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Classical Mechanics, Second Edition, 2nd Edition
book

Classical Mechanics, Second Edition, 2nd Edition

by Tai L. Chow
May 2013
Intermediate to advanced content levelIntermediate to advanced
639 pages
23h 12m
English
CRC Press
Content preview from Classical Mechanics, Second Edition, 2nd Edition
530 Classical Mechanics
© 2010 Taylor & Francis Group, LLC
Solution:
The operator T has the form
T
H
pq
H
pp t
p
mq
mq
p
ii ii
=
+
=
ω
2
.
Then,
(Tq)
t=0
= p
0
/m, (T
2
q)
t=0
= −ω
2
q, (T
3
q)
t=0
= −ω
2
p/m, (T
4
q)
t=0
= −ω
4
q
and so on. Substituting these into Equation 15.42, we obtain
qt qt
p
m
t
q
tp
m
q
()
!
()
!
...=+ +− +−
+
=
0
0
2
2
0
3
2
0
23
ωω
00
22
0
33
0
1
23
−+
+−+
=
ω
ω
ω
ωtp
m
t
t
q
!
...
!
...
ccossinω
ω
ωt
p
m
t+
0
with a similar development for p(t).
15.4 H-J THEORY AND WAVE MECHANICS
There is a formal analogy between the trajectory of a particle in a conservative force eld and
the path followed by a light ray in a region of space where the index of refraction does ...
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Publisher Resources

ISBN: 9781466569980