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Lecture 3
9
Oct
9, 2002
3
Special Relativity
Before going any further, we quickly review the evidence and kinematics of special relativity.
3.1
A Short Review of Classical Mechanics
All classical mechanics is based on
F
=
m
a
≡
m d
v
/
dt
≡
m d
2
x
/
dt
2
, where
F
denotes the vector sum of
all forces acting on the object of mass
m
. The result is an acceleration
a
(
t
) of the object in the direction
of the net force. The goal of classical mechanics is to deduce/predict the full
trajectory
of the object
over time
x
(
t
), i.e. predict where it will be at any time
t
in the future based on full knowledge of the
forces at play and
initial conditions
(position
x
0
and velocity
v
0
at a starting time
t
=0). Vice versa,
given a detailed measurement of the trajectory
x
(
t
) of an object, we want to deduce the details of the
force(s) that played a role.
In general, the force
F
may depend on location
x
and velocity
v
of the object, as well as depend on in
trinsic properties like orientation, charge, spin, color charge, mass, and so on:
F
=
F
(
x
,
v
=
d
x
/
dt
,…,
t
).
Only in the simplest cases for
F
can we solve the equations of motion analytically, in all other cases
computer solutions must be found. A few simple cases are listed Table 3 below.
Table 3. Some common types of forces and resulting analytical trajectories
Force type
Equations of Motion
Shape
Constant;
e.g. near
Earth gravity
F
= constant
a
= constant
0
0
00
0
0
22
0
0
11
()
t
tt
d
dt
d
t
t
dt
d
dt
t dt
d
dt
≡⇒ = →=
−→
=
+
≡⇒ = +
=
→+
=
−→=++
∫∫
∫
v
v
x
x
v
aa
v
a
v
v
v
v
x
vv
v
ax
vax
xx
x
va
a
Parabolic trajectory
Proportional;
e.g. spring
force
F
= –
kx
2
max
2
max
0
0
0
2
max
0
0
2
0(
) s
i
n
( )
,
with:
,
where and
are determined by
and
:
if
0; else:
tan
, and
tan
1
dx
mk
x x
t
x
t
dt
k
m
x
xv
x
x
v
x
ωφ
ω
φ
+=⇒
=
+
≡
=
==
+=
+
Harmonic motion
Inverse square
e.g. Coulomb
or Gravity
2
/
kr
=−
Fr
±
Central force problem: (Keppler equations)
Elliptical or hyper
bolic trajectories
The conservation of momentum and energy is a consequence of Newton’s Laws:
1.
if
0 then
constant
Momentum conservation
net
i
net
i
d
dt
==⇒ =
=
⇒
∑
p
FF
F
p
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View Full DocumentLecture 3
10
Oct
9, 2002
2.
22
2
2
11
1
1
()
21
2 1
RHS of
1
(ork)
(
)
2
Fm
a
d
m
d
m
d
m
d
m
v
v
K
K
dt
=
≡⋅=
⋅=
⋅
=
−≡−=
∆
∫∫
∫
∫
xx
v
x
v
v
x
v
v
Fx
x
vv
±²³²´
Wd
(I.15)
The possibility for integration of the LHS of
F
=
m
a
depends on the nature of the force
F
: many “wellbehaved” forces can be written as a
K
gradient of a potential
:
F
(
x
) = –
dU
/
d
x
. Only for this type of “conservative” force the integral can be done, and as result
total
mechanical
energy is conserved:
x
1
∫
x
2
F
⋅
d
x
=
−
U
(
x
2
) +
U
(
x
1
)
⇒
−∆
U
=
∆
K
⇒
∆
U
+
∆
K
= 0. Thus, with total energy defined as
E
≡
U
+
K
we have (for conservative
forces):
∆
E
= 0, i.e. Energy conservation. For nonconservative forces the integral can
not be done: that means that such forces cannot be written as a gradient of a potential
energy function. An example of a nonconservative force is simple friction: the total
work done by friction depends on the detailed trajectory taken moving from position
x
1
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 Fall '01
 Rijssenbeek
 Physics, mechanics, Special Relativity

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