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