Lecture 16
8.251 Spring 2007
Lecture 16 - Topics
• Light-cone fields and particles
Reading: Sections 9.5-10.1
Space-filling D-brane
xµ (τ, σ) = xµ0 +
�1
√
√
2α� α0µ τ + i 2α�
αµ e−inτ cos(nσ)
n n
n=0
�
α0µ =
√
2α� pµ
√
µ
α−n
= naµn
√
αnµ = nanµ
n · x(τ, σ) = βα� (n · p)τ
n · P τ constant along string.
ẍµ − xµ�� = 0
1 ∂xµ
Pµσ = −
2πα� ∂σ
1 ∂xµ
Pµτ =
2πα� ∂τ
(ẋ ± x� )2 = 0
ẋµ (τ, σ) =
�
√
√
2α� α0µ + 2α�
aµn e−inτ cos(nσ)
n=0
�
�
√
= 2α�
αnµ e−inτ cos(nσ)
n∈Z
�
√
x�µ (τ, σ) = −i 2α�
αnµ e−inτ sin(nσ)
√
n=0
�
= −i 2α�
�
αnµ e−inτ sin(nσ)
n∈Z
ẋµ ± x�µ =
�
√
2α�
αnµ e−in(τ ±σ)
n∈Z
√
√
Let ηµ = (1/ 2, 1/ 2, 0, 0, . . .)
1
Lecture 16
8.251 Spring 2007
1
n · x = √ (x0 + x1 ) = x+
2
n · p = p+
x+ (τ, σ) = βα� p+ τ
Open Strings:
x+ = 2α� p+ τ
x+
0 =0
+
αn+ = α−n
for n = 1, . . . , ∞
Coordinates of string: (x+ , x− , xI ) I = 2, 3 . . . , d. Each xI has associated xI0 ,αnI .
Remarkable statement: Can generate fulll solution of wave equation using the
constraints.
Recall: a · b = −a+ b− − a− b+ + aI bI , (ẋ ± x� )2 0
−∂(ẋ+ ± x+� )(ẋ− ± x−� ) + (ẋI ± xI � )2 = 0
Here x− appears linearly with number coefficient, so we can solve for x− and
get string motion.
(ẋ− ± x−� ) =
1 1
(ẋI ± xI � )2
2 2α� p+
Determines for you ẋ− and x−� . If you know x− (P ) can get x− (Q) where P
and Q are any two points.
∂σ ẋ− = ∂τ x−�
For closed strings, paths from P to Q not deformable. Think about? Is true an
extra constraint?
Let’s solve this finally:
ẋ− ± x−� =
√
2α�
�
αn− e−in(τ ±σ)
n∈Z
�
1 1
+
�
=
1p
(2α
)
αpI αqI e[−i(p+q)(τ ±σ)]
2 2α�
p,q∈Z
1 �1� I I
= +
αp αn−p
e(−in(τ ±σ))
p
2
n∈Z
p∈Z
2
Lecture 16
8.251 Spring 2007
√
1 � I I
αp αn−p
2α� αn− = +
2p
p∈Z
Now solved motion of open string for any constraints.
Transverse Virasora Mode
L⊥
n =
1� I I
αp αn−p
2
p∈Z
√
1
2α� αn− = + L⊥
p n
All of 2D field theory based on Virosora algebra. Show up everywhere in math
and physics. Here we’ve seen the origins.
√
2α� p− = α0−
α−
2p+ √ � −
1
2p+ p− = 2p+ √ 0 =
( 2α α0 ) = � L⊥
0
�
�
2
α
α
2α
1
M 2 = −p2 = 2p+ p− − pI pI = � L+
− pI pI
α 0
1
M = �
α
2
� �
�
1
I I
αp α−p −p
I pI
2
p∈Z
∞
�
1
I I
=
(α α +
αI αI ) − pI pI
2α� 0 0 n=1 n −n
α0I =
M2 =
√
2α� pI
∞
1 � I∗ I
na a
α� n=1 n n
If no constants of oscillation, then M = 0, string collapses to a point. Classical
string theory solved! Most complete solution of equation of motion of string.
But need other things with no mass - photons, gluons, gravitons.
Prepare ground for quantization of strings.
3
Lecture 16
8.251 Spring 2007
EM fields ⇒ “photon”, spin-1 field, Aµ (x)
Gravity ⇒ “graviton states”, spin-2 field, hµν (x)
Scalar fields just have Φ(x) ∈ �. Simplest of all! But might not exist in nature.
(“Scalar” under Lorentz symmetry)
For E&M fields:
� B
� ) ⇒ Aµ (x) ⇒ F µ ⇒ S =
(E,
1
4
�
F2
This took a lot of work! Will be very easy for scalar fields Φ(t, �x)
1 ∂φ ∂φ
2 ∂x0 ∂x0
1
= M 2 φ2
2
KEdensity =
PEdensity
L=
1 ∂φ ∂φ
1 ∂φ ∂φ
1
−
− M 2 φ2
2 ∂x0 ∂x0
2 ∂xi ∂xi
2
4