x 2 = (ct) 2 +1, x 2 = (ct ) 2 +1, Major disadvantage of geometric approach: scaling according to hyperbolic laws.

Transcription

1 Scaling of axes The Lorentz transformation S S changes the scale of the axes. Choose a length unit by s 2 = 1. Then x 2 = (ct) 2 +1, ie hyperbola, all points on it have length s 2 = 1. In particular t = 0 cuts x-axis at x = 1. But as s 2 is a Lorentz invariant, then this is equivalent to t = 0 cuts x -axis at x = 1. Analogously s 2 = 1 gives the time unit x 2 = (ct ) 2 +1, (ct) 2 = x 2 +1 (ct ) 2 = x 2 +1, again a hyperbola x = 0 gives time unit in S frame; x = 0 in S frame. (Not shown on plot.) Major disadvantage of geometric approach: scaling according to hyperbolic laws. 93

2 Chapter 8 Covariant formulation of Special Relativity Must be possible to write laws of physics in a covariant form ie one that under Lorentz transformations remains form invariant The mathematics we need is tensor analyis (in four dimensions), as we shall know how they transform under (Lorentz) transformations. So form invariant equations tensor equations Develop in analogy with our three dimensional experience. 8.1 Covariant and Contravariant Tensors Three dimensional rotations leave r 2 = x 2 1 +x2 2 +x2 3 invariant; Lorentz transformations of the coordinates leaves invariant. The sign is a problem. Define (x 0, x), where x 0 = ct, s 2 = x 2 0 (x2 1 +x2 2 +x2 3 ), x i = (ict,x 1,x 2,x 3 ) i Then Euclidean metric (irrelevant overall sign) and Lorentz transformations become cosθ sinθ 0 0 x i = l ij x j, with L = sinθ cosθ

3 So ict = ictcosθ +xsinθ x = ictsinθ+xcosθ So comparing gives cosθ = γ, sinθ = iγv/c, or iθ = φ, tanhφ = v c, (Again φ is the rapidity variable.) This makes Minkowski diagrams apparently look simpler(just rotations in a 4-dimensional Euclidean space), but eventually must return to t Modern approach is not to use this, but to use covariant/contravariant vector formalism because for General Relativity essential Quantum Field Theory already has an i which can easily get mixed up with the additional i here The disadvantage is that it is more complicated The three dimensional experience Orthogonal axes e2 x e1 e3 Orthogonal co-ordinate system: e i e j = δ ij. Co-ordinates x i are given by x = x i e i x i = e i x. Rotation to a new co-ordinate system, e i = l ip e p e i = l pi e p e p l pi. Vectors: three cpts transforming according to a i = l ipa p Tensors: (second rank) nine cpts transforming according to T ij = l ip l jq T pq. 95

4 Oblique axes e2 x e3 e1 Base vectors not orthogonal e i e j = g ij. Then x i and x i defined by x = x i e i, and x i = e i x, are different x i = e i x = e i (x j e j ) = e i e j x j = g ij x j x i. Consider an arbitrary transformation e i = β p i e p. Note the index position and spaces in β p i. The Einstein summation convention is now applied to summing over one upper and one lower index. Define inverse transformation as e i = e pα p i. Again note the index position and spaces in α p i. Then relation between α, β is given by where e i = (βp j e j )α p i = α p i β p j = δ j i e i = β p i ( e j αj p ) = β p i αj p = δj i, δ j i = { 1 if i = j 0 otherwise 96

5 (no need to worry about position of indices here). Then x i = e i x = β p i e p x or x i = β p i x p. Also x = x p e p = xp e p = x p ( e j αj p ) giving x j = x p α j p or x i = α i p xp. Can also find inverse from x i = α i p xp or β j i x i = β j i αi p xp = δ j p xp = x j or Note that we also have Raising and lowering indices x i = x j β i j. x x = (x i e i ) (x j e j ) = g ij x i x j x i x i x i x i. For any vector a the a i (where a = a i e i ) and a i (defined by e i a) are called contravariant or covariant components of a respectively. They are related by a i = g ij a j. We might wish to invert this equation. Define g ij by g ij g jk = δk i. Then g ij a j = g ij g jk a k = δk iak = a i or a i = g ij a j. Then we can define e i = g ij e j, so a = a i e i = g ij a j e i = a i e i, ie g ij lowers indices and g ij raises indices. Transformation properties of g ij are given by g ij = e i e j = (β p i e p) (β q j e q), ie g ij = β p i β q j g pq. 97

6 Reciprocal basis As e i e j = (g ik e k ) e j = g ik g kj = δ i j, we can find an explicit form for e i the reciprocal basis. Now e 1 must be to e 2, e 3, so e 1 = c e 2 e 3. Now e 1 e 1 = 1 = c e 1 ( e 2 e 3 ) which yields c. Similarly for the others, so e 1 = e 2 e 3 e 1 ( e 2 e 3 ), e2 = e 3 e 1 e 1 ( e 2 e 3 ), e3 = e 1 e 2 e 1 ( e 2 e 3 ). Now e i = g ij e j gives e i e j = g ik e k e j = g ik δ j k = gij, and hence g ip = ( e j e k ) ( e q e r ) e 1 ( e 2 e 3 ) 2. where ijk and pqr are cyclic. Compare to the simplicity of g ip = e i e p. Note that obviously for an orthonormal basis, e 1 = e 1 etc... Example Define an oblique co-ordinate system by e 1 = 2 e x +3 e y + e z e 2 = e x 3 e y + e z e 3 = e y + e z. Find the expansion of a = 5 e x +6 e y +7 e z in this basis. From above the reciprocal basis is given by so e 1 = 1 6 (2 e x + e y e z ) e 2 = 1 6 (2 e x 2 e y +2 e z ) e 3 = 1 6 ( 4 e x + e y +5 e z ), a 1 = e 1 a = 3/2, a 2 = e 2 a = 2, a 3 = e 3 a = 7/2, and a 1 e 1 +a 2 e 2 +a 3 e 3 = 5 e x +6 e y +7 e z = a as expected. 98

7 8.1.2 Generalisation to four dimensions Straightforward, using previously developed machinery. In preparation for future use we introduce the conventions Greek indices run over spatial and time co-ordinates, ie 0, 1, 2, 3 Latin indices run over spatial co-ordinates, ie 1, 2, 3 Scalars A scalar is any quantity specified by a single number at each point, which has the same value in all co-ordinate systems. Vectors, tensors Any quantity specified specified by components such that a µ = α µ ν aν contravariant vector a µ = β ν µ a ν covariant vector with Second rank tensor: T µν = α µ λ αν ρt λρ T µν = βµ λ βν ρ T λρ T µ ν = α µ λ β ν ρ T λ ρ contravariant tensor covariant tensor mixed tensor α λ µ β ν λ = δν µ, β λ µ αν λ = δν µ. Observations If φ is invariant, then φ/ x µ is a covariant vector, for [x ν = x µ β ν µ ] ( φ x µ ) = φ φ x ν = x µ x ν x = β ν µ µ δ µ ν is a mixed (isotropic) tensor for δµ ν = β λ ν αµ ρ δρ λ φ x. ν Scalar products in form a b a µ b µ are invariant (a µ b µ, a µ b µ not invariants, will not arise) More generally, any vector or tensor combination has repeated indices in upper and lower positions 99

8 Can write in a matrix notation, for example T µν = (αtα T ) µν Quotient rule: If a set of components T µν is such that for all covariant vectors a µ the product T µν a ν is a contravariant vector then T µν is a tensor. The metric The co-ordinate transformation leaves invariant ds 2 = g µν dx µ dx ν. g µν is now a (symmetric) tensor called the metric tensor. Use g µν and g µν to raise/lower indices (as before). 8.2 The Special Relativity case Define 4-vectors and x µ = (x 0,x 1,x 2,x 3 ) (ct,x,y,z) (ct, r), ds 2 = c 2 dt 2 dx 2 dy 2 dz 2 = η µν dx µ dx ν, where instead of the more general symbol g µν we use η µν, the Minkowskian or West Coast metric with η µν = = ηµν The other metric in common use is diag( 1,1,1,1), the East Coast metric, so be careful. [a strong opinion topic] As η µν is the same in all frames, then we have the further constraint giving ds 2 = η µν dx µ dx ν = η µν (α µ λ dxλ )(α ν ρ dxρ ) η µν dx µ dx ν, η µν α µ λ αν ρ = η λρ = α λ ν α ν ρ = δ λ ρ. Thus comparing with previous result gives β = α Λ say. So we have η µν Λ µ λ Λν ρ = η λρ, as fundamental constraint on Lorentz transformations Λ. 100