Lecture 9
8.251 Spring 2007
Lecture 9 - Topics
• Change of variable
1. Change of Variables, 1 Variable
�
�
dx(u)
f�(u)
du
du
dx
f (x)dx = f�(u) du
du
u = u(x). Assume invertible function x(u)
f (x)dx =
f�(u) = f (x(u))
2 Variables of Integration
�
Let Mij =
f (ξ 1 , ξ 2 )dξ 1 ξ 2
∂ξ i
∂ξ j
dv1i =
∂ξ i �1
dξ
∂ ξ�1
dv2i =
∂ξ i �2
dξ
∂ξ�2
f (ξ 1 , ξ 2 )dξ 1 ξ 2 = f (ξ 1 , ξ 2 )dA
dA = |d�v1 ||d�v2 | sin θ
�
= |dv1 ||dv2 | − (d�v1 · d�v2 )
��
�� j j � � i i �2
∂ξ i ∂ξ i
∂ξ ∂ξ
∂ξ ∂ξ
=
−
dξ�1 dξ�2
1
2
�
�
�
�
1
2
∂ξ ∂ξ
∂ξ ∂ξ
∂ξ�1 ∂ξ�2
1
Lecture 9
8.251 Spring 2007
dA =
�
(Mi1 · Mi1 )(Mj2 · Mj2 ) − Mi1 Mi2 Mj1 Mj2 dξ�1 ξ�2
M1i = (M T )i1
dA =
�
(M T M )11 (M T M )22 − (M T M )212 dξ�1 dξ�2 =
�
det(M T M )dξ�1 dξ�2
det(M T M ) = det(M T )det(M )
dA = |det(M )|dξ�1 dξ�2
So:
i
∂ξ
f (ξ 1 , ξ 2 )dξ 1 dξ 2 = f�(ξ�1 , ξ�2 )|det( j )|dξ�1 dξ�2
∂ξ
�
√
The goal is to verify: A = dξ 1 dξ 2 g where g = det(gij ) is reparam. invariant.
gij (ξ) · dξ i dξ j = g�pq (ξ�)dξ�p dξ�q
� �p �� �q �
∂ξ
∂ξ
= g�pq (ξ�)
dξ i dξ j
i
∂ξ
∂ξ j
�ij =
Let M
∂ξ�i
∂ξj
�pi M
�pj
gij (ξ) = g�pq M
�T )ip g�pq M
�
qj
= (M
�T gM
= (M
� �
)ij
�T )gdet
�) = g|det(
�)|2
g = det(gij ) = det(M
� (M
�
M
�T ) = det(M
�)
det(M
�
A=
√
dξ 1 dξ 2 g =
�
�
�)
dξ�1 dξ�2 det(u) gdet(
�
M
�)ij = Mik M
�ki =
(M M
∂ξ i ∂ξ�k
∂ξ i
= j
j
∂ξ
∂ξ�k ∂ξ
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�1 dξ�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: (ξ 1 , ξ 2 ) → (τ, σ) where τ is “like time” and σ is “like time”.
Target space:
xµ = (x0 , x1 , . . . , xd )
D = d + 1 = space time dimension. d = spatial dimension.
Mapping:
xµ (τ, σ) = X µ (τ, σ)
X µ sometimes called “string coordinates”. σ has a finite range. For a closed
string, periodic. τ can have an infinite range.
Area constructed from: dv1µ =
By analogy: dA =
�
dX µ
dτ dτ ,
dv2µ =
dX µ
dσ dσ.
(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:
X 0 (τ, σ) = cτ
X i (τ, σ) = f i (σ)
dv1µ has only µ = 0 component.
� 0 components.
dv2µ has only µ =
dv1 · dv1 < 0
dv2 · dv2 > 0
dv1 · dv2 = 0
Therefore:
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Lecture 9
8.251 Spring 2007
dA =
√
<0
So instead:
��
�2 �
��
�
∂X ∂X
∂X ∂X
−
·
·
∂τ ∂τ
∂σ ∂σ
��
�
�
��
�
2
∂X µ ∂Xµ
∂X µ ∂Xµ
−
···
= dτ dσ
∂τ ∂σ
∂τ ∂τ
dA = dτ dσ
∂X ∂X
·
∂τ ∂σ
Consider worldline xµ (τ ):
dxµ
dτ
is timelike. Particle moves slower than light.
Consider point P on worldsheet of a string. Worldsheet described by τ and σ
so:
Xp = X µ (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.
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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
∂X µ
∂X µ
∂τ (P ), ∂σ (P ).
Consider 1 parameter family of vectors.
∂X µ
∂X µ
+λ
∂τ
∂σ
Linear combination of ∂X µ /∂τ and ∂X µ /∂σ with coefficients 1&λ. Most agreed
would have 2 arbitrary coefficients, but here only care about direction.
v µ (λ) =
2
v (λ) = λ
2
�
∂X
∂σ
�2
�
� �
�2
∂X ∂X
∂X
·
+∂x
+
∂σ ∂τ
∂τ
Get quadratic equation for λ
λ=
−b ±
√
b2 − 4ac
2a
5
Lecture 9
8.251 Spring 2007
If b2 − 4ac ≤ 0 will get complex (not real) roots. So b2 − 4ac > 0.
6