ALL 9 CHAPTERS COVERED
SOLUTIONS MANUAL
,Solution Manual for Advanced Mechanics and
General Relativitỵ
Joel Franklin
Draft date August 25, 2010
,2
, Chapter 1
Newtonian Gravitỵ
Problem 1.1
From the Euler-Lagrange equations of motion:
µ ∂
d ∂L ∂L
— =0 (1.1)
dt ∂x˙ ∂x
we have
m ẍ + A = 0 −→ m ẍ = −A, (1.2)
and this corresponds to motion under the influence of a constant force A (for
example, A = m g). The constant B provides an energỵ offset to the potential
U = Ax + B, and has no dỵnamical effect.
Problem 1.2
t
a. For x(t) = xf
T
, we have:
Z T Z T ≥ x ¥2 1 x2f
1 1
S= m ẋ 2 dt = m f dt = m . (1.3)
0 2 0 2 T 2 T
P ≥ ¥
x t
∞ jπt
b. For x(t) = f
T
+ j=1 αj sin the action is:
T
õ !
Z T X∞
jπ ∂ ∞ µ ∂
x2 kπt
S = 1 m f +1m jπt X kπ
αj cos α cos dt.
2 T 2 0 j=1
T T k
T T
k=1
(1.4)
Noting that Z µ ∂ µ ∂
T jπt kπt 1
cos cos dt = T δjk, (1.5)
0 T T 2
3
SOLUTIONS MANUAL
,Solution Manual for Advanced Mechanics and
General Relativitỵ
Joel Franklin
Draft date August 25, 2010
,2
, Chapter 1
Newtonian Gravitỵ
Problem 1.1
From the Euler-Lagrange equations of motion:
µ ∂
d ∂L ∂L
— =0 (1.1)
dt ∂x˙ ∂x
we have
m ẍ + A = 0 −→ m ẍ = −A, (1.2)
and this corresponds to motion under the influence of a constant force A (for
example, A = m g). The constant B provides an energỵ offset to the potential
U = Ax + B, and has no dỵnamical effect.
Problem 1.2
t
a. For x(t) = xf
T
, we have:
Z T Z T ≥ x ¥2 1 x2f
1 1
S= m ẋ 2 dt = m f dt = m . (1.3)
0 2 0 2 T 2 T
P ≥ ¥
x t
∞ jπt
b. For x(t) = f
T
+ j=1 αj sin the action is:
T
õ !
Z T X∞
jπ ∂ ∞ µ ∂
x2 kπt
S = 1 m f +1m jπt X kπ
αj cos α cos dt.
2 T 2 0 j=1
T T k
T T
k=1
(1.4)
Noting that Z µ ∂ µ ∂
T jπt kπt 1
cos cos dt = T δjk, (1.5)
0 T T 2
3
