Intrinsic derivatives). The intrinsic derivatives 1 of the base vectors are c e b = e a ω a, c e b = Γ a bc e a (1.15) c ω a = ω b c ω a, e b = Γ a bc ωb (1.16) Definition 1.9 (Covariant derivative). Let c be any index in 1... n. derivative of a scalar function S is equal to the partial derivative: The partial covariant c S = c S (1.17) The partial covariant derivative c T of any tensor T, say of type (q, r) as in equation (1.12), is obtained by using linearity and the Leibniz product for the partial derivatives of the components, and the intrinsic partial derivatives of the base vectors. Remark. Suppose a specific embedding M R N of the manifold is given. Then the following situation occurs: The intrinsic part of any vector X R N is given by the projection Pro j TP X = e a ω a, X (1.18) a=1...n As explained in lemma 1.12 below, the connection turns out to be symmetric: Γ a bc = Γa cb holds for all indices a, b, c. In traditional Gaussian differential geometry the dimensions are n = 2 and N = 3. Here the normal parts can be calculated, and yield the Weingarten formulas. One needs to carefully distinguish partial derivatives, which refer to the embedding R 3, from the intrinsic derivatives used in this exposition. Lemma 1.9. The following are the two simplest cases for a covariant derivative. Let u = v s e s be a contravariant vector and c be any index in 1... n. The c-th partial covariant derivative has the components c v a = c v a + Γ a scv s (1.19) Let f = f b ω b be a covariant vector. The c-th partial covariant derivative has the components c f b = c f b Γ s bc f s (1.20) 1 In this exposition to differentiable manifolds, the intrinsic derivatives get no own symbol.

10 10 F. Rothe Lemma 1.10 (Rule to get covariant derivatives). In the general case of a tensor of type (q, r), the components of the partial covariant derivative are denoted by c T a 1...a q b 1...b r or even simpler T a 1...a q b 1...b r ;c They are obtained by the following rule: Each such components is the sum of 1 + q + r terms. The first term is the partial derivative c T a 1...a q b 1...b r The remaining terms are all products of the tensor components with a Christoffel symbol. The next q terms are added. For each term, a Christoffel symbol has robbed a different one of contravariant indices a 1... a q and replaced this index by a contravariant summation index s. The Christoffel symbol gets the robbed index, the covariant summation index s, and c as last index. The last r terms are subtracted. Once more, for each term, a Christoffel symbol "has robbed" a different one of covariant indices b 1... b r, and replaced it by a covariant summation index s. The Christoffel symbol gets as a contraction the corresponding contravariant summation index s, the robed index, and c as last index. Proof for a contravariant vector. Let u = v s e s be a contravariant vector and c be any index in 1... n. The c-th partial covariant derivative is c v = e a ω a, Dv Dx c where the capital D means taking into account the derivatives of the base vectors e s, too. But because of the projection along the tangent plane, only the tangential part of these derivatives incorporated into the connection coefficient is taken into account. Hence c v = e a ω a, D(vs e s ) Dx c = e a v a x c + e au s ω a, e s x c = e a = e a ω a, vs x c e s + v s e s x c [ ] v a x c + Γa scv s Lemma For any product or contraction of tensors, the covariant derivatives are formed following the Leibniz product rule. Proof for the simplest case. Take the contraction v a f a of a contravariant vector v = v a e a and a covariant vector f = f b ω b. Since the contraction is a scalar, its covariant derivative is just the partial derivative, and clearly satisfies the Leibniz product rule. c (v a f a ) = c (v a f a ) = ( c v a ) f a + v a ( c f a ) = ( c v a + Γ a scv s ) f a + v b ( c f b Γ s bc f s) = ( c v a ) f a + v b ( c f b ) One can add and subtract the Christoffel terms to get the corresponding covariant derivatives. Thus one sees that the covariant partial derivative satisfies the Leibniz product rule. One sees that formula (1.17), together with lemma 1.9 and 1.11 already uniquely determine the covariant derivatives of tensors of types (0, 0), (1, 0) and (0, 1). One may proceed inductively from tensors of type (q, r) to those of types (q + 1, r) and (q, r + 1). Indeed, take any tensor [T a 1...a q b 1...b r ] of type (q, r), and [v s ] of types (1, 0), as well as [ f t ] of type (0, 1).

11 Topics from Relativity 11 The outer product [v s T a 1...a q b 1...b r ] is of type (q+1, r). Similarly, the outer product [T a 1...a q b 1...b r f t ] is of type (q, r + 1). There covariant derivatives are to be obtained via the stipulated Leibniz rule. One obtains: c (v s T a 1...a q b 1...b r ) = ( c v s )T a 1...a q b 1...b r + v s ( c T a 1...a q b 1...b r ) (1.21) where the right-hand side may already be calculated using formula (1.19), and the rule for tensors of type (q, r). Thus one may proceed inductively to tensors of any type, and check that the rules given by definition 1.9 are always valid. Indeed, with a bid of additional work, one obtains following result. Theorem 1.1. Once a connection is specified, and the Leibniz rule rule is stipulated, the covariant derivatives are uniquely determined for all tensors of arbitrary types. Indeed, they are obtained by the rule from lemma 1.10, and moreover have the following properties: The covariant derivative of any contraction equals the contraction of the covariant derivative. As an example we take a tensor [t ab d f ] of type (2, 2). If v ab d f c = ct ab d f and one contracts to y a f = t ab b f then c y a f = v ab b f c In the end of course, both sides are called c t ab b f. The Leibniz product rule holds for all possible products of tensors. As an example take a tensor [t ab ] of type (2, 0) and a tensor [ f d ] of type (0, 1). c (t ab f d ) = ( c t ab ) f d + t ab ( c f d ) Problem 1.3. Apply the rules from lemma 1.10 to get the covariant derivative d t c ab. Answer. d t c ab = dt c ab Γs ad t c sb Γs bd t c as + Γ c sd t s ab Proposition 1.1 (Transformation of an affine connection). For any C 2 -smooth point transformation between two coordinate systems, say (x 1,..., x n ) and (x 1,..., x n), the Christoffel symbols are transformed following the rule Γ a bc = x a x f x g x d x b x c Γd f g + x a 2 x d x d x c x b = x a x d x f x b x g x c Γd f g x f x b x d x c 2 x a x d x f Proof. The connection coefficients in the S -system are defined following the rule (1.13) Γ a bc = ω a, e b x c Now the point transformation gives ( ) Γ a bc = x f x a x d ωd, x c x e b f ( = x a x f e f x d ωd, x b x c + 2 x f ) x b x e c f [ = x a x f x d x b ωd, xg e f x c x g + 2 x f ] x b x c ωd, e f [ = x a x f x g x d x b x c Γd f g + 2 x d ] x b x c (1.22)

12 12 F. Rothe thus proving the first formula. Inverse point transformations have inverse Jakobi matrices. Hence x b x a x d x d x c = δa c ( x a x d ) x d x c = 0 2 x a x f x d x a 2 x d + x d x f x b x c x d x c x = 0 b Thus we get the second formula from the first one. Corollary 1. For a C 2 -smooth manifold, the variation of the connection symbols δγ a bc is a tensor. Definition 1.10 (Torsion tensor). For a C 2 -smooth manifold, the antisymmetric part T a bc = Γ a bc Γa cb is a tensor, called the torsion tensor. Definition 1.11 (Symmetric connection). A connection is called symmetric iff Γ a bc = Γa cb holds for all indices a, b, c. Problem 1.4. Convince convince yourself that a C 2 -smooth point transformation takes a symmetric connection to a symmetric one. Lemma Suppose there exists a C 2 -smooth embedding M R N of the manifold. Then the connection coefficients are symmetric: Γ a bc = Γa cb holds for all indices a, b, c. Proof. With the embedding given by formulas (1.8), the tangent space basis at any point P = (x 1,..., x n ) has the vectors [ ] X 1 T e b = x,..., XN = X b x b x b for b = 1,..., n. Hence the connection coefficients are Γ a bc = ωa, c e b = ω a, 2 X x b x c = Γa cb since the order taking partial derivatives can be exchanged for C 2 -smooth functions. Proposition 1.2. Assume that the connection is symmetric. For any given point, there exists a point transformation which makes all connection coefficients zero at this point. Proof. To transform the connection coefficients at point P to zero, the quadratic transformation x a = x a x a (P) + Γa bc (P) 2 [ x b x b (P) ] [ x c x c (P) ] for a = 1... n (1.23) will do. Since the connection is symmetric the derivatives of the above transformation are x a x = b δa b + Γa bc (P)(xc x c (P)) 2 x a x b x = c Γa bc (P)

13 Topics from Relativity 13 Now the second formula from equation (1.22) to transform the connection coefficients yields Γ a bc = x a x d x f x b x g x c Γd f g x f x b x d x c 2 x a x d x f = x f x b x g x c [ δ a d + Γa dc (P)(xc x c (P)) ] Γ d f g x f x b x d x c Γa d f (P) = x f x b x g x c Γa f g + x f x b x g x c Γa dc (P)(xc x c (P))Γ d f g x f x d x b x c Γa d f (P) = x f x b x g x c [ Γ a f g Γa g f (P)] + x f x b x g x c Γa dc (P)(xc x c (P))Γ d f g Both terms are zero at point P. If the connection coefficient are C 1 -smooth, both terms are small near the point P. Indeed Γ (x ) = O( x x(p) ) = O( x x (P) ) = O( x ) Problem 1.5. If the connection is not assumed to be symmetric, convince yourself that one may at least achieved by the above quadratic transformation that Γ a bc (P) = Γ a cb (P). Problem 1.6. Assume that the connection is symmetric and the connection coefficient are C 1 - smooth. Use the first formula from equation (1.22) and the quadratic transformation x x x d = x d (P) + x d Γd bc (P) x b x c for d = 1... n (1.24) 2 to achieve that the transformed connection Γ (P) = 0 and moreover Γ (x ) = O( x ) near to the point P. Proof. Since the connection is symmetric the derivatives of the above transformation (1.24) are x d x = b δa b Γd bc (P)x c 2 x d x b x = c Γd bc (P) Now the first formula from equation (1.22) to transform the connection coefficients yields [ Γ a bc = x a x f x g x d x b x c Γd f g + 2 x d ] x c x b = x a x d [ (δ f b Γ f bc (P)x c )(δ g c Γ g ce(p)x e )Γ d f g Γd bc (P)] = x a x d [ δ f b δg cγ d f g δ f b Γg ce(p)x e Γ d f g Γ f bc (P)x c δ g cγ d f g + Γ f bc (P)x c Γ g ce(p)x e Γ d f g Γd bc (P)] = x a x d [ Γ d bc Γg ce(p)x e Γ d bg Γ f bc (P)x c Γ d f c + Γ f bc (P)x c Γ g ce(p)x e Γ d f g Γd bc (P)] = x a x d [ Γ d bc Γd bc (P)] + O( x ) = O( x ) Theorem 1.2. The partial covariant derivatives ( c T)e c of any tensor T of type (q, r), is a tensor of type (q, r + 1).

14 14 F. Rothe Proof for the case (q, r) = (1, 0). Let v = v a e a be a contravariant vector. The c-th partial covariant derivative has the components c v a = va x c + Γa scv s obtained from formula (1.19). Under any point transformation x a = x a(x 1,..., x n ) the contravariant components [v b ] have the transformation law The partial derivatives are v a xc v a = x e x e x c But we need the covariant derivatives v a = x a x b vb ( ) = xc x a x e x c x b vb = xc 2 x a x e x b x c vb + xc x a v b x e x b x c ev a = v a x e + Γ a be v b The connection coefficient are transformed by the second form of the rule (1.22) Γ a be = x a x f x g x d x b x e Γd f g x f x d 2 x a x b x [ e x d x f x Γ a a x g be v b = x d x e Γd f g xd 2 x a ] x f x e x d x f [ x a x c = x d x e Γd f c xb 2 x a ] x e v f x b x f v b x b and addition of the formulas yields the covariant derivatives, and the terms with the second derivative cancel. ev a = v a x e + Γ a be v b = xc 2 x a x e x b x c vb + xc x a v b x e x b x c + x a x c x b = xc x e x a x b [ v b x c + Γb f c v f ] = xc x e x a x b cv b x e Γb f c v f xb 2 x a x e x b x v f f Proof for the case (q, r) = (0, 1). Let f = f a ω a be a covariant vector. The c-th partial covariant derivative has the components c f a = f a x c Γs ac f s obtained from formula (1.20). Under any point transformation x a = x a (x 1,..., x n) the transformed covariant components [ f b ] are f b = xs x f b s and taking the partial derivatives one gets ( ) x s = 2 x s f b x c = x c x b f s x b x c f s + xa x b f a x d x d x c

15 Topics from Relativity 15 But we need the covariant derivatives c f b = f b x c Γ s bc f s The connection coefficient are transformed by the first form of the rule (1.22) Γ s bc = x s x h x g x d x b x c Γd hg + x s 2 x d [ x d x c x b x Γ s bc f s h x g = x b x c Γd hg + 2 x d ] x s x c x b x f d s [ x h x g = x b x c Γd hg + 2 x d ] f x c x b d and subtraction of the formulas yields the covariant derivatives. derivative cancel. Some index shoveling is needed, and one gets c f b = f b x c Γ s bc f s = xa x d f a x b x c x xh x g d x b x c Γd hg f d = xa x d [ ] fa x b x c x d Γs ad f s = xa x d x b x c c f a The terms with the second End of the proof of theorem 1.2. One now proceeds by induction. Because of formula (1.17) the covariant derivatives of tensors of type (0, 0) have type (0, 1). In other words, the derivative of a scalar is a covariant vector. By the proof for the case (q, r) = (1, 0), the covariant derivative [ c v a ] of a tensor [v a ] of type (1, 0) has type (1, 1). The Leibniz product rule from equation (1.3) implies that the covariant derivatives of tensors of type (0, 1) has type (0, 2). Indeed, for any contravariant vector [v a ] and covariant vector [ f a ] holds c (v a f a ) ( c v a ) f a = v b ( c f b ) Both terms on the left-hand side have type (0, 1). The contravariant vector [v b ] has type (1, 0), and let the tensor [ c f b ] have type (q, r). The contraction operation lowers the type from (q, r) to (q, r 1) for the tensor [v b c f b ]. Since types on both sides are equal one concludes (0, 1) = (q, r 1), and hence tensor [ c f b ] has type (0, 2). Based on the fact that all covariant derivatives are to be obtained via the stipulated Leibniz rule (1.21) c (v s T a 1...a q b 1...b r ) = ( c v s )T a 1...a q b 1...b r + v s ( c T a 1...a q b 1...b r ) one may proceed inductively from tensors of type (q, r) to those of types (q + 1, r) and (q, r + 1), and check that indeed the tensor [ c T a 1...a q b 1...b r ] is of type (q, r + 1). Definition 1.12 (Intrinsic derivative of a vector along a curve). For a field v = v a e a of contravariant vectors, the intrinsic derivative along the curve x = x(u) with any real parameter u is Dv a du = dva du + Γa scv s dx c du Dv du = ( cv a dx c )e a du (1.25) (1.26)

16 16 F. Rothe For a field f = f b ω b of covariant vectors, the intrinsic derivative along the curve x = x(u) with any real parameter u is Remark. The first formula (1.25) yields Dv a du = dva du + Γa scv s dx c du = D f b du = d f b du Γs bc f dx c b du Df du = ( c f b )ω b dxc du [ ] v a dx x c + Γa scv s c Hence after putting a basis the second formula (1.26) is obtained. Dv du = D(va e a ) = Dva Du du e a = ( c v a dx c )e a du du = cv a dxc du (1.27) (1.28) Remark. Suppose a specific embedding M R N of the manifold is given. The intrinsic part of any vector X R N is given by the projection Pro j TP X from equation (1.18). The intrinsic derivative of the vector v along a curve x c = x c (u) gets Indeed Dv du = Pro j dv T P du dv Pro j TP du = Pro j d(v a e a ) T P du = dva du e a + v s e s dx c Pro j TP x c du = dva [ ] dv a = du + vs Γ a dx c sc e a = Pro j TP [ dv a du e a + v s e s dx c ] x c du du e a + v s e a ω a, e s x c dxc du du = [ c v a] dx c e a du (1.29) Definition 1.13 (Intrinsic derivative along a curve). Given is a curve x c = x c (u) and a tensor field T = T(x) of type (q, r). The intrinsic derivative DT Du of the tensor T along this curve is defined by the identity DT du = ( ct a 1...a q b 1...b r )e b 1...b r a 1...a q Under the assumption that specific embedding M R N of the manifold is given DT du = e b 1...b r a 1...a q dx c du (1.30) ω a 1...a q b 1...b r, dt(x(u)) (1.31) du Corollary 2. We assume an embedding M R N of the manifold exists. Then the intrinsic derivative of the tensor T of type (q, r) is again a tensor of the same type (q, r). DT Du Proof. The equation (1.31) is a coordinate free definition since the projection involved is coordinate free. 1 Corollary 3. We assume an embedding M R N of the manifold exists. The covariant derivative of the tensor T of type (q, r) T = ( c T) ω c is a tensor of the same type (q, r + 1). 1 Helmholtz ants can feel tensors but no coordinates.

17 Topics from Relativity 17 Proof. The contraction DT du = ( ct) dxc du is of type (q, r) and the tangential vector dxc du is of type (1, 0). Hence the tensor ct is of type (q, r + 1) Riemannian manifold Definition 1.14 (Riemannian manifold). A Riemannian manifold M is a differentiable manifold with an additional metric structure. At every point and for every differential dx = e a dx a at that point, the length ds is given by the Riemannian metric The symmetric matrix [g ab ] is assumed to be nonsingular ds 2 = g ab dx a dx b (1.32) g ab = g ba and g := det g ab 0 and have the same number of positive and negative eigenvalues at all points of the manifold. 1 Definition 1.15 (Dot product). By putting e a e b = g ab for all a, b = 1... n and extending by linearity, a commutative dot product is defined on a Riemann manifold. Postulate. For a Riemannian manifold, the tangent space T P and the cotangent space T P are identified by the requirement that the inner product equals the bilinear form: b a = b, a for all b T P and a T P (1.33) Lemma The postulate to identify the tangent plane to the cotangent plane is equivalent to the rules for lifting and lowering an index: ω a = g ab e b and e a = g ab ω b Here the matrices [g ab ] and [g ab ] are inverse of each other. The rules extend to any of the indices of any tensor of any type. Moreover, under this postulate hold both ω a e c = δ a c e a e c = g ac (1.34) for all a, c = 1... n. The postulate is compatible with the identification T P = T P from lemma 1.4. Proof. For the base vectors, the requirement (1.33) gives ω a e c = ω a, e c = δ a c for all a, c = 1... n (1.35) Since [e b ] is a basis of the tangent plane, any identification T P T P gives formulas ωa = g ab e b with a, b = 1... n. From equations (1.35) one obtains g ab e b e c = g ab g bc = δ a c (1.36) Hence the matrices [g ab ] and [g ab ] are inverse to each other, and one has obtained the rule to lift the index. The rule to lower the index is now easy to check. 1 The last requirement can be proved under the assumption that the manifold is connected.

18 18 F. Rothe Conversely, the rules to lower and lift indices imply the requirement (1.33) that the inner product equals the bilinear form. The identifications TP = id(t P) = T P from equation (1.10) and TP = T P from equation (1.33) work together to produce b a = b, a = id(a), b = a, b = a b for all b TP and a T P. One puts.,. =.,.. No contradiction arises since the dot product is commutative. Corollary 4. Corresponding rules to lift and lower indices apply to tensors of any type (q, r). Problem 1.7. Apply the rules for lifting and lowering the indices from lemma 1.13 to get the components t a bc in terms of the components t f de. Answer. t a bc = gad g c f t Lemma For any C 1 -smooth Riemann manifold, the metric tensor has covariant derivatives zero, and indeed satisfies c g ab = Γ bac + Γ abc and c g ab = 0 (1.37) Proof. The connection coefficients are defined by equation (1.13). By the rule from lemma 1.13, one may lower the first indices on both sides and obtain f db Γ a bc = ωa, c e b Γ abc = e a, c e b Taking the intrinsic partial derivative c on both sides of the second equation (1.34) and using Leibniz rule and commutativity of the dot product, we get e a e b = g ab e b ( c e a ) + e a ( c e b ) = c g ab From the postulate to identify the tangent space T P and the cotangent space TP, the dot products are bilinear forms and e b, c e a + e a, c e b = c g ab For the partial and the covariant derivatives of the metric tensor, one gets c g ab = Γ bac + Γ abc c g ab = c g ab g sb Γ s ac g as Γ s bc = cg ab Γ bac Γ abc = 0 Theorem 1.3. For a C 2 -smoothly embedded Riemann manifold, the metric tensor determines the connection symbols. Γ abc = 1 2 ( bg ca + c g ab a g bc ) (1.38) Γ a bc = gad 2 ( bg cd + c g db d g bc ) (1.39)

19 Topics from Relativity 19 Proof. By lemma 1.12 the connection coefficients are symmetric: Γ a bc = Γa cb and hence Γ abc = Γ acb. Since the equations Γ bac + Γ abc = c g ab and Γ abc = Γ acb holds for all indices a, b, c = 1... n and especially their permutations, they imply identity (1.38). The second identity (1.39) follows by the rule to lift an index. Problem 1.8. A metric is called diagonal if g ab = 0 for all a b. Convince yourself that for a diagonal metric and symmetric connection, the connection coefficients Γ a bc are zero for a, b, c all three different. Check the formulas where no summation is implied. Γ a ac = cg aa 2g aa for all a, c = 1... n; Γ a bb = ag bb 2g aa for all a, b = 1... n with a b; Given a point P on the pseudo-riemannian manifold, as used in general relativity. We know that there exists a point transformation such that g ab (P) = η ab (P) at this one point P. Problem 1.9. Explain, using linear algebra, how to get such a point transformation, it is even a linear one. In the above situation, the cotangent and tangent base vectors satisfy ω 0 = e 0 and ω i = e i for i, k = 1, 2, 3 and α, β = 0, 1, 2, 3. e 0 e 0 = 1, e 0 e i = 0, e i e k = δ ik e α e β = η αβ Definition 1.16 (Tetrad). The base vectors satisfying ê α ê β = η αβ for α, β = 0, 1, 2, 3 are called a tetrad. They are denoted by carots. I call an orthonormal basis in three dimension a tetrad, too, and use the same notation. Problem Suppose the metric g ab is diagonal and positive definite, as occurs for example for spherical coordinates in R 3. Write down, in terms of g aa : the relations of the bases e a for the tangent space and ω a for the cotangent space; and the relations to the corresponding tetrad ê a. Answer. ω a = g ab e b = e a ê a = g aa e a gaa = g aa ω a

20 20 F. Rothe In the case of orthogonal coordinates, one obtains the same tetrad from the tangent basis as the cotangent basis. Hence one may further simplify the notation. For example, take spherical coordinates (r, θ, φ) in R 3. It is customary to denote the unit vector by the boldface name of the coordinate with a carot put above it: r = e r = ω r θ = r 1 e θ = r ω θ φ = (r sin θ) 1 e φ = r sin θ ω φ Problem Assume at some point, the electrical field has the covariant components [E r, E θ, E φ ] = [2, 3, 5]. Calculate the components for the orthonormal basis r, θ, φ. Problem Assume at some point, the velocity of a particle has the contravariant components [v r, v θ, v φ ] = [2, 3, 5]. Calculate the components for the orthonormal basis r, θ, φ. Definition 1.17 (Christoffel symbols). For any smooth Riemann manifold, the Christoffel symbols are defined by M abc = 1 2 ( bg ca + c g ab a g bc ) (1.40) { } a = gad b c 2 ( bg cd + c g db d g bc ) = gad 2 M dbc (1.41) Lemma For any C 2 -smooth Riemann manifold { } a Γ a bc = + 1 b c 2 { } a Γ a bc = + 1 b c 2 1 ( Γ a bc 2 + ) { } a Γa cb = + 1 b c 2 where T a bc = Γa bc Γa cb is the torsion tensor. ( T a bc + T a cb + T a bc ( T a bc T a c b + T ) a bc ( T a bc + T ) a cb Proof. Since the torsion is a tensor, it is sufficient to prove Γ abc = M abc (T bca + T cba + T abc ) Γ abc = M abc (T abc T cab + T bca ) 1 2 (Γ abc + Γ acb ) = M abc (T bca + T cba ) Since c g ab = Γ bac + Γ abc by equation (1.37), the definition (1.40) yields ) (1.42) (1.43) (1.44) M abc = 1 2 ( bg ca + c g ab a g bc ) = 1 2 (Γ acb + Γ cab + Γ bac + Γ abc Γ cba Γ bca ) M abc Γ abc = 1 2 (Γ acb + Γ cab + Γ bac Γ abc Γ cba Γ bca ) = 1 2 (T acb + T cab + T bac ) Γ abc M abc = 1 2 (T bca + T cba + T abc ) = 1 2 (T abc T cab + T bca ) (Γ abc + Γ acb ) 2M abc = 1 2 (T bca + T cba + T abc + T cba + T bca + T acb ) = T bca + T cba which checks formulas (1.42) and (1.43) and (1.44).

21 Topics from Relativity Lie derivative Any vector field [v a ] defines infinitesimal point transformations with x a = x a + εv a x a = x a εv a + O(ε 2 ) (1.45) x a x c = δa c + ε c v a x c x a = δc a ε a v c + O(ε 2 ) (1.46) One may even define on the manifold a (local) flow ε x = Φ(ε, x) as the (local) solution of the initial value problem Φ a (ε, x) = v a (x) (1.47) ε For any tensor field one gets the induced flow t = Φ(ε, t). Here the tensor t has been transformed, but is still evaluated at the same coordinates x. 1 t (x ) = Product o f Jakobians t(x) (1.48) Φ(ε, t)(x) = t (x) = Product o f Jakobians t(φ( ε, x)) (1.49) Definition 1.18 (Lie derivative). The Lie derivative L v t of any tensor t along the vector field v is determined by the transformation of the tensor under the above flow. Either in terms of infinitesimal point transformations (1.45), one defines t(x ) t (x ) = εl v t + O(ε 2 ) t (x) t(x) = εl v t + O(ε 2 ) (1.50) or the flow x = Φ(ε, x) and its induced flow t = Φ(ε, t) Φ(ε, t) t lim = L v t (1.51) ε 0 ε Lemma 1.16 (Rule to get the Lie derivative). In the general case of a tensor T of type (q, r), the components of the Lie derivative are denoted by L v T a 1...a q b 1...b r They are obtained by the following rule: Each component is the sum of 1 + q + r terms. The first term is the directional derivative v s s T a 1...a q b 1...b r The remaining terms are all products of the tensor components with some partial derivative of components from the vector field v along which the Lie derivative is taken. The terms corresponding to the q upper indices are subtracted. For each term, the vector field has robbed a different one of contravariant indices a 1... a q and replaced this index by a contravariant summation index s. The partial derivative s is taken along the robbed index, of the vector component with the robbed index. The last r terms are added. Once more, for each term, a different one of covariant indices b 1... b r has been "robbed" and is replaced by a covariant summation index s. The partial derivative of the vector field components v s is taken along the robbed index. Problem Convince yourself that for any scalar function S = S (x), the Lie derivative is the directional derivative: L v S = v a a S (1.52) 1 The minus sign appears naturally, well motivated by throughout use of passive transformations.

22 22 F. Rothe Answer. Since a scalar function transforms by the rule S (x ) = S (x) under any point transformation, one gets from the first formula (1.5) and from any Taylor expansion S (x ) S (x) = S (x ) S (x ) = εl v S + O(ε 2 ) S (x ) = S (x) + (x a x a ) a S + O( x x 2 ) = S (x) + εv a a S + O(ε 2 ) Thus comparison give the formula (1.52). Hence the Lie derivative of a scalar is the directional derivative. Problem Apply the rules for the Lie derivative, given by lemma 1.16, to get L v t c ab of tensor t along the vector field v. Answer. Proof of validity. The first term comes from partial derivatives: L v t c ab = vd d t c ab ( av s )t c sb ( bv s )t c as + ( s v c )t s ab t(x ) t (x ) = t(x ) t(x) [ t (x ) t(x) ] t c ab (x ) t c ab (x) = εvs s t c ab + O(ε2 ) The second term comes from the transformation flow: t c ab(x ) = xd x a x e x b x c x f t f de (x) = ( δ d a ε a v d) ( δ e b ε bv e) ( δ c f + ε f v c) t f de (x) + O(ε2 ) = t c ab (x) ε av d t c db ε bv e tae c + ε f v c t f ab + O(ε2 ) t ab(x c ) t c ab (x) = ε av s t c sb ε bv s tas c + ε s v c t s ab + O(ε2 ) Subtraction of the second from the first yields t c ab (x ) t c ab(x ) = εv s s t c ab + ε av s t c sb + ε bv s t c as ε s v c t s ab + O(ε2 ) The second formula is obtained since t c c ab (x) t ab(x) = t c ab (x ) t ab(x c ) + O(ε 2 ) The third formula (1.51) is obtained from the definition (1.48). Corollary 5. The Lie derivative L v of a tensor t is a tensor of the same type, provided the vector field v is transformed, too. Proposition 1.3. The Lie derivative L v applied to tensors of any type with the vector field v fixed, obeys the Leibniz product rule. Proof of the most simple case. For any contravariant vector [h a ] and covariant vector [k a ] holds (L v h a )k a + h a L v (k a ) = [ (v s s h a ) h s ( s v a ) ] k a + h a [ (v s s k a ) + k s ( a v s )k s ] = (v s s h a )k a + h a (v s s k a ) = v s s (h a k a ) = L v (h a k a )

23 Topics from Relativity Special Relativity 2.1. Relativity of time and length (ct, x, y, z) be the coordinates for any event, measured in the inertial system S. Let (ct, x, y, z ) be the coordinates for the same event, measured in the inertial system S. Assume that the origins of systems S and S are equal, and that the system S moves with velocity v in +x direction relative to system S. We want to determine the linear transformation t = At + Bx x = Dt + Ex y = y and z = z (2.1) In special relativity, it is customary to introduce the dimensionless parameters β := v c and γ := 1 1 β 2 Problem 2.1. From the relative velocity of the two systems, and from the postulate of constancy of the velocity of light c, one gets the three assumptions x = vt if and only if x = 0; x = ct if and only if x = ct ; x = ct if and only if x = ct. Use these three assumptions to determine the constants B, D and E in terms of A and the relativity parameters β and γ. Answer. x = vt x = 0 yields D = Ev; x = ct x = ct yields ca D + c 2 B ce = 0 since 0 = ct x = (ca D)t + (cb E)x = (ca D + c 2 B ce)t x = ct x = ct yields ca D + c 2 B + ce = 0 since 0 = ct + x = (ca D)t + (cb E)x = (ca D c 2 B + ce)t Subtracting the last two relations yields 2cA 2cE = 0, hence A = E. Adding them yields 2D + 2c 2 B = 0, hence D = c 2 B. After eliminating B, D and E one gets t = A t Avx/c 2 x = Av t + A x y = y and z = z In 4d-matrix notation: ct x y z = A βa 0 0 βa A ct x y z Definition 2.1 (Isochronic Lorentz-, proper Lorentz transformation). An isochronic Lorentz transformation maps the cone of light rays which point into the future into itself. A proper Lorentz transformation maps the cone of light rays which point into the future into itself, and has the determinant +1.

24 24 F. Rothe Problem 2.2. In many texts and lectures, it is customary to begin with the stronger postulate of invariance of the Minkowski metric: c 2 t 2 x 2 y 2 z 2 = c 2 t 2 x 2 y 2 z 2 (2.2) instead of the weaker postulate of constancy of the velocity of light. Use the invariance of the Minkowski metric and the fact that x = vt x = 0, to determine the constants A, B, D and E in the transformation (2.1), in terms of the relativity parameters β and γ. There are four solutions, with different signs of these constants. Write down all four solutions. What is the meaning of these solutions? Which one of these four solutions is a proper Lorentz transformation. Answer. Invariance of the Minkowski metric implies c 2 t 2 x 2 = c 2 (At + Bx) 2 (Dt + Ex) 2. We compare the coefficients of t 2, tx and x 2 and get c 2 = c 2 A 2 D 2 0 = 2c 2 AB 2DE 1 = c 2 B 2 E 2 We still use that D = Ev for a boost with relative velocity v in +x direction. Hence c 2 AB = ve 2 from the second relation. Squaring and plugging in the first and third relation yields v 2 E 4 = (c 2 A 2 )(c 2 B 2 ) = (c 2 + v 2 E 2 )(E 2 1) 0 = c 2 + (c 2 v 2 )E 2 c E = ± c2 v = ±γ 2 A 2 = 1 + c 2 D 2 = 1 + c 2 v 2 E 2 = 1 + First consider the solution with A = E = γ. transformation A = ±γ t = γ t γvx/c 2 x = γv t + γ x y = y and z = z which is the Lorentz boost with relative velocity +v. The solution with A = γ and E = +γ is t = γ t + γvx/c 2 x = γv t + γ x y = y and z = z which is the Lorentz boost followed by time reversal. The solution with A = +γ and E = γ is t = γ t γvx/c 2 x = +γv t γ x y = y and z = z v 2 c 2 v = c 2 2 c 2 v 2 One gets B = ve 2 /(c 2 A) = vγ/c 2 and the

25 Topics from Relativity 25 which is the Lorentz boost followed by space reflection. For the forth solution t = γ t + γvx/c 2 x = +γv t γ x both the time is reversed and the space reflected. y = y and z = z Here is another way to determine the constant A left open in problem 2.1. We begin with the assumptions (i) The proper Lorentz transformations are a group. (ii) Among the Lorentz transformations are the boosts as well as the usual 3-dimension rotations. (iii) The rotations without boost leave the time invariant. (iv) Any isochronic Lorentz transformation that is diagonal leaves the time invariant. Thus both the boost and the rotation L = S = A βa 0 0 βa A about the z-axis by 180 are Lorentz transformations. Problem 2.3. Calculate the matrix products S LS and LS LS. Give a convincing argument that A 2 (1 β 2 ) = 1 holds in the physically meaningful case. Determine the sign of A occurring for a proper Lorentz transformation. Once more, write down the matrix for a boost in +x direction. Answer. S LS = A βa 0 0 βa A and LS LS = A 2 (1 β 2 ) A 2 (1 β 2 ) This last matrix is a proper Lorentz transformation, and is diagonal, too. By the above assumption item (iv), it leaves the time invariant. Hence A 2 (1 β 2 ) = 1 and A = ±γ. Only the case A = +γ is an isochronic Lorentz transformation. For a graphic representation of the usual boost ct = γct βγx x = βγct + γx (2.3) one uses the same scale for the lengths ct and x, and a right angle between the ct-axis and the x-axis.

26 26 F. Rothe Problem 2.4. Convince yourself that the same angle occurs between the ct- and ct -axis as between the x- and x -axis; the angle between the x - and the ct -axis is acute. Draw the future light ray x = ct, t 0. Draw the hyperbola of all points which have the invariant space-like squared distance 1 from the origin. To illustrate the Lorentz contraction, we imagine a rigid tube of (comoving) length d with mirrors on both ends. Now the tube is moving with velocity v relative to the S -system. Thus the tube is at rest in the comoving S system. I calculate at first with coordinates from the S -system. The mirrors at the end of the tube move along the lines x = vt and x = vt + L Let a light ray OA be sent from the mirror at the right end to the mirror at the left end, and reflected into a light ray AB from the mirror at the left end to the mirror at the right end. The equations of these light rays are x = ct and x = ct + a. Problem 2.5. Determine constant a. Determine the S -system coordinates of the reflection events A and B. Answer. Point A is the intersection of light ray x = ct with the world line of the right mirror x = vt + L. Hence t = L c v and the coordinates are ( cl (ct, x) A = c v, cl ) ( ) L = c v 1 β, L 1 β and a = 2ct A = 2cL c v. Point B is the intersection of light ray x = a ct with the world line of the left mirror x = vt. Hence t = a c+v and the coordinates are ( 2c 2 L (ct, x) B = c 2 v, v 2 c 2c 2 ) ( ) L 2L = c 2 v 2 1 β, 2βL 2 1 β 2 Problem 2.6. Use the Lorentz transformation (2.3) to determine the S -system coordinates of the reflection events A and B. Answer. From the equations (2.3) for the boost one gets the S -coordinates of point A to be ct A x A and the S -coordinates of point B to be L = γct βγx = γ(1 β) 1 β = γ L L = βγct + γx = γ( β + 1) 1 β = γ L ct B = γct βγx = γ(1 2L β2 ) 1 β = 2γL 2 x B Problem 2.7. Convince yourself that 2L = βγct + γx = γ( β + β) 1 β = 0 2 L = 1 β 2 d < d (2.4) which is the famous Lorentz-FitzGerald contraction. There are no forces of any kind involved.

27 Topics from Relativity 27 Answer. In the comoving system S, the coordinates of points A and B are, by definition of the proper distance d, (ct, x ) A = (d, d) and (ct, x ) B = (2d, 0) Comparing with the result of the previous problem 2.6 yields the relation (2.4) for the Lorentz contraction. To illustrate the time dilation, we imagine a rigid tube of (comoving) length d with mirrors on both ends. Again the tube is moving with velocity v relative to the S -system, but turned by 90. Thus the tube is at rest in the comoving S system, and lying on the y -axis. I calculate at first with coordinates (ct, x, y ) from the comoving S -system. The mirrors at the end of the tube move along the lines x = 0, y = 0 and x = 0, y = d Let a light ray OA be sent from the mirror at the right end to the mirror at the left end, and reflected into a light ray AB from the mirror at the left end to the mirror at the right end. The light goes on to be reflected forth and back, and each reflection give a tick of this mirror-clock. In the comoving system, the equations of these light rays are OA : x = 0, y = ct AB : x = 0, y = 2d ct The emission and reflection events have S -coordinates O = (0, 0, 0), A = (cd, 0, d), B = (2cd, 0, 0). Thus cd is the proper time interval between the ticks of the mirror clock. Problem 2.8. Determine the S -system coordinates (ct, x, y) of the reflection events A and B. Determine the equations of the light rays OA and AB. Answer. One needs the inverse of the boost (2.3). The S -coordinates of reflection event A are The S -coordinates of reflection event B are ct A = γct + βγx = γcd x A = βγct + γx = βγd y A = y = d ct B = γct + βγx = 2γcd x B = βγct + γx = 2βγd y B = y = 0 We determine the equations of the light rays OA and AB. In the S -system, the equations of light ray OA are ct = γct + βγx = γct x = βγct + γx = γvt y = y = ct Remark. One has to keep some parameter along the light ray. This cannot be the proper time. Such a parameter is also called an affine parameter. I have chosen t for that role.

28 28 F. Rothe In the S -system, the equations of light ray AB are ct = γct + βγx = γct x = βγct + γx = γvt y = y = 2d ct Problem 2.9. Convince yourself that times of the reflection events A and B in the S -system and γcd and 2γcd. Since γ > 1, we get longer time intervals between the ticks of the clock. The moving clock is slowing down by the relativity parameter γ. Answer. From the expression for ct A and ct B calculated in the previous problem 2.8, we found already that γcd is the time interval between the ticks of the clock, as observed in the S -system. In spite of the slowing down of the clock rate, the velocity of light remains the same. Using the theorem of Pythagoras, we find which distance the light has travelled from event O to A to be x 2A + y2a = β 2 γ 2 d 2 + d 2 = γd So the moving clock slows down since the light has to travel a longer distance between the reflection events Discovery of Aberration and Parallax In 1725, James Bradley, who held a position at Oxford as astronomer and natural philosopher, began observations of γ Draconis at the home of a friend, Samuel Molyneux. Using a telescope affixed to a chimney, so that it pointed nearly vertically, he changed the position of the telescope very slightly, and very accurately measured its change in position, using a screw and plumb-line; and over the course of a year or so, found that the star did indeed vary in position during the course of the year by 40 arc-seconds, just like Polaris. Figure 2.1. Aberration. Stellar aberration produces an elliptical motion, circular at the Ecliptic poles, and linear at the Ecliptic plane, whose semi-major axis equals a constant, regardless of the distance or angular

29 Topics from Relativity 29 Figure 2.2. Parallax. position of the star, equal to one radian multiplied by the ratio of the Earth s orbital velocity, to the speed of light. This ratio is about The first figure gives radian measure, the second figure gives the angle in angular seconds multiply by /π. As the Earth moves, the apparent positions of any star are shifted in the direction of the velocity of Earth s motion. Exactly like already had been known for Polaris, the change in motion was in the wrong direction for stellar parallax. Parallax was really observed more than hundred years later, for the first time by F. W. Bessel and W. Struve in They observed a shift of for the star 61 Cyni. This star has a distance of about 11 light years from the earth, one of the stars nearest to the earth. Roughly spoken, parallax is about 1/100 of aberration, or less. Parallax produces a shift reciprocal to each star s distance in parsecs, and is used to measure the distance of the stars nearest to the earth. Parallax produces an elliptical motion of the star, circular at the Ecliptic poles, and linear at the Ecliptic plane, whose semi-major axis equals the reciprocal of each star s distance in parsecs, which is of course different for different stars. As the Earth moves, the apparent positions are shifted in the direction of the radius from the earth to the sun. Parallax occurs a quarter cycle of the circular motion later than of the aberration shift Aberration and the Doppler effect Here is the simple non-relativistic reasoning. Assume raindrops fall with the velocity c vertically. A pedestrian moves with velocity v. In which angle shall he observe the rain? Seen in a frame at rest, the rain gives a right triangle with hypothenuse c and horizontal leg c cos α, vertical leg c sin α. In the moving frame, the right triangle has same vertical leg c sin α and (non-relativistic!) horizontal leg c cos α + v. Hence tan α = c sin α c cos α + v Remark. Part of the information above is taken from website of Courtney Seligman, Professor of Astronomy.

30 30 F. Rothe Figure 2.3. Aberration of the rain and the light. A second source is dtv Atlas der Astronomie. Figure 2.4. The one-dimensional Doppler effect The one-dimensional Doppler effect Let the observer O be stationary in the inertial S - frame, with world-line CD. Let the light source or emitter E be stationary in the inertial S -frame, with world-line AB. Let the S -frame move with uniform velocity v along the positive x-axis.

31 Topics from Relativity 31 Suppose a light ray is sent in the negative x-direction. Let rays AC and BD be two successive crests of the light wave. I calculate in the S -system. A figure is provided on page 30. Subtracting (c + v)t A + k = ct A + x A = ct C + x C and (c + v)t B + k = ct B + x B = ct D + x D = ct D + x C yields (1 + β) t AB = t CD To obtain the frequency shift, one needs to calculate with the proper times, both for the emitter and observer. Since the phase of the wave is invariant, and differs by 2π for successive crests of the light wave ν E τ AB = ν O τ CD = 2π and taking into account the time dilation τ AB = 1 β 2 t AB and τ CD = t CD For the ratio of the frequencies we obtain ν O ν E = τ AB τ CD = 1 β 2 t AB t CD = 1 β β = 1 β 1 + β For β > 0, we see that ν O < ν E. Thus a receding light source is redshifted, as expected Four-vectors and Minkowski metric The approach to special relativity we use here, goes back to a lecture by Minkowski from I use a modern notation as in Hobson s book [7] General Relativity. I shall use space-time with dimensions. Space-time vectors are either denoted by their contravariant components with upper Greek indices and put into square brackets, or by an invariant four-vector and written in bold face. To begin with, to denote any position in space-time the four-vector x, or [x µ ] is used. Here x 0 = ct and x 1 = x, x 2 = y, x 3 = z are the components of the three-dimensional space vector x. We assume t, x, y, z to be real. I anyways make an endeavor that all components of a four-vector have the same dimension. Any two spacetime vectors x = (ct, x, y, z) and p = (s/c, p, q, r) have a Lorentz-invariant scalar product With the matrix notation from linear algebra this means that η = x T η p = ts xp yq zr (2.5) x p := x T η p (2.6) for the Lorentz-invariant scalar product. In spite of its surprising properties, the Lorentz-invariant scalar product has still many properties in common with the ordinary scalar product. It is easy to check that the scalar product is commutative: x p = p x non-degenerate: If x p = 0 for all vectors p, then x = 0. Definition 2.2 (Perpendicular subspace). For any subspace U R 4, the Minkowski-perpendicular space is defined as U = {y R 4 : y u = 0 for all u U}

32 32 F. Rothe Definition 2.3 (Present, future, future light-cone).. The future cone F uture = {(ct, x, y, z) : c 2 t 2 x 2 y 2 z 2 > 0 and t > 0}. A vector such that x F uture or x F uture is called time-like. The future light-cone Light + = {(ct, x, y, z) : c 2 t 2 x 2 y 2 z 2 = 0 and t 0}. A vector x Light + is called future light-like. A vector such that x Light + or x Light + is called light-like. The present Present = {(ct, x, y, z) : c 2 t 2 x 2 y 2 z 2 < 0}. A vector x Present is called space-like. The proper time along a time-like vector x = (ct, x, y, z) F uture or x F uture is the Lorentz invariant quantity x = x x = c 2 t 2 x 2 y 2 z The relativistic Doppler effect Now I give the correct dimensional relativistic argument. The amplitude of a pure plane sinus wave is given by a space-time function is described by plane waves cos(ωt k x), and their superposition. We consider any material- or light-wave but disregard polarization. The vector k R 3 is called wave vector and ω > 0 is the angular frequency. The wave length λ is related to the wave vector by 2π λ = k. Concerning (passive) transformation by the Lorentz group, we know that (ct, x) is a four-vector and the phase (ωt k x) is invariant. Hence [k µ ] = [ω/c, k] is a four-vector, too. Thus the phase has turned out to be the invariant scalar product x k of two four-vectors. So far, the arguments hold for any type of waves. We now restrict ourselves to light waves in vacuum. In that case, the wave equation implies k k ω 2 /c 2 = 0 and hence the four-vector (ω/c, k) is light-like. The wave length λ, the frequency ν and the circular frequency ω are related by k = 2π λ = ω c = 2πν c Suppose a light ray lies in the xy-plane, and let θ be the angle between the direction of propagation of light and the positive x-axis. In this setting, the wave four-vector is (ω/c, k) = 2π (1, cos θ, sin θ, 0) λ Let the observer O be stationary in the inertial S -frame. Let the light source or emitter E be stationary in the inertial S -frame. Let the S -frame move with uniform velocity v along the positive x-axis. The corresponding quantities referring to a S -frame are denoted by primes. The relative velocity is β = tanh λ L with the Lobashevskij parameter λ L. We use the abbreviation, common in relativity β = v c and γ = The Lorentz transformation between the two frames is ct = γct βγx x = βγct + γx 1 1 β 2 = cosh λ L y = y and z = z

33 Topics from Relativity 33 or in matrix notation ct x y z = γ βγ 0 0 βγ γ ct x y z Exactly the same Lorentz transformation applies to (ω/c, k) since this is a the space-time vector, too. Hence we conclude ω /c γ βγ 0 0 ω/c k x βγ γ 0 0 k k y = x k y k z k z So far, this transformation even holds for any type of waves. We specialize to light rays in the xy-plane, and use polar coordinates to get 1 γ βγ π cos θ λ sin θ = 2π βγ γ 0 0 cos θ λ sin θ Hence the Doppler shift of the wave length, respectively frequency, is ν E = ν ν O ν = λ 1 β cos θ = γ(1 β cos θ) = λ 1 β 2 The transformation of the spatial components give the change of direction called aberration. We obtain the formulas cos θ = cos θ β 1 β cos θ sin θ = γ 1 sin θ 1 β cos θ tan θ sin θ = γ cos θ γβ For the motion of the earth around the sun β Even motion through the milky way gives relative velocity of the same magnitude. And even the recently measured motion of the milky way against the average position of many spiral nebula gives β in the order of Hence in the case of astronomical aberration, the deflection angle θ θ is small. After some calculation we get sin(θ θ) = sin θ cos θ cos θ sin θ = β sin θ + sin θ cos θ(γ 1 1) 1 β cos θ = β sin θ + O(β 2 ) For the approximation in first order of β, we have confirmed the common formula θ = θ + β sin θ + O(β 2 ) Remark. With a "wave" vector pointing from the observer, we need to do the substitutions: θ α and θ α Hence the different sign of β occurs in the intuitive argument above.

34 34 F. Rothe 2.7. Four-velocity For a particle moving under arbitrary forces, one uses the world-line x µ = x µ (τ) for components µ = 0, 1, 2, 3 and the proper time τ as parameter. For the worldline of a light ray x µ = x µ (p), one has to use any other arbitrary parameter p since the proper time is constant along the light path. The four-velocity [u µ ] is defined as the derivative whereas the Newtonian velocity is u µ = dxµ dτ v = d x dt Because of time dilation dt = γdτ, they are related by 1 [u µ ] = γ[c, v] = γ[c, v x, v y, v z ] Lemma 2.1. The four-velocity for a material particle is a time-like vector pointing to the future, and has the Lorentz-invariant length u = c. Reason. and u 0 = γc c > 0. u u = γ 2 (c 2 v 2 ) = c 2 γ 2 (1 β 2 ) = c The energy-momentum vector For any particle of rest-mass m, the energy-momentum vector [p µ ] is defined by p µ = mu µ = m dxµ dτ The space components of this four vector are p = γm d x dt This turns out to be the momentum occurring in Newton s law 2 F = d p dt (Newton) What is the meaning of the component p 0? To find out, one needs to use the relation the relation W = F x (work) for the work W done by a force F during a motion over a distance x. We imagine that a particle is moving along some path, under the influence of forces. These may be for example forces from electric and magnetic fields, among other possibilities. The path is denoted as x µ = x µ (τ), with the proper time τ as parameter. From the property of the four-velocity, checked in lemma 2.1, one gets (p 0 ) 2 p 2 = p p = m 2 c 2 As long as the forces do not charge the rest-mass, which is true for electric forces, anyway, we have found the constant of motion m 2 c 2. One differentiates by the parameter τ used on the particle path and obtains p 0 dp0 dτ = p d p dτ 1 To avoid upper indices 1, 2, 3 for the components of a common vector, I use instead the lower indices x, y, z. 2 I have to ask the reader to accept this statement at face-value.

35 Topics from Relativity 35 The definition of the four-momentum implies One uses this identity and cancels p 0 p = γm v = p0 c v cp 0 dp0 dτ = p0 v d p dτ c dp0 dτ Now Newton s law (Newton) is used to obtain d p = v dτ = d x dt d p dτ = d x dτ d p dt c dp0 dτ = d x dτ F The right-hand side is the rate at which work is done by the force F. This work is only used to increase or decrease the kinetic energy T of the particle. Hence c dp0 dτ = dt dτ Integration over the particle path yields cp 0 = T + R with some constant of integration R. The constant is determine by referring to a spot p µ (τ 0 ) along the path, where the particle is at rest. For the parameter τ 0 we obtain p 0 = mc and T = 0, and hence conclude that R = mc 2. Since R is constant, one has obtained cp 0 = T + mc 2 Following Einstein, the term mc 2 is interpreted as the energy equivalent of the rest mass, and E := T + mc 2 as the total energy of the particle. Theorem 2.1. The four-momentum of a particle moving in any field of forces is p µ = [E/c, p ] The total energy E and the rest mass m are related by E = m 2 c 4 + c 2 p 2 (2.7) The momentum and the velocity by p = E v (2.8) c2 These formulas retains their meaning for m = 0, as does occur for a photon, or other massless particle. Problem A particle is non-relativistic if c p mc 2, or equivalently β 1. Use the power expansion x 1 + x = x2 8 ±... to get the approximation of a kinetic energy for a non-relativistic particle.

36 36 F. Rothe Answer. From the relation (2.7), the total energy E = T + mc 2 and the kinetic energy T are E = m 2 c 4 + c 2 p 2 = mc p2 m 2 c 2 [ T = mc p2 m 2 c 1 2 = p 2 ] mc2 2m 2 c p4 2 8m 4 c ±... 4 T p2 2m as well known from the basics of classical mechanics. Theorem 2.2. The four-momentum of a massive particle moving in a field of forces is, in terms of its velocity p µ = [E/c, p ] = [γmc, γm v] The total energy E and the total mass γm are related by E = γmc 2 These formulas do not retain their meaning for m = 0. Figure 2.5. The principal setup of Compton s experiment The Compton effect If light consists of photons, collisions between photons and particles of matter should be possible. For photons and electrons, this quantum effect was discovered in 1922 by A.H. Compton from the university of St. Louis in Missouri. The figure on page 36 shows the principle setup of Compton s experiment. The monochromatic Mo-K α -rays are scattered by a graphite crystal. The wave-length spectrum dw/dλ of the scattered x-radiation is measured, by means of a Bragg crystal, for different scattering angles θ. Additionally to photons of the incident wavelength λ = m = 70 pm, photons with a longer wavelength λ occurs. The shift λ(θ) = λ (θ) λ is an increasing function of the scattering angle θ. For example, λ(90 ) = 2.4 pm. Compton already gave the correct interpretation of his results as an elastic scattering of photons by the quasi-free electrons inside the graphite.

37 Topics from Relativity 37 Actually one finds that the scattered photons have two different wavelengths. One set of photons gets their wavelength shifted, by a shift depending on the scattering angle. The shift comes out as predicted below, by the scattering from electrons. A second set of photons has unshifted wavelength. This set is due to scattering from the positively charged ions. The mechanism of scattering is the same for both sets, except that for the ions the electron mass has to be replaced by the ion mass, which is many thousand time larger. So the shift is tiny. Figure 2.6. The kinematics of the collision of a photon with an electron, initially at rest. In short-hand, the scattering process is written as γ + e γ + e. The situation in the lab system is shown in the figure on page 37. For a relativistic calculation, we assume that the four-momenta of the incident photon and electron are and for the scattered photon and electron 1 are q = ( ω/c)[1, 1, 0, 0] and p = [mc, 0, 0, 0] q = ( ω /c)[1, cos θ, sin θ, 0] and p = (E /c)[1, β cos φ, β sin φ, 0] and use the conservation of the total four-momentum. Thus one gets four equations q + p = q + p (2.9) from which we want to eliminate the less interesting quantity p, by means of the known relation (2.7), which hold both for the incident and the scattered electron. From here on, it is perfectly possible to proceed just by elementary means. 2 I prefer to take advantage of the invariant scalar products as follows: (q + p) 2 = (q + p ) 2 q q p + p 2 = q q p + p 2 We use now q 2 = q 2 = 0 and p 2 = p 2 = m 2 c 4, and next eliminate the variable p : q p = q p = q (p + q q ) = q (p + q) 1 I prefer to use unwound angles from polar coordinates. The drawing on page 37 is an example with φ > 0 and θ < 0. 2 A up-to-date presentation along these lines is given for example in the textbook [6] p.114

38 38 F. Rothe We put the assumed vectors into the last equation to get ωmc 2 = ω ( ω + mc 2 ) 2 ωω cos θ λ λ = ω ω(1 cos θ) + mc2 = ω mc 2 λ ωλ(1 cos θ) λ = = h (1 cos θ) mc 2 mc The factor h mc = λ C = pm is called the Compton wavelength of the electron. A photon of the wavelength λ C has the energy mc 2 = ev, equivalent to the rest mass of the electron. Thus we have obtained the shift of the wavelength to be λ λ = λ C (1 cos θ) (2.10) Problem Check that the kinetic energy of the scattered electron is T = mc 2 λ2 C (1 cos θ) λλ Take as an example the monochromatic Mo-K α -rays from A.H. Compton s experiment of The incident wavelength is λ = 70 pm. Calculate the kinetic energy of the scattered electron. Answer. From the 0-component of the four-momentum conservation (2.9), the kinetic energy of the electron is T = cq 0 cq 0 and converted to wavelengths becomes ( 1 T = cq 0 cq 0 = ω ω = hc λ 1 ) ( λ = mc 2 λc λ λ ) C λ The equation (2.10) for the shift of the wave length yields T = mc 2 λ2 C (1 cos θ) λλ With the data from Compton s experiment of 1922, one obtains T = (1 cos θ)635.9 ev. Problem Determine the direction of the scattered electron, from the y-component of the fourmomentum conservation (2.9). (a) Write down the y-component of the four-momentum conservation. (b) Get sin φ in terms of sin θ, and the wave length λ and relativity parameter γ for the scattered electron. (c) From the kinetic energy T obtained in the previous problem see that γ 1 = λ2 C (1 cos θ) λλ Use this expression to simply. After a bid lengthy calculation one gets sin φ = A cos(θ/2) with a factor A < 1 which is approaching one for non-relativistic electron. (d) Express the result sin φ = cos(θ/2), to be expected in the experiments, in simple geometric terms.

39 Topics from Relativity 39 Answer. (a) The y-components q y = ( ω /c) sin θ and p y = mcγ β sin φ add up to zero. One gets cp sin φ + ω sin θ = 0. (b) (c) sin φ = ω cp sin θ = h λ p sin θ = h mcλ β γ sin θ = λ C λ (γ 2 1) sin θ λ C sin φ = λ γ + 1 γ 1 sin θ = λ γ + 1 λ 2 = λ γ + 1 cos(θ/2) λ C λλ λ C sin θ (1 cos θ) (d) The non-relativistic electron is scattered along the ray opposite to the angle bisector of the incident and scattered photon. The deviation from this rule is of first order in v/c. Remark. We see that the kinetic energy of the scattered electron increases when using harder x- rays. The scattered electrons have been visible by means of a Wilson cloud chamber, for the first time by Compton in Moreover Bothe and Geiger in 1925, have used an electronic coincidence circuit connecting two Geiger counters, and have demonstrated the scattered electron and x-ray to appear simultaneously. Decreasing the intensity of the incoming x-ray, one obtains in the counters no continuous intensity, but discrete pulses. Similar to observations with a photon multiplier, these observations cannot be explained by considering light as a wave. Light acts, in these experiments, as a flow of discrete particles with all mechanical properties of a particle. During the same experiment, in the interaction with the Bragg crystal, diffraction is used for the measurement of the wave length. Here the same light acts as a wave! Part of the information above is taken from the script Wellen und Quanten, by K.R. Schubert (2004/5), professor of physics. Problem Rotate the lab-system by the angle φ such that the scattered electron moves along the positive x-axis. Transform the components of the four-momenta for the incoming and scattered photon and electron to this system. Convince yourself that sin(φ θ) sin φ = ω ω = 1 + λ C (1 cos θ) λ Conclude that for an experiment with λ λ C, one gets indeed φ θ/2 90. Answer. One applies the rotation matrix R = cos φ sin φ 0 0 sin φ cos φ to get transformed components of four-momenta. For the incident photon and electron ( ω/c)r[1, 1, 0, 0] T = ( ω/c)[1, cos φ, sin φ, 0] T (mc)r[1, 0, 0, 0] T = (mc)[1, 0, 0, 0] T and

40 40 F. Rothe and for the scattered photon and electron ( ω /c)r[1, cos θ, sin θ, 0] T = ( ω /c)[1, cos(θ φ), sin(θ φ), 0] T (E /c)r[1, β cos φ, β sin φ, 0] T = (E /c)[1, β, 0, 0] T and Only the photon has a momentum along the new y-axis. By conservation of momentum one gets ω sin φ = ω sin(θ φ). Together with the formula for the shift of the wave length sin(φ θ) sin φ = ω ω = 1 + λ C (1 cos θ) λ For an experiment with λ λ C, one gets indeed sin(φ θ) sin φ. Hence φ θ 180 φ, φ θ/ and φ 90 θ / Collision of particles Problem Inverse Compton scattering 1 occurs whenever a photon scatters off a particle moving with almost speed of light. Suppose that a particle with rest mass M and total energy E collides head on with a photon of energy E γ. For simplicity, assume that the scattered particles move back along the same axis. (i) Convince yourself that the conservation of the four-momentum implies E γ (E + cp) = E γ(e cp) + 2E γ E γ (ii) Solve for E γ and use that Mc 2 E and hence E + cp 2E holds with any accuracy needed. Show that under this assumption the scattered photon has the energy E γ = 4E 2 E γ M 2 c 4 + 4E E γ (iii) Such a situation may occur for example for a collision of an ultra-relativistic cosmic ray with a photon from the cosmic background radiation. How much energy can such a cosmic ray proton transfer to a microwave background photon? Energies up to E = ev may occur in cosmic rays. The proton has rest energy Mc 2 = MeV. A typical energy of photon from the cosmic background radiation is E γ = 2.7 K ev/k. Answer. The four-momenta of the incident photon and proton are cq = E γ [1, 1, 0, 0] and cp = [E, cp, 0, 0] the four-momenta for the scattered photon and proton are cq = E γ[1, 1, 0, 0] and cp = [E, cp, 0, 0] From the conservation of four-momentum (2.9) we eliminate the less interesting quantity p. I prefer to take advantage of the invariant scalar products and get, with the same calculation as above q p = q p = q (p + q q ) = q (p + q) 1 See Hobson [7] p.132 problem 5.14

41 Topics from Relativity 41 We put the assumed vectors into the last equation and obtain E γ (E + cp) = E γ(e cp) + 2E γ E γ E γ = E γ(e + cp) E cp + 2E γ = E γ (E + cp) 2 E 2 c 2 p 2 + 2E γ (E + cp) = E γ (E + cp) 2 M 2 c 4 + 2E γ (E + cp) We use that Mc 2 E and hence E + cp 2E holds with any accuracy needed to simplify E γ = 4E γ E 2 M 2 c 4 + 4E γ E Definition 2.4 (Center of mass system). For any set of colliding particles, the center of mass system or CMS, is the inertial system for which the total momentum of the incoming particles is zero. Thus their four-momentum is [Mc, 0, 0, 0] where M is the total rest mass. Problem Consider the Compton scattering process in the CMS system, with incoming photon moving in the positive x-direction. Write down the four-momenta of the incident photon and electron, and the possible four-momenta for the scattered photon and electron. Answer. The four-momenta of the incident photon and electron are q = ( ω/c)[1, 1, 0, 0] and p = (γmc)[1, β, 0, 0] with ω = mc 2 βγ, and the possible four-momenta for the scattered photon and electron are with any angle θ. q = ( ω/c)[1, cos θ, sin θ, 0] and p = (γmc)[1, β cos θ, β sin θ, 0] Problem The Bevatron at Berkeley was built with the idea of producing antiprotons, 1 by the reaction p+ p p+ p+ p+ p. It was designed to give about 6.2 GeV kinetic energy to the protons it accelerates. Thus one intended to let a high-energy proton strike a proton at rest; at that point of history, it was not yet possible to send two proton rays against each other. By known conservation laws, it was clear that only an additional pair of proton and antiproton could be expected. Find the threshold for the energy E, and kinetic energy T, of the incoming proton at which this reaction becomes possible. (i) Calculate the total energy-momentum four-vector p tot of the incoming particles in the lab system. (ii) At the threshold, the created four particles cannot have any additional kinetic energy. Thus they need to be at rest in the CMS-system. Calculate the total energy-momentum four-vector p tot of the scattered particles in the CMS-system. (iii) From the conservation of the four-momentum and Lorentz invariance, we know that (iv) Determine E and c p from the two equations p tot p tot = p tot p tot (E + Mc 2 ) 2 c 2 p 2 = (4Mc 2 ) 2 E 2 c 2 p 2 = M 2 c 4 1 See also R. Feynman [10], vo. II, chpt. 25 and D. Griffiths [5], p.106

42 42 F. Rothe (v) Determine the kinetic energy T and the speed of the incoming protons at threshold. Answer. (i) p tot = [(E + Mc 2 )/c, p, 0, 0] is the total four-momentum for the incoming two protons, in the lab system. (ii) p tot = [4Mc, 0, 0, 0] is the total energy-momentum four-vector of the four scattered particles, in the CMS system. (iii) The Lorentz invariant of the total four-momenta are p tot p tot = (E/c + Mc) 2 p 2 p tot p tot = 16M 2 c 2 (iv) From the conservation of the four-momentum and Lorentz invariance, we get the first equation. The second one refers to the incoming proton from the ray. One gets E = 7Mc 2 and c p = 48Mc 2. (E + Mc 2 ) 2 c 2 p 2 = (4Mc 2 ) 2 E 2 c 2 p 2 = M 2 c 4 (v) T = E Mc 2 = 6Mc 2. Indeed the antiprotons were discovered when the machine reached about 6000 MeV. At threshold, the speed of the proton is β = ( 48)/7 times the speed of light The motion of particles Definition 2.5 (Four-force). is called the four-force. f µ = [ (γ/c) v F, γ F ] Proposition 2.1. We imagine that a particle is moving along some path, under the influence of forces of whatever origin. Newton s law F = d p (Newton) dt together with the relation W = F x (work) for the work W done by a force F during a motion over a distance x are equivalent to the equation of motion f µ = d pµ (2.11) dτ Under the assumptions made above, the four-force is a four-vector. Reason. The relation (work) gives the rate at which work is done on the particle, which is equal to the rate of energy increase for the particle. Hence de dt = F d x dt

43 Topics from Relativity 43 Now the definition of the four-force and Newton s law (Newton) yield [ f µ = γ d x ] [ d ct F, γ F = γ de d ct, γ d p ] d t We convert the derivatives by coordinate time back to derivatives by the proper time, and finally use the definition of the four-momentum. Hence [ de f µ = d cτ, d p ] = d pµ d τ dτ The converse statement is checked in the same way. Remark. The conservation of the four-momentum both for the free motion, and in collision processes is well confirmed experimentally, and justified theoretically. Already for these reason, it is sound to formulate Newton s equation of motion together with the rate of work, and in terms of a differential equation for the four-momentum p µ. Moreover, the actual studies about the motion of electrons and other particles under the influence of electromagnetic forces confirm the equation of motion d p dt = q[ E + v B] (Lorentz) for a charged particle with charge q, both in the non-relativistic as well the relativistic regions. Based on these facts, the occurrence of the acceleration in the traditional form of Newton s law seems for me to be not more than an artifact. The complicated distinguishment of the parallel and transverse mass in the corrected form of the Newton s law is a further hint supporting that point of view. Definition 2.6 (Four-accelaration). For any particle with four-velocity [u µ ], and moving along any path x µ = x µ (τ), with the proper time τ as parameter, the four-accelaration vector [a µ ] is defined by a µ = d uµ dτ Lemma 2.2. In the Minkowski metric, the four-velocity and the four-acceleration are perpendicular. For a material particle, the four-acceleration is a space-like vector. Reason. We know from lemma 2.1 that the four-velocity has length c. Differentiating by the proper time yields u u = c 2 0 = d( u u ) dτ = 2 u a For a material particle the four-velocity is a time-like vector. As shown in lemma 3.3, any vector perpendicular to a time-like vector is space-like. Proposition 2.2. We assume the equations of motion (2.11) to be valid. Equivalent are: (i) The equations of motion have the form f µ = ma µ ; (ii) the four-force [ f µ ] does not change the rest mass m; (iii) the four-force is perpendicular to the four-velocity: f u = f µ u µ = 0.

44 44 F. Rothe Proof. The equations of motion (2.11) imply f µ = d muµ dτ = d m dτ uµ + ma µ (2.12) ( f µ ma µ )u µ = d m dτ uµ u µ = c 2 d m dτ (2.13) Assume item (i) holds. We conclude 0 = ( f µ ma µ )u µ = c 2 d m dτ and thus item (ii) holds. Too, item (i) and lemma 2.2 yield f µ u µ = ma µ u µ = 0 and thus item (iii) holds. Conversely assume item (iii) to hold. Hence lemma 2.2 and equation (2.13) imply 0 = f µ u µ = ( f µ ma µ )u µ = c 2 d m dτ and thus item (ii) holds. Assume now item (ii) to hold. Hence we get the equation of motion in the form f µ = d muµ dτ = d m dτ uµ + ma µ = ma µ of item (i). Lemma 2.3. In terms of the velocity v, acceleration a and relativity parameter γ, the fouracceleration has the components a 0 = cγ dγ dt, ai = γ 2 a i + γ dγ dt v i for i = 1, 2, 3

45 Topics from Relativity The Lorentz Group 3.1. Different aging of twins Problem 3.1 (Twin paradox or travelling keeps young). For this problem, I put c = 1 and consider only one space dimension. We do a series of simplifying calculations, and prove the conjecture at first in dimensions. Confirm for any two time-like vectors x = (t, x) F uture and p = (E, p) F uture, the reversed triangle inequality x + p x + p holds even with proper inequality > unless they are proportional. Answer. All the following inequalities are equivalent: x + p x + p x 2 + p x p x + p 2 t 2 x 2 + E 2 p (t 2 x 2 )(E 2 p 2 ) (t + E) 2 (x + p) 2 2 (t 2 x 2 )(E 2 p 2 ) 2tE 2xp (t 2 x 2 )(E 2 p 2 ) [ te xp ] 2 t 2 p 2 x 2 E 2 2tExp 0 (tp xe) 2 Problem 3.2. At least the calculation in problem 3.1 above is a guide, as I go now back to dimensions. Check that any two vectors x, p F uture Light + have Minkowski scalar product x p 0. Answer. Let x = (t, x, y, z) and p = (E, p, q, r) be the two vectors in F uture Light +. We calculate the Minkowski scalar product, we use the common Cauchy-Schwarz inequality xp + yq + zr (x2 + y 2 + z 2 )(p 2 + q 2 + r 2 ). Since the vectors are assumed to be in the future or light cone, 0 x 2 + y 2 + z 2 t and 0 p 2 + q 2 + r 2 E. We obtain as claimed x p = te + xp + yq + zr (x 2 + y 2 + z 2 )(p 2 + q 2 + r 2 ) te 0 (3.1) Problem 3.3. Assume that two nonzero vectors x, p F uture Light + \ {0} have Minkowski scalar product x p = 0. Check that x = αp Light + holds with α > 0. Thus they are light-like and linearly dependent. Answer. The assumption that the vectors are nonzero implies t > 0, E > 0 and te > 0. Hence (x, y, z) 0 and (p, q, r) 0. We get equality everywhere in formula (3.1). Hence the vectors (x, y, z) and (p, q, r) are proportional, as can be seen from the following calculation: (xp + qy + rz) 2 = (x 2 + y 2 + z 2 )(p 2 + q 2 + r 2 ) 2xpyq + 2xprz + 2qyrz = (x 2 q 2 + y 2 p 2 ) + (x 2 r 2 + z 2 p 2 ) + (y 2 q 2 + z 2 q 2 ) (xq yp) 2 + (xr zp) 2 + (yr zq) 2 = 0 xq = yp, xr = zp, yr = zq

46 46 F. Rothe Since (x, y, z) 0 and (p, q, r) 0, we conclude that x = αp, y = αq, z = αr with some α 0. Furthermore, (x 2 + y 2 + z 2 )(p 2 + q 2 + r 2 ) = te, x 2 + y 2 + z 2 t and p 2 + q 2 + r 2 E imply x 2 + y 2 + z 2 = t and p 2 + q 2 + r 2 = E. Hence x, p Light + and t = αe with α > 0. Together, we have confirmed x = αp Light + as conjectured. Problem 3.4. Assume the sum of any two light-like vectors is light-like. Prove that they are linearly dependent. Answer. We assume x, p Light and x + p Light. Calculate the Minkowski scalar products 0 = x + p x + p = x x + p p + 2 x p = 2 x p Hence the two vectors are orthogonal. By the last problem, they are linearly dependent. Lemma 3.1 (Reversed inequalities in the future cone). For any two future time-like or future light-like vectors x, p F uture Light + we get the inequalities 0 ( x x )( p p ) ( x p ) 2 (3.2) x p x p (3.3) x + p x + p (3.4) Equality occurs in any one of these formulas if and only if the vectors x and p are linearly dependent. Note that these inequalities in the future cone are the reversed versions of the corresponding inequalities from Euclidean geometry. Proof. Take vectors x = (t, x, y, z) and p = (E, p, q, r) F uture Light + and p 0. ( x x )( p p ) ( x p ) 2 = (t 2 x 2 y 2 z 2 )(E 2 p 2 q 2 r 2 ) (te xp yq zr) 2 [ (t 2 x 2 y 2 z 2 )(E 2 p 2 q 2 r 2 ) te (x 2 + y 2 + z 2 )(p 2 + q 2 + r 2 ) = t 2 E 2 + (x 2 + y 2 + z 2 )(p 2 + q 2 + r 2 ) t 2 (p 2 + q 2 + r 2 ) (x 2 + y 2 + z 2 )E 2 t 2 E 2 (x 2 + y 2 + z 2 )(p 2 + q 2 + r 2 ) + 2tE (x 2 + y 2 + z 2 )(p 2 + q 2 + r 2 ) = t 2 (p 2 + q 2 + r 2 ) E 2 (x 2 + y 2 + z 2 ) + 2 t 2 (p 2 + q 2 + r 2 ) E 2 (x 2 + y 2 + z 2 ) = [ t 2 (p 2 + q 2 + r 2 ) E 2 (x 2 + y 2 + z 2 )] 2 0 We have confirmed formula (3.2). To check under which assumptions equality occurs, assume formula (3.2) holds with equality. Equality occurs in the common Cauchy-Schwarz inequality xp + yq + zr = (x 2 + y 2 + z 2 )(p 2 + q 2 + r 2 ), hence x = αp, y = αq, z = αr with some factor α. The last line implies t 2 (p 2 + q 2 + r 2 ) = E 2 (x 2 + y 2 + z 2 ) and hence t = αe and α 0. Together, we see that x = αp. Formula (3.3) follows by taking roots since x p 0 by Problem 3.2. Finally we check the reversed triangle inequality (3.4) x + p 2 + ( x + p ) 2 = x + p 2 + x 2 + p x p = x + p x + p x x p p + 2 x p = 2 x p + 2 x p 0 ] 2

47 Topics from Relativity 47 Lemma 3.2 (Convexity).. The convex combination of any two independent future light-like vectors is in the future cone. The future cone F uture is convex. The union F uture Light + of the future cone and the future light cone is convex. Proof. For given linearly independent vectors x, p Light + the inequality (3.4) holds in the strict form. Hence with any 0 < α < 1 0 = αx + (1 α)p < αx + (1 α)p and αx + (1 α)p F uture. The other claims follow similarly. Proposition 3.1 (Twin paradox). For any two future time-like or future light-like vectors x, p F uture Light + which are not linearly dependent holds the reversed strict triangle inequality 1 x + p > x + p Lemma 3.3 (Orthogonal complement of a one-dimension space). Let e 0 be any vector. The orthogonal complement e is always three-dimensional. If e 0 is time-like, the orthogonal complement e consists only of space-like vectors. If e 0 is light-like, the orthogonal complement e is spanned by e itself and two space-like vectors. If e 0 is space-like, the orthogonal complement e contains both space-like and time-like vectors. There is a basis of two light-like, and a space-like vector. +?????????? We use the abbreviation Imagine that a radar signal is sent from a station on earth to the moon and the reflected signal is received. The duration between the sending time t 1 and the receiving time t 2 is measured. In which sense can we conclude the distance from the station on earth to the reflector on the moon is c(t 2 t 1 )/2? I assume for simplicity that both the station on earth and the reflector on the moon move without acceleration, and do not discuss the effects of gravity. So I deal with the following simplified picture: In space-time, there is a sender active at space-time ŝ, a reflector on the moon hit at spacetime m, and a receiver getting the signal back at space-time r. The vectors m ŝ and r m along the radar signal are light-like. The proper time r ŝ is measured. Question. What is the distance between the world line ŝ r and the reflector m? How can the spatial distance be specified precisely? Answer. Indeed we can say the distance to be moon seen in the frame with sender and receiver on the same spot is r ŝ /2. Let ĥ = ŝ + r 2 be the midpoint of the segment ŝ r from sender to receiver. We get the following three statements specifying a spatial distance: (a) The vectors m ĥ and r ŝ are perpendicular: 1 That is why traveling keeps young! m ĥ, r ŝ = 0

48 48 F. Rothe (b) The space-like distance from the midpoint to the moon is greater than the distance from any other point on the word line sender-receiver to the moon: m ĥ, m ĥ m p, m p for any point p on the word line ŝ r. (c) The space-like distance from the midpoint to the moon is the (negative) half the reflection time: 3.2. The Lorentz transformations m ĥ, m ĥ = 1 ŝ r, ŝ r 4 Definition 3.1. A Lorentz-transformation is a linear mapping from space-time to space-time that leaves the Minkowski metric (2.5) invariant. Let A be the matrix for the Lorentz transformation. Any two vectors x, ŷ and their images A x, Aŷ satisfy A x, Aŷ = x y (3.5) In other words, this means that the Lorentz transformations are the isometries for the Minkowski metric. We convert requirement to a matrix equation characterizing the Lorentz transformations. For all vectors x, ŷ x T A T GA ŷ = (A x) T G Aŷ = x T G ŷ yields the matrix equation A T G A = G (3.6) Proposition 3.2. The set of all Lorentz transformations is a group. Definition 3.2 (orthochronous and time-reversing transformations).. A Lorentz transformation which maps the future cone to itself is called orthochronous. A Lorentz transformation which maps the future cone to the past cone is called antichronous or time-reversing. Problem 3.5. Give a reason why any Lorentz transformation either maps future and past cones to themselves, or exchanges them. Reason. The invariance of the Minkowski metric implies A(F uture) F uture ( F uture) Since F uture is a convex set, the image A(F uture) is convex, too. There exist linear convex combinations of a time-like vector in the future and the past which lie in the present. Hence any convex subset of the union F uture F uture is either a subset of F uture or a subset of F uture. Since a Lorentz transformation is a bijection, and the inverse is Lorentz, too, we conclude that either A(F uture) = F uture or A(F uture) = ( F uture). These two cases lead to the orthochronous and time-reversing transformations. From the matrix equation (3.6), we see that (det A) 2 = 1 and hence det A = +1 or det A = 1.

49 Topics from Relativity 49 Definition 3.3. An orthochronous Lorentz transformation with determinant one is called proper. The subgroup of proper Lorentz transformations is the proper Lorentz group. Problem 3.6. Fix an arbitrary vector â. Show that the transformation x = x + x x â N N = 1 2 x a + a a x x (3.7) is its own inverse. It leaves the light-cone invariant. This is a nonlinear conformal transformation. Problem 3.7. Prove that any linear conformal transformation x Ax such that x x = A x, A x for all vectors x is a Lorentz transformation. Lemma 3.4. If a subspace U R 3 is invariant, then the orthogonal complement is Lorentz invariant, too. Especially, the orthogonal complement of an eigenvector is invariant. Proof. Assume the subspace U R 3 is invariant for Lorentz transformation A. This means by definition A(U) U, and hence A(U) U A 1 A(U) A 1 (U). Since the Lorentz transformation is invertible, these spaces have equal finite dimension and hence A(U) = U = A 1 (U). Given is any vector x U. Hence x u = 0 for all û U. The Lorentz invariance implies A x, Aû = x u = 0 for all û U. Hence as to be shown. A x A(U) = U Lemma 3.5. Assume a Lorentz transformation A has two linearly independent light-like eigenvectors l, m Light +. Then there eigenvalues ν, µ have product νµ = 1. The vector b orthogonal to their span is a space-like eigenvector, indeed A b = (det A) b. Proof. Let U = span( l, m) be the linear span of the two light-like eigenvectors. The orthogonal complement space U is spanned by one space-like vector b. Since the orthogonal complement of an invariant space is invariant, this is an eigenvector. A l = ν l, A m = µ m and A b = β b The Lorentz invariance implies A l, A m = l m νµ l m = l m Since the two light like vectors are independent, we know by Lemma 3.1, inequality (3.3) that l m < 0, and hence νµ = 1 and ν = µ 1. Since det A = νµβ, we get β = det A. Question. Explain these facts geometrically in Klein s model. Lemma 3.6. Assume an orthochronous Lorentz transformation A has a space-like eigenvector. Then there exist (i) either two linearly independent light-like eigenvectors; (ii) or a time-like eigenvector.

50 50 F. Rothe If both (i) and (ii) happen together, the transformation is the identity A = I, or det A = 1, and these three vectors lie in a plane and the transformation A is a reflection across this plane. Proof. Let b Present be the eigenvector and A b = β b. The orthogonal complement b is invariant, too. It contains two different linearly independent light-like eigenvectors l, m Light +. Now the invariance of b together with the Lorentz invariance leaves us with two possibilities: (i) They are both eigenvectors: A l = ν l (ii) They are switched: A l = ν m and A m = µ l and A m = µ m Since the Lorentz transformation A is orthochronous, in both cases ν, µ > 0. Here are the further conclusions for case (ii). Indeed A( µ l + ν m) = νµ ( µ l + ν m) is an eigenvector, which is a convex combination of vectors in the future light-cone, and hence by Problem?? it is time-like. Now we assume that both (i) and (ii) happen together. Hence ν = µ > 0 and, by Lemma 3.5, νµ = 1, which implies ν = µ = 1. By Lemma 3.5, the orthogonal complement U = span( l, m) is a space-like eigenvector satisfying A b = (det A) b. Altogether, we see that either det A = 1 and A = I, or det A = 1 and the transformation A is a reflection across the plane U = span( l, m). Theorem 3.1 (Structure of a orthochronous Lorentz transformation). Each proper Lorentz transformation A has an eigenvector with eigenvalue 1. Except for the identity, this eigenvector ê is unique. There are three mutually exclusive cases: Rotation If the eigenvector ê is time-like, the orthogonal complement is an invariant plane spanned by two space-like vectors. The restriction of A to this plane is a rotation by some angle α. There exists a proper Lorentz transformation S such that cos α sin α 0 S 1 AS = sin α cos α and hence TrA = cos α. Lorentz boost If the eigenvector ê is space-like, the orthogonal complement has an orthonormal basis of a space-like and a time-like unit vector. The restriction of A to this plane is a Lorentz boost with some Lobachevskij parameter λ. There exists a proper Lorentz transformation S such that cosh λ 0 sinh λ S 1 AS = 0 det A 0 sinh λ 0 cosh λ and hence TrA = det A + 2 cosh λ. Rotation about a light ray If the eigenvector ê is light-like, the orthogonal complement has an orthonormal basis of a space-like unit vector and a light-like vector. The restriction of A to this plane is sheering map. The transformation A has the eigenvalue one with geometric multiplicity one and algebraic multiplicity three. There exists a proper Lorentz transformation S such that 1 α2 α S 1 2 α 2 2 AS = α 1 α (3.8) α2 2 α 1 + α2 2 Indeed, the parameter α > 0 or α < 0 can be chosen arbitrarily except for its sign!

51 Topics from Relativity 51 Identity The existence of three linearly independent light-like eigenvectors implies that the transformation is the identity. Each orthochronous Lorentz transformation A with det A = 1 has an eigenvector with eigenvalue 1. This eigenvector b is unique, and it is always space-like. The orthogonal complement b has an orthonormal basis of a space-like and a time-like unit vector. The restriction of A to this plane is a Lorentz boost with some Lobachevskij parameter λ. Only the Lorentz boost case can occur, where λ = 0 yields a common reflection. Proof. Given is an orthochronous Lorentz transformation A. Since dimension 3 is odd, there exists a real eigenvalue and eigenvector. It is obvious from logic to state the following mutually exclusive cases: (a) (b) (c) (d) (e) There exists a time-like eigenvector. There does not exist any time-like eigenvector, but there exists two light-like eigenvectors. There exist no time-like and only one light-like eigenvector. There exist three linearly independent light-like eigenvectors. There exist only space-like eigenvectors. Consider case (a) and let ê F uture be a time-like eigenvector, hence Aê = αê. The Lorentz invariance (3.6) implies α 2 = 1. Because the transformation is orthochronous, we know that α = +1. By Lemma 3.4, the orthogonal complement ê is invariant, too. By Lemma 3.3, the orthogonal complement is spanned by two space-like vectors â and b. We may choose â, b, ê all to be orthogonal unit vectors. The matrix S for the change of basis has the new basis vectors as columns: S = [ â, b, ê ] Question. Check that our choice of the basis makes S an orthochronous Lorentz transformation, and indeed the sign of â can be chosen to make it a proper one. Too, a straightforward matrix calculation that each student needs to do at least ten times tells that the above explanations mean n 11 n 12 0 AS = S N with the normal form N = n 21 n Since the proper Lorentz transformations are a group, the normal form is an orthochronous Lorentz transformation and det N = det A. Because of the zeros and one occurring in N, the 2 2 matrix on the upper left corner is a two-dimensional rotation or reflection. In the latter case, we can choose the space-like base vectors along and perpendicular to the axis of reflection, and get even case (b) with λ = 0. Now we consider case (b). Let l, m Light + be two linearly independent light-like eigenvectors. The eigenvalues are µ > 0, since the transformation is orthochronous and µ 1 by Lemma 3.5: A m = µ m and A l = µ 1 l

52 52 F. Rothe Any linear combination of l and m with positive coefficients is in the future cone, as shown in Problem??. Hence the span U = span( l, m) contains a normalized time-like vector ê F uture, from which we get an orthogonal space-like vector â U. Here is the most simple choice: â = 2 l 2 m l m ê = 2 l + 2 m l m The action of transformation A on these vectors is A(â + ê) = µ 1 (â + ê) A(â ê) = µ(â ê) Aâ = µ + µ 1 2 Aê = µ 1 µ 2 â + µ 1 µ 2 â + µ + µ 1 2 In terms of the Lobachevskij parameter λ, the matrix elements are ê ê µ + µ 1 2 = cosh λ and µ 1 µ 2 = sinh λ The orthogonal complement space U is spanned by one space-like vector b. Since the orthogonal complement of an invariant space is invariant, this is an eigenvector with eigenvalue det A. The matrix S = [â, b, ê] for the change of basis has these new basis vectors as columns. Our choice of an orthonormal basis makes S an orthochronous Lorentz transformation. Indeed, switching the sign of â if needed, makes det S = +1 and hence S a proper Lorentz transformation. The above calculations are summarized in a matrix equation cosh λ 0 sinh λ AS = S N with the normal form N = 0 det A 0 sinh λ 0 cosh λ Consider case (d) and assume there exist three linearly independent light-like eigenvectors. Using just any two of them, we argue as in case (b). The existence of the third eigenvector implies µ = 1 and hence Lobachevskij parameter λ = 0. Hence the transformation is the identity. Finally we discuss the case (e), with the only purpose to rule it out. Assume there exist only space-like eigenvectors. Let b Present and A b = β b. By Lemma 3.6 two cases can occur: (i) either two linearly independent light-like eigenvectors exist which has already been considered in case (b); (ii) or there exists a time-like eigenvector which has already been considered in case (a). The case (c) is possible for det A = 1, and interesting. The argument is continued in the next section.

53 Topics from Relativity Infinitesimal generators Let A(x) = e xb (3.9) in the sense of the matrix exponential function. The set of A(x) for all real x is called a oneparameter group and B is called the generator. Lemma 3.7. Formula (3.9) generates a one-parameter group of Lorentz transformations if and only if B T G + G B = 0 (3.10) Proof. We need to confirm that Differentiating by x yields e xbt G e xb = G (3.11) d dx exbt G e xb = e xbt (B T G + G B) e xb Equation (3.11) holds for all x if and only if the derivative of the left-hand side is zero which happens if and only the relation (3.10) holds for the generator. In dimensions with the Minkowski metric as above, relation (3.10) holds if and only if 0 b 3 b 2 B = b 3 0 b 1 = b 1S 1 + b 2 S 2 + b 3 S 3 (3.12) b 2 b 1 0 where I have defined S 1 = We need to assume b S 2 = S 3 = Lemma 3.8. The eigenvectors of the Lorentz transformation e xb are eigenvectors of the generator B, too. If generator B has the eigenvalue β, the Lorentz transformation A = e xb has the eigenvalue e xβ with the same eigenvector. The characteristic polynomial of B is det(b λi) = λ 3 + (b b2 2 b2 3 )λ It is easy to see that B has the eigenvalue 0 with eigenvector b = (b 1, b 2, b 3 ) T. Hence b is the eigenvector with eigenvalue one for the Lorentz transformation e xb. The other two eigenvalues of B are ± b b2 2 b2 3. For the evaluation of the exponential series, one needs the powers b 2 B b2 2 b 1 b 2 b 1 b 3 = b 1 b 2 b b2 1 b 2 b 3 (3.13) b 1 b 3 b 2 b 3 b b2 2 B 3 = (b b2 2 b2 3 )B

54 54 F. Rothe The last formula can be obtained from the Hamilton-Cayley theorem. By the Hamilton-Cayley theorem, one gets zero by plugging the matrix into its own characteristic polynomial. I now consider the case (c) of Theorem 3.1. In this case b is a light-like vector. Hence B 3 = 0. This implies that the exponential series of formula (3.9) has only three terms: e xb = I + xb + x2 2 B2 Problem 3.8. Calculate the example replacing b by f = (1, 0, 1) T and matrix B by F = = S 1 + S Confirm you get back the right-hand side of formula (3.8) in Theorem 3.1, the Special case. Hence N = e αf Proof of Theorem 3.1 for case (c). Given any other proper Lorentz transformation A in the case (c) of Theorem 3.1. Let f = (1, 0, 1) T be the eigenvector for the normal form N in formula (3.8). There exists a proper Lorentz transformation R such that b = R f The matrices N and R 1 AR have the eigenvector f in common. Are the matrices N and R 1 AR equal? No, one cannot expect such a coincidence, because of the free parameter α appearing in formula (3.8). But there is a similarity transformation to adjust this parameter. Let cosh λ 0 sinh λ L(λ) = sinh λ 0 cosh λ With a bid of calculations one sees that Hence the exponential series implies L(λ) 1 FL(λ) = e λ F L(λ) 1 e αf L(λ) = e αeλ F Thus we get a transformation α αe λ. We choose λ such that the matrices L(λ) 1 NL(λ) and R 1 AR have equal restrictions to the invariant subspace f. After all the work, these two matrices have the following common properties: Both are proper Lorentz transformations. Both have the eigenvector f with eigenvalue 1. Both have equal restrictions to the invariant subspace f. Both have the same characteristic polynomial. Both have the same Jordan normal form with triple eigenvalue 1, and only a one-dimensional eigenspace spanned by f. Question. Give a detailed reasoning with linear algebra to prove that any two matrices with all the common properties mentioned above are equal. Question. Given are two automorphic collineations, and of both one knows They preserve the orientation.

55 Topics from Relativity 55 They have the same fixpoint on the line of infinity. Neither one of them has a second fixpoint Give a simple geometric argument why these two automorphic collineations are equal. In the end, with the appropriate choice of λ, we have confirmed that R 1 AR = L(λ) 1 NL(λ). Hence S 1 AS = N holds with the proper Lorentz transformation S = RL(λ) 1. Problem 3.9. Convince (at least) yourself that rotation angle α mod 2π, and the Lobacheskij parameter λ in Theorem 3.1, and the shift α along a horocycle are unique, once you require that the transformation S is a proper Lorentz transformation.

56 56 F. Rothe 4. The Poincaré Half-Plane Model In the first subsection, we construct the half-plane model via an isometric mapping of the disk to the half plane. We obtain the hyperbolic lines and distances as expected. Both the distance of two points and the Riemann metric of the Poincaré s half-plane are calculated via the isometry. The next section explains the Euler-Lagrange equation from the calculus of variations. In the following sections, we reconstruct all features of the Poincaré s half-plane model, taking the Riemann metric as starting point. At first, the curve of minimal distance between any two given points is calculated from the Euler-Lagrange equation. It turns out to be a circular arc with center on the boundary line of the half-plane, or in a special exceptional case a Euclidean line perpendicular to the boundary line. These minimizing lines specify the hyperbolic line connecting the two given points. The minimum of the hyperbolic length of any connecting curve determines the hyperbolic distances between these two points. Using the Riemann metric, the length of the hyperbolic segment is obtained by integration. The hyperbolic distance turns out to be the logarithm of the cross ratio of the two endpoints of the given segment, and the ideal endpoints of the hyperbolic line through these two points Poincaré half-plane and Poincaré disk The open unit disk is denoted by D = {(x, y) : x 2 + y 2 < 1}, and its boundary is D = {(x, y) : x 2 + y 2 = 1}. We denote the upper open half-plane by H = {(u, v) : v > 0}. Its boundary is just the real axis H = {(u, v) : v = 0}. We shall construct the half-plane model via an isometric mapping of the disk to the half plane. It is convenient to use the complex variables z = x + iy and w = u + iv. We use the notation w = u iv for the conjugate complex. Proposition 4.1 (Isometric mapping of the half-plane to the disk). The linear fractional function w = i 1 z 1 + z is a conformal mapping, and a bijection from C { } to C { }. The inverse mapping is z = i w i + w These bijective mappings preserves angles, the cross ratio, the orientation, and map generalized circles to generalized circles. The unit disk D = {z = x + iy : x 2 + y 2 < 1} is mapped bijectively to the upper half-plane H = {w = u + iv : v > 0}. Especially, one easily checks that Problem z = 1 w =, z = i w = 1, z = 1 w = 0, z = 0 w = i (a) We check whether the mapping (4.1) maps indeed the boundaries D H and find the restriction of the mapping to the boundary. Confirm that a point z = e iθ is mapped to w = tan θ 2. Do not separate real and imaginary parts. (b) Only now separate now real and imaginary parts. Use the mapping (4.1) to confirm the identities tan θ 2 = sin θ 1 + cos θ = 1 cos θ sin θ (4.1) (4.2)

57 Topics from Relativity 57 (c) Use the inverse mapping (4.2) to confirm the identities cos θ = 1 tan2 θ tan 2 θ 2 and sin θ = 2 tan θ tan 2 θ 2 Solution of part (a). We plug z = e iθ into formula (4.1) and get w = i 1 z 1 + z = i 1 eiθ 1 + e iθ = i e iθ/2 e iθ/2 2 sin(θ/2) = e iθ/2 + eiθ/2 2 cos(θ/2) = tan θ 2 The Poincaré half-plane model of hyperbolic geometry is constructed from the disk model via this isometry. One translates the definitions from the section on the Poincaré disk model to the half-plane model and arrives at the following conventions: The points of H are the "points" for Poincaré s model. The points of H are called "ideal points" or "endpoints". Those are not points for the hyperbolic geometry. The "lines" for Poincaré s model are circular arcs, or in a special case Euclidean lines perpendicular to H. The "angles" for Poincaré s model are the usual Euclidean angles between tangents to the circular arcs. Remark. The inversion image of any point P = (u, v) obtained by reflection across the real axis is P = (u, v). In complex notation, reflection by the real axis is complex conjugation: point P = w = u + iv is reflected to P = w = u iv. Remark. The one-point compactification C = C { } of the complex plane is well-known and useful in complex analysis, especially it is possible to define regularity and power series of functions in a neighborhood of. Only a linear fractional mapping as for example mapping (4.1) and its inverse are naturally extended to bijective continuous mappings 1 C { } C { } Hence the half-plane of Poincaré s model gets just one point. This point can occur as ideal end of a line, circle, equidistance lines or horocyle. Especially for the half-plane, this state of affairs is a bid contrary to the common imagination. Indeed, we have a totally different definition and usage of improper points in the projective completion from projective geometry. Given are any two points A and B. Let P and Q be the ideal endpoints of the hyperbolic line through A and B. These points are named in a way that A, B, P, Q occur in this order during an entire turn around the circle. For the definition of a hyperbolic "distance" and of "congruence of segments", the Definition?? and the preservation of the cross ratio are used as starting points. Thus we arrive at the following Definition 4.1 (Distance and Congruence). The hyperbolic distance of points A and B is given by AP BQ s(a, B) := ln(ab, PQ) = ln (4.3) AQ BP Two segments AB and XY are called "congruent" iff s(a, B) = s(x, Y). 1 They are indeed the only analytic mappings with such an extension

58 58 F. Rothe Since the mapping (4.2) provides an isometry between the half-plane and the disk model, we can directly calculate the Riemann metric for Poincaré s half-plane: Proposition 4.2 (Riemann Metric for Poincaré s half-plane). In the Poincaré half-plane, the infinitesimal hyperbolic distance ds of points with coordinates (u, v) and (u + du, v + dv) is Proof. The metric is determined by the requirement that (ds H ) 2 = du2 + dv 2 v 2 (4.4) z = i w i + w provides an isometry from the half-plane to the disk: Hence we need to convert the known metric (4.2) ds D = ds H (4.5) ds 2 = 4 dx2 + dy 2 (1 x 2 y 2 ) 2 (??) of the Poinaré disk model to a metric in the half-plane. We calculate the denominator 1 z 2 = i + w 2 i w 2 (w + i)(w i) (i w)( i w) = = i + w 2 w + i 2 and the derivative of the mapping (4.2): Putting these two results into formula (??) yields dz dw = 2 (w + i) 2 ds 2 = 4(dx2 + dy 2 ) 4 dz 2 = (1 x 2 y 2 ) 2 (1 z 2 ) 2 = 4 dz 2 ( ) w + i dw dw = 4 2 4v (w + i) 2 = dw 2 = du2 + dv 2 v 2 v 2 2 ( ) w + i dw v Thus formula (4.4) arises from the isometry (4.5) of the half-plane and the disk. 4v w + i The Euler-Lagrange equation The basic problem of the calculus of variations is to determine the curve y = y(x) between two given points (a, y(a)) and (b, y(b)) for which the prescribed functional L[y] := b a F(x, y, y ) dx assumes an extremum (minima or maxima), or simply becomes stationary. 1 It turns out that the stationary curves for the functional L[y] satisfy the Euler-Lagrange equation d F dx y F y = 0 1 In physical applications, the functional is obtained from first principles of physics.

59 Topics from Relativity 59 To derive the Euler-Lagrange equation, we take a pencil of connecting curves y = y(x, p) depending smoothly on a parameter p, and differentiate the functional L[y(., p)] by the parameter p. It is customary to denote the derivative of any quantity by p with the symbol δ and call it the variation of this quantity. One obtains b δ L[y] = d dp L[y] = d F(x, y, y ) dx a dp b [ F(x, y, y ) = y a y p + F(x, y, y ) y ] y dx p [ F(x, y, y ) = y y ] b b [ F(x, y, y ) + y p y p d ( F(x, y, y ) ) dx y y ] p a a The boundary terms vanish for a problem with prescribed endpoints (a, y(a)) and (b, y(b)) of the curve. Hence we obtain b [ F δ L[y] = y d ( )] F dx y δy dx a Since the variation δy of the curve can be chosen to be any smooth function of x, the Lemma of the calculus of variations shows that the expression in the bracket has to vanish identically. Thus we obtain the Euler-Lagrange differential equation. Lemma 4.1 (Lemma of the calculus of variations). Let g(x) be a piecewise continuous function and suppose that 1 0 η(x)g(x)dx = 0 for all functions η C. Then the function g is identically zero. Proof. We show that for a continuous function g 0 the assertion 1 0 η(x)g(x)dx = 0 does not hold for all functions η C. Assume g 0 and g continuous. Hence there exists 0 < a < 1 such that g(a) > 2ε > 0. There are cases where you need to go with the negative g and get the following reasoning for the negative function g. Because of the continuity of g there exists δ > 0 such that x a < 2δ implies g(x) g(a) < ε and hence g(x) > g(a) ε > ε. There exists a continuous, and even C function η 0 such that η(x) = 0 for x a > 2δ and η(x) = 1 for x a < δ. Hence 1 Hence the assumption that 0 η(x)g(x)dx = a+2δ a 2δ η(x)g(x)dx 1 0 a+δ a δ η(x)g(x)dx = 0 η(x)g(x)dx = 2δε > 0 for all functions η C does not hold. As a contrapositive, the assumptions that a continuous function g 0 satisfies 1 for all functions η C imply g(x) = 0 for all x. 0 η(x)g(x)dx = 0 dx

60 60 F. Rothe 4.3. The curve of minimal hyperbolic length We want to find the curve of minimal hyperbolic length connecting two given points. In this problem, it turns out to be more convenient to use the right half plane {(x, y) : x > 0} as model of hyperbolic geometry. The corresponding Riemann metric is dx2 + dy ds = 2 (4.6) x The hyperbolic length of any curve y = y(x) between two given points (a, y(a)) and (b, y(b)) is given by the functional b dx2 + dy L[y] := 2 a x Choosing the variable x as independent, we get b dx2 + dy 2 b 1 + y 2 = dx x x a In the variational problem occurs the function 1 + y F(x, y, y 2 ) = x The Euler-Lagrange equation becomes particularly simple. Since the variable y does not occur in the functional F, we can immediately perform one integration and get d dx 1 + y 2 y x = y 2 y = c x y x 1 + y 2 a = c y 2 = (1 + y 2 )c 2 x 2 y 2 (1 c 2 x 2 ) = c 2 x 2 y = cx 1 c2 x 2 Here c denotes a constant independent on x. Of course the value of c can still depend on the coordinates of the endpoints. The last line is a first order differential equation. If c = 0, we get the solution y = const. The minimizing curve is a Euclidean line perpendicular to the boundary. If c 0, we do a further integration and obtain y = y 0 + cx dx 1 c2 x 2 We substitute v = 1 c 2 x 2 and dv = 2c 2 xdx to obtain y = y 0 1 dv 2c v v = y 0 c = y 0 c 2 x 2 This is the equation of an circular arc with center (0, y 0 ) and radius c 1.

61 Topics from Relativity The minimum of hyperbolic length I go now back to the more commonly used upper half-plane. For the convenience of the reader, I use variables x and y. The upper half plane is {(x, y) : y > 0} and has the metric dx2 + dy ds = 2 (4.7) y The minimum of the hyperbolic length of any connecting curve determines the hyperbolic distances between two points. Given are two points A with Euclidean coordinates (x A, y A ) and B with coordinates (x B, y B ). In the case x A x B, the minimizing curve of connection is a circular arc with center on the x-axis. 1 The equation of such an arc is y = + r 2 (x x 0 ) 2 (4.8) The radius r > 0 and the center (x 0, 0) are to be determined from the coordinates of the two points A and B. Problem 4.2. We can check directly that the function (4.8) is a solution of the Euler-Lagrange equation for the functional 1 + y F(x, y, y 2 ) = y (a) Confirm that d F dx y F y = 1 + y 2 + y y y 2 (1 + y 2 ) 1 + y 2 (b) Check that the derivatives of function (4.8) for the upper half-circle satisfy y 2 + yy + 1 = 0. The hyperbolic length of the arc is s(a, B) = xb x A dx2 + dy 2 y = xb x A 1 + y 2 dx y We need to differentiate the square root composite function (4.8) occurring inside the integral and get y = 1 + y 2 = We plug into the distance functional and obtain s(a, B) = 1 I leave the special case x A = x B as an exercise. = = x x 0 r2 (x x 0 ) 2 r 2 r 2 (x x 0 ) 2 = r2 y 2 xb x A xb x A xb x A 1 + y 2 dx y r dx y 2 r dx r 2 (x x 0 ) 2

62 62 F. Rothe This integral can be calculated by means of the partial fraction decomposition r r 2 (x x 0 ) = 1 2 2(r x + x 0 ) + 1 2(r + x x 0 ) s(a, B) = = xb x A dx 2(r + x x 0 ) + xb x A dx 2(r x + x 0 ) [ 1 2 ln(r + x x 0) 1 2 ln(r x + x 0) x A It remains to check that this result agrees with formula (4.3). We calculate the logarithm of the cross ratio of the two endpoints A and B of the given segment, and the ideal endpoints P and Q of the hyperbolic line through these two points. We assume x A < x B. The points A, B, P, Q occur during a clockwise turn around the circle. The Euclidean coordinates of the endpoints are (x P, y P ) = (x 0 + r, 0) and (x Q, y Q ) = (x 0 r, 0). Hence the cross ratio, and its logarithm are (AB, PQ) = AP BQ AQ BP ln(ab, PQ) = 1 AP 2 ln 2 AQ 1 BP 2 ln 2 2 BQ 2 = 1 [ln (r x + x 0) 2 + y 2 2 (r + x x 0 ) 2 + y 2 = 1 [ln 2(x x 0)r + 2r 2 2 2(x x 0 )r + 2r 2 = 1 [ ln x + x ] xa 0 + r 2 x x 0 + r x B in agreement with the result (??). In the special case x A = x B, the minimizing curve is a Euclidean line perpendicular to the x-axis. We leave the calculation of the distance in the special case as an exercise Some useful reflections in the half-plane Problem 4.3. Check how the mapping w = i 1 z 1 + z maps the boundaries D H. Confirm that a point z = e iθ is mapped to w = tan θ 2. Use the inverse mapping z = i w i + w to confirm the identities cos θ = 1 tan2 θ tan 2 θ and sin θ = 2 tan θ tan 2 θ 2 Problem 4.4. Let S α denote the reflection across the line with ends α R and. Confirm that for any α R. Use this result for an easy check that holds for any α, β R. ] xa x B ] xa x B ] xb (4.1) (4.2) S α (w) = 2α w (4.9) S α+β = S α S 0 S β (4.10)

63 Topics from Relativity 63 Problem 4.5. Let M γ denote the reflection across the line with ends γ > 0 and γ. Confirm that for any γ > 0. Use this result for an easy check of Problem 4.6. Confirm that S 0 M γ = M γ S 0 and for any α, γ R. M γ (w) = γ2 w w 2 (4.11) M γδ = M γ M 1 M δ (4.12) M γ M 1 S α M 1 M γ = S αγ 2 (4.13)

64 64 F. Rothe 5. Equation of motion The motion of a particle or photon in a gravitational field can in principle be determined by either one of the following three principles: (A) "The trajectory is even on itself". (B) "The geodesic is an extremal for the proper time or distance." (C) The quadratic Lagrangian integral is stationary Affine geodesic Alternative (A) leads to the equation of parallel transport to be postulated for the tangent of the trajectory. With any parameter u, consider the trajectory u x a (u) with tangent t a = dxa du. Using the definition 1.12 of the intrinsic derivative of a vector along a curve, postulate (A) gives the equation of motion Dt = λ(u)t (5.1) du The arbitrary function λ(u) depends on the choice of the parameter. Using equation (1.26), we may write out (5.1) in coordinates: d 2 x a du 2 + dx b dx c Γa bc du du = λ(u)dxa du Definition 5.1. The solutions of the equation of motion (5.2) with [ dxa du ] 0 everywhere define the affine geodesics. A parameter with λ(u) 0 is called an affine parameter. Lemma 5.1. There exist affine parameters. Any two affine parameters v and v are related by v = Av + B with any constants A 0 and B. One may find the bijective substitution (reparametrization) v = v(u) for which λ(v) = 0 holds along the entire curve as a solution of the linear differential equation d 2 v dv = λ(u) (5.3) du2 du Proof. Putting v = v(u) into the equation (5.2) for the affine geodesic, one obtains 0 = d2 x a du [ 2 d 2 x a = dv 2 + dx b Γa bc du + Γa bc dx b dv dx c du λ(u)dxa du dx c ] ( dv dv du ) 2 + dxa dv [ d 2 ] v dv λ(u) du2 du A nonconstant solution of the linear differential equation (5.3) exists and has dv du 0 everywhere. Hence one obtains d 2 x a dv + dx b dx c [ ] dx a 2 Γa bc dv dv = 0 and 0 dv as expected. Lemma 5.2. For an affine parameter, the affine equation of motion (5.2) has the following equivalent forms (5.2) ẍ a + Γ a bcẋb ẋ c = 0 (5.4) d (g ab ẋ b ) Γ c du abẋcẋ b = 0 (5.5) where the Christoffel symbol is defined by equation (1.41). In the second form, we use the covariant velocity (or momentum) ẋ c = g cd ẋ d.

65 Topics from Relativity Metric geodesic Following alternative (B) the geodesic, more precisely metric geodesic, is defined to be an extremal for the proper time or distance. The Euler-Lagrange equation for this variational problem is the geodesic equation. Let x a = x a (u) be a parametric equation for the curve using any, not necessarily affine parameter u. For fixed endpoints P and Q, we have to solve the variational problem δ Q P ds = 0 or δ Q with the line element ds = Ldu or cdτ = Ldu and the Lagrangian P dτ = 0 (5.6) L := g ab ẋ a ẋ b (5.7) Here the derivative by any arbitrary parameter u is denoted by a dot. Without any further assumption, the quantity ṡ = ds du = L may depend on the parameter u. We see immediately that only trajectories may be considered for which the tangent vector is either everywhere time-like (material particles), or everywhere spacelike. In the second case one needs to put ṡ = L. Only the first alternative will be pursued here. The Euler-Lagrange equation for the variational problem (5.6) is d L du ẋ a L x a = 0 (5.8) Lemma 5.3. Even without any further assumption on the parameter u, the Euler-Lagrange equation (5.8) has the following equivalent forms ( d gab ẋ b ) ( ag bc )ẋ b ẋ c du L 2 = 0 (5.9) L d g ab ẋ b 1 du 2 ( ag bc )ẋ b ẋ c = ṡ s g abẋ b (5.10) { } a ẍ a + ẋ b ẋ c = ṡ b c s ẋa (5.11) where the Christoffel symbol is defined by equation (1.41). Checking the calculations. d du d du (g abẋ b )ṡ (g ab ẋ b ) s ṡ 2 ( gab ẋ b L ) (δ ag bc )ẋ b ẋ c 2 L ( ag bc )ẋ b ẋ c 2ṡ = 0 d g ab ẋ b 1 du 2 ( ag bc )ẋ b ẋ c = ṡ s g abẋ b = 0 from L = ṡ and the quotient rule; ( c g ab )ẋ c ẋ b + g ab ẍ b 1 2 ( ag bc )ẋ b ẋ c = ṡ s g abẋ b from product and chain rule; g ab ẍ b ( cg ab + b g ac a g bc )ẋ b ẋ c = ṡ s g abẋ b { } d ẍ d + ẋ b ẋ c = ṡ b c s ẋd multiplying by g da one gets

66 66 F. Rothe Let the rest mass be m > 0 and let c denote the speed of light. We assume the tangent vector is everywhere time-like along the geodesic. This case occurs for the motion massive particles. Here the most convenient choice of curve parameter is the proper time τ defined by to hold. L := g ab dx a dτ dx b dτ = c2 (5.12) Lemma 5.4. There exists parameters for which s 0 and hence the Lagrangian is constant along the geodesic. Any two such parameters v and v are related by v = Av + B with any constants A 0 and B. One may find the bijective substitution (reparametrization) τ = τ(u) by integrating and thus introduce the proper time τ as curve parameter. c dτ du = L (5.13) 5.3. The quadratic Lagrangian We now consider alternative (C), which turns out to be a very attractive possibility. For fixed endpoints P and Q, we have to solve the variational problem δ Q for the quadratic Lagrangian One obtains the Euler-Lagrange equation P Ldu = 0 (5.14) d L du ẋ a L x a = 0 (5.15) Lemma 5.5. The Euler-Lagrange equation (5.15) has the following equivalent forms d (g ab ẋ b ) 1 du 2 ( ag bc )ẋ b ẋ c = 0 (5.16) { } a ẍ a + ẋ b ẋ c = 0 (5.17) b c { } dẋ a c du ẋ c ẋ b = 0 (5.18) a b where the Christoffel symbol is defined by equation (1.41). In the third form, we use the covariant velocity (or momentum) ẋ c = g cd ẋ d. Proposition 5.1. The Euler-Lagrange equation (5.15) is equivalent to the affine geodesic equation (5.4) if and only if the torsion tensor satisfies T a bc + T a cb = 0 (5.19) Especially, for a symmetric connection, the Euler-Lagrange equation (5.15) is equivalent to the affine geodesic equation (5.4). Proof. From formula (1.44) one concludes ẍ a + Γ a bcẋb ẋ c = ẍ a + { } a ẋ b ẋ c + T a bc b c ẋb ẋ c

67 Topics from Relativity 67 Lemma 5.6. Along the trajectories of equation (5.16), the quadratic Lagrangian L g ab ẋ a ẋ b = Const (5.20) is a constant of motion. Under the assumption that this constant of motion L > 0 is positive, the parameter u is automatically u = As + B or u = Aτ + B with some constants A 0 and B. Thus we have obtained as solution for equation (5.10) with s 0, too. Proof. We differentiate the quantity L by the parameter u, denoting any derivatives by u with a dot. The Leibnitz product rule yields d g ab ẋ a ẋ b du = d g ab du ẋa ẋ b + g ab ẍ a ẋ b + g ab ẋ a ẍ b = 2 d g abẋ b ẋ a d g ab du du ẋa ẋ b = ( a g bc )ẋ b ẋ c ẋ a ( c g ab )ẋ a ẋ b ẋ c = 0 and confirms that L is a constant of motion. The situation with L > 0 holds always for a positive definite metric, as usually assumed in classical differential geometry. Too, in general relativity one gets L > 0 for the time-like geodesics, which are the paths of massive particles. In all these cases the parameter is automatically the arclength or proper time, up to a linear transformation. Indeed, under the assumption that L > 0, the definition of arc-length, respectively proper time, gives d 2 s du = d L 2 du = 1 2 L and hence u = As + b, or u = Aτ + B, respectively. Theorem 5.1. Each solution of the variational problem δ Q P d L du = 0 L du = 0 (5.14) with the quadratic Lagrangian L has the Lagrangian as a constant of motion. Under the additional assumption that this constant of motion L > 0 is positive, the curve parameter u is automatically proportional to the arc-length or proper time; and the trajectory satisfies the variational problem δ Q P ds = 0 (5.6) for the arc-length or proper time, too. Conversely, from solution of the variational problem (5.6) by introducing the arclength or proper time as parameter, one obtains a solution of the variational problem (5.14) for the quadratic Lagrangian L. The geodesic equation has the two following two equivalent forms, which I nickname the "physical form" and the "Christoffel form": d ( gab ẋ b) 1 dλ 2 ( ag bc )ẋ b ẋ c = 0 (5.21) dẋ a dλ + Γa bcẋb ẋ c = 0 (5.22)

68 68 F. Rothe Problem 5.1. As a first benefit, one can use the equivalence of the two forms (5.21) and (5.22) for the geodesic equation in order to calculate the connection coefficients Γ a bc of a symmetric connection. As an example, we take the spherical coordinates (r, θ, φ) of the Euclidean R 3. They have the metric ds 2 = dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2 Get all non-zero connection coefficients. Answer. The r-component of the physical form of the geodesic equation yields d dλ (g rrṙ) 1 2 ( rg θθ ) θ ( rg φφ ) φ 2 = 0 r r θ 2 r sin 2 θ φ 2 = 0 Γ r rr = 0, Γ r θθ = r, Γr φφ = r sin 2 θ The θ-component of the physical form of the geodesic equation yields d ( gθθ θ ) 1 dλ 2 ( θg φφ ) φ 2 = 0 r 2 θ + 2rṙ θ r 2 sin θ cos θ φ 2 = 0 Γ θ θr + Γθ rθ = 2r 1, Γ θ φφ = sin θ cos θ one has to use that the connection is symmetric to end up with Γ θ θr = r 1. The φ-component of the physical form of the geodesic equation yields d ( gφφ φ ) = 0 dλ 2r sin 2 θṙ φ + 2r 2 sin θ cos θ θ φ + r 2 sin 2 θ φ = 0 Γ φ rφ = r 1, Γ φ θφ = cot θ again one has used that the connection is symmetric to cancel the factor Null geodesics In general relativity the null-geodesics with L 0 are the paths of photons. Here are two among several possibilities to find and justify the equation of motion for the photon: The affine geodesic equation (5.4) may be applied, and an affine parameter is still available; In the variational problem δ Q P dτ = 0 (5.6) for the proper time, one takes the limit rest mass to zero. I define the physical affine parameter as λ = τ m where m denotes the rest mass. The equations governing the four-momentum p a = dxa and p 0 = dx0 dλ dλ = E c = c dt dλ and pa p a = m 2 c 2 (5.23) hold independently of the rest mass. In the limit m 0, they remain valid and are still physically meaningful. Indeed, in this way one obtains correct dynamical equations for the photon. The affine geodesic equation (5.4) may be written as the system dx a dλ = pa dp a dλ = Γc ab p dx b c dλ (5.24)

69 Topics from Relativity 69 which is an initial value problem for the functions x a (λ), p a (λ). Quite similarly, at least for m > 0, the Euler-Lagrange equation (5.15) may be written as dx a dλ = pa dp a dλ = 1 2 ( ag bc )p c dxb dλ (5.25) Proposition 5.2 (Local smoothness). The solutions of both initial value problems (5.24) and (5.25) depend smoothly on the rest mass and the physical affine parameter. Especially, both problems are well posed for rest mass m = 0. More precise: assume the respective initial value problem has a solution for some mass m 0 and some initial value [x a ], [p a ] such that p a p a = m 2 c 2 and [p a ] 0, existing on an open interval including the closed interval [Λ 1, Λ 2 ] 0 including zero, and having everywhere nonzero fourmomentum [p a ] 0. Then there exist open neighborhoods for this initial value, this mass m and the interval [Λ 1, Λ 2 ] on which the initial value problem is solvable. Moreover, the solution depends smoothly on these parameters and the four-momentum [p a ] 0 is nonzero everywhere. Remark. Still a blow-up of the solution after a finite "parameter-lifetime" is possible. The end of existence may occur for a photon even slightly sooner than for a massive particle. Proof. Because of the conservation law g ab p a p a = m 2 c 2 = Const and the signature of the metric, one obtains bounds cp 0 [E min, E max ] > 0 for all the energy values occurring within parameters m [0, M] and λ [ Λ, +Λ]. Hence dt dλ = p0 /c [c 2 E min, c 2 E max ] which allows to use the coordinate time as independent variable. With this substitution, the Euler- Lagrange equation (5.15) are transformed into dx a dt dp a dt = c pa p 0 = 1 2 ( ag bc )p c dxb dt We have set up a system which remain meaningful for rest mass m = 0. Choose T c 2 E max. The smooth dependence of the solution for m [0, M] and t [ T, +T] is a standard theorem from differential equations. Hence smooth dependence for the solution of equation (5.25) and the Euler-Lagrange equation (5.15) holds within any bounded parameter ranges m [0, M] and λ [ Λ, +Λ], under the only assumption that p a 0 holds for all four-momentum values occurring. The proof for the affine geodesic equation (5.4) is similar The method The equation of motion can be derived either from (A) parallel transport; (B) the extremum for proper time; (C) the variational principle for the quadratic Lagrangian integral.

70 70 F. Rothe Methods (A) and (C) lead to well-posed initial value problems, with solutions which depend smoothly on the rest mass m 0 and the physical affine parameter λ. Moreover, under the additional assumption of a symmetric connection, we have shown that methods (A) and (C) lead to the same initial-value problem. For the (somehow most attractive) method (B) we have obtained a well-posed initial value problem only for massive particles. Indeed, for positive rest mass m > 0, the solution of the initial value problem depends smoothly on m > 0 and the proper time τ. Moreover, under the restriction of positivity of mass, methods (B) and (C) turn out to be essentially equivalent. The method (C) allows to include the case of zero rest mass continuously. On the other hand, method (B) may be physically more justified because of its relation to the principles of superposition of waves. These principles of interference are well-known to hold both for massive particles and light. Hence there are reasons to believe that the method (B) can be extended smoothly, to include the case of massless particles, too. In case, one accepts such a postulate, the unique smooth extension to rest mass m 0 obtained from method (C), is the physically meaningful extension of method (B) to rest mass m 0, too, since for positive mass, methods (B) and (C) are essentially equivalent. Corollary 6 ("The method"). Assuming the connection is symmetric, the same equation of motion are obtained either from parallel transport, the extremum of proper time, or the variational principle for the quadratic Lagrangian. Both for photons and massive particles, the equation of motion may be obtained as solution of the variational problem δ Q P L dλ = 0 (5.26) with the quadratic Lagrangian dx a dx b L := g ab (5.27) dλ dλ In place of u, we have to use the physical affine parameter λ = τ/m; and denote derivative by λ with a dot. The physical momentum of the particle or photon is p a = ẋ a, and may be used, and indeed shows up, both in its original contravariant, as well as the covariant form p a = g ab p b. One obtains (and easily checks) the equations of motion dx a dλ = pa dp a dλ = 1 2 ( ag bc )p c p b (5.28) Automatically, the Lagrangian is a constant of motion. Because of the choice of the physical affine parameter, its value is fixed to be L = p a p a = m 2 c 2 (5.29) 5.6. Killing vector Definition 5.2 (Isometry). A point transformation x = x (x) on a Riemannian manifold is called an isometry iff g ab (x) = g ab(x) holds for all coordinates x and indices a, b. Definition 5.3 (Killing vector). A vector field [K a ] is a Killing vector field or simply Killing vector iff the flow Φ a (ε, x) = K a (x) (5.30) ε produces isometries.

71 Topics from Relativity 71 Proposition 5.3. A vector field K is a Killing vector field if and only if the Lie derivative of the metric along the vector field vanishes: L K g ab = 0 (5.31) K c c g ab + ( a K s )g sb + ( b K s )g as = 0 (5.32) Independent proof. The flow defined by the initial value problem (5.30) gives for small ε the infinitesimal transformations x a = Φ a (ε, x) = x a + εk a (x) + O(ε 2 ) Under these point transformations the metric is transformed by g ab (x ) = xd x e x a x g de(x) = g b ab (x) ε( a K d )g db ε( b K e )g ae + O(ε 2 ) At the same coordinate x one gets as transformation of the metric g ab (x) g ab(x) = [g ab (x) g ab (x )] + [g ab (x ) g ab (x)] = εk c c g ab ε( a K d )g db ε( b K e )g ae + O(ε 2 ) Assuming that the flow (5.30) of the vector field K induces isometries for all ε, we conclude that the first order term is zero and get equation (5.32). Conversely, let us assume that equation (5.32) holds identically on the entire manifold. Then the Killing flow induces transformations of the metric such that Φ(ε, g)(x) = g (x) = Product o f Jakobians g(φ( ε, x)) = g(x) ε [ K c c g ab + ( a K d )g db + ( b K e )g ae ] ω a e b + O(ε 2 ) = g(x) + O(ε 2 ) for all ε and hence g = g for all ε on the entire manifold. This means that the Killing flow induces isometries. Proposition 5.4. Given is the vector field [K a ] on a Riemannian manifold. The scalar product S = K a ẋ a is a constant of motion along all geodesics λ x a (λ) if and only if the Killing equation holds: a K b + b K a = 0 (5.33) Proof. The covariant derivative of the scalar product S = K a ẋ a = K a ẋ a along the geodesic is calculated. Since Leibniz rule holds for covariant derivatives, and the geodesic equation is assumed Ṡ = ds dλ = ( bk a ẋ a )ẋ b = ( b K a )ẋ a ẋ b + K a ( b ẋ a )ẋ b (5.34) = ( b K a )ẋ a ẋ b + K a Dẋ a dλ = 1 2 ( ak b + b K a )ẋ a ẋ b (5.35) Hence the scalar product S = K a ẋ a is a constant of motion along all geodesics if and only if the Killing equation (5.33) holds for the vector field [K a ]. Remark. Note that for the velocities and corresponding momenta, the index is lowered and lifted by the rules even if the metric is not constant. ẋ a = g ab ẋ b and ẋ a = g ab ẋ b p a = g ab p b and p a = g ab p b

72 72 F. Rothe Corollary 7. Assume the metric does not depend on the index c, and the connection is symmetric. Under that assumption, the vector field K a := δa c is a Killing vector. Hence the corresponding covariant velocity u c = g ca u a and momentum p c = g ca p a is constant of motion. Proof. a K b = Γ b sak s = Γ b ca a K b = Γ bca a K b + b K a = Γ bca + Γ acb = Γ bac + Γ abc = δ c g ab = 0 Corollary 8. Assume the metric satisfies K c δ c g ab = 0 for a constant vector [K c ], and the connection is symmetric. Under that assumption, the product S = g ab K a u b and the corresponding momentum is a constant of motion. Proof. a K b = Γ b sak s a K b = Γ bsa K s = Γ bas K s a K b + b K a = K s δ s g ab = 0 Hence K is a Killing vector and the scalar product S = g ab K a u b is a constant of motion. Theorem 5.2. Let K be a vector field on a Riemannian manifold with symmetric connection. Equivalent are (i) K is a Killing vector, which means the flow produces isometries. Φ a (ε, x) ε = K a (x) (5.30) (ii) The infinitesimal point transformations x a = x a + εk a (x) + O(ε 2 ) are isometries up to order O(ε 2 ). (iii) The Lie derivative of the metric along the vector field vanishes L K g ab = 0. (iv) K c c g ab + g bc a K c + g ac b K c = 0 (v) The Killing equation a K b + b K a = 0 holds identically. (vi) The scalar product S = K a ẋ a = K a ẋ a is constant along all geodesics. Proof. We need still to check that the Killing equation is equivalent to item (iv). a K b + b K a = a (g bc K c ) + b (g ac K c ) = g bc a K c + g ac b K c = g bc ( a K c + K s Γ c sa) + g ac ( b K c + K s Γ c sb ) = g bc a K c + K s Γ bsa + g ac b K c + K s Γ csb = g bc a K c + g ac b K c + K s (Γ bas + Γ cbs ) = K s s g ab + g bc a K c + g ac b K c

73 Topics from Relativity 73 Problem 5.2. Check that the metric has the Killing vector K = r 2 sin φ dθ + r 2 cos θ sin θ cos φ dφ. ds 2 = dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2 (??) Answer. We may directly check the Killing equation (5.33). The non-zero Christoffel symbols have been calculated in problem 5.1 To check whether a K b + b K a = 0 calculate Γ r rr = 0, Γ r θθ = r, Γr φφ = r sin 2 θ Γ θ θr = r 1, Γ θ φφ = sin θ cos θ Γ φ rφ = r 1, Γ φ θφ = cot θ r K θ + θ K r = r K θ Γ s θr K s + θ K r Γ s θr K s = r K θ 2Γ θ θr K θ = 2r sin φ 2r 1 r 2 sin φ = 0 r K φ + φ K r = r K φ 2Γ s φr K s = 2r cos θ sin θ cos φ 2r 1 r 2 cos θ sin θ cos φ = 0 θ K θ = θ K θ Γ s θθ K s = θ K θ = 0 φ K φ = φ K φ Γ θ φφ K θ = r 2 cos θ sin θ sin φ + sin θ cos θr 2 sin φ = 0 θ K φ + φ K θ = θ K φ + φ K θ 2 cot θk φ = 0 Problem 5.3. Find a more inspired solution of problem 5.3 without use of Christoffel symbols. Problem 5.4. Fear Schopenhauer s mousetraps and do no mousetrap proofs! Give the reason behind the last calculation. Why does the Killing flow maps geodesics into geodesics, and even why do these have the same two constants of motion: both K a ẋ a as well as g ab ẋ a ẋ b. Lemma 5.7. For any vector field [K a ] holds L K K a = 0. For a Killing vector field holds L K K a = 0, too. Proof. L K K a = (K s s )K a K s ( s K a ) = 0 L K K a = L K (g ab K b ) = (L K g ab )K b + g ab L K K b = 0 The first term is zero by proposition 5.3. The second term is zero by first line above. Lemma 5.8. Let x a = x a (λ, ε) be a family of curves satisfying x a (λ, ε) ε = v a (x) (5.36) In other words, the curve x a = x a (λ, 0) is transported by the flow of vector field [v a ]. Then L v ẋ a = 0, where the dot denotes partial derivative by curve parameter λ. Proof. L v ẋ a = (v s s )ẋ a ẋ s ( s v a ) = 2 x a ε λ ẋs ( s v a ) = dva dλ xs λ ( sv a ) = 0

74 74 F. Rothe Proposition 5.5. Let x a = x a (λ, ε) be a family of curves satisfying x a (λ, ε) ε = K a (x) (5.37) Moreover, assume the curve x a = x a (λ, 0) a geodesic and it is transported by the flow of Killing field [K a ]. (i) The curves λ x a (λ, ε) are geodesics for all ε. (ii) The Lagrangian L = g ab ẋ a ẋ b is a independent of both λ and ε. (ii) The quantity S = g ab K a ẋ b is a independent of both λ and ε. Proof of item (i). This is left to the reader. Proof of item (ii). L ε = L K(g ab ẋ a ẋ b ) = (L K g ab )ẋ a ẋ b + 2g ab ẋ a L K ẋ b = 0 The first term is zero by proposition 5.3. The second term is zero by lemma 5.8. Proof of item (iii). S ε = L K(g ab K a ẋ b ) = (L K g ab ) K a ẋ b + g ab (L K K a )ẋ b + g ab K a L K ẋ b = 0 The first term is zero by proposition 5.3. The second term is zero by lemma 5.7. The third term is zero by lemma 5.8.

75 Topics from Relativity Geodesics in the Schwarzschild metric The Schwarzschild metric is the solution of Einstein s field equation for a mass M at the center. ( ) ) 1 ds 2 = c 2 1 r dt 2 (1 r dr 2 r 2 dθ 2 r 2 sin 2 θdφ 2 (6.1) r r Here r = 2GM (6.2) c 2 is the Schwarzschild radius. For the sun r 2.96 km. Problem 6.1. Write down the "physical form" of the geodesic equations in the Schwarzschild metric, with proper time as parameter. ) ] d [(1 r dt = 0 (6.3) dτ r dτ ) 1 d dτ (1 r dr r dτ = 1 2 ( rg bc )ẋ b ẋ c (6.4) [ d r 2 dθ ] ( ) 2 dφ = r 2 sin θ cos θ (6.5) dτ dτ dτ [ d r 2 sin 2 θ dφ ] = 0 (6.6) dτ dτ Remark. The right-hand side of the radial equation is a nuisance, but can be avoided. The orbits lie in plane. For simplicity I may choose the plane θ 90. Since the Schwarzschild metric does not depend on time t nor angle φ, one gets two constants of motion ) (1 r dct r dτ = E esc const (6.7) mc r 2 dφ dτ = l const (6.8) m The physical meaning of these integration constants can be determined from the limit r where special relativity holds. It turns out that E esc is the energy of the particle or planet escaped to infinity. The relativity parameter is γ = E esc /(mc 2 ). One gets the slow-down of proper time t = γτ as known from special relativity. Let p esc be the momentum of the escaped particle. From special relativity we know the important formula Eesc 2 = m 2 c 4 + c 2 p 2 esc It turns out that l is the angular momentum of the particle or planet around the z-axis. Let us introduce the impact parameter b to be the perpendicular distance of the line of straight motion of the particle from the center mass. Suppose that the particle escapes along a line x(τ)e x + be y. With x = cos φ, y = r sin φ, one gets the angular momentum as expected. l = mr 2 φ = xmẏ + ymẋ bmẋ = bp esc (6.9)

76 76 F. Rothe Remark. For bounded orbits we may still use the parameters p esc i[0, ) and b = lp esc i[0, ), but with imaginary values. The boundary case of parabolic motion has p esc = 0 and any value A [0, ) for the semilatus rectum and l [0, ). The integration of the equations of motions becomes possible since the Lagrangian is the further constant of motion. With the proper time τ as curve parameter dx a dx b L := g ab dτ dτ = c2 (5.12) is the constant value of the Lagrangian. For the Schwarzschild metric this identity reduces to ( ) ( ) 2 ) 1 ( ) 2 ( ) 2 ( ) 2 c 2 1 r dt (1 r dr dθ dφ r 2 r 2 sin 2 θ = c 2 r dτ r dτ dτ dτ and is further simplified by use of θ 90. Too, I use from now on the simpler dot notation. ( ) ) 1 c 2 1 r ṫ 2 (1 r ṙ 2 r 2 φ 2 = c 2 (6.10) r r A bid of arithmetic is still needed. Then I further simplify by use of the constants of motion from equations (6.7) and (6.8). ( ) 2 ) ( ) c 2 1 r ṫ 2 ṙ 2 r (1 2 r φ 2 = c 2 1 r r r r ( ṙ 2 + r 2 1 r r ṙ 2 + (1 r r ) ) φ 2 c2 r r l 2 m 2 r c2 r 2 r = c 2 ( 1 r r ) 2 ṫ 2 c 2 = E2 esc m 2 c 2 c2 p2 esc m 2 The last equation may be multiplied by m/2 to get the energy balance for the Kepler motion, with just one extra term! Too, we shall need the angular equation of motion. ) m 2 ṙ2 + (1 r l 2 r 2mr GMm = p2 esc (6.11) 2 r 2m mr 2 φ = l bp esc (6.12) For the discussion of motion of photons, I use the physical affine parameter λ = τ/m, for which as expected, one gets equations which are meaningful in the limit m 0. Equation (6.11) is multiplied by 2m and the affine parameter λ is introduced. Too, we need the equation of motion (6.7)for the time and the angular equations of motion (6.8). Altogether we obtain as equations of motion for photons and massless particles ( ) 2 ) dr + (1 r l 2 dλ r r = 2 p2 esc l2 (6.13) b ) (1 2 r dct r dλ = E esc p esc (6.14) c r 2 dφ dλ = l bp esc (6.15) Problem 6.2. Write a paragraph on the rotation of the perihelion of mercury. How does one proceed from equations of motion (6.11) and (6.12) to get an equation about the shape of the orbit.

77 Topics from Relativity 77 One does an expansion in powers of r, respectively c 2, which is a small parameter. What is the zeroth order term obtained with r = 0. One puts r 1 =: u = u 0 + u Which equation does one get for u 1. Which formula for the perihelion rotation is obtained? 6.1. The equation for the shape of relativistic orbits For the equation about the shape of the orbit, we use the variable u := r 1 and need an equation of motion for the function u = u(φ). Since du dφ = 1 dr r 2 dφ = ṙ r 2 φ = mṙ l (6.16) we multiple equation (6.11) by 2ml 2 and substitute to obtain ( ) 2 du + (1 r u)u 2 2GMm2 u = p2 esc dφ l 2 l 2 This step excludes the case l = 0 of radial motion. As a convenient geometric quantity, one may introduce the semilatus rectum l 2 A := GMm 0 (6.17) 2 and gets ( ) 2 du + u 2 2u dφ A r u 3 = p2 esc (6.18) l 2 Remark. Equation (6.18) is valid both for unbounded as well as bounded orbits, as long as l 0. Note that in the latter case, one needs to use imaginary values p esc i[0, ) for the escape momentum. For hyperbolic motion the impact parameter b = lp esc from equation (6.9) and the escape momentum p esc are convenient parameters. Since equations (6.17), (6.9) and (6.2) imply one obtains A = l 2 GMm = b2 p 2 esc 2 GMm = 2b2 2 r ( pesc ) 2 mc ( ) 2 ( ) 2 du + u 2 r mc u r u 3 = 1 (6.19) dφ b 2 p esc b 2 Remark. Again the equation (6.19) is valid both for unbounded orbits with p esc > 0, as well as bounded orbits. But in the latter case, one needs to use imaginary values p esc i(0, ) and b = lp esc i(0, ). Remark. The boundary case of parabolic motion occurs for p esc = 0 and any value of the angular momentum l [0, ). The semilatus rectum from equation (6.17) still exists and takes any value A [0, ). For A 0, one get an impact parameter formally b =. For the parabolic motion and A 0, one gets the equation of shape ( ) 2 du + u 2 2u dφ A r u 3 = 0 (6.20)

78 78 F. Rothe 6.2. Kepler s classical nonrelativistic orbits One does an expansion in powers of r, respectively c 2, which is a small parameter. The zeroth order term is obtained by putting r = 0 into equation (6.18). One obtains the classical approximation of Kepler motion ( ) 2 du + u 2 2u dφ A = Const p2 esc (6.21) l 2 from which the orbits are obtained to be conic sections. The equation of motion (6.21) is also called Binet equation. All cases lead to the orbits u(φ) = 1 + e cos φ A (6.22) with the eccentricity e [0, ) and the semilatus rectum A (0, ) as convenient parameters. Only the purely radial motion with l = 0 needs to be treated separately. Plugging the solution (6.22) into the equation of motion (6.21), the reader should check that the equation of motion holds. Moreover one obtains e 2 1 A 2 = p2 esc l 2 = 1 b 2 (6.23) to be the relation between the gets geometrical and physical parameters and the impact parameter. The first equation holds in all cases with l > 0. For e 1 the major half-axis a > 0 satisfies b 2 = with the upper minus sign for the ellipsis. Hence the equation A2 e 2 1 = Aa = (e2 1)a 2 (6.24) b = a e 2 1 (6.25) holds wonderfully in all cases. These are the different shapes of orbits for l > 0: The case p 2 esc < 0 and u const gives circles. Eccentricity is e = 0 and the radius is 1/u r = A = l p esc The case p 2 esc < 0 but u not constant gives ellipses. The eccentricity lie in the range e (0, 1). Reusing relation (6.23) and the almost magical semilatus rectum from equation (6.17), the major half axis turns out to be a = A 1 e 2 = l2 Ap 2 esc = GMm2 p 2 esc = r 2 ( pesc ) 2 mc into which formula even the Schwarzschild radius from (6.2) fits! The case p 2 esc = 0 gives parabolas, here e = 1. The impact parameter is formally b =. The case p 2 esc > 0 gives hyperbolas, here the eccentricity lies in the range e (1, ). the major half axis is a = A e 2 1 = l2 Ap 2 esc = r 2 ( pesc ) 2 mc

79 Topics from Relativity Scattering in Newtonian dynamics From the equation (6.25), we see that the impact parameter is the minor half axis of the hyperbola. The scattering angle θ is supplementary to the angle between asymptotes of the hyperbola. Hence the Cartesian equation implies y 2 b 2 = x2 a 2 1 cot θ 2 = b a I have shall use spherical coordinates with the incoming beam in +z direction. One wants to express the ray properties in terms of impact parameter b and escape momentum p esc. To this end, we need b 2 = aa from good old conics (6.24), the semilatus rectum from equation (6.17), and l = bp esc and produce cot θ 2 = b a = A b = l 2 GMm 2 b = bp2 esc GMm 2 For small angles, we get the first order approximation θ = 2 tan θ 2 = 2GMm2 bp 2 esc = r m 2 c 2 b p 2 esc which is only half of the physical value calculated in equation (6.32) below. For arbitrary scattering angles, but nonrelativistic dynamics, we may obtain the Rutherford scattering formula. The argument below works both for attractive as well as repulsive inter interactions. For an incoming ray of intensity I, some fraction di is scattered into the space angle dω = sin θdθ dφ, and is measured at large distance R from the scatterer as an intensity IR 2 dσ. Thus one defines the differential cross section dσ. These scattered particles correspond to impact parameter [b, b + db] and rotation angles [φ, φ + dφ]. Hence Simply differentiate dσ = b db dφ dσ dω = b db sin θ dθ b = GMm2 cot θ p 2 esc 2 db dθ = GMm2 sin 2 θ 2p 2 esc 2 b db dθ = M 2 m 4 G2 cos θ θ 2p 4 esc 2 sin 3 2 dσ dω = b db sin θ dθ = M 2 m 4 G2 sin 4 θ 4p 4 esc Perturbation expansion for relativistic bounded orbits One expands in powers of the small parameter r and puts u = u 0 + u into the equation of shape (6.18). The zeroth order

80 80 F. Rothe term is the classical Kepler ellipse, given by equation (6.22). For the next term of order r we get a linear the equation, which one may solve 2 du 0 du 1 dφ dφ + 2u 0u 1 2u 1 A = r u 3 0 e sin φ du 1 dφ + e cos φ u 1 = r (1 + e cos φ) 3 2A 2 u 1 = 3r eφ sin φ 2A 2 + periodic terms with only sin φ, cos φ, sin 2 φ, cos 2 φ One needs only the secular term proportional to φ sin φ. The period terms with sin φ, cos φ, sin 2 φ, cos 2 φ are below the accuracy of observation. Problem 6.3. Since expansions work best for linear equation, some people may want to see the following approach. We differentiate the equation of shape (6.18) by the angle φ and obtain d 2 u dφ 2 + u 1 A 3r 2 u2 = 0 (6.26) One expands in powers of the small parameter r and puts u = u 0 + u The zeroth order term is the classical Kepler ellipse, given by equation (6.22). Determine and solve the first order equation. Get once more the secular term of first order and thus check the above calculation The mercury perihelion rotation The solution u = 1 + e cos φ + 3r eφ sin φ A 2A 2 has a tiny perihelion advance ε per rotation. Successive maxima of the solution take place for φ = 0 and φ = 2π + ε. u = e sin φ A + 3r e(sin φ + φ cos φ) 2A 2 A e u (2π + ε) = sin ε + 3r (sin ε + (2π + ε) cos ε) 2A ε + 3r 2A [ε + (2π + ε)] + O(ε2 ) We need to put this derivative to zero and solve for ε. The terms of order r 2 are neglected, once more. One gets ε = 3πr A = 3πr a(1 e 2 ) This value is approximately ε = 3r 2A Next I introduce the major half axis and the eccentricity, easily to be obtained from astronomical tables. For the sun the Schwarzschild radius is approximately r 2.96 km. But we may instead use the known velocity c of light, and obtain r from Kepler s third law. Finally, one needs to convert to arcseconds per century and gets 4π 2 a 3 = GM = c2 r T 2 2 3πr ε = a(1 e 2 ) = 6πr 4π 2 a 3 a(1 e 2 ) c 2 T = 24π 2 a 2 2 (1 e 2 )c 2 T T century 24π 2 a 2 ε = π T (1 e 2 )c 2 T 2 which you may now check with easily obtainable data. The famous value is ε 43 century 1.

81 Topics from Relativity Perburtation expansion for the angle of deflection We start from the equation (6.19) ( ) 2 ( ) 2 du + u 2 r mc u r u 3 = 1 (6.19) dφ b 2 b 2 p esc for the shape of the orbit. We assume that the Schwarzschild radius r b is a small quantity compared to the impact parameter, and the particle is either a photon, or at least moving fast enough such that say mc/p esc 10. To start an perturbation expansion, we put r := 0 and obtain for the zeroth order term the equation of motion ( du dφ ) 2 + u 2 = 1 b 2 which has the solution u 0 = cos φ b This is simply a straight line with impact parameter b. The goal is to set up the expansion u = u 0 + u by powers of r. We need the first order term u 1 in order to calculate the deflection angle. From the terms linear in r, we get a linear differential equation for u 1, which one has to solve. 2 du ( 0 du 1 dφ dφ + 2u 0u 1 = r mc b 2 p esc ( mc 2 sin φ du 1 dφ + 2 cos φu 1 = r b 2 p esc ) 2 u 0 + r u 3 0 ) 2 cos φ + r b 2 cos3 φ I take the Ansatz u 1 = A + B cos 2 φ and compare coefficients of A and B. The reader should check the following result: ( ) 2 (2A + 4B) cos φ 2B cos 3 φ = r mc cos φ + r b 2 b 2 cos3 φ u = cos φ b u = cos φ b p esc + A + B cos 2 φ ( + r b + r mc 2 2b 2 p esc ) 2 r 2b 2 cos2 φ The ingoing and outcoming rays correspond to r and hence u = 0. They occur for the polar angles ±φ = 90 + ε/2, where ε is the total deflection. Since cos(90 + ε/2) ε/2 for small ε 1, one gets 0 = ε ( ) r b + r mc + O(ε 2 ) 2 2b 2 p esc and finally the first order approximation for the small bending angle ε 2r b ( ) 2 mc 2 + O(r 2 ) (6.27) 6.7. The bending of light The equation of shape (6.19) is valid for photons, too. One may simply put m = 0 and obtains a "Binet-type" equation for the inverse radius u = r 1 as a function of φ ( ) 2 du + u 2 r u 3 = 1 (6.28) dφ b 2 p esc

82 82 F. Rothe The bending of light is obtained from the perturbation expansion, similarly as in the last section. One obtains the small bending angle in first approximation to be ε 2r b + O(r 2 ) (6.29) For illustration of this important result, I give the entire reasoning independently, once more. Clearly we may start with equations of motion (6.13) and (6.15) and eliminate the proper time by means of du dφ = 1 dr r 2 dφ = dr ( r 2 dφ ) 1 = 1 dλ dλ l dr dλ = 1 dr bp esc dλ (6.30) Thus one arrives at the same equation (6.28) about the shape of the orbit. An expansion in terms of powers of the small parameter r is needed. The zeroth order term is obtained putting r = 0 ( ) 2 du + u 2 = 1 dφ b 2 which equation has the solution u 0 = cos φ b As to be expected, this is simply a straight line with impact parameter b. We set up the expansion u = u 0 + u by powers of r. We need only the first order term u 1 in order to calculate the small deflection angle. I illustrate a variant method which is sometimes useful since expansions work best for linear equations. We differentiate the equation of shape (6.28) by the angle φ and obtain d 2 u dφ + u 3r 2 2 u2 = 0 (6.31) From the terms linear in r, we get a linear differential equation for u 1, which one has to solve. To this end, I make the Ansatz u 1 = A + B cos 2 φ and compare coefficients of A and B. d 2 u 1 dφ 2 + u 1 = 3r 2 u2 0 2B(1 2 cos 2 φ) + A + B cos 2 φ = 3r 2b 2 cos2 φ u = cos φ b u = cos φ b + A + B cos 2 φ + r b 2 r 2b 2 cos2 φ The reader should check the above calculation. We may now calculate the total deflection ε of the light ray. The ingoing and outgoing rays correspond to r and hence u = 0. They occur for the polar angles ±φ = 90 + ε/2. Since cos(90 + ε/2) ε/2 for small ε 1, one has to solve 0 = ε 2 + r b 2 + O(ε2 ) and finally the first order approximation for the small bending angle ε 2r b + O(r 2 ) (6.32)

83 Topics from Relativity Gauss Differential Geometry and the Pseudo-Sphere 7.1. Introduction Through the work of Gauss on differential geometry, it became clear after a painfully slow historic process that there is a model of hyperbolic geometry on surfaces of constant negative Gaussian curvature. One particularly simple such surface is the pseudo-sphere. According to Morris Kline, it is not clear whether Gauss himself already saw this non- Euclidean interpretation of his geometry of surfaces. Continuing Gauss work, Riemann and Minding have thought about surfaces of constant negative curvature. Neither Riemann nor Minding did relate curved surfaces to hyperbolic geometry (Morris Kline III, p.888 etc). But, independently of Riemann, Beltrami finally recognized that surfaces of constant curvature are non-euclidean spaces. Due to the ideas forwarded by Gauss, mathematicians have in the end advanced to the concept of a curved surface as a space of its own interest. Gauss work implies that there are non- Euclidean geometries on surfaces regarded as spaces in themselves. An obvious and important idea is finally spelled out! As we explain in detail below, Beltrami shows that one can realize a piece of the hyperbolic plane on a rotation surface of negative constant curvature. This surface is called a pseudo-sphere. But this new discovery comes with a disappointment: by a result of Hilbert, there is no regular analytic surface of constant negative curvature on which the geometry of the entire hyperbolic plane is valid (see Hilbert s Foundations of Geometry, appendix V). Concerning models of hyperbolic geometry, the final outcome turns out to be a trade off between the pseudo-sphere and the Poincaré disk. Both have their strengths and weaknesses. The pseudo-sphere is a model for a limited portion of the hyperbolic plane. Both angle and length are represented correctly. The arc length of a geodesic is the correct hyperbolic distance. Furthermore, because of the constant Gaussian curvature, on the pseudo-sphere a figure may be shifted about and just bending will make it conform to the surface. The situation is similar to the more familiar case of Euclidean geometry on a circular cylinder or cone. As everybody knows, on a circular cylinder, a plane figure can be fitted by simply bending it, without stretching and shrinking. On the other hand, only the Poincaré disk is a model for the entire hyperbolic plane. Here only angles are still represented correctly, but the price one finally has to pay is that hyperbolic distances are distorted. The hyperbolic lines become circular arcs, perpendicular to the ideal boundary. One can see the distortion easily in Esher s superb artwork, based on tiling of the hyperbolic plane with congruent figures. The trade off just explained makes the isometry between the pseudo-sphere into the Poincaré disk especially interesting. One such isometric mapping is explicitly constructed below. Hilbert s result gets rather natural, too. As explained below, in the sense of hyperbolic geometry, the boundary of the pseudo-sphere turns out to be an arc of a horocyle About Gauss differential geometry Karl Friedrich Gauss had devoted an immense amount of work to geodesy and map making, starting This stimulus leads to his definitive paper in differential geometry of 1827: "Disquisitiones Generales circa Surperficies Curvas". In this work, Gauss introduces the basics of curved surfaces, and goes far beyond. The real benefit is that, due to the ideas forwarded by Gauss, mathematicians have in the end advanced to the concept of a curved surface as a space of its own interest. To begin with, one imagines a curved surface to be embedded into three dimensional space R 3, and given by some parametric equations x = x(u, v), y = y(u, v), z = z(u, v) (7.1) The distance ds of neighboring points on the surface with parameters (u, v) and (u + du, v + dv) is given by the first fundamental form ds 2 = Edu 2 + 2Fdudv + Gdv 2 (7.2)

84 84 F. Rothe The first fundamental form is straightforward to calculate from the parametric equations (7.1) since E = x 2 u + y 2 u + z 2 u F = x u x v + y u y v + z u z v (7.3) G = x 2 v + y 2 v + z 2 v follows from elementary vector calculus. The geodesics on curved surfaces are defined to be the shortest curves lying on the given surface, connecting any two given points. Gauss work sets up the differential equation for the geodesics. Gauss introduces the two main curvatures, called c 1, c 2. They turn out to be simply the extremal curvatures of normal sections of the surface. A new important feature is the Gaussian curvature, called K. Gauss shows that K = LN M2, the quotient of the determinants of the EG F 2 second and first fundamental form. But, even simpler, the Gaussian curvature turns out to be the product of the two principle curvatures: K = c 1 c 2 (7.4) Gauss shows the remarkable fact that this curvature is preserved during the process of bending the curved surface inside a higher dimensional space, without stretching, contracting or tearing it. On the contrary, the two main curvatures are changed by flexing the surface. There are actually at least two different proofs for this fact contained in Gauss work. The first one depends on Gauss characteristic equation K = 1 2H u [ F E EH v 1 H G u ] + 1 2H [ 2 F v H u 1 E H v where H = EG F 2. Obviously, any such equation implies that the Gaussian curvature depends only on the first fundamental form. The first fundamental form is preserved, if one bends the curved surface in three space, without stretching, contracting or tearing it. Therefore the functions E, F, G, H which determine the first fundamental form depend only on the parameters (u, v), but do not depend at all on how or even whether at all the surface lies in a three dimensional space. Because of the Gauss characteristic equation (7.5), the same is true for the Gaussian curvature K. Because of all that, one says that the Gaussian curvature is an intrinsic property of the curved surface Riemann metric of the Poincaré disk Proposition 7.1 (Riemann Metric for Poincaré s Model). In the Poincaré model, the infinitesimal hyperbolic distance ds of points with coordinates (x, y) and (x + dx, y + dy) is F EH ] E u (7.5) (ds D ) 2 = 4(dx2 + dy 2 ) (1 x 2 y 2 ) 2 (7.6) Reason. The fact that angles are measured in the usual Euclidean way implies that ds 2 = E(x, y)(dx 2 + dy 2 ). The rotational symmetry around the center O implies that E(x, y) = E( x 2 + y 2, 0). Hence ds 2 = E( x 2 + y 2, 0)(dx 2 + dy 2 ) (7.7) Now it is enough to calculate the distance of the points (x, 0) and (x + dx, 0). The hyperbolic distance of a point (x, 0) from the center (0, 0) is 2 tanh 1 x, as we have derived in Proposition??

85 Topics from Relativity 85 in the section on the Poincaré disk model. See formula (??) there, which is of course the primary origin of the hyperbolic distance! Taking the derivative by the variable x yields 4 Hence E(x, 0) = and (1 x 2 ) 2 ds dx = d dx (2 tanh 1 x) = d dx ln 1 + x 1 x = x x = 2 1 x 2 (7.8) ds 2 4 = (1 x 2 ) 2 dx2 E( x 2 + y 2, 0) = Now formulas (7.7) and (7.9) imply the claim (7.14). 4 (1 x 2 y 2 ) 2 (7.9) Problem 7.1 (Hyperbolic circumference of a circle). Calculate the circumference of a circle of hyperbolic radius R. We use the Poincaré disk, put the center of the circle at the center O of the disk. In polar coordinates, the Riemann metric is ds 2 = 4 dx 2 + dy 2 (1 x 2 y 2 ) = 4 dr2 + r 2 dθ 2 2 (1 r 2 ) 2 (a) Calculate the hyperbolic length R of a segment OA with Euclidean length OA = r < 1. (b) Get the circumference C = ds of the circle around O, at first in terms of the Euclidean radius OA = r < 1. (c) Get the circumference C of this circle in terms of the hyperbolic radius R. Solution. We take O as center of the circle, and point A on the circumference. Let r = OA denote the Euclidian radius, and R = s(o, A) be the hyperbolic radius. The hyperbolic radius R can be found directly for the Riemann metric (7.14). One needs partial fractions to do the integral. R = r 0 ds = = 2 tanh 1 r r 0 2dr 1 r 2 = r 0 [ 1 1 r r ] dr = [ ln(1 r) + ln(1 + r)] r 0 Remark. Of course, we can go back once more to Proposition??, formula (??) from the section on the Poincaré disk model and get R = s(o, A) = 2 tanh 1 OA = 2 tanh 1 r. We solve R = 2 tanh 1 r for the Euclidean radius and get r = tanh R 2 = er/2 e R/2 e R/2 + e R/2 = er 1 e R + 1 For the usual Euclidean polar coordinates (r, θ) we get the Euclidean arc length: 2π 2π 2π L Eucl = dx 2 + dy 2 = dr2 + r 2 dθ 2 = r dθ = 2πr (7.10) 0 0 The first line holds for any smooth curve. In the second line, we go to the special case of a circle. For a circle, the coordinate r is constant and hence dr = 0, and the factor r can be pulled out of the integral. 0

86 86 F. Rothe Now the distance along the circumference is measure in the hyperbolic metric (7.14) from Proposition 7.1. Hence the calculation above is modified to L hyp = 2π 0 = 2r 1 r 2 dx2 + dy x 2 y = 2 2π 0 dθ = 2π 0 4πr 1 r 2 2 dr2 + r 2 dθ 2 1 r 2 (7.11) We have found the correct hyperbolic arc length. But still, one needs to use the formula r = er 1 e R +1 to express r in terms of the hyperbolic distance R. L hyp = 4πr 1 r = 4π(e R 1) 2 [1 ( er 1 e R +1 )2 ](e R + 1) = 4π(eR 1)(e R + 1) (e R + 1) 2 (e R 1) = 4π(e2R 1) 2 4e R = π(e R e R ) = 2π sinh R (7.12) Proposition 7.2 (The circumference of a circle). In hyperbolic geometry, the circumference of a circle of hyperbolic radius R is 2π sinh R. Problem 7.2. The hyperbolic circumference of a circle is much larger than the Euclidean circumference. Let R = 1, 2, 5, 10 and estimate how many times the radius fits around the circumference of a circle of that radius. Answer. A simple calculation yields R (2πsinh R)/R Problem 7.3 (Hyperbolic area of a circle). For a circle of hyperbolic radius R, calculate the area A. Again, we use the Poincaré disk, put the center of the circle at the center O of the disk. For the area, we use the formula from differential geometry A = 2π r 0 0 EG F2 dr dθ The first fundamental form is the Riemann metric. It has been already given by formula (7.14), and transformed to polar coordinates in the previous problem. ds 2 = 4 dx2 + dy 2 (1 x 2 y 2 ) 2 = 4 dr2 + r 2 dθ 2 (1 r 2 ) 2 = Edr 2 + 2Fdrdθ + Gdθ 2 (7.13) (a) Get the area of the circle, at first in terms of the Euclidean radius OA = r < 1. (b) Get the area A of this circle in terms of the hyperbolic radius R.

87 Topics from Relativity 87 (c) Check that da dr = C (a) The first fundament form (7.13) yields H = EG F 2 = 4r (1 r 2 ) 2 and hence the hyperbolic area of a circle of Euclidean radius r is A = 2π r EG F2 dr dθ = 2π r 0 4r dr (1 r 2 ) 2 This integral is solved with the substitution u = r 2 and du = 2rdr. u [ ] u 2 du A = 2π (1 u) = 4π 4π 4πr2 = 4π = 2 (1 u) (1 r 2 ) (1 r 2 ) This is the area in terms of the Euclidean radius OA = r < 1. 0 (b) The hyperbolic radius R has already been calculated in the previous problem. R = 2 tanh 1 r for the Euclidean radius and get We solve r = tanh R 2 = er/2 e R/2 e R/2 + e R/2 = er 1 e R + 1 r 2 = (er 1) 2 (e R + 1) 2 and 1 r 2 = (er + 1) 2 (e R 1) 2 (e R + 1) 2 = Now plug this formula into the result from part (a) and get A = An alternative formula is 4e R (e R + 1) 2 4πr2 (1 r 2 ) = π(er 1) 2 e R = π(e R 1 + e R ) = 2π(cosh R 1) A = π(er 1) 2 e R = π(e R/2 e R/2 ) 2 = 4π sinh 2 R 2 (c) We have obtained in Proposition 7.2 from the section on the Poincaré disk model, that the hyperbolic circle of radius R has the circumference C = 2π sinh R. On the other hand, differentiating the result of (b) gives as to be shown. da cosh R = 2πd = 2π sinh R = C dr dr Problem 7.4. Use the fundamental form for the Poincaré disk model to calculate its Gaussian curvature. (ds D ) 2 = 4(dx2 + dy 2 ) (1 x 2 y 2 ) 2 (7.14)

88 88 F. Rothe Answer. Formula (7.14) implies that the functions in the first fundamental form are E = G = H = 4(1 x 2 y 2 ) 2 and F = 0. Hence, with x = u and v = y, we get from formula (7.5) K = 1 [ 1 ] E + 1 [ 1 ] E = 1 ( ) 2 2E x E x 2E y E y 2E x + 2 ln E 2 y 2 = + 1 ( ) 2 E x + 2 ln(1 x 2 y 2 ) = + 1 ( 2x 2 y 2 E x (1 x 2 y 2 ) + ) 2y y (1 x 2 y 2 ) = 2 ( 1 x 2 y 2 + 2x x2 y 2 + 2y 2 ) = ( 2) 2 = 1 E (1 x 2 y 2 ) 2 (1 x 2 y 2 ) 2 4 By the way, the result K = 1 motivates the annoying factor 4 in formula (7.14) Riemann metric of Klein s model Proposition 7.3 (Hilbert-Klein Metric). In the Klein model, the infinitesimal hyperbolic distance ds of points with coordinates (X, Y) and (X + dx, Y + dy) is ds 2 = dx2 + dy 2 (XdY YdX) 2 (1 X 2 Y 2 ) 2 (7.15) Proof. We shall derive this metric using the transformation from the Poincaré to the Klein model. As stated in Proposition??, the mapping from a point P in Poincaré s model to a point K in Klein s model is OK = 2 OP (??) 1 + OP 2 requiring that the rays OP = OK are identical. We use Cartesian coordinates and put P = (x, y) for Poincaré s model and K = (X, Y) for the points in Klein s model. Finally we put r 2 = x 2 + y 2 and R 2 = X 2 + Y 2. From the mapping (??), we get X = 2x and Y = 2y (7.16) 1 + r r 2 The Riemann metric for Poincaré s model has been derived in Proposition 7.1 to be ds 2 = 4 dx2 + dy 2 (1 x 2 y 2 ) 2 = E dx2 + 2F dxdy + G dy 2 Here E, F, G denotes the fundamental form for the Poincaré model in terms of (x, y). following we shall use the matrix [E ] [ ] F = F G (1 x 2 y 2 ) In the (7.17) From the fact that the transformation from Poincaré s to Klein s model is a passive coordinate transformation, we know that the infinitesimal hyperbolic distance ds of points is left invariant. Because of the invariance, the fundamental form E, F, G for the Klein model has to satisfy ds 2 = E dx 2 + 2F dxdy + G dy 2 = E dx 2 + 2F dxdy + G dy 2 We take for now (x, y) as independent variables. From calculus, we know that by means of the chain rule X Y [ ] x x X Y X X [ ] E F x y E F F G Y Y = (7.18) F G y y x y

89 Topics from Relativity 89 It now remains to carry out the arithmetic. The superscript T denotes transposition of matrices and the superscript 1 denote inversion of matrices. As usual, we use X X D X Dx = x y Y Y x y as shorthand for the Jacobi matrix of the transformation (7.16). We need to solve the equation (7.18) for the new fundamental form E, F, G to obtain [ D X ] T [ ] [ E F D X ] [ ] E F = Dx F G Dx F G [ ] [ E F D X ] T, 1 [ ] [ E F D X ] 1 = F G Dx F G Dx The Jacobi matrix of the transformation (7.16) is obtained explicitly from equations (7.16) to be [ ] D X Dx = 2 1 x 2 + y 2 2xy (1 + x 2 + y 2 ) 2 2xy 1 + x 2 y 2 The determinant is Det D X Dx = 4 (1 + r 2 ) 4 [ 1 (x 2 y 2 ) 2 4x 2 y 2] = 4 [ ] 1 r 4 = 4(1 r2 ) (1 + r 2 ) 4 (1 + r 2 ) 3 Hence the inverse turned out to be [ D X ] 1 = (1 + r2 ) 3 [ ] x 2 y 2 2xy Dx 4(1 r 2 ) (1 + r 2 ) 2 2xy 1 x 2 + y 2 = 1 + [ ] r2 1 + x 2 y 2 2xy 2(1 r 2 ) 2xy 1 x 2 + y 2 With the fundamental form from formula (7.17) and the inverse Jacobi matrix just obtained plugged into equation (??), we calculate [ ] E F = (1 + r2 ) 2 [ ] x 2 y 2 2 2xy F G 4(1 r 2 ) 2 (1 r 2 ) 2 2xy 1 x 2 + y 2 = (1 + r2 ) 2 [ ] 1 + 2x 2 2y 2 + (x 2 y 2 ) 2 + 4x 2 y 2 4xy (1 r 2 ) 4 4xy 1 2x 2 + 2y 2 + r 4 = (1 + r2 ) 2 (1 r 2 ) 4 [ ] (1 + r 2 ) 2 4y 2 4xy 4xy (1 + r 2 ) 2 4x 2 This is the new fundamental form. We need still to introduce the new coordinates (X, Y). We use the short hands r 2 = x 2 + y 2 and R 2 = X 2 + Y 2. By means of equation (7.16) we get 1 X 2 = (1 + r2 ) 2 4x 2 (1 + r 2 ) 2, 1 Y 2 = (1 + r2 ) 2 4y 2 (1 + r 2 ) 2, XY = 4xy, 1 R 2 = (1 r2 ) 2 (1 + r 2 ) 2 (1 + r 2 ) 2 Thus the new fundamental form miracously simplifies to be [ ] [ ] E F 1 1 Y 2 XY = F G (1 R 2 ) 2 XY 1 X 2 (7.19)

90 90 F. Rothe For the line element we get from this fundamental form ds 2 = E dx 2 + 2F dxdy + G dy 2 = (1 Y2 )dx 2 + 2XYdXdY + (1 X 2 )dy 2 (1 R 2 ) 2 = dx2 + dy 2 (XdY YdX) 2 (1 X 2 Y 2 ) 2 Problem 7.5 (Gaussian curvature of the Hilbert-Klein metric). Use Gauss characteristic equation (7.5) and check directly that the Hilbert-Klein metric (7.15) from proposition 7.3 has constant Gaussian curvature K = 1. We use polar coordinates X = r cos θ, Y = r sin θ and convert formula to ds 2 = dr2 + r 2 (1 r 2 )dθ 2 (1 r 2 ) 2 (7.20) since this simplifies the calculation considerably. Answer. We have to put u = r and v = θ. The first fundamental form and its coefficients become ds 2 = Edr 2 + 2Fdrdθ + Gdθ 2 E = (1 r 2 ) 2, F = 0, G = r 2 (1 r 2 ) 1 H = EG F 2 = r(1 r 2 ) 3/2 Hence we get for the Gaussian curvature K from the characteristic equation 2HK = [ ] 1 G H = [ r 1 (1 r 2 ) 3/2 r 2 ] r u r r 1 r 2 = [ r 1 (1 r 2 ) 3/2 2r(1 r2 ) r 2 ] ( 2r) = [ 2(1 r 2 ) 1/2] r (1 r 2 ) 2 r = ( 2)( 1/2)(1 r 2 ) 3/2 ( 2r) = 2r(1 r 2 ) 3/2 = 2H We get the constant Gaussian curvature K = 1, as expected. Problem 7.6 (Distortion of angles by the Hilbert-Klein metric). We use polar coordinates to simplify the calculation, and the corresponding contravariant components for the tangent vectors. At a point K with polar coordinates X = r cos θ, Y = r sin θ are attached the radial tangent vector (a r, a θ ) = (1, 0) and any other tangent vector (b r, b θ ). Hence the apparent angle α satisfies cos α = b r b 2 r + r 2 b 2 θ and tan α = rb θ b r Check with the Hilbert-Klein metric (7.20) that the angle ω between the two vectors in Klein s model satisfies tan ω = tan α 1 r 2 (7.21)

91 Topics from Relativity 91 Answer. The apparent angle α and the hyperbolic angle ω between the two tangent vectors (a r, a θ ) and (b r, b θ ) satisfy cos α = a r b r + r 2 a θ b θ (a 2r + r 2 a 2θ ) (b 2 r + r 2 b 2 θ ) cos ω = a r b r + r 2 (1 r 2 )a θ b θ a 2 r + r 2 (1 r 2 )a 2 θ b 2 r + r 2 (1 r 2 )b 2 θ In the given example with a r = 1, a θ = 0 the expressions simplify cos α = b r b 2 r + r 2 b 2 θ and tan 2 α = r2 b 2 θ b 2 r cos ω = b r b 2 r + r 2 (1 r 2 )b 2 θ and tan 2 ω = r2 (1 r 2 )b 2 θ b 2 r Hence tan ω = tan α 1 r A second proof of Gauss remarkable theorem The most enlightened proof that the Gaussian curvature is an intrinsic property of the surface uses Gauss notion of integral curvature. For any domain G on a given curved surface, the integral curvature is defined as the integral KdA, where da denotes the area element of the surface. G Take a geodesic triangle ABC. Let T denote the region bounded by the geodesics between any three given points A, B, C on the surface. Let α, β, γ be the angles (between tangents) to the geodesics at the three vertices. Gauss proves KdA = α + β + γ π (7.22) T where angles are to be measured in radians. The quantity on the right hand side is the deviation of the angle sum α + β + γ from the Euclidean value 180, respectively π. 1 The quantity α + β + γ π is called the excess of the triangle ABC. For the hyperbolic case, the excess is negative. In that case, one calculates using the excess times 1, which is called defect. In words, Gauss theorem tells the following: For a geodesic triangle, the integral curvature equals the excess of its angle sum. This theorem, Gauss says, ought to be counted as a most elegant theorem. I discuss a few immediate, but important consequences of (7.22). First of all, instead of the complicated characteristic equation (7.5), one has a simple property of a geodesic triangle from which to derive the Gaussian curvature in a limiting process. Secondly, as an immediate implications of (7.22), the Gaussian curvature is an intrinsic property of a curved surface. Recall that both geodesics, as well as measurement of area depend only on the first fundamental form. Hence, because of (7.22), the same is true for the Gaussian curvature. Another easy consequence of (7.22) is obtained from the special case of a sphere. For this surface, the Gaussian curvature is constant, and equal to K = R 2 where R is the radius of the sphere. Hence one obtains for the area of a spherical triangle A = (α + β + γ π)r 2. as was already known before Gauss, e.g. to Lambert. 1 In radian measures, the Euclidean angle sum is π.

92 92 F. Rothe Problem 7.7. Tile a sphere by equilateral triangles. It can be done in three ways: (i) Four triangles with α = β = γ = 120. (ii) Eight triangles with α = β = γ = 90. (iii) N triangles with α = β = γ = 72. Explain and draw these tilings. To which Platonic bodies do the vertices correspond? Determine the surface area of the sphere from (i) and (ii), then get the number N in (iii). Answer. From item (i), the area of the sphere is A = 4 (3 2π3 ) π R 2 = 4πR 2 The vertices of the four triangles form a tetrahedron. Similarly, item (ii) yields A = 8 (3 π ) 2 π R 2 = 4πR 2 The vertices of the eight triangles form a octahedron. We can now calculate the number N of triangles in the tiling (iii). Because of A = N (3 2π5 ) π = 4πR 2 one gets N = 20. The vertices of the twenty triangles form an icosahedron. Here is a further important consequence of equation (7.22). Corollary 9 (A common bound for the area of all triangles). On a surface with negative constant Gaussian curvature K < 0, the area of any triangle is less than π K. On December 17, 1799, Gauss wrote to his friend, the Hungarian mathematician Wolfgang Farkas Bolyai ( ): As for me, I have already made some progress in my work. However, the path I have chosen does not lead at all to the goal which we seek [deduction of the parallel axiom], and which you assure me you have reached. It seems rather to compel me to doubt the truth of geometry itself. It is true that I have come upon much which by most people would be held to constitute a proof; but in my eyes it proves as good as nothing. For example, if we could show that a rectilinear triangle whose area would be greater than any given area is possible, then I would be ready to prove the whole of Euclidean geometry absolutely rigorously. Most people would certainly let this stand as an axiom; but I, no! It would indeed be possible that the area might always remain below a certain limit, however far apart the three angular points of the triangle were taken. From about 1813 on Gauss developed his new geometry. He became convinced that it was logically consistent and rather sure that it might be applicable. His letter written in 1817 to Olbers says: I am becoming more and more convinced that the physical necessity of our Euclidean geometry cannot be proved, at least not by human reason nor for human reason. Perhaps in another life we will be able to obtain insight into the nature of space, which is now unattainable. Until then we must place geometry not in the same class with arithmetic, which is purely a priori, but with mechanics.

93 Topics from Relativity 93 Problem 7.8. (a) Find two further enlightening statements of Gauss, and comments on all four statements. (i) Are they courageous? (ii) Are they to the benefit of the scientific community? (iii) Are they helpful for the person he addresses? (iv) Are they just against other people? (v) What would you have done in Gauss place? (b) Choose two of Gauss comments. Write a letter as you imagine you would have written in place of Gauss. Problem 7.9. To test the applicability of Euclidean geometry and his non-euclidean geometry, Gauss actually measured the sum of the angles of the triangle formed by three mountain peaks in middle Germany: Broken, Hohenhagen, and Inselberg. the sides of the triangle were 69, 85, 197 km. His measurement yielded that the angle sum exceeded 180 by (a) Use Herons formula A = s(s a)(s b)(s c) to calculate the area of the triangle, in a very good approximation. (b) Take R = 6378 km as radius of the earth. Calculate the angle excess for a spherical triangle between the three mountain peaks. You need to convert angular measurements! 1 radian equals =. π π (c) Is the triangle that Gauss measured actually a spherical triangle. Why or why not? (d) Reflect on the motives why Gauss did his measurement. Find and read some further sources. Think of the following and further motives and possibilities. Did Gauss really just want to (i) check accuracy? (ii) check geometry? (iii) It was just a theoretical thought experiment, not really performed. Answer. (a) Herons formula give the area A = s(s a)(s b)(s c) = km 2. (b) Take R = 6378 km as radius of the earth. The angle excess for a spherical triangle between the three mountain peaks is α + β + γ π = A R 2 = (7.23) This is the value in radian measure. Converted to degrees, we get which is (c) Of course the triangle one measures is not a spherical triangle, since light rays do not follow the curvature of the earth.

94 94 F. Rothe 7.6. Principal and Gaussian curvature of rotation surfaces Before introducing the pseudosphere, we need some facts about the curvature of general rotation surfaces. We take the graph of an arbitrary function y = f (x), and rotate it about the y-axis to produce a rotation surface in three dimensional space. The first principle curvature of a rotation surface in the xy plane is y c 1 = (1 + y 2 ) 3 2 (7.24) This is just the curvature of the graph y = f (x). Recall that the perpendicular to the tangent of a curve is called the normal of the curve. The second principal curvature occurs for a section of the surface by a plane P 2, which intersects the xy plane along the normal of the curve y = f (x), and is perpendicular to the xy plane. The second principal curvature is c 2 = y x(1 + y 2 ) 1 2 (7.25) Proposition 7.4. The Gaussian curvature of the rotation surface produced by rotating the graph of y = f (x) around the y-axis, is the product K = y y x(1 + y 2 ) 2 (7.26) The formula (7.24) for a curvature of a plane curve is standard. Finally, since K = c 1 c 2, formulas (7.24) and (7.25) imply the claim (7.26). Here is an argument to justify (7.25): Let tan β = y be tangent of the slope angle for y = f (x), as usual. Calculate sin β. Calculate the hypothenuse AB of the right ABC, with vertex A = (x, y) on the curve, leg AC parallel to the x-axis, leg BC on the axis of rotation, and hypothenuse AB perpendicular to the curve. One can show that point B is the center of the best approximating circle in the plane P 2. Hence c 2 = 1 AB. Use this idea to get the main curvature c 2. Answer. tan β = AC BC = y Hence sin β = AC AB = c 2 = 1 AB = sin β = x AC BC 2 + AC 2 = y x(1 + y 2 ) 1 2 y 1 + y 2 A second proof of Proposition 7.4. On the surface of rotation, we choose as parameters u = φ the rotation angle, and v = r the distance from the rotation axis. Since the y-axis is the axis of rotation, the surface of rotation gets the parametric representation x = v cos u y = f (v) z = v sin u The derivatives by the parameters are x u = v sin u x v = cos u y u = 0 y v = f (v) z u = v cos u z v = sin u (7.27)

95 Topics from Relativity 95 Figure 7.1. Curvature of a rotation surface. From these derivatives, one gets the first fundamental form. We use the general formulas E = x 2 u + y 2 u + z 2 u F = x u x v + y u y v + z u z v (7.3) G = x 2 v + y 2 v + z 2 v valid for any surface, and specialize to the surface of rotation given above. Now calculate E, F, G from (7.3) and (7.27), to get the first fundamental form (7.2). Next get the root of the determinant H = EG F 2. Finally calculate the Gaussian curvature from the characteristic equation (7.5). Problem Use the approach as indicated to confirm formula (7.26). Answer. One gets E = v 2, F = 0 and G = 1 + f 2 and ds 2 = v 2 du 2 + (1 + f 2 )dv 2 = r 2 dφ 2 + (1 + f (r) 2 )dr 2 The root of the determinant is H = v 1 + f 2. Because all four quantities E, F, G, H depend only on v, the partial derivatives by u all vanish. Thus Gauss characteristic equation (7.5) can be simplified to yield K = 1 [ F E 2H u EH v 1 ] G + 1 [ 2 F H u 2H v H u 1 E H v F ] E EH u = 1 [ 1 ] E = 1 [ ] d 2v 2H v H v 2H dv H

96 96 F. Rothe Now use H = v(1 + f 2 ) 1 2 and go on. We arrive at K = 1 2H d dv [ ] 2v H 1 = 2v 1 + f 2 = 1 v 1 + f 2 d(1 + f 2 ) 1 2 dv [ ] (1 + f 2 ) f f = f f v(1 + f 2 ) 2 This result is equivalent to formula (7.26), since x = v = r is the distance from the axis of rotation and y = f (x). Problem Calculate the Gaussian curvature of a three dimensional sphere of radius a. Answer. The sphere is provided by rotating the graph of x 2 + y 2 = a 2 about the y-axis. Implicit differentiation yields 2x + 2yy = 0 and hence y = x y y = 1 y xy y 2 K = = xy y y 2 = x2 /y y y 2 = a2 y 3 y y x(1 + y 2 ) = a 2 2 y 4 ( 1 + x 2 /y 2) = a 2 2 (x 2 + y 2 ) = 1 2 a The pseudo-sphere The issue is now to find a rotation surface of constant negative Gaussian curvature K = a 2. Such a surface is called pseudo-sphere. Problem Use the formula (7.26) for the Gaussian curvature, and get a differential equation of first order for the function u := y 2. You may begin by getting the derivative du dx. Answer. The derivative of the function u := y 2 is du dx = 2y y. Next, I put the requirement K = a 2 into the formula (7.26). One gets Problem Solve the differential equation y y x(1 + y 2 ) 2 = 1 a 2 (7.28) 2y y = 2x(1 + y 2 ) 2 a 2 (7.29) du + u)2 = 2x(1 dx a 2 (7.30) du + u)2 = 2x(1 (7.30) dx a 2 by separation of variable. For simplicity, we use the initial data u(a) = 0, and get a curve through the point x = a, u = 0.

97 Topics from Relativity 97 Answer. u a du + u)2 = 2x(1 dx a 2 du (1 + u) = 2xdx 2 a 2 du x (1 + u) du = 2xdx dx 2 0 a 2 [ 1 ] u ] x = [ x2 1 + u a a u + 1 = x2 a u = a2 x 2 0 u = a2 x 2 x 2 Problem Check that the differential equation y a2 x = 2 x with x a, has the solution y = a ln a + a 2 x 2 x Too, find the general solution of the equation a 2 x 2 + C (7.31) Answer. y = a ln y = + a a 2 x 2 x a2 x 2 x + a 2 x 2 + C Definition 7.1 (Pseudo-sphere). The rotation surface with constant negative Gaussian curvature is called a pseudo-sphere. With curvature K = a 2 and the y-axis as axis of rotation, its equation is y = a ln a + a 2 x 2 z 2 a 2 x 2 z 2 x2 + z 2 Problem Check, once more, that the Gaussian curvature of the specified surface is K = a 2. Answer. Problem Check the following fact: The segment on the tangent to the curve (7.31), between the touching point T, and the intersection S of the tangent with the y-axis has always the same length a. For that reason, the curve (7.31) is called tractrix.

98 98 F. Rothe Figure 7.2. The tractrix has a segment on its tangent of constant length. Answer. Take the right TS C, formed by the segment TS on the tangent at point T, and the perpendicular from T onto the y-axis. 1 We know that y = tan α = S C TC TS 2 = TC 2 + S C 2 = x 2 (1 + y 2 ) = a 2 Problem The surface area of of rotation surface, made by rotating y = f (x) about the y-axis, for x 1 x x 2 is x2 S = 2πx 1 + y 2 dx Calculate the surface of the pseudo-sphere for bounds 0 < x a. Answer. Because of 1 + y 2 = 1 + u = a2, we get x 2 S = x 1 a 0 2πx a dx = 2πa2 x We introduce now (φ, r) as two convenient coordinates on the pseudo-sphere. As first coordinate, we choose the angle of rotation φ about the y-axis. The second coordinate is the radius 1 This triangle is different from triangle ABC in the figure on page 95.

99 Topics from Relativity 99 r = x 2 + z 2 measured from the axis of rotation, Up to now it was called x, but now I choose to name it r. The three parameters r, φ, y are cylindrical coordinates of three dimensional space. The first two of them are convenient parameters on the pseudo-sphere. Proposition 7.5 (Riemann Metric for the Pseudo Sphere). The infinitesimal distance ds of points with coordinates (φ, r) and (φ + dφ, r + dr) is ds 2 = r 2 dφ 2 + a2 r 2 dr2 (7.32) Proof. The distance on the pseudo-sphere is calculated from the usual Euclidean distance for points of the three dimensional space into which the surface is embedded. At first, I convert the distance from Cartesian to cylindrical coordinates. Because the y-axis is the rotation axis, its coordinate stays, but the pair (x, z) is converted to polar coordinates. Hence one gets ds 2 = dx 2 + dy 2 + dz 2 = dr 2 + r 2 dφ 2 + dy 2 (7.33) We restrict to points on the pseudo-sphere. Hence the coordinates r and y are related in the same way as x and y before. Thus y = a 2 x 2 gets x dy dr = a2 r 2 Now we use (7.34) to eliminate y from (7.33) and get r (7.34) ds 2 = dr 2 + r 2 dφ 2 + dy 2 = dr 2 + r 2 dφ 2 + = dr 2 + r 2 dφ 2 + a2 r 2 r 2 ( ) 2 dy dr 2 dr dr 2 = r 2 dφ 2 + a2 r 2 dr2 as to be shown. As an alternative, we can use the first fundamental form calculated above. Since 1 + f (r) 2 = 1 + a2 r 2 = a2, one gets again r 2 r 2 ds 2 = r 2 dφ 2 + (1 + f (r) 2 )dr 2 = r 2 dφ 2 + a2 r 2 dr Poincaré half-plane and Poincaré disk Throughout, we denote the upper open halfplane by H = {(u, v) : v > 0}. Its boundary is just the real axis H = {(u, v) : v = 0}. The open unit disk is denoted by D = {z = x + iy : x 2 + y 2 < 1}, and its boundary is D = {z = x + iy : x 2 + y 2 = 1}. The following isometric mapping of the half-plane to the disk is used in this section. It differs from the one used in the previous section by a rotation of the disk by a right angle. We repeat for convenience. Proposition 7.6 (Isometric Mapping of the Half-plane to the Disk). The linear fractional function z = iw + 1 (7.35) w + i is a conformal mapping and a bijection from C { } to C { }. The inverse mapping is w = 1 iz z i (7.36)

100 100 F. Rothe These mappings preserves angles, the cross ratio, the orientation, and map generalized circles to generalized circles. The upper half-plane H = {w = u + iv : v > 0} is mapped onto the unit disk D = {z = x + iy : x 2 + y 2 < 1}. Especially w = 0 z = i, w = 1 z = 1, w = z = i, w = i z = 0 Proposition 7.7 (Riemann Metric for Poincaré s half-plane). In the Poincaré half-plane, the infinitesimal hyperbolic distance ds of points with coordinates (u, v) and (u + du, v + dv) is (ds H ) 2 = du2 + dv 2 v 2 (7.37) The mapping (7.35) provides an isometry between the half-plane and the disk: ds D = ds H (4.5) Proof. The metric of the half plane is calculated from the known metric of the Poinaré disk model. The mapping (ds D ) 2 = 4(dx2 + dy 2 ) (1 x 2 y 2 ) 2 (7.14) z = iw + 1 w + i provides an isometry from the half-plane to the disk. The denominator is 1 z 2 = The derivative of the mapping (7.35) is Putting the last two formulas into (7.14) yields (w + i)(w i) (iw + 1)( iw + 1) 2iw 2iw = = w + i 2 w + i 2 dz dw = 2 (w + i) 2 ds 2 = 4(dx2 + dy 2 ) 4 dz 2 = (1 x 2 y 2 ) 2 (1 z 2 ) 2 = 4 dz 2 ( ) w + i dw dw = 4 2 4v (w + i) 2 2 4v w + i 2 ( ) w + i dw v = dw 2 = du2 + dv 2 v 2 v 2 Hence formula (7.37) arises from the isometry (7.35) between half-plane and the disk Embedding the pseudo-sphere into Poincaré s half-plane (7.35) Proposition 7.8. The mapping w = φ + i a (7.38) r transforms the line element ds PS of the pseudo-sphere to the line element ds H of the half-plane such that ds PS = a (ds H ) (7.39) For a = 1, we get an isometry. This is just the case with Gaussian curvature K = a 2 = 1. Because an isometry conserves the Gaussian curvature, this shows that the Poincaré half-plane has Gaussian curvature 1.

101 Topics from Relativity 101 Proof. We separate equation (7.38) into its real- and imaginary part to get u = φ, v = a r (7.40) Using its derivatives, we plug into One gets Now comparing with one concludes ds H 2 = dφ2 + ( ar 2) 2 dr 2 a 2 r 2 ds H 2 = du2 + dv 2 v 2 (7.37) ) = a (r 2 2 dφ 2 + a2 r 2 dr2 ds 2 PS = r2 dφ 2 + a2 r 2 dr2 (7.32) ds H 2 = a 2 (ds PS ) 2 and hence equation (7.39) holds Embedding the pseudo-sphere into Poincaré s disk The next goal is to construct an isometric mapping from the pseudo-sphere to the Poincaré disk. It is convenient to get this mapping as composition of a mapping from the pseudo-sphere to the half-plane, and the conformal mapping from the half-plane to the disk. Proposition 7.9. We take a pseudo-sphere with a = 1. This normalizes the Gaussian curvature to be K = 1. The mapping z = r 1 + irφ (7.41) rφ + i(r + 1) maps the pseudo-sphere isometrically into the Poincaré disk. Proof. The mapping (7.41) is constructed as a composition of two mappings PS H D. Take the mapping PS H given by equation (7.42), and the mapping H D given by equation (7.35). The composition of the mapping w = φ + i 1 r from the pseudo-sphere to the Poincaré half-plane, with the mapping z = iw + 1 w + i from the Poincaré half-plane to the Poincaré disk is the required mapping. following mapping (7.42) (7.35) z = i ( ) φ + i r + 1 φ + i r + i = r 1 + irφ rφ + i(r + 1) from the pseudo-sphere to the Poincaré disk. Both mappings (7.42) and (7.35) are isometries, as stated by formulas (7.39) with a = 1 and formula (??). Hence their composition (7.41) conserves the line element: ds PS = ds H = ds D.

102 102 F. Rothe About circle-like curves We now go back to the Poincaré disk model. At first, here are a few remarks about circle-like curves. In hyperbolic geometry, there exist three different types of circle-like curves. I define as a circle-like curve a curve which appears to be a circle in the Poincaré model. Recall that D is the boundary circle of the Poincaré disk. Take any second circle C. I call its Euclidean center M the quasi-center. The meaning of C for the hyperbolic geometry of the Poincaré disk depends on the nature of the intersection of the two circles C and D. There are three important cases: (i) The circle C lies totally in the interior D. In that case, it is a circle for hyperbolic geometry. This circle has a center A in hyperbolic geometry. Note that the quasi-center M is different from the center A of C as an object of hyperbolic geometry. (ii) The circle C touches the boundary D from inside, say at endpoint E. In that case, it is a horocycle for hyperbolic geometry. A horocycle has no hyperbolic center, instead it contains an ideal point E. Hence it is unbounded. The hyperbolic circumference of a horocycle is infinite, as follows from part (c) below. (iii) The circle C intersects the boundary D at two endpoints E and F. In that case, the circular arc inside the disk D is either an equidistance line or a geodesic for hyperbolic geometry. A geodesic intersects D perpendicularly. In the case of non perpendicular intersection of D and C, one gets an equidistance line. Actually all points of that equidistance line have the same distance from the hyperbolic straight line with ends E and F. Problem Take points Y + = (1, 0) and O = (0, 0). Find the analytic equation for a horocycle H with apparent diameter OY +. Answer. In complex notation, point Y + is i. The quasi-center is M = i 2, and the apparent radius is 1 2. Hence one gets the equation z i 2 2 = 1 4 ( x 2 + y 1 ) = 0 x 2 + y(y 1) = 0 We need another more convenient parametric equation for the horocycle H. Let Z = (x, y) be any point on H and define the circumference angle d OY + Z. Calculate tan d in terms of (x, y). Then express x and y in terms of the central angle 2d OMZ. Use double angle formulas, and finally express x and y in terms of tan d. Answer. tan d = x = y = x 1 y = y x sin 2d = sin d cos d = 2 1 cos 2d 2 tan d 1 + tan 2 d = sin 2 d = tan2 d 1 + tan 2 d (7.43) Problem Confirm that the hyperbolic arc length of the arc OZ on the horocycle H is just s = 2 tan d.

103 Topics from Relativity 103 Figure 7.3. Measuring an arc of a horocycle. Answer. Let t = tan d. Differentiation yields x = t 1 + t 2, dx dt = y = t2 1 + t, dy 2 dt = ( ) 2 ( ) 2 1 x 2 y 2 1 dx dy = 1 y = + = 1 + t 2 dt dt Hence the hyperbolic metric (7.14) implies ( ) 2 ( ) 2 ds dx = 4(1 x 2 y 2 ) 2 dt + dt Hence by elementary integration s = 2t = 2 tan d. ( ) 2 dy dt = 4 1 t2 (1 + t 2 ) 2 2t (1 + t 2 ) 2 1 (1 + t 2 ) 2 Problem Give the representation of the horocycle H with this arc length as parameter. Explain in a drawing, how to measure the arc length on this horocycle. Answer. We get the parametrization x = 2s 4 + s 2 (7.44) y = s2 4 + s 2 The hyperbolic arc length of OZ on the horocycle H is the Euclidean length Y Z since s = 2 tan d = Y Z.

104 104 F. Rothe Figure 7.4. Isometry of the sliced pseudo-sphere to a half infinity strip Mapping the boundaries There cannot exist an isometry of between the pseudo-sphere and the half-plane, since they have different topologies. A corresponding problem already arises in Euclidean geometry, for the construction of for an isometry between the cylinder and the plane. At least, there exists a non-invertible homomorphism from the plane onto the cylinder. This homomorphism can be restricted to on isomorphism between a strip of the plane and the sliced cylinder. We return to the hyperbolic case. By slicing the pseudo-sphere, we get an isomorphism of the sliced pseudo-sphere into a strip of the half-plane, and furthermore into part of the disk, too. The pseudo-sphere is sliced along the geodesic in the negative (x, y)-plane, restricting the rotation angle to the half-open interval π φ < π. The mapping w = φ + i 1 r (7.42) maps the sliced pseudo-sphere onto a half open rectangular domain PS H in the upper half-plane. The boundary of PS H consists of a segment AB with the endpoints A = π + i, B = π + i, and two unbounded rays A and B with vertices A and B parallel to the positive v axis. Furthermore, we map the sliced pseudo-sphere to the Poincaré disk via the isometry (7.41). The image PS D of the pseudo-sphere is a part of the interior of the horocycle H with apparent diameter 0 to i. Problem On the pseudo-sphere, we use as parameters the cylindrical coordinates r and φ. The boundary of the sliced pseudo-sphere is given by r = 1, and π < φ < π. To which curve in the disk D is the boundary mapped by the isometry (7.41)? Answer. The boundary PS D of image PS D consists of three circular arcs. A segment AB of a horocycle H with endpoints A = iπ π + 2i, B = iπ π 2i as well as two geodesic rays AY+ and BY+ with vertices A and B pointing to the ideal endpoint Y + = i.

105 Topics from Relativity 105 Figure 7.5. Isometric image of the sliced pseudo-sphere in the Poincaré disk. Problem Give a parametric equation for the boundary, with parameter φ, at first in complex notation for z = x + iy. Then separate into real and imaginary parts to get equations for x and y. Answer. Simply put r = 1 into equation (7.41). One gets z = iφ φ + 2i To separate real and imaginary parts, one needs a make the denominator real: z = iφ φ + 2i = iφ(φ 2i) (φ + 2i)(φ 2i) x + iy = iφ2 + 2φ φ x = 2φ φ y = φ2 φ Problem Check that your parametric equation is a circle with center i 2. Answer. This is a parametric equation of a circle with center i 2 because (d) Compare (7.46) with the result z i 2 = iφ φ + 2i i i(φ 2i) = 2 2(φ + 2i) z i 2 = 1 2 tan d x = 1 + tan 2 d y = tan2 d 1 + tan 2 d (7.45) (7.46) (7.47) (7.48)