8.251 2005 Midterm Solutions
Question 1
z = x³
σ=0
y = x²
σ = σ₁
x = x¹
⎫
X 0 (τ, σ1 ) ⎬
X 1 (τ, σ1 )
all “Free”
⎭
X 2 (τ, σ1 )
X 0 (τ, 0),
X 1 (τ, 0),
X 2 (τ, 0),
X 3 (τ, 0) = a
X 3 (τ, σ1 ) = 0
�
Dirichlet
Question 2
y
i
x
(a)
z ∼z+i
F:
0≤y<1
M:
infinite cylinder
(b)
z ∼ 2z
1
This scales z, keeping the argument unchanged.
F : 1 ≤ |z| < 2
M:
torus
Technically we should also include the point z = 0.
1
2
Question 3
�
S=
�
δS =
�
=
�
δS =
d3 x (A0 F12 + A1 F20 + A2 F01 )
d3 x (δA0 F12 + A1 (∂2 δA0 ) − A2 ∂1 (δA0 ))
d3 x (δA0 F12 − δA0 ∂2 A1 + (∂1 A2 )δA0 )
d3 xδA0 (2F12 )
EOM:
F12 = 0
Question 4
δ = 8πGT0
δ should have no units with suitable c’s and �’s
(a) Units of GT0 ?
F =
GM 2
r2
[G] = [F ]
L2
L L2
=M 2 2
2
M
T M
L3
T 2 M
M L
[T0 ] = 2
T
[G] =
2
→ [GT0 ] =
δ=
L4
T4
thus need a c4
8πGT0
c4
(b) Mass density µ0 =
T0
c2
T0
c2
=
δ
·
=
c2
8πG
�
�
2π
�
0.5
360 · 60 · 60
�
�2
3 × 108
� · 4 · (6.67 × 10−11 )
2π
�
�
T0
kg
= 1.30 × 1020 !
c2
m
Question 5
r
dr
(a)
v = ωr
(dm)ωr
dp = �
2 2
1 − ωc2r
but dm =
dp =
dE
T0 dr
= 2
2
c
c
T0 ω
rdr
�
2
2 2
c
1− ω r
c2
dJ = rdp =
T0 ω
r2 dr
�
2
2 2
c
1− ω r
c2
(b)
�
J =2
0
l/2
T0 ω
r2 dr
�
2 2
c2
1− ω r
c2
ωr
c
=x r= x
c
ω
� 1 � �3
T0 ω
c
x2 dx
√
J =2 2 ·
c
ω
1 − x2
0
let:
3
1
x2 dx
π T0 c
√
=
2
2 ω2
1−x
� 0 ��
�
π
4
l
2c
ω =c ω=
2
l
2c
πT0 c
ω= 2 E =
E
πT
J =2
T0 c
ω2
�
0
π T0 cE 2
E2
J=
=
2 π 2 T02 c2
2πT0 c
Question 6
�
�
� (t, σ) = 1 F� (u) + G
� (�v )
X
2
(a)
� (t, σ + σ1 ) = X
� (t, σ)
X
� (v − σ1 ) = F� (u) + G
� (v)
F� (u, +σ1 ) + G
� (v) − G
� (v − σ1 )
F� (u + σ1 ) − F� (u) = G
(b)
�
�
� − �g (v − σ1 ) + β
� (v − σ1 )
f�(u + σ1 ) + α
� (u + σ1 ) − f�(u) − α
� (u) = �g (v) + βv
�
→
α
� =β
�
�
� (t, σ) = 1 f�(u) + αu
� + �g(u) + αv
�
X
2
�
1
1 ��
= α
� (u + v) +
f (u) + �g (v)
2
2
�
�
1
� (t, σ) = αct
� +
f�(u) + �g (v)
X
2
ασ
� 1 = β σ�1
(c)
T0 ∂�x
P�τ = 2
c ∂t
�
T
T0 � ��
0
P�τ =
α
�+
f (ct + σ) + g�� (ct − σ)
2c
� σ1 c
�
�
T0
T 0 σ1 ∂ � �
P�
P�τ dσ =
σ1 α
�+
f (ct + σ) − �g (ct − σ) dσ
c
2c 0 ∂σ
0
�
��
�
0, by periodicity of f and g
T0 σ1
P� =
α
�
c
Since σ1 =
E
T0 ,
we can also write:
E
P� = α
�
c
4
Question 7
E₂
O
E₁
P
Consider an arbitrary point P on the string. If each crawler travels L/4, the two endpoints E1
and E2 are separated along the ellipse by L/2. Moreover, they are opposite, and the midpoint of
E1 E2 is the origin. Thus P will go to the origin at t = L4 1c . Since P is arbitrary, the ellipse collapses
� �
to the origin after time L4 1c .
Comment: You may convince yourself that any curve C with inversion symmetry in the plane (if
�a ∈ C, then (−�a) ∈ C) will collapse to zero size.
5