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RELATIVISTIC MOMENTUM AND ENERGY
We have derived the addition of velocity
equation for motion parallel to the motion of
the moving frame
u—v
x
ux—
vux
2
c
Now we need the equation for motion
perpendicular to the direction of motion of the
moving frame.
From the Lorentz-Einstein Equation we have
y,=y
=y(t-)
1
Differentiate to get
dy’=dy
vdx\
dt’=y(dt—
j
2
c
Divide to get
dy’
dt’
dy
y(dt_
vdx’
c2)
Or
uy
y(1
—“‘
C2J
2
w
Di
(‘I
r)
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ae sJaAJasqo OM atu ‘IeD!UaP! ae seq
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Or using the railroad car example.
Looking down on the railroad tracks:
0’ ystcin
[hrow
uL
C
(utdi
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C’
+
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i12
lb
()
‘rtirow arid catch
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=
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Momentum is not conserved if u *
Butz4 is
uy
y(1x)
ux=o
Thus
U
y
Y
Not
Momentum is not conserved.
7
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We will show it from a different point of view
and do what some other books do.
Consider the equation
p=ymu
We can interpret this as the mass increasing
with velocity.
p
=
(relativistic mass)u
=
(ym)u
Then
9
0’ %y%tc;n
I hrow
(utch
:E:::
g
:z::
!:If;:
I
,
LJ/2
0
fhrow dnd Lat(.h
-
I
(a)AsseenbyO
For the observer in rest frame
Change in momentum for ball in rest frame
2 (mestuy)
Change in momentum for ball in moving frame
10
2 (mojngu)
=
2 mmovjng
(i
ux=o
2 (mmojgu)
=
2 [mmoving
2 (mmojngu)
=
0)1
2
m
movjng
Conservation of momentum the two changes
must be equal
11
2 (mrestuy)
mmo
-
=
2
m
mov ing
=
ymres t
This states that a moving mass has a larger
mass than the same object when moving.
The conservation of momentum holds if we
have a different definition of momentum
p=ymu
Or
Have a different definition for mass
mmovjng
=
ymrest
12
Either interpretation is proved with the
experiment of accelerating electrons and
measuring the momentum.
5
E 4
V
3
E
C 2
E
V
1
0
0
0.2 0.4 0.6 0.8 1.0 1.2
v/c
2GtJ6 Brooks!Ce
13
KINETIC ENERGY
WHAT WE DID IN BASIC MECHANICS
KE
WORK TO BRING OBJECT FROM REST
TO STATE OF MOTION WITH VELOCITY
=
K
=
CALCULATE THIS WORK AND IT IS THE KINETIC
ENERGY
WORK
=
FORCE X DISTANCE
Fds
AND
14
dv
F=m
dt
so
dv
I ds
K=m
ds= I m
dv
dt
dt
J
OR
K =fmvdv
If the object goes from 0 to some speed v
K
=
Vmvdv
0
f
=
mfVvdv
=
m
15
-
z
-n
-n
rn
rn
z
m
o
-
rn
rn
II
—1
C
z
rn
0
r)
-
rn
C
-n
-
C
—1
0
-‘
START WITH NEWTON’S SECOND LAW
d
F=
p
dt
Relativistic momentum
p=ymu
d
F=
ymu
dt
Kinetic Energy, K, then is force times distance
K=W= fFds=(ymu)ds
s=ut
17
ds=udt
So
K
=
f
=
K
=
d(yu)
mfu(yu)dt
mfud(yu)
=
udy + ydu
1
I
dy=d 1—
3
dy=
if
\
2
u
2C2)
2u
)du
2
(_
18
2
1_
d=
)
2
d(yu)=u
u
(
\
2
C
u /
2
2
u
1_
d(u)=
)
2
du
2
u
C2)
—3’
/2
dul+Ydu
u
2
)
2
du+(1_
_1/2
Find common denominator
19
d(yu)
du
=
(
C)
Put in KE Equation
du
K=mf U
(‘\
3
K=mc
I
F
I
3
K=mc
C)
udu
2
(c
—
1
—
)
2
u
3’
/2
i
2j
20
Put in the limits
1
3
K=mc
1
2
2
e
2
K=mc
[C
[2
2
K=mc
C
—
C
2
U
1
F
—1
2
me
—
/
C
2
me
C
K= 2
ymc 2
—
me
Or
K= (y—1)mc
2
21
2
K=(y—1)mc
K
=
2
ymc
2
ymc
=
—
2
mc
2
K + mc
ALL ENERGY UNITS
TOTAL ENERGY = KE
TOTAL ENERGY
=
E
=
+
MASS ENERGY
2
ymc
2
)fmc