Lecture 14
8.251 Spring 2007
Lecture 14 - Topics
• Momentum charges for the string
• Lorentz charges for the strings
• Angular momentum of the rotating string
• Discuss α� and the string length �s
• General gauges: Fixing τ and natural units
Reading: Section 8.4-8.6 and 9.1
�
S=
dξ 0 dξ 1 . . . dξ p L(φa , ∂α φa )
ξ α : coordinates, φa (ξ): fields, ∂α =
∂
∂ξ α
α: coord. index α = 0, 1, . . . , p
a: field index a = 1, . . . , m
i: index for various symmetries
δφa (ξ) =i hai (φ(ξ))
Leaves L invar. to first order.
1.
∂L a
∂L
δφ +
∂α (δφa ) = 0
∂φa
∂(∂α φa )
3.
J µ → Jiα → i Jiα =
Similar to mechanics:
∂L
δφa
∂(∂α φa )
∂L
∂ q̇ δq
Claim: Given this transformation leaves L invar. to first order then:
2.
∂α Jiα = 0 ∀i Conserved Current
Check this yourself using 1. and the E − L equations of motion. Done in book
as well.
Conserved charge too:
�
Qi =
JiO (ξ)dξ 1 dξ 2 . . . dξ p
Answer independent of time.
1
Lecture 14
8.251 Spring 2007
dQ
=0
dξ O
Nambu-Gotta action:
�
dξ 0 dξ 1 L(∂0 xµ , ∂1 xµ )
���� ����
S=
dτ
a
dσ
µ
This means α = 0, 1. φ = x ⇒ a = 0, . . . , d = spatial dimension
Let’s look for asymmetry. A variation of the field that leaves the field invar.
δxµ =µ = constant
Constant translations of a worldsheet should by asymmetric. Why would Nambu-
Gotta action care if rigidally moved worldsheet through time or space?
So:
δ(∂0 xµ ) = ∂0 (δxµ ) = 0
δ(∂1 xµ ) = 0
So δxµ =µ indeed asymmetric.
Apply (3)
µ
Jµα =
∂L µ
∂(∂α xµ )
∂L
∂(∂α xµ )
�
�
∂L ∂L
0
1
(Jµ , Jµ ) =
,
= (Pµτ , Pµσ )
∂ẋµ ∂x�µ
Jµα =
Conservation law: ∂α J α = 0 gives us collection of conservation laws for µ.
∂Pµτ
∂Pµa
+
∂τ
∂a
�σ
Pµ (τ ) = 0 1 Pµτ (τ, σ)dσ conserved quantity indexed by spacetime index µ.
Pµ (τ ): conserved momentum for the string not dependent on τ since conserved.
∂α J α = 0 =
Check Pµ is conserved
∂Pµτ
(τ, σ)dσ
∂τ
0
� σ1
∂Pµσ
=−
dσ
∂σ
0
= [−Pµσ ]σ0 1
dPµ
=
dτ
�
σ1
2
Lecture 14
8.251 Spring 2007
This yields the free BCs.
This is the hardest part of the course. After this, it gets easier.
A momentum is in general a variation of a Lagrangian with respect to a velocity
eg ∂∂ẋLµ conserved, has units of momentum. We will see this is indeed the relative
momentum of a piece of string.
When we had (ρ, J�):
[ρ] = LQ3
[J�] = TQL2
Now we have (Pµτ , Pµσ ):
P
[Pµτ ] = Lµ
P
[Pµσ ] = Tµ
Call Pµτ momentum density, and Pµσ momentum current.
Okay, we have:
dPµ
=0
dτ
But would like:
dPµ
=0
dt
Conserved for Lorentz observer. Is this the case? (Yes).
dP
Sure, could work in static gauge. τ = t ⇒ dtµ = 0
But what about an arbitrary τ curve on worldsheet?
Look for a generalization formula (clue from divergence theorem)
A =flux of vector field
3
Lecture 14
8.251 Spring 2007
�
x
� �
y
(A dy − A dx) =
R
�
�
[Pµτ dσ − Pµσ dτ ] =R
Γ
�
∂Ax
∂Ay
+
dxdy
∂x
∂y
�
∂P τ
∂P σ
+
dτ dσ = 0
∂τ
∂σ
Given an arbitrary curve γ, claim momentum given by:
�
[Pµτ dσ − Pµσ dτ ]
Pµ (γ) =
γ
γ = α → γ2 → β → −γ1
�
(
�
�
+
γ2
�
+
α
+
) (Pµτ dσ − Pµσ dτ ) = 0
�
��
�
−γ1
β
�
�
κ
κ=
κ=0
α
β
Pµ (γ1 ) = Pµ (γ2 )
Usually will use Pµ (τ ) =
general formulation.
� σ1
0
Pµτ (τ, σ)dτ with constant τ , but nice to have this
4
Lecture 14
8.251 Spring 2007
Lorentz Transformation
xµ = Lµν xν
Leaves ηµν xµ xν invar. Vary xµ subject to xν
δxµ =µα xα
µα
δ(ηµν xµ xν ) = 2ηµν
xα x ν
= 2µα xα xµ
where ηµν xν = xµ
If we want δ(ηµν xµ xν ) = 0, we make antisymmetric.
µν
= −νµ
Claim:
δ(ηµν ∂α xµ ∂β xν ) = 0
So Nambu-Gotta action invar and get new set of symmetries:
δxµ (τ, σ) =µν xν (τ, σ)
µν
α
Jµν
=
µν
∂L
δxµ = Pµα δxµ = Pµαµν xν
∂(∂α xµ )
α
Jµν
=−
1 µν
(xµ Pνα − xν Pµα )
2
No physical relevance to − 12
So define:
α
α
mα
µν (τ, σ) = xµ Pν − xν Pµ
α
Conserved currents: ∂α mµν
=0
σ1
�
m0µν dσ
Mµν =
0
Conserved Charge
�
Mij =
�
mij dσ =
0
123
σ
σ1
(xi Pjτ − xj Piτ )dσ =ijk Lk
0
= +1, totally antisymmetric. eg:
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Lecture 14
8.251 Spring 2007
�
M12 =
σ1
(x1 P2τ − x2 P1τ )dσ =12l Lk = L3
0
� = �r × p�
L
So Mij =angular momentum, conserved.
Angular momentum of rotating string:
σ1
�
(x1 P2τ − x2 P1τ )dσ
M12 = L3 = J =
0
�
σ1
πσ
�x(t, σ) =
cos
π
σ1
��
�
�
�
��
πct
πct
cos
, sin
σ1
σ1
Parametrized String
�
�
�
�
��
� τ = T0 ∂�x = T0 cos πσ − sin πct , cos πct
P
c2 ∂t
c
σ1
σ1
σ1
� �2
� �
σ1 T0
πσ
x1 P2 − x2 P1 =
cos2
π
c
σ1
J=
1
E2
2πT0 c
E
= σ1
T0
6