Read
Rohlf,
P99‐111
Galilean
Transforma.on
x ′ = x − vt y ′ = y z ′ = z
t′ = t
x = x ′ + vt ′ y = y ′ z = z ′
t = t′
Lorentz
Transforma.on
x′ = γ (x − vt) y′ = y z′ = z
⎛
⎞
t ′ = γ ⎜ t − v2 x ⎟
⎝
c ⎠
x = γ ( x′ + vt ′) y = y′ z = z′
⎛
⎞
t = γ ⎜ t ′ + v2 x′⎟
⎝
c ⎠
Velocity
Transforma9on
Rohlf
P105
u x′ =
(u x − v)
v ⎞
⎛
1−
⎜⎝
2 ux ⎟
⎠
c
u y′ =
uy
v ⎞
⎛
γ ⎜ 1− 2 u x ⎟
⎝
⎠
c
uz′ =
v
Low velocity limit: u << c, v << c. , → 0 γ → 1.
c
u x′ = u x − v
u y′ = u y
uz′ = uz
uz
v ⎞
⎛
γ ⎜ 1− 2 u x ⎟
⎝
⎠
c
3‐Dim.
Lorentz
Transforma9on
S
Rohlf,
P111
1
S’
β
2
Two
events
happen
at
the
same
9me
in
S.
What
is
the
9me
interval
in
S’
⎛
t2 = t1 t1′ = γ ⎜ t1 −
⎝
⎛
v ⎞
v ⎞
x
t
=
γ
t
−
x2
′
1
2
2
⎜
⎝
c2 ⎟⎠
c2 ⎟⎠
t2′ − t1′ = γ (t2 − t1 ) − γ
Δt ′ = γ 0 − γ
v
( x2 − x1 )
c2
v
v
Δx
=
−
γ
Δx
c2
c2
Δt ′ = γΔt − γ
v
Δx
c2
Exercise
Two
events
happen
simultaneously
at
x1=0
m
and
x2=1000
m
Suppose
S’
is
moving
to
the
right
at
0.9
c.
Which
event
occurs
first
and
by
how
much?
Event
1
occurs
at
t=0
µs,
and
event
2
occurs
at
t=2
µs.
When
does
event
2
occur
in
S’?
Solu9on
Two
events
happen
simultaneously
at
x1=0
m
and
x2=1000
m
Suppose
S’
is
moving
to
the
right
at
0.9
c.
Which
event
occurs
first
an
by
how
much?
Δt ′ = γΔt − γ
v
0.9c
−6
Δx
=
−
2.3
1,
000
=
−
6.9
×
10
s = − 6.9 µs
c2
c2
Event
1
occurs
at
t=0
µs,
and
event
2
occurs
at
t=2
µs.
When
does
event
2
occur
in
S’?
Δt ′ = γ ⎛⎜ Δt −
⎝
⎛
⎞
v ⎞
−6 0.9c
Δx
=
2.3
2
×10
−
1,000
= 2.3( 2 ×10 −6 − 3 ×10 −6 ) = −2.3µs
⎟
⎜
⎟
2
2
⎠
⎝
⎠
c
c
The
order
of
the
events
is
reversed.
Does
this
violate
causality?
No.
Events
can
be
causally
connected
only
if
they
can
communicate.
The
fastest
communica9on
is
at
the
speed
of
light.
For
event
2
to
be
influenced
by
event
1,
a
signal
from
event
1
must
travel
at
a
speed
at
least
Δx
1000
9
=
−6 = 0.5 × 10 m/s ≈ 1.5c
Δt 2 × 10
v
Δt ′ = γ ⎛ Δt − 2 Δx ⎞ > 0
⎝
⎠
c
Δt >
v
Δx (timelike)
c2
Δt <
v
Δx (spacelike)
c2
Space
‐
Time
Since
space
and
9me
are
not
independent,
define
a
new
4‐dimensional
space‐9me,
with
coordinates:
x0 ≡ ct, x1 ≡ x, x2 ≡ y, x3 ≡ z, or simply xµ
This
is
called
Minkowski
space‐9me.
The
evolu9on
of
a
body
is
a
con9nuum
of
points
which
form
a
curve,
called
a
world
line,
in
space
9me.
ct
For
1‐space
and
9me
dimension:
x=‐ct
9melike
world
line
light
speed
world
line
x=ct
x
Minkowski
Space‐Time:
2
spa9al
and
1
9me
dimensions.
World
line
Lorentz
Transforma9on
x ′ = γ (x − vt) y′ = y, z ′ = z
β≡
v
c
γ ≡
t ′ = γ (t −
v
x)
c2
1
1− β 2
Define
4
dimensional
space‐9me.
All
vectors
become
4‐vectors.
(Rohlf
P123‐126)
⎛ ct ⎞
⎜ x⎟
R = x0 , x1 , x2 , x3 = t, x, y, z = ⎜ ⎟ = xµ
⎜ y⎟
⎜⎝ z ⎟⎠
Define
4
dimensional
Lorentz
transforma9on.
Λ = Λ µν
⎛ γ
⎜ −γβ
=⎜
⎜ 0
⎜⎝ 0
−γβ
γ
0
0
0
0
1
0
0⎞
0⎟
⎟
0⎟
1 ⎟⎠
Transform
from
S
to
S’
frame
3
R′ = Λ R = ∑ Λ µν Rν
ν =0
⎛ ct ′ ⎞ ⎛ γ
⎜ x ′ ⎟ ⎜ −γβ
⎜ ⎟ =⎜
⎜ y′ ⎟ ⎜ 0
⎜⎝ z ′ ⎟⎠ ⎜⎝ 0
−γβ
γ
0
0
0 0 ⎞ ⎛ ct ⎞ ⎛ γ ct − βγ x ⎞
0 0 ⎟ ⎜ x ⎟ ⎜ − βγ ct + γ x ⎟
⎟⎜ ⎟ =⎜
⎟
1 0⎟ ⎜ y ⎟ ⎜
y
⎟
⎟⎠
0 1 ⎟⎠ ⎜⎝ z ⎟⎠ ⎜⎝
z
Exercise:
Verify
that
above
transforma9on
is
equivalent
to
the
Lorentz
transforma9on.
Lorentz
transforma9on
wri[en
in
matrix
nota9on
3
R′ = Λ R = ∑ Λ µν Rν
ν =0
⎛ ct ′ ⎞ ⎛ γ
⎜ x ′ ⎟ ⎜ −γβ
⎜ ⎟ =⎜
⎜ y′ ⎟ ⎜ 0
⎜⎝ z ′ ⎟⎠ ⎜⎝ 0
ct ′ = γ ct − βγ x
−γβ
γ
0
0
0
0
1
0
0 ⎞ ⎛ ct ⎞ ⎛ γ ct − βγ x ⎞
0 ⎟ ⎜ x ⎟ ⎜ − βγ ct + γ x ⎟
⎟⎜ ⎟ =⎜
⎟
0⎟ ⎜ y ⎟ ⎜
y
⎟
⎟⎠
1 ⎟⎠ ⎜⎝ z ⎟⎠ ⎜⎝
z
v
⎛
t′ = γ ⎜ t − 2
⎝ c
x ′ = − β cγ t + γ x = γ ( x − vt )
⎞
x⎟
⎠
Specialize
to
2‐space/9me
dimensions
⎛ ct ⎞
ct, x = ⎜ ⎟
⎝ x⎠
⎛ ct ′ ⎞ ⎛ γ
⎜⎝ x ′ ⎟⎠ = ⎜ −γβ
⎝
−γβ ⎞ ⎛ ct ⎞
⎟
γ ⎠ ⎜⎝ x ⎟⎠
Inverse
Lorentz
Transforma9on.
β → −β
⎛ ct ⎞ ⎛ γ
⎜⎝ x ⎟⎠ = ⎜ +γβ
⎝
+γβ ⎞ ⎛ ct ′ ⎞
⎟
γ ⎠ ⎜⎝ x ′ ⎟⎠
Inverse
Lorentz
Transforma9on.
⎛ ct ′ ⎞ ⎛ γ
⎜⎝ x ′ ⎟⎠ = ⎜⎝ −γβ
Let β-β,
then
−γβ ⎞ ⎛ ct ⎞
γ ⎟⎠ ⎜⎝ x ⎟⎠
⎛ γ γβ ⎞ ⎛ γ
−γβ ⎞ ⎛ I 0⎞
Λ −1
Λ
=
µδ δν
⎜⎝ γβ γ ⎟⎠ ⎜⎝ −γβ γ ⎟⎠ = ⎜⎝ 0 I ⎟⎠
Exercise:
−1
Λ µδ Λδν
Verify:
Λ −1
µδ Λδν
⎛ γ γβ ⎞ ⎛ γ
=⎜
⎝ γβ γ ⎟⎠ ⎜⎝ −γβ
−γβ ⎞ ⎛ γ 2 − γ 2 β 2
=
γ ⎟⎠ ⎜⎝ γ 2 β − γ 2 β
⎛ I 0⎞
=⎜
⎝ 0 I ⎟⎠
⎞ ⎛ 1 0⎞
− γ 2 β + γ 2 β ⎞ ⎛ γ 2 (1 − β 2 )
0
=
=
−γ 2 β 2 + γ 2 ⎟⎠ ⎜⎝
0
γ 2 (1 − β 2 )⎟⎠ ⎜⎝ 0 1⎟⎠
Time dilation (Exercise)
Two
events
occur
at
the
same
place
Δx=0
at
different
9mes
Δt.
How
much
9me
has
elapsed
according
to
the
non‐rest
observer,
and
where
does
she
calculate
it
took
place?
That
is,
find
Δx’
and
Δt’
.
Time dilation
Two
events
occur
at
the
same
place
Δx=0
at
different
9mes
Δt.
How
much
9me
has
elapsed
according
to
the
non‐rest
observer,
and
where
does
she
calculate
it
took
place?
That
is,
find
Δx’
and
Δt’
⎛ cΔt ′ ⎞ ⎛ γ
⎜⎝ Δx ′ ⎟⎠ = ⎜ −γβ
⎝
or Δt ′ = γτ
−γβ ⎞ ⎛ cΔt ⎞ ⎛ γ
=⎜
⎟
γ ⎠ ⎜⎝ Δx ⎟⎠ ⎝ −γβ
−γβ ⎞ ⎛ cΔτ ⎞ ⎛ cγτ ⎞
=⎜
⎟
⎟
γ ⎠ ⎜⎝ 0 ⎟⎠ ⎝ −cγβτ ⎠
τ ≡ Δt proper time
and Δx ′ = −cγβτ = −γ vτ
Since
γ
>
1
Δt’
>
τ.
This
is
called
9me
dila9on.
When
two
events
occur
at
the
same
place,
then
that
system
is
called
the
proper
frame,
and
the
9me
between
the
events
is
called
the
proper
9me.
That
is
τ=
Δt.
Proper
9me
is
the
shortest
9me
interval
of
any
reference
frame.
Length
Contrac.on.
S’
frame
obtain
length
by
measuring
2
events
at
the
same
place
x2’=x1’=0’.
t1’
t2’
L = vt 2′ − vt1′ = vΔt ′
(note, Δt’ is
proper
9me)
S
frame
on
rocket
with
Δt = 0.
Find
the
length
Δx’
in
the
S’
frame.
x2
x1
Proper frame: L0 = x2 − x1
L0 = x2 − x1 = γ ( x2′ + vt 2′ ) − γ ( x1′ + vt1′ ) = γ ( x2′ − x1′ ) + γ v ( t 2′ − t1′ )
L0 = γ v ( t 2′ − t1′ ) = γ vΔt ′ = γ L
L=
L0
γ
Comment
on
Ch.
4
problem
11.
S
u2 x = 0
c
u2 y =
2
u2′ x = ?
u2′ y = ?
c
2
u1x =
⎫
⎪
⎪
u2 y
⎬ u2′ =
u2′ y =
v
⎛
⎞⎪
γ ⎜ 1− 2 u2 x ⎟ ⎪
⎝
⎠⎭
c
(u x − v)
v ⎞
⎛
⎜⎝ 1 − 2 u x ⎟⎠
c
u1x = 0
u1y = 0
u1y = 0
u2′ x = −v = −
u x′ =
S’
=
(0 − v)
v ⎞
⎛
⎜⎝ 1 − 2 (0)⎟⎠
c
1
v2
1
2
= 1− 2 = 1− =
γ
4
c
3
c
2
= −v = −
c
2
uy c 3
u y′ =
=
γ
4
(u2′ x )
2
u y′ =
( )
+ u2′ y
2
uy
⎛
⎝
γ ⎜1 −
v ⎞
ux
c2 ⎟⎠
u ′ = u x′ + u y′ =
2
 c
| v |=
2
2
=
uy
⎛
⎝
γ ⎜1 −
v ⎞
(0
c2 ⎟⎠
=
uy
γ
c 2 3c 2 c
+
=
7 = .66c
4
16
4
Class
exercises:
The
earth
has
a
diameter
of
about
12.8X103
km.
It
has
a
mass
about
6
X1024kg.
1.What
is
the
density
in
kg/m3?
A space cosmic neutrino with velocity 0.999c is directed toward the earth.
2. What is the earthʼs thickness and diameter as seen by the neutrino?
3. What is it the earthʼs density in kg/m3? 4. How long does it take for the neutrino to pass through the earth
as seen by the earth?
5. How long does it take for the neutrino to pass through the earth
as seen by the neutrino?
Class
exercises:
The
earth’s
thickness
Δx ′ = 1 Δx =
γ
as
seen
by
the
neutrino?
The
earth’s
density
as
seen
from
the
earth.
The
earth’s
density
as
seen
by
the
neutrino.
(
)
1 − β 2 Δx = 1 − .9992 6 × 106 = .27 × 106 m = 270 km
m
6 × 10 24
ρ= =
= 5.5 × 10 3 kg/m 3 = 5.5 g/cm 3
4
V
π (6.4 × 10 6 )3
3
ρ = γρ0 = 22.4(5.5 × 10 3 )kg/m 3 = 123 × 10 3 kg/m 3 = 123g/cm 3
The
9me
to
pass
through
the
earth
as
seen
by
the
earth?
13 × 10 6
Δt =
= 0.04 s
.999 × 3 × 10 8
Δt
The
9me
to
pass
through
Δt ′ = Δτ =
= .002 s
γ
the
earth
as
seen
by
thneutrino?
Δt ′ = Δx ′ / v
(Note,
you
could
have
used
.)
Example:
An event occurs in coordinate system S at x=0, and t=0.
A second event occurs at x=0 and time t= 10 ns.
An observer in S’ moving with velocity β'= 0.9 relative to S synchronizes his
clock so that he also sees the first event at x’=0 and time t’= 0.
a. What will be the time t’ for the second event, as measured in S’?
b. A second observer in S’’, moving with a velocity β' '= 0.9 relative to S’
synchronizes her clock so that she also sees the first event at x’’=0 and
t’’=0. At what time t’’ will she see the second event?
Procedure:
Take x’ and t’’ as the starting point, and perform a second Lorentz
transformation from S’ to S’’ with S’’ moving at β'' = 0.9 relative to S’ .
S’
β'
S’’
β''
S
β ′ = β ′′ = 0.9,
S’
S’’
β'
γ ′ = γ ′′ = 2.3,
β’’
τ = 1 × 10−8 s =10 ns,
cτ = 3m
⎛ ct ′ ⎞ ⎛ γ ′
⎛ 1
−γ ′β ′ ⎞ ⎛ ct ⎞
−0.9⎞ ⎛ 3⎞ ⎛ 6.9 ⎞
=
=
2.3
=
⎜⎝ x ′ ⎟⎠ ⎜⎝ −γ ′β ′
⎜⎝ −0.9
γ ′ ⎟⎠ ⎜⎝ x ⎟⎠
1 ⎟⎠ ⎜⎝ 0⎟⎠ ⎜⎝ −6.2⎟⎠
⎛ ct ′′ ⎞ ⎛ γ ′′
⎛ 1
−γ ′′β ′′ ⎞ ⎛ ct ′ ⎞
−0.9⎞ ⎛ 6.9 ⎞
=
=
2.3
⎜⎝ x ′′ ⎟⎠ ⎜⎝ −γ ′′β ′′
⎜⎝ −0.9
γ ′′ ⎟⎠ ⎜⎝ x ′ ⎟⎠
1 ⎟⎠ ⎜⎝ −6.2⎟⎠
t ′′ = 29 / 3 × 108 ≈ 10 × 10−8 s = 100 ns,
x ′′ ≈ −29 m
⎛ 29 ⎞
≈⎜
⎝ −29⎟⎠
Successive
matrix
opera9ons.
Rµ′ = ∑ Λ ′µν Rν
ν
Rµ′ = Λ ′µν Rν
Rα′′ = ∑ Λαµ
′′ Rµ′ =
µ
Λαµ
′′ Λ ′µν Rν
∑µ Λαµ′′ ∑ν Λ′µν Rν = ∑
µν
Rα′′ = Λαµ
′′ Λ ′µν Rν
⎛ ct ′ ⎞ ⎛ γ ′
−γ ′β ′ ⎞ ⎛ ct ⎞
=
⎜⎝ x ′ ⎟⎠ ⎜⎝ −γ ′β ′
γ ′ ⎟⎠ ⎜⎝ x ⎟⎠
⎛ ct ′′ ⎞ ⎛ γ ′′
−γ ′′β ′′ ⎞ ⎛ ct ′ ⎞
=
⎜⎝ x ′′ ⎟⎠ ⎜⎝ −γ ′′β ′′
γ ′′ ⎟⎠ ⎜⎝ x ′ ⎟⎠
⎛ ct ′′ ⎞ ⎛ γ ′′
−γ ′′β ′′ ⎞ ⎛ γ ′
−γ ′β ′ ⎞ ⎛ ct ⎞
=
⎜⎝ x ′′ ⎟⎠ ⎜⎝ −γ ′′β ′′
γ ′′ ⎟⎠ ⎜⎝ −γ ′β ′
γ ′ ⎟⎠ ⎜⎝ x ⎟⎠
S
S’
β’
S’’
β’’
Relativistic Dynamics
(Rohlf P112-116)

 dp
dv

F=
= ma = m
dt
dt
 1 t
v = ∫ Fdt → ∞ as t → ∞
m 0
It implies that the speed will increase indefinitely as F is applied.
If we want to maintain this equation we need to redefine momentum.


dr
dr dt

p≡m
=m
dτ
dt dτ

dr

= m γ = γ mv
dt
 dp


F=
p = γ mv
dt
Homework
6
Due
Tues.
Oct.
22.
• Relativistic Mechanics: P112-121, Problems 13,18,23
• Four Vectors: P121-125, Problems 25,29,31
Mass‐Energy
Rela9on
Change
in
kine9c
energy
Show:

   
 
 
2
ΔE = K = ∫ Fidl = ∫ md (γ v )iv = mc ∫ (γ β id β + β iβdγ )
 dp
Hint: use F =
,
dt


p = γ mv,
γ =
1
1− β
2
(
)
, γ 2 1− β2 = 1
Mass-Energy Relation: E=mc2
Work=change in kinetic energy

 





d (γ mv )
dp
 dl

ΔK = ∫ Fidl = ∫ idl = ∫
idl = m ∫ d (γ v )i = m ∫ d (γ v )iv dt
dt
dt
   
 
v dv v v
= ∫ md(γ v)iv = mc 2 ∫ (γ i + i dγ )
c c c c
   
= mc ∫ (γ β id β + β iβdγ )
2
Mass-Energy Relation
Change in kinetic energy
   
ΔK = mc ∫ (γ β id β + β iβdγ )
2
Show:
γ
ΔE = ΔK = mc 2 ∫1 d γ = γ mc 2 − mc 2
ΔE = E − E0 ⇒
E = γ mc 2 and
Hint:
γ =
1
1− β
2
(
)
, γ 2 1− β2 = 1
E0 = mc 2
Mass-Energy Relation
ΔK
= mc2
   
∫ (γ β id β + β iβdγ )
 
First γβ id β term:
Starting with γ 2 (1− β 2 ) = 1
Show:
 
2
2
2γ dγ 1− β − γ 2β id β = 0
(
)
1− β )  
(
dγ
= β id β
2
and
γ
Find the energygained in going from rest, γ =1, to finite velocity γ =1/ (1− v2 / c2 ).
By direct substitution show
ΔK = mc2 ∫ dγ
= mc2
γ
∫
γ =1
dγ =γ mc2 − mc2
Finally
ΔE = K
γ
2
= mc ∫1
dγ = γ mc2 − mc2
ΔE = E − E0 ⇒ E = γ mc2 and E0 = mc2
Rela9onship
between
momentum
and
energy
Show:
E 2 = p2 c2 + m 2 c 4
Hint:
p = γ mv = γ mβ c
1
2
2 2
2
=
γ
⇒
β
γ
=
γ
−1
2
1− β
Rela9onship
between
momentum
and
energy
p = γ mv = γ mβ c
1
2
=
γ
1− β2
(
p2 = β 2γ 2 m 2 c2
(
)
1 = 1 − β 2 γ 2 = γ 2 − β 2γ 2
)
β 2γ 2 = γ 2 − 1
p2 c2 = γ 2 − 1 m 2 c 4 = γ 2 m 2 c 4 − m 2 c 4 = E 2 − m 2 c 4
E 2 = p2 c2 + m 2 c 4
Non
rela9vis9c
limit
of
kine9c
energy.
Starting with E 2 = p 2 c 2 + m 2 c 4
Show that in the low speed limit (β <<1)
the non relativistic expression for kinetic energy K =
Hint: For small x x 2 << x,
and
1+x 2 ≈ 1+
1 2
x
2
1 2
mv is retrieved.
2
Non
rela9vis9c
limit
of
kine9c
energy.
E 2 = p2 c2 + m 2 c 4
E=
p c + m c = mc
2 2
2
4
2
γ 2 m 2 v2 c2
p2 c2
2
+ 1 = mc
+1
m2 c4
m2 c4
= mc
2
γ 2 v2
β2
2
2 2
2
2
2
2
+
1
=
mc
1
+
γ
β
=
mc
1
+
≈
mc
1
+
(1
+
β
)
β
c2
1− β2
≈ mc
2
1 + β + β ≈ mc
2
4
E − mc = E − E0 = K
2
2
1 2⎞
mc 2 β 2
⎛
2
1 + β ≈ mc ⎜ 1 + β ⎟ = mc +
⎝
2 ⎠
2
2
2
1 2 2
K = mc β
2
1 2
K = mv
2


Rela9vis9c
Equivalent
to
F = ma
a
is
not
invariant
in
all
frames.
Rohlf,
p118,
shows
that
the
rela9vis9c
equivalent
is

a=
   
F − β Fiβ
( )
mγ
Note,
they
are
not
in
the
same
direc9on.
Homework
6
‐
hints.
Chapter 4, Problems 13,18,23,25,29,31
13. E = E0 + K
pc = E 2 − (mc 2 )2
18. ΔE = mC c 2 − mN c 2 − me c 2
mC c 2 = 13040.976 MeV
mN c 2 = 13040.309 MeV me c 2 = 0.511 MeV
23. u ′ = speed in cm. u = 0.9c = speed in lab.
To go from cm to lab, inverse boost by v=-u ′ to
sit on one electron.
25. E = γ mc 2
γ =
29. Ee+ + Ee− = EZ
u=
(u ′ − v)
v′ ⎞
⎛
1−
u
⎜⎝
c2 ⎟⎠
1
1− β2
=
1
(1 − β ) (1 + β )
β = 1 − 10 −24
Mini
Review

 dp
d (γ v )
Force: F =
=m
dt
dt


Momentum: p = γ mv
Energy:

 


 
d (γ v )
 dl
 
K = ∫ Fidl = m ∫
idl = m ∫ d (γ v )i = m ∫ d (γ v )iv
dt
dt
= γ mc 2 − mc 2 = E − E0
E = γ mc 2
Energy-Momentum:
(
E0 = mc 2
)
p 2 = γ 2 m 2 v 2 = m 2 c 2 γ 2 β 2 = m 2 c 2 γ 2 − 1 ⇒ p 2 c 2 = E 2 − E0 2
E 2 = p2 c2 + m 2 c 4
Low velocity limit:
⎛
⎞
1
1 2 ⎞
1 2
2⎛
2⎛ 1
2⎞
K ⎯β⎯⎯
→
mc
γ
−
1
=
mc
−
1
≈
mc
1+
β
−
1
=
mc
β
=
mv
(
)
⎜
⎟
⎜⎝
⎟⎠
⎜⎝
⎟⎠
→0
2
2
2
2
⎝ 1− β
⎠
2
2
Class
Exercise
Suppose
one
wants
to
travel
to
the
nearest
star,
Alpha
Centauri
within
a
9me
span
of
10
years
(proper
9me).
The
distance
is
4.3
ly,
with
1
ly~1x1016
m.
1
yr
=3.2x107
s.
1. How
fast
would
one
have
to
go?
2. How
much
energy
in
Joules
would
have
to
be
expended
to
get
a
10
ton
vehicle
going
that
fast?
3. How
many
1
GW
power
plants
running
for
1
year
would
produce
the
needed
energy?
Class
Exercise
Suppose
one
wants
to
travel
to
the
nearest
star,
Alpha
Centauri
within
a
9me
span
off
10
years
(proper
9me).
The
distance
is
4.3
ly,
with
1
ly~1x1016
m.1
yr
=3.2x107
s.
1. How
fast
would
one
have
to
go?
x = vt = vγτ = γβ cτ =
β (c τ + x ) = x
2
2
2
2
2
β cτ
1− β2
β=
(
)
x 2 1 − β 2 = β 2 c 2τ 2
x 2 = β 2 c 2τ 2 + x 2 β 2
x2
4.32 ly 2
8
=
=
0.4
v
=
β
c
=
1.2
×
10
m/s
c 2τ 2 + x 2
10 2 ly 2 + 4.32 ly 2
2. How
much
energy
in
Joules
would
have
to
be
expended
to
get
a
10
ton
vehicle
going
that
fast?
1
1
E − E0 = (γ − 1) mc 2 γ =
=
= 1.09
2
2
1− β
1 − .4
E − E0 = 0.09mc 2 = 0.09(10 × 10 3 )(3 × 10 8 )2 = 8 × 1019 J
3.How
many
1
GW
power
plants
running
for
1
year
would
produce
the
needed
energy?
1 GW provides 3.2 × 10 J/yr.
16
8 × 1019
≈ 2.5 × 10 3
16
3.2 × 10
Op9onal
Graphical
Representa9on
of
Lorentz
Transforma9on
2‐D
Lorentz
transforma9on
x ′ = γ (x − β ct)
⎛ ct ′ ⎞ ⎛ γ
⎜⎝ x ′ ⎟⎠ = ⎜⎝ −γβ
−γβ ⎞ ⎛ ct ⎞
γ ⎟⎠ ⎜⎝ x ⎟⎠
ct ′ = γ (ct − β x)
S’
coordinates
seen
from
the
observer
in
S.
1
ordinate : x ′ = 0 ⇒ ct = x
β
1
straight line with slope
β
ct’
ct
⎛ 1⎞
tan −1 ⎜ ⎟
⎝ β⎠
abscissa : t ′ = 0 ⇒ ct = β x
straight line with slope β
x’
tan −1 ( β )
x
http://www.univie.ac.at/future.media/moe/galerie/struct/struct.html
Op9onal
ct
ct E′′ = ct D′′ +
Twin
Paradox
Revisited
cT cT
=
γ 2
γ
1
ct ′′
ct D′′ = ct B′ =
lightcone
cT
1
x′
γ 2
ct B =
cT
2
x ′′
cT
c
1
γ
γ 2
ct B′ = t B =
ct ′
lightcone
ct A = ct A′
x
Return
to
4‐Vectors
All
the
rules
of
vector
manipula9on
can
be
applied
to
4‐vectors
A = A µ = A 0 , A1 , A 2 , A 3
A + B = C = C µ = Aµ + Bµ
B = B µ = B 0 , B1 , B 2 , B 3
Rela9vis9c
invariants
The
dot
product,
or
scalar
product
in
4
dim.
space‐9me
3
 
Ai B → ∑ gµν A µ Bν = gµν A µ Bν
1
This
is
a
scalar
invariant
since
which
has
the
same
value
in
all
reference
frames.
gµν
is
called
the
metric
tensor
and
describes
the
intrinsic
proper9es
of
the
space
9me.
For
a
flat
space‐9me
in
Cartesian
coordinates
it
is
given
by:
gµν
⎛ −1
⎜ 0
=⎜
⎜ 0
⎜
⎝ 0
0
1
0
0
0
0
1
0
0⎞
0⎟⎟
0⎟
1⎟⎠
 
Ai B = −A 0 B 0 + A1 B1 + A 2 B 2 + A 3 B 3
Example:
space‐9me
vector
Ri R ≡ ΔS 2 = ∑ x µ gµν xν = ( ct x
µν
⎛ −1
⎜ 0
y z )⎜
⎜ 0
⎜
⎝ 0
0
1
0
0
0
0
1
0
0⎞ ⎛ ct ⎞
0⎟⎟ ⎜ x ⎟
⎜ ⎟ = −c 2 t 2 + x 2 + y 2 + z 2
0⎟ ⎜ y ⎟
1⎟⎠ ⎜⎝ z ⎟⎠
Space‐9me
intervals.
ΔS 2 = −c 2 Δt 2 + Δx 2 + Δy 2 + Δz 2
Example:
space‐9me
vector
The
space
9me
interval
ΔS2
has
the
following
proper9es:
ΔS2
>
0
:
spacelike
separa9on
of
events;
meaning
you
can
always
find
a
frame
with
v
<
c
where
the
events
take
place
at
the
same
9me
in
different
spa9al
loca9ons.
The
events
cannot
be
causally
connected.
ΔS2
<
0
:
9melike
separa9on
of
events;
meaning
you
can
always
find
a
frame
with
v
<
c
where
the
events
take
place
at
the
same
place
at
different
9mes.
The
events
may
be
causally
connected.
ΔS2
=
0
:
Lightlike
separa9on
of
events;
events
are
separated
by
a
light
ray‐
causally
related
events
have
a
9melike
separa9on,
at
most
a
lightlike
one
(ex.
E+M
interac9ons)
Energy
Momentum
4
Vector

pµ = (E, pc)= (E, px c, py c, pz c)
⎛ E′ ⎞ ⎛ γ
⎜ p ′ c ⎟ ⎜ −γβ
⎜ x ⎟ =⎜
⎜ py′ c ⎟ ⎜ 0
⎜⎝ p ′c ⎟⎠ ⎜⎝ 0
z
−γβ
γ
0
0
0
0
1
0
0 ⎞ ⎛ E ⎞ ⎛ γ E − βγ px c ⎞
0 ⎟ ⎜ px c ⎟ ⎜ − βγ E + γ px c ⎟
⎟ =⎜
⎟
⎟⎜
py c
0 ⎟ ⎜ py c ⎟ ⎜
⎟
⎟⎠
pz c
1 ⎟⎠ ⎜⎝ pz c ⎟⎠ ⎜⎝
The
magnitude
of
the
energy‐momentum
4‐vector
is
invariant
in
all
reference
frames.
That
is,
P2
is
conserved
in
all
frames,
whereas
non‐rela9vis9cally
they
are
separately
conserved
in
one
frame,
and
are
different
and
separately
conserved
in
another
frame.
Rela9vis9cally
the
combina9on
is
conserved
in
all
frames.
Charge‐current
4‐vector.
⎛ ρ ′c⎞ ⎛ γ
⎜ j ′ ⎟ ⎜ −γβ
⎜ x ⎟ =⎜
⎜ j y′ ⎟ ⎜ 0
⎜⎝ j ′ ⎟⎠ ⎜⎝ 0
z
j ′ µ = ∑ Λνµ j ν
ν
−γβ
γ
0
0
0
0
1
0
0 ⎞ ⎛ ρ c ⎞ ⎛ γρ c − βγ j x ⎞
0 ⎟ ⎜ j x ⎟ ⎜ − βγρ c + γ j x ⎟
⎟
⎟⎜ ⎟ =⎜
jy
0⎟ ⎜ jy ⎟ ⎜
⎟
⎟⎠
jz
1 ⎟⎠ ⎜⎝ jz ⎟⎠ ⎜⎝
Op9onal
Example
of
scalar
invariance
in
par9cle
physics.
A
photon
beam
is
incident
on
a
proton
target.
When
a
photon
is
absorbed,
it
produces
a
proton
in
an
excited
quantum
state.
The
lowest
mass
state
is
called
a
Δ.
The
masses
of
a
proton
and
Δ
are
mpc2
=.938
GeV
and
mΔc2
=1.233
GeV,
respec9vely.
Find
the
energy
of
thephoton
necessary
to
produce
a
Δ.
Op9onal
In
the
laboratory
frame
of
reference.
Define
the
invariant
S
as:
(
)(
) (
S = P = PiP = Pγ + Pp i Pγ + Pp = Eγ + E p
2
(
= Eγ + E p
⎛ −1 0 ⎞ ⎛ Eγ + E p ⎞
pγ c + p p c ⎜
⎝ 0 1 ⎟⎠ ⎜⎝ pγ c + p p c ⎟⎠
⎛ −Eγ − E p ⎞
pγ c + p p c ⎜
= − Eγ + E p
⎝ p c + p c ⎟⎠
)
γ
p
(
)
) + ( p c + p c)
2
γ
2
p
The proton is at rest: p p = 0 and E p =m p c 2 .
The photon has mγ =0, so that Eγ = pγ c
(
S = − Eγ + m p c 2
) + ( p c)
2
γ
2
= −Eγ 2 − 2Eγ m p c 2 − m p 2 c 4 + Eγ 2 = −2Eγ m p c 2 − m p 2 c 4
Op9onal
In
the
Delta
rest
frame,
which
is
the
CM
frame:
S = PΔ′ i PΔ′ = ( EΔ′
⎛ −1 0 ⎞ ⎛ EΔ′ ⎞
pΔ′ c ) ⎜
= ( EΔ′
⎝ 0 1 ⎟⎠ ⎜⎝ pΔ′ c ⎟⎠
= − EΔ′ EΔ′ + pΔ′ pΔ′ c 2 = − EΔ′ 2 + pΔ′ 2 c 2
The Delta is at rest: EΔ′ ' = mΔ c 2 and pΔ′ = 0
S = −mΔ 2 c 4
⎛ − EΔ′ ⎞
pΔ′ ) ⎜
⎝ pΔ′ c ⎟⎠
Op9onal
By
conserva9on
of
energy‐momentum,
that
is,
conserva9on
of
4‐momentum,
we
can
equate
S
before
and
aner.
Equate
S
in
the
Lab
and
CM
frames:
Lab: S = −2Eγ m p c 2 − m p 2 c 4
CM : S = −mΔ 2 c 4
Equate:
−mΔ 2 c 4 = − 2Eγ m p c 2 − m p 2 c 4
Eγ =
mΔ 2 c 4 − m 2 p c 4
2m p c 2
1.232 2 − .938 2
=
= 0.34 GeV = 340 MeV
2(.938)
Op9onal
“contravariant”
and
“covariant”
4‐vectors

A = (A 0 , A1 , A 2 , A 3 ) = (A 0 , A)
The
4‐vector
used
here
is
a
“contravariant
4‐vector:
Examples: R = x 0 x1 x 2 x 3 = ct, x, y, z = x µ
P = p 0 p1 p 2 p 3 = E, cpx , cpy , cpz = p µ

There
is
also
a
“covariant”
4‐vector
Bµ = ( B0 , B1 , B2 , B3 ) = gµν ( −B 0 , B1 , B 2 , B 3 ) = ( −B 0 , B )
(
Examples: R = ( −ct, x, y, z ) = xµ
)
P = −E, cpx , cpy , cpz = pµ
 
Ai B = ∑ Aµ B µ = ∑ A µ Bµ = ∑ A µ gµν Bν = A 0
So
that
(
 
Ai B = −A 0 B 0 + A1 B1 + A 2 B 2 + A 3 B 3
S = −c Δt + Δx + Δy
Space‐9me
interval:
S = −E 2 + p 2x + p 2y + p 2z
4‐momentum
interval:
2
2
2
2
+ Δz 2
1
A
A
2
⎛ −1
⎜0
3 ⎜
A
⎜0
⎜0
⎝
)
0⎞ ⎛ B0 ⎞
0
0
1
0
0 ⎟ ⎜ B1 ⎟
0
1
0⎟ ⎜ B2 ⎟
0
0
⎟⎜
⎟
1 ⎟⎠ ⎜⎝ B 3 ⎟⎠
Op9onal
Nota9on
used
in
par9cle
physics.
⎛1 0 0 0 ⎞
⎜ 0 −1 0 0 ⎟
⎟
Define different metric gµν = ⎜
⎜ 0 0 −1 0 ⎟
⎜ 0 0 0 −1⎟
⎝
⎠
⎛ 1 0 0 0 ⎞ ⎛ B0 ⎞ ⎛ B0 ⎞
⎜ 0 −1 0 0 ⎟ ⎜ B1 ⎟ ⎜ −B1 ⎟
ν
⎟⎜ 2⎟ = ⎜ 2⎟
Bµ = ∑ gµν B = ⎜
0
0
−1
0
⎜
⎟ ⎜ B ⎟ ⎜ −B ⎟
v
⎜ 0 0 0 −1⎟ ⎜ B 3 ⎟ ⎜ −B 3 ⎟
⎠
⎝
⎠⎝ ⎠ ⎝
 
Ai B = ∑ Aµ B µ = ∑ A µ Bµ
(
= (B ,
Bν = B 0
Bν
0
−B1
−B 2
B1 ,
B2 ,
) (
)
) = ( B , B) contravariant 4-vector
−B 3 = B 0 , −B covariant 4-vector
B3
0
 
Ai B = ∑ A µ Bµ = A 0 B 0 − A1 B1 − A 2 B 2 − A 3 B 3
S = c 2 Δt 2 − Δx 2 − Δy 2 − Δz 2
S = P 2 = E 2 − p 2x − p 2y − p 2z = m 2 c 4
since S is positive,
S = mc 2
Op9onal
Exercise
A
π0
has
a
rest
energy
mc2=135
MeV.
In
the
lab
it
has
a
momentum
of
pc=1
GeV
(=1000
MeV).
It
decays
into
2 photons
k1
and
k2
In
the
cm.
frame,
k1
and
k2
emerge
back‐to‐back,
each
having
an
equal
and
opposite
momentum.
π0
β
k1
k2
1.
Find
the
momentum
and
energy
of
k1
and
k2
in
the
CM
frame.
(The
cm
frame
is
the
one
in
which
the
π0
is
at
rest.
The
mass
of
the
photon
is
zero.)
2.
Find
the
velocity
β
of
the
π0.
3.
By
making
a
Lorentz
transforma9on
find
their
momentum
and
energy
in
the
lab
frame.
(The
cm
frame
is
moving
with
a
velocity
equal
to
that
of
the
pion.
Therefore,
the
transforma9on
from
the
CM
to
the
lab
involves
a
velocity
‐β
)
4.
By
making
a
Lorentz
transforma9on
find
their
veloci9es
in
the
lab
frame.
Op9onal
In pion rest frame:
Before the pion decays:
SI = ( Eπ
⎛ −Eπ ⎞
2
2 2
2 4
=
−
E
+
p
c
=
−m
c
π
π
⎝ pπ c ⎟⎠
pπ c ) ⎜
After the pion decays:
SF = ( E1 + E2
⎛ −E1 − E2 ⎞
2
2
=
−
E
+
E
+
p
c
+
p
c
= SI = −m 2 c 4
(
)
(
)
1
2
1
2
⎟
⎝ p1 c + p2 c ⎠
p1 c + p2 c ) ⎜
The pion rest frame is the 2-photon cm frame p1 c = − p2 c E1 = E2 and m1 = m2 = 0
Thus ( 2E1 ) = m 2 c 4
2
2E1 = mc 2 E1 = E2 = mc 2 / 2 p1 c = E1 = − p2 c
In Lab: need Lorentz transformation with velocity − βπ
′ ⎞ ⎛ γ
⎛ E L1,2
⎜⎝ p ′ c ⎟⎠ = ⎜⎝ −γβ
L1,2
−γβ ⎞ ⎛ E1,2 ⎞ ⎛ γ E1,2 − βγ p1,2 c ⎞
=
γ ⎟⎠ ⎜⎝ p1,2 c ⎟⎠ ⎜⎝ − βγ E1,2 + γ p1,2 c ⎟⎠
Therefore we need γ , β , E1,2 , p1,2
p1,2 = E1,2 = mc 2 / 2 = 67.5 MeV
Evaluate γ and β :
(
Eπ L = γ mπ c
2
2
)
2
= ( pπ L c) + (mπ c )
2
2 2
γ =
2
( pπ L c)2
(m c )
2
2
+ 1 Op9onal
π
γ =
( pπ L c)2
(m c )
2
2
+1=
π
(1000)2
135 2
+1=
(1000)2
135 2
+1 =
56 = 7.4746
β2 = 1 −
1
γ2
= 0.9821 β = −0.9910
Perform Lorentz Transformation:
′ ⎞ ⎛ γ
⎛ E L12
⎜⎝ p ′ c ⎟⎠ = ⎜⎝ −γβ
L12
p1 = positive
−γβ ⎞ ⎛ E12 ⎞ ⎛ γ E12 − βγ p12 c ⎞
⎛ E12 − β p12 c ⎞
=
=
γ
⎜⎝ − β E + p c ⎟⎠
γ ⎟⎠ ⎜⎝ p12 c ⎟⎠ ⎜⎝ − βγ E12 + γ p12 c ⎟⎠
12
12
E L1 = γ E1 (1 − β ) ⎫
⎛ E1 − β E1 ⎞
⎛ 1− β ⎞
γ⎜
=
γ
E
⎬ 7.4746 ( 67.5 )(1 + .991) = 1004.5 MeV
1 ⎜
⎝ − β E1 + E1 ⎟⎠
⎝ − β + 1⎟⎠
pL1 c = γ E1 (1 − β )⎭
⎛ E2 + β E2 ⎞
⎛ 1+ β ⎞
p2 = negative γ ⎜
=
γ
E
2 ⎜
⎝ − β E2 − E2 ⎟⎠
⎝ − (1 + β )⎟⎠
Check : Eπ L =
E L 2 = γ E2 (1 + β ) = 7.4746 ( 67.5 )(1 − .991) = 4.5 MeV
pL 2 c = −γ E2 (1 + β ) = −4.5 MeV
( pπ L c)2 + (mπ c 2 )2 = 1000 2 + 135 2 = 1009 MeV