String Theory- Change of Variable

� � � Lecture 9 8.251 Spring 2007 Lecture 9 -Topics • Change of variable 1. Change of Variables, 1 Variable dx(u)f(x)dx = f�(u) du du dx f(x)dx = f�(u) du du u = u(x). Assume invertible function x(u) f�(u)= f(x(u)) 2 Variables of Integration f(ξ1,ξ2)dξ1ξ2 ∂ξi Let Mij = ∂ξj dv1 i = ∂∂ξξ�1 i dξ�1 dv2 i = ∂ξi dξ�2 ∂ξ�2 f(ξ1,ξ2)dξ1ξ2 = f(ξ1,ξ2)dA dA = |d�v2sin θ � v1||d�| = |dv1||dv2− (d�v1 · d�v2)�� | �� �� �2∂ξi ∂ξi ∂ξj ∂ξj ∂ξi ∂ξi dξ�1dξ�2= ∂ξ�1 ∂ξ�1 ∂ξ�2 ∂ξ�2 − ∂ξ�1 ∂ξ�2 1 � � � � �� � � �� � Lecture 9 8.251 Spring 2007 dA = (Mi1 · Mi1)(Mj2 · Mj2) − Mi1Mi2Mj1Mj2dξ�1ξ�2 M1i = (MT )i1 dA =(MT M)11(MT M)22 − (MT M)2 dξ�1dξ�2 = det(MT M)dξ�1dξ�2 12det(MT M)= det(MT )det(M) dA = |det(M)|dξ�1dξ�2 So: ∂ξi f(ξ1,ξ2)dξ1dξ2 = f�(ξ�1 ,ξ�2)|det( ∂ξj )|dξ�1dξ�2 The goal is to verify: A = � dξ1dξ2√g where g = det(gij ) is reparam. invariant. gij (ξ) dξidξj = g�pq (ξ�)dξ�pdξ�q · ∂ξ�p ∂ξ�q = g�pq (ξ�) dξidξj ∂ξi ∂ξj Let �= ∂ξ�iMij ∂ξj gij (ξ)= �gpqMpiMpj=( ��MT )ipg�pq MqjMT ��=( �gM)ij g = det(gij )= det(M�T )gdet�( �g|det( �|2M)= �M)det( �M)MT )= det( �A = dξ1dξ2√g = dξ�1dξ�2det(u) ��M)gdet( �∂ξi ∂ξ�k ∂ξi (M ��==M)ij = MikMki ∂ξ�k ∂ξj ∂ξj If i =�j, this equals 0. If i = j, this equals 1. Therefore, we have δji . 2 � � � Lecture 9 8.251 Spring 2007 det(M)= det( �M) A = dξ�1dξ�2 �g Goal: Write area functional for spacetime surface. Just did this for a surface in Euclidean space. Now do for a surface in Mintowsk space (so there’s a negative sign instead of all positive signs). Change of notation: (ξ1,ξ2)(τ,σ) where τ is “like time” and σ is “like time”. → Target space: xµ =(x 0 ,x 1 ,...,x d) D = d + 1 = space time dimension. d = spatial dimension. Mapping: xµ(τ,σ)= Xµ(τ,σ) Xµ sometimes called “string coordinates”. σ has a finite range. For a closed string, periodic. τ can have an infinite range. dXµ dXµArea constructed from: dv1 µ = dτ dτ , dv2 µ = dσ dσ. By analogy: dA =(dv1 · dv1)(dv2 · dv2) − (dv1 · dv2)2? The problem is that the number under the square root is less than 0, and we don’t want an imaginary dA! Static String: X0(τ,σ)= cτ Xi(τ,σ)= fi(σ) dv1 µ has only µ = 0 component. dv2 µ has only µ = 0 components. �dv1 · dv1 < 0 dv2 · dv2 > 0 dv1 · dv2 =0 Therefore: 3 �� Lecture 9 8.251 Spring 2007 dA = √< 0 So instead: ��2 � �� � ∂X ∂X ∂X ∂X ∂X ∂X dA = dτdσ ∂τ · ∂σ − ∂τ · ∂τ ∂σ · ∂σ ��2 � ��� ∂Xµ ∂Xµ ∂Xµ ∂Xµ= dτdσ ∂τ ∂σ − ∂τ ∂τ ··· Consider worldline xµ(τ): dxµ is timelike. Particle moves slower than light. dτ Consider point P on worldsheet of a string. Worldsheet described by τ and σ so: Xp = Xµ(τP ,σP ) If all points P on string ∃ a point P � on string at time Δt. uP �,P is timelike then string moving slower than light. The worldsheet is the area swept out by the string over time. 4 Lecture 9 8.251 Spring 2007 1. ∀P ∃ spacelike tangent 2. Just saw ∀P ∃ timelike tangent too. Will use 1 and 2 to show dA = √> 0. Consider tangent vectors at P spanned by ∂Xµ (P ), ∂Xµ (P ).∂τ ∂σ Consider 1 parameter family of vectors. ∂Xµ ∂Xµ vµ(λ)= + λ ∂τ ∂σ Linear combination of ∂Xµ/∂τ and ∂Xµ/∂σ with coefficients 1&λ. Most agreed would have 2 arbitrary coefficients, but here only care about direction. ��2 � ���2 v 2(λ)= λ2 ∂X +∂x ∂X ∂X + ∂X ∂σ ∂σ · ∂τ ∂τ Get quadratic equation for λ −b ±√b2 − 4ac λ = 2a 5 Lecture 9 8.251 Spring 2007 If b2 − 4ac ≤ 0 will get complex (not real) roots. So b2 − 4ac > 0. 6