Вы находитесь на странице: 1из 7

# Classical Mechanics - Homework Assignment 10

## Alejandro Gomez Espinosa

December 5, 2012
Goldstein, Ch.10, 6 A charged particle is constrained to move in a plane under the inuence of a central
force potential (nonelectromagnetic) V =
1
2
kr
2
, and a constant magnetic eld B perpendicular to the
plane, so that
A =
1
2
Br
Set up the Hamilton-Jacobi equation for Hamiltons characteristic function in plane polar coordinates.
Separate the equation and reduce it to quadratures. Discuss the motion if the canonical momentum p

is zero at time t = 0.
Let us start with the potential energy due to the constant magnetic eld. We can write down A
explicitly:
A =
1
2
Br =
1
2
(B z) (x x + y y + z z) =
B
2
(x y y x)
Then, the potential due to this eld is given by:
V
B
=
q
c
( r A)
=
qB
2c
( yx xy)
=
qB
2c
((r cos )( r sin + r

cos ) (r sin )( r cos r

sin ))
=
qB
2c
r
2

(1)
Hence, the Lagrangian of this system in planar polar coordinates is:
L =
m
2
( r
2
+ r
2

2
) +
qB
2c
r
2

1
2
kr
2
(2)
Let us use (2) to found the Hamiltonian of the system:
p
r
=
L
r
= m r r =
p
r
m
p

=
L

= mr
2

+
qBr
2
2c

=
p

mr
2

qB
2mc

gomez@physics.rutgers.edu
1
H = p
r
r + p

L
= m r
2
+ mr
2

2
+
qBr
2

2c

m
2
( r
2
+ r
2

2
)
qB
2c
r
2

+
1
2
kr
2
=
m
2
( r
2
+ r
2

2
) +
1
2
kr
2
=
m
2
_
_
p
r
m
_
2
+ r
2
_
p

mr
2

qB
2mc
_
2
_
+
1
2
kr
2
=
p
2
r
2m
+
1
2mr
2
_
p

qBr
2
2c
_
2
+
1
2
kr
2
Now, using the Hamilton-Jacobi formalism:
H
_
q
1
, ..., q
n
,
S
q
1
, ...,
S
q
n
_
+
S
t
= 0
where our old momenta are p
i
=
S
q
k
. Therefore:
1
2m
_
S
r
_
2
+
1
2mr
2
_
S

qBr
2
2c
_
2
+
1
2
kr
2

S
t
= 0 (3)
From equation (1) is easy to see that is a cyclic coordinate, i.e., p

## must be constant. In addition, as

our system is conservative, the energy must also be a constant. Thus, the function S has to have the
form S = W(r, E, ) + Et according to eq. (10.38). Now, equation (3) can be written as:
1
2m
_
W
r
_
2
+
1
2mr
2
_

qBr
2
2c
_
2
+
1
2
kr
2
= E (4)
where we can solve for W:
W =
_
dr

2mE mkr
2

1
r
2
_

qBr
2
2c
_
2
(5)
and S:
S =
_
dr

2mE mkr
2

1
r
2
_

qBr
2
2c
_
2
+ Et (6)
Since at t = 0, the canonical momentum is zero, i.e., = 0, and equation (5) becomes:
W =
_
dr
_
2mE mkr
2

qBr
2
2c
=

2mE
_
dr

1
_
k
2E

qBr
4mEc
_
r
2
that is similar to equation (10.22) for the simple harmonic oscillator, where we can think about the
term inside the parenthesis as the frequency of the particle due to the nonelectromagnetic potential
and the constant magnetic eld.
2
Goldstein, Ch.10, 7 (a) A single particle moves in space under a conservative potential. Set up the
Hamilton-Jacobi equation in ellipsoidal coordinates u, v, dened in terms of the usual cylindrical
coordinates r, z, by the equations
r = a sinh v sin u, z = a cosh v cos u
For what forms of V (u, v, ) is the equation separable?
T =
m
2
_
r
2
+ r
2

2
+ z
2
_
(7)
where r and z in terms of ellipsoidal coordinates described above are:
r = a cosh v sin u v + a sinh v cos u u
z = a sinh v cos u v a cosh v sin u u
Plugging this velocities and the coordinates into equation (7):
T =
m
2
_
(a cosh v sin u v + a sinh v cos u u)
2
+ r
2

2
+ (a sinh v cos u v a cosh v sin u u)
2
_
=
m
2
_
a
2
cosh
2
v sin
2
u v
2
+ a
2
sinh
2
v cos
2
u u
2
+ r
2

2
+ a
2
sinh
2
v cos
2
u v
2
+ a
2
cosh
2
v sin
2
u u
2
+ r
2

2
_
=
m
2
_
a
2
cosh
2
v sin
2
u( v
2
+ u
2
) + a
2
sinh
2
v cos
2
u( u
2
+ v
2
) + a
2
sinh
2
v sin
2
u

2
_
=
ma
2
2
_
( v
2
+ u
2
)((1 + sinh
2
v) sin
2
u + sinh
2
v cos
2
u) + sinh
2
v sin
2
u

2
_
=
ma
2
2
(( v
2
+ u
2
)(sin
2
u + sinh
2
v) + sinh
2
v sin
2
u

2
)
The Lagrangian for this system can now be written as:
L =
ma
2
2
(( v
2
+ u
2
)(sin
2
u + sinh
2
v) + sinh
2
v sin
2
u

2
) V (u, v, ) (8)
Hence, the momenta are:
p
v
=
L
v
= ma
2
(sin
2
u + sinh
2
v) v
2
p
u
=
L
u
= ma
2
(sin
2
u + sinh
2
v) u
2
p

=
L

= ma
2
sin
2
usinh
2
v v
and the Hamiltonian is given by:
H =
p
2
v
2ma
2
(sin
2
u + sinh
2
v)
+
p
2
u
2ma
2
(sin
2
u + sinh
2
v)
+
p
2

2ma
2
sin
2
usinh
2
v
+ V (u, v, )
In the Hamilton-Jacobi formulation, the function S would be S = W
u
+ W
v
+ W

## Et, and the

Hamilton-Jacobi equation is:
_
Wu
u
_
2
2ma
2
(sin
2
u + sinh
2
v)
+
_
Wv
v
_
2
2ma
2
(sin
2
u + sinh
2
v)
+
_
W

_
2
2ma
2
sin
2
usinh
2
v
+ V (u, v, ) = E (9)
3
Let us nd out the way to make equation (9) separable:
1
2ma
2
_
W
u
u
_
2
+
1
2ma
2
_
W
v
v
_
2
+
(sin
2
u + sinh
2
v)
2ma
2
sin
2
usinh
2
v
_
W

_
2
+V (sin
2
u+sinh
2
v) = E(sin
2
u+sinh
2
v)
1
2ma
2
_
W
u
u
_
2
+
1
2ma
2
_
W
v
v
_
2
+
1
2ma
2
_
1
sin
2
u
+
1
sinh
2
v
__
W

_
2
+V (sin
2
u+sinh
2
v) = E(....)
(10)
where if we take the potential as:
V (u, v, ) =
V

()
sin
2
usinh
2
v
the Hamilton-Jacobi equation becomes separable:
1
2ma
2
_
W
u
u
_
2
+
1
2ma
2
_
W
v
v
_
2
+
1
2ma
2
_
1
sin
2
u
+
1
sinh
2
v
__
W

_
2
+V

_
1
sin
2
u
+
1
sinh
2
v
_
= E(....)
1
2ma
2
_
W
u
u
_
2
+
1
2ma
2
_
W
v
v
_
2
+
1
2ma
2
_
_
W

_
2
+ V

_
_
1
sin
2
u
+
1
sinh
2
v
_
= E(....)
1
2ma
2
_
_
W

_
2
+ V

_
_
1
sin
2
u
+
1
sinh
2
v
_
= E(....)
1
2ma
2
_
W
u
u
_
2

1
2ma
2
_
W
v
v
_
2
1
2ma
2
_
_
W

_
2
+ V

_
=
_
1
sin
2
u
+
1
sinh
2
v
_
1
_
E(....)
1
2ma
2
_
W
u
u
_
2

1
2ma
2
_
W
v
v
_
2
_
(b) Use the results of part (a) to reduce to quadratures the problem of a point particle of mass m moving
in the gravitational eld of two unequal mass points xed on the z axis a distance 2a apart.
To solve this part, let us compute rst the distance between one of the xed mass points and the
mass m:
(r + a z) = r
2
+ (z + a)
2
= a
2
sinh
2
v sin
2
u + a
2
(cosh v cos u + 1)
2
= a
2
_
sinh
2
v sin
2
u + cosh
2
v cos
2
u + 2 cosh v cos u + 1
_
= a
2
_
sinh
2
v sin
2
u + (1 + sinh
2
v)(1 sin
2
u) + 2 cosh v cos u + 1
_
= a
2
_
sinh
2
v sin
2
u + 1 + sinh
2
v sin
2
u sinh
2
v sin
2
u + 2 cosh v cos u + 1
_
= a
2
_
sinh
2
v + 1 + 1 sin
2
u + 2 cosh v cos u
_
= a
2
_
cosh
2
v + cos
2
u + 2 cosh v cos u
_
= a
2
(cosh v + cos u)
2
Similar with the other xed mass point, and now the potential can be written as:
V =
Gmm
1
|r + a z|

Gmm
2
|r a z|
= Gm
_
m
1
a (cosh v + cos u)

m
2
a (cosh v cos u)
_
=
Gm
a
_
m
1
(cosh v cos u) m
2
(cosh v + cos u)
cosh
2
v cos
2
u
_
=
Gm
a
_
(m
1
m
2
) cosh v (m
1
+ m
2
) cos u
sinh
2
v + sin
2
u
_
4
where replacing in the results of part (a), we found:
1
2ma
2
_
_
W

_
2
+ V

_
=
_
1
sin
2
u
+
1
sinh
2
v
_
1
_
E(....)
1
2ma
2
_
W
u
u
_
2

1
2ma
2
_
W
v
v
_
2
_
1
2ma
2
_
_
W

_
2
+
Gm
a
_
(m
1
m
2
) cosh v (m
1
+ m
2
) cos u
sinh
2
v + sin
2
u
_
_
=
=
_
1
sin
2
u
+
1
sinh
2
v
_
1
_
E(....)
1
2ma
2
_
W
u
u
_
2

1
2ma
2
_
W
v
v
_
2
_
where we can see that this equation will be reduce to quadratures if we arrange correctly the
terms.
Goldstein, Ch.10, 15 A particle of mass m moves in one dimension subject to the potential
V =
a
sin
2
_
x
x
0
_
Obtain an integral expansion for Hamiltons characteristic function. Under what conditions can action-
angle variables be used? Assuming these are met, nd the frequency of oscillation by the action-angle
method. (The integral for J can be evaluated by manipulating the integrand so that the square root
appears in the denominator.) Check your result in the limit of oscillations of small amplitude.
Let us nd out the Hamiltons characteristic function. Using the proposed potential, the Lagrangian
can be written as:
L =
m x
2
2

a
sin
2
_
x
x
0
_
and the Hamiltonian:
H = p x L =
p
2
2m
+
a
sin
2
_
x
x
0
_
Then, Hamilton principal function must be S = W(x, ) + t. Hence:
1
2m
_
W
x
_
2
+
a
sin
2
_
x
x
0
_ = (11)
Using this result we can nd the integral solution of this problem:
W =

2m
_

a
sin
2
_
x
x
0
_dx
The condition to use action-angle variables, and to have a solvable W, is a/ sin
2
(x/x
0
). Then, the
action-variable is given by:
J =
_
pdx =

2m
_

a
sin
2
_
x
x
0
_dx
By solving this complicated integral we can found the frequency of the system.
5
Goldstein, Ch.10, 17 Solve the problem of the motion of a point projectile in a vertical plane, using the
Hamilton-Jacobi method. Find both the equation of the trajectory and the dependence of the coordinates
on time, assuming the projectile is red o at time t = 0 from the origin with the velocity v
0
, making
an angle with the horizontal.
Let us write the Lagrangian as:
L =
m
2
( x
2
+ z
2
) mgz
and the Hamiltonian:
H =
p
2
x
2m
+
p
2
z
2m
+ mgz
The Hamiltons principal function can be written as S = W
z
+ x
x
t and therefore:

2
x
2m
+
1
2m
_
W
z
z
_
2
+ mgz =
Then, the integral solution of this problem is given by:
W =
_
_
2m
2
x
2m
2
gzdz =
(2m
2
x
2m
2
gz)
3/2
3m
2
g
and the principal function:
S =
x
x
(2m
2
x
2m
2
gz)
3/2
3m
2
g
t
Hence, the constants are:
=
S

=
(2m
2
x
2m
2
gz)
1/2
mg
t

x
=
S

x
= x +
(2m
2
x
2m
2
gz)
1/2
m
2
g
Solving for x and z:
mg + (2m
2
x
2m
2
gz)
1/2
+ mgt = 0 (12)
m
2
g
x
m
2
gx (2m
2
x
2m
2
gz)
1/2
= 0 (13)
From (12):
(mg + mgt)
2
= 2m
2
x
2m
2
gz
z =

mg
+

2
x
2m
2
g
g( + t)
2
(12) - (13):
mg + mgt m
2
gx + m
2
gx = 0
x =
x

m

t
m
With this results, let us nd the velocities:
x =

m
z = 2g( + t)
6
Using initial conditions:
x(0) =

m
= v
0
cos z(0) = 2g = v
0
sin
x(0) = 0 = x

m
z(0) = 0 =

mg
+

2
x
2m
2
g
g
2
we can found the relations for the constants:
=
tan
2gm
= v
0
mcos

x
=

m

2
x
=
tan
4
2
gm
2
+
v
0
cos
g
7

## Нижнее меню

### Получите наши бесплатные приложения

Авторское право © 2021 Scribd Inc.
Авторское право © 2021 Scribd Inc.
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9325910806655884, "perplexity": 10784.017448264449}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988775.80/warc/CC-MAIN-20210507090724-20210507120724-00068.warc.gz"}