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