Lecture 5
8.251 Spring 2007
Lecture 5 - Topics
• Nonrelativistic strings
• Lagrangian mechanics
Reading: Zwiebach, Chapter 4
Non-Relativistic Strings
Study nonrelativistic strings first to develop intuition and math notation before
moving to the relativistic strings that we actually care about.
Non-relativistic string:
Characterized by:
Tension, T0 : [T0 ] = [Force] = [Energy/Length] =
Mass/Length: µ0
T0 ≈ µ0 v
2
�
Natural velocity: v =
T0 /µ0
M
L
[v 2 ]
Transverse Oscillation: Mark point P on string and see it moving up and down:
y(P, t), x(P, t) = x(P )
(x not dependent on t)
Small Oscillation:
�
�
� ∂y
�
� (t, x)�<< 1
� ∂x
�
Consider small section of string:
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Lecture 5
8.251 Spring 2007
Approximate tensions on endpoints as equal (good for transverse waves, terrible
for longitudinal)
∂y
∂y
(t, x + dx) − T0 (t, x)
∂x
∂x
∂ 2 y
= T0 2
(t, x)dx
∂x
∂ 2 y
≈ µ0 dx 2
∂t
dFν = T0
∂2y
1 ∂2y
−
=0
2
∂x
T0 /µ0 ∂t2
The Wave Equation! t, x are parameters. Motion described by y(t, x). (If had
motion in more than 1 dimension �y (t, x))
Stretching of string:
�
dx2 + dy 2 − dx
�
= dx( 1 + (dy/dx)2 − 1)
1
= dx(dy/dx)2
((small))
2
Δl =
General form of wave equation:
∂2f
1 ∂2f
−
=0
∂x2
v 2 ∂t2
v: velocity of wave, v =
�
T0 /µ0
General Solution:
y(x, t) = h+ (x − v0 t) + h− (x + v0 t)
Note: the h’s are function of 1 variable (x ± v0 t) not 2 variables x and t inde­
pendently.
Boundary Conditions: Behavior of endpoints at all times (special points at all
times)
Open string:
y(t, x = 0) = 0
(Dirichlet condition - for fixed end point)
∂y
(t, x = 0) = 0
∂x
(Free BD, Neumann condition)
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Lecture 5
8.251 Spring 2007
For free endpoint (hoop on string), means string must be perp. here
Initial Conditions: All points on string at some t0 (all points at special time)
y(λ, t = 0)
∂y
(x, t = 0)
∂x
Example:
Fixed Endpoints:
y(t, 0) = h+ (−v0 t) + h− (v0 t) = 0
Let u = v0 t
= h+ (−u) + h− (u)
h− (u) = −h+ (−u)
y(t, x = a) = 0 = h+ (a − v0 t) + h− (a + v0 t)
h+ (a − v0 t) = −h− (a + v0 t) = h+ (−a − v0 t)
Let u = −a − v0 t
h+ (u + 2a) = h+ (u)
Variational Principle
Consider point mass m doing 1D motion x(t).
Assume x(ti ) = xi , x(tf ) = xf . Under the influence of potential V (x)
Know:
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Lecture 5
8.251 Spring 2007
Possible motions:
Not possible:
Given a path:
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Lecture 5
8.251 Spring 2007
Functional: S : x(t) ⇒ � (not a function of time)
Hamilton’s Principle: Principal path makes S stationary.
Call true path x(t). Consider new path x(t) + δx(t)
S[x(t) + δx(t)] = S[x] + θ[(δx)2 ]
Assume δx(ti ) = 0, δx(tf ) = 0
Lagrangian:
L(t) = Kinetic Energy - Potential Energy
t2
t2 �
�
1
2
S=
L(t)dt =
m(ẋ(t)) − V (x(t)) dt
2
t1
t1
�
� tf �
1
2
S[x + δx] =
m(ẋ + δẋ) − V (x + δx) dt
2
ti
�
� tf �
� tf
∂V
1
1
2
= S[x] +
mxδ
˙ x˙ −
(x(t)δx(t)) dt +
m(δ x(t))
˙
− V �� (δx)2
2
∂x
2
ti
t
� i
��
�
�
�
θ(δx2 )
Need to eliminate second term.
� tf
[mxδ
˙ x˙ −
be true.
ti
∂V
∂x
(x(t)δ(x(t)))]dt must go away for S[x + δx] = S[x] + θ[(δx)2 ] to
Call this the variation δS
�
� tf �
d
δS =
dt
(mxδx)
˙
− m¨
xδx − V � (x(t))δ(x(t))
dt
ti
Integrate by parts
� tf
dS = mẋ(tf )δx(tf ) − mẋ(ti )δx(ti ) +
dtδx(t)[−mẍ − V � (x(t))]
ti
δx(tf ) = δ(ti ) = 0 from before.
The integral
� tf
ti
dtδx(t)[−mẍ − V � (x(t))] must be 0 too, so:
mẍ = −V � (x(t))
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Lecture 5
8.251 Spring 2007
String Lagrangian
�
T : Kinetic energy =
Potential Energy =
1
2 µ0 dx
�
L=
0
a
�2
ΔlT0 =
string
�
∂y
∂t
�
�a
1
dx
0 2
∂y
∂x
�2
T0
�
�
1
1
2
2
dx µ0 (∂y/∂t) − T0 (∂y/∂t)
2
2
� tf
S=
L(t)dt
ti
Call L: Lagrangian Density
L=
1
∂y
1 ∂y
µ0 ( )2 − ( )
2
∂t
2 ∂t
So:
�
tf
S=
a
�
dt
ti
0
�
�
∂y ∂y
dxL
,
∂t ∂x
δy(ti , x) = 0
δy(tf , x) = 0
Don’t know δy(x = 0, t) or δy(x = a, t)
�
� tf � a �
∂L
∂L
δS =
dt
dx
δẏ + � δy �
∂ẏ
∂y
ti
0
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Lecture 5
8.251 Spring 2007
Let:
P t = ∂L/∂ẏ
P x = ∂L/∂y �
�
tf
δS =
ti
�
tf
δS =
a
�
dt
ti
0
�
� a�
t ∂(δy)
x ∂(δy)
P
+P
∂t
∂x
0
�
� t
�� � a
� tf
∂P
∂P x
t
P x [δy]x=a
dx −δy(x, t)
+
+
dxP t [δy]tfi +
x=0
∂t
∂x
0
ti
δy(ti ) = δy(tf ) = 0
Must have:
∂P t
∂P x
∂2y
∂2y
+
= 0 = µ0 2 − T0 2
∂t
∂x
∂t
∂x
Some kind of conservation law like ∂µ J µ = 0
�
tf
dtP
x
[δy]x=a
x=0
=
ti
�
tf
dt[P x (t, x = a)δy(t, x = a) − P x (t, x = 0)δy(t, x = 0)]
ti
For ∗ ∈ 0, a:
P x (t, x∗ )δy(t, x∗ )
Dirichlet condition:
y(t, x∗ ) = fixed, δy(t, x∗ ) = 0
Free boundary condition:
P x (t, x∗ ) = 0, ∂y/∂x = 0 (Neumann condition)
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