Mass, Energy, and Momentum
•In Relativity, Momentum = mu (u = speed in frame)
•Where m = mo
u
u’
V
S’ S
Relative velocity
• Why the change in mass?
• Consider the relative velocities between frames:
u
S’S
u’
V
)('
)''(
'
2
'
c
vxtt
tvxxdt
dt
dt
dx
dt
dxu
Velocity transformation
dt
dt
dt
dx
dt
dxu
''
)()''(
2' c
vxt
dt
dvtx
dt
d
22
( ' )(1 )uv
u u vc
-v-- v,u'--u :, transforminverseor 1
)'(
2
cuvvu
u
'
1
)('
2cvuvu
u
u)for (solve
Mass, Energy, and Momentum
u
u’
VS’S
21
)'(
cuvvu
u
If u’ = .9c, and v = .9c,
Using Galilean Relativity, u = 1.8c…not allowed!
then u = .994c !
Now Consider a two body collision•View Collision from Frame S:
V
SfinalSinitial
m mo M
0
(0)
,omu m Mv
Also
m m M
Combining gives:omm
muv
u
Consider a two body collision•View Collision from Frame S’ moving with velocity V=u’ relativie to S, so that M is at rest.
S’finalS’initial
m’ m’ M’
u’ u’Vm mo M
21'
cuvvu
u
v
cvv
ufor
2
)0(1
0' :righton mass
mass on left: u' v or else M' not at rest!for
Proof that m=m0
S’fin
al
S’initia
l
m’ m’ M’
u’ u’
V
Sfina
l
Sinitial
m mo M
2
'and
'1
u' = v
u vu
u vc
where
Now that we have our velocities properly tranformed, lets combine the results of momentum conservation in frame S with the velocity transform equation between S and S’:
2
2
2 that u =
1
vso
vc
Namely V = o
mu
m m
Proof that m=m0
S’fin
al
S’initia
l
m’ m’ M’
u’ u’
V
Sfina
l
Sinitial
m mo M
22
22
2( )
11 ( )
11 ( ) 2( )
o
o
o o
mum m
umum m c
mu m
m m c m m
let v =o
mu
m m
Simplifying:
21o
o
mm m m
S’fin
al
S’initia
l
m’ m’ M’
u’ u’
V
Sfina
l
Sinitial
m mo M
2 2 2(1 )om m
2 2( ) ( ) 2 ( )o om m m m m m
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