Relativistic Momentum and Energy We have derived the addition...

39

Relativistic Momentum and

Energy

We have derived the addition of

velocity equation for motion

parallel to the motion of the

moving frame

'

21

xx

x

u vu vuc

−=

−

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Now we need the equation for

motion perpendicular to the

direction of motion of the

moving frame.

From the Lorentz-Einstein

Equations we have

2

'

' ( )

y y

vxt tc

γ

=

= −

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Differentiate to get

2

'

' ( )

dy dy

vdxdt dtc

γ

=

= −

Divide to get

2

'

2

'' ( )

(1 )

yy

x

dy dyvdxdt dtc

or

uu vu

c

γ

γ

=−

=−

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For z the same

'

2(1 )

zz

x

uu vuc

γ=

−

We will use these equations to

see how mass depends on the

speed of an observer relative to

an object.

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Railroad car from above with

Observer carrying BB and

Observer on ground with BB

The railroad car is moving

down the tracks (in x direction).

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Two individuals, one on the car

the other standing by the tracks,

each throw a basketball

perpendicular to the tracks. The

balls travel the dashed paths and

collide. The collision is elastic

so the speed of each stays the

same but the direction is

reversed as shown.

From the view of the observer

on the ground an equation for

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the change in momentum from

conservation of momentum will

be

' '

02( ) 2( )y ym u mu=

0m is the mass of the ball in the

frame of the observer at rest

beside the track. 'm is the mass

of the ball in the frame of the

car for the observer on the

ground.

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(Remember 0m is m in the

book.)

From the transformation

equations we have

'

2(1 )

yy

x

uu vu

cγ

=−

So

'

0

2(1 )

yy

x

um u m vu

cγ

=−

But

0xu =

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So

'

0y

y

mum u

γ=

Or

'

0m mγ=

The mass of the ball in the

moving frame is γ times the

mass of the ball in the stationary

frame.

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We have chosen to call the mass

of an object m0 if the observer

and the object are in the same

frame. If the observer observes

an object in the moving frame

the mass is m and we write

0m mγ= . This is different from

the symbols used in your book.

I prefer this because it reminds

you that 0m is the rest mass.

49

On the other hand we can do as

your book does and define

momentum as

2

21

mupuc

=

− .

Then we always treat m as a

constant but have a new

equation for defining

momentum.

50

Mass-Energy

Remember from first year

physics KE = K = work to bring

object from rest to state of

motion with velocity v.

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Then

K Fds

anddvF mdt

sodv dsK m ds m dvdt dt

=

=

= =

∫

∫ ∫

In classical physics m is

constant.

21

2K m vdv mv= =∫

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But if we treat m as a variable

or we use the new equation for

momentum we get something

else.

53

Start with

122

2

2

2

3 12 22 2

2 2

322

2

( )

(1 )

(1 ) (1 )

(1 )

d dpF mudt dt

d uF mudt c

u du dum mc dt dtFu uc c

dumdtFuc

−

= =

⎡ ⎤= −⎢ ⎥

⎣ ⎦

= +− −

=−

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Then

322

2

322

2

(1 )

(1 )

dumdtK Fds udtuc

uduK muc

= =−

=−

∫ ∫

∫

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Intergrating to get

122

2 20

22

122

2

122

2

11( )( 1)

(1 )

1

(1 )

u

K mu

c c

mcK mcuc

and

uc

γ

⎡ ⎤⎢ ⎥

= −⎢ ⎥⎢ ⎥− +⎢ ⎥⎣ ⎦

= −−

=−

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So 2 2K mc mcγ= −

Let

2

20

E mc

and

E mc

γ=

=

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Then

0

0

K E E

or

E K E

= −

= +

We see that mass is a form of

energy.

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We also have

2E mcγ=

Einstein’s Equation for the

equivalence of mass and

energy.

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Energy-Momentum

Relationship

We want to derive a useful

relationship between energy and

momentum. We start with

2

21

mup muuc

γ= =

−

Square and multiply by 2c

2 2 2 2 2 2p c m u cγ=

Or

60

22 2 2 2 4

2

up c m cc

γ=

Then starting with

2

2

1

1 uc

γ =−

Squaring and rearranging to get

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2

2 2

11uc γ

= −

Putting into the equation above

2 2 2 2 42

2 2 2 2 4 2 4

1(1 )p c m c

p c m c m c

γγ

γ

= −

= −

Or

2 2 4 2 2 2 4m c p c m cγ = +

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And

2 2 4 2m c Eγ =

So

2 2 2 2 4E p c m c= +

or

2 2 2 2

0E p c E= +

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If a particle has zero mass like

a photon then

2 2 2 2 4 2 2 0E p c m c p c

E pc

= + = +

=

or

Epc

=

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A mass less particle (for

example a photon) will have

momentum.

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Simultaneous Events

Consider the equation we

obtained for time dilation

2 12' ( ( )vt t x xc

γ ⎡ ⎤∆ = ∆ − −⎢ ⎥⎣ ⎦

For two events to be

simultaneous in the rest frame

the interval between the events

will be zero.

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2 1 0t t t− = ∆ =

For the same two events to be

simultaneous in the moving

frame

' '2 1 ' 0t t t− = ∆ =

Putting these two values back

into the time dilation equation

gives

2 120 0 ( ( )v x xc

γ ⎡ ⎤= − −⎢ ⎥⎣ ⎦

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or

2 1x x=

The only way for the events to

be simultaneous in both the rest

frame and the moving frame is

for event to be at a position

where

2 1x x=.

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Simultaneity in the moving

train example.

The twin paradox

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