Special Relativity 3
•
4-momenta
– Scalar derivative of a 4 vector
interval
– Use of proper time
– Interpretation of zero component
– Non-relativistic limits of energy and
momenta
•
Rest mass as the 4-vector invariant
“length”
– 4 momenta from boosts
•
Subatomic energy units
– Calculating velocity from kinetic
energy
– What determines the energy of a
photon?
•
•
Doppler shifts using photon 4momenta
Conservation of 4-momenta
– Deducing the mass of a parent from
the energy and momenta of its
daughters.
– Analysis of Compton Scattering
•
Center of momentum (cm) frame
– Computing the cm energy using
invariants
– Computing the energy threshold for
subatomic processes
– Describing collisions in the cm frame
• Computing the cm energies from s
and m
1
What will we do in this chapter?
In this chapter we discuss the concept of 4
momentum. We convert Newtonian 3momentum into a 4 vector using a velocity
based on differentiating the interval with
respect to proper time rather than time. In the
process we obtain a 4-vector with a zero
component of energy and momenta as the 1,2,3
components. The relativistic energy includes a
rest energy (mc2) as well as a kinetic energy
term which goes over to the Newtonian form (T
= mv2/2) as v/c → 0.
We show that dot product of the 4 momentum
with itself is the square of the particle’s rest
mass and show that a Lorentz boost of the rest
mass creates the relativistic energy and
momentum of the particle.
We discuss eV based units and give the rest
energy of the some common subatomic
particles. We show how the velocity of a
particle can be computed from its kinetic
energy.
We introduce the idea of a photon with an
energy proportional to its frequency and use
this idea to derive a more general form for the
Doppler shift by transforming the photon
energy in an arbitrary frame using our invariant
trick.
We discuss the conservation of 4-momenta in a
particle collision and discuss how the mass of a
short-lived subatomic parent can be deduced
from the energy and momenta of its daughters.
We analyze Compton scattering or the
scattering of x-rays from electrons in metal.
We introduce the concept of the center of
momentum frame and use it determine the
minimum (threshold) energy required to allow
subatomic collisions to take place.
Finally we show how to boost into the cm
frame and compute the incident and final
particle cm energies and momenta from the cm
energy and the particle masses.
2
4-momentum
r
r
r
A natural candidate for 4-momenta is p = mv ⇒ p% = m η% → p% = ( γ mc γ mv )
This result should raise some questions. Does p% have anything to do with
conventional momentum? And what's up with the zero component p0 = γ mc ?
We can answer both by expanding the answer to second order in β .
1
−1/ 2
We begin with γ . Recall (1 − x )
≈ 1+ x / 2 → γ =
≈ 1+ β 2 / 2
1-β 2
mv 2
E
⎛E
p0 = γ mc ≈ (1 + β / 2 ) mc ≈ mc +
+ L Defining p0 = → p% = ⎜
2c
c
⎝c
2
r⎞
p⎟ ,
⎠
r
r
r
mv 2
we have E = cp0 ≈ mc +
and p =γ mv ≈ mv
2
2
We see the usual Newtonian 3 momentum. The cp0 component contains
the usual kinetic energy plus an “energy” of the form E=mc2. This is
called the rest energy. We will show later that one can indeed convert
rest energy into kinetic energy.
3
What is ther length of 4-momentum?
What is p% • p% if p% = ( γ mc γ mv ) ?
As a 4 vector, p% can be transformed to another frame
using the Lorentz boost. As an example, consider
transforming p% from its rest frame to the lab frame.
p% • p% = ( γ mc ) − ( γ mc ) β =
2
2
2
p% • p% = ( mc ) γ 2 (1 − β 2 ) = ( mc )
2
2
Let the primed frame be the rest frame of mass m.
r
% mc 0 since
In the mass's rest frame we have p'=
(
Hence the squared length of the momentum
is the squared rest energy/c of the particle
in this frame the mass has no momentum but only rest
energy.
carrying the momentum. In this formulation
of relativistic kinematics, mass is the
important invariant.
2
For cp << mc 2
E = mc 2
( cp )
2
+ ( mc 2 )
⎛ cp ⎞
1+ ⎜ 2 ⎟
⎝ mc ⎠
2
⎛
⎞
cp
1
p2
⎛
⎞
2
2
E ≈ mc ⎜1 + ⎜ 2 ⎟ ⎟ ≈ mc +
⎜ 2 ⎝ mc ⎠ ⎟
2m
⎝
⎠
2
v
x'
x
We will usually write the 4 momentum as
r
2
p% = ( E / c p ) and p% • p% = ( mc ) implies
(E / c) 2 − p 2 = ( mc ) or E =
)
2
⎛ p 0 ⎞ ⎛ E/c ⎞ ⎛ γ βγ ⎞ ⎛ po ' ⎞
⎜ ⎟=⎜
⎟=⎜
⎟⎜ p '⎟ =
p
p
βγ
γ
⎠⎝ x ⎠
⎝ 1⎠ ⎝ x ⎠ ⎝
⎛ γ βγ ⎞ ⎛ mc ⎞ ⎛ γ mc ⎞ ⎛ γ mc ⎞
⎜
⎟⎜ ⎟ = ⎜
⎟=⎜
⎟
⎝ βγ γ ⎠ ⎝ 0 ⎠ ⎝ γβ mc ⎠ ⎝ γ mv ⎠
r
r
In general E / c =γ mc → E =γ mc 2 and p=γ mv
We thus have obtained exactly the same form
for 4-momentum by boosting the particle’s rest4
mass to the lab.
Units and velocity calculation
Relativity is often used to describe subatomic
particles which often travel at relativistic
velocities (i.e. velocities approaching the speed
of light). It is much more convenient to use the
rest energy (m c2) for these particles rather than
their mass and specify this energy in units
based on electron volts or eV. An electron (or
any charged particle with a charge of |e| =
1.6×10-19 C acquires a kinetic energy of 1 eV
when accelerated through a 1 volt electrical
potential difference. 1eV = 1.6×10-19 Joules. If
an electron with m c2 = 0.511 MeV is
accelerated through 1 million Volts it acquires a
lab energy of 1.511 MeV.
Accelerating an electron from
rest through 1 million volts.
mc 2 = 0.511× 106 eV
V
E = T + mc 2 = eV + mc 2
If V=1 MeV = 1×106 eV
T = 1 MeV and E = 1.511 MeV
We illustrate a typical kinematics calculation
for the velocity of a 1 MeV electron (i.e. an
electron w/ a kinetic energy of 1 MeV).
How fast does the 1 MeV electron travel?
p=γ mv E =γ mc 2 →
v cp
=
c E
E 2 − ( cp ) = ( mc 2 ) → cp = E 2 − ( mc 2 )
2
2
E − ( mc
v cp
=
Thus =
E
c E
2
)
2 2
2
⎛ mc 2 ⎞
= 1− ⎜
⎟
⎝ E ⎠
For the case of E = 1.511 MeV
2
v
⎛ 0.511 ⎞
= 1− ⎜
⎟ = 0.941
c
⎝ 1.511 ⎠
It is clear from our formula for v/c that an
infinite amount of energy is required to
accelerate a massive particle up to v=c.
5
2
More particles
The electron is a real light weight. We have
been introduced to the muon which has a rest
energy of mμ c2 = 105 MeV. Another common
particle is the proton with a rest energy of
mp c2 = 940 MeV. At present protons and
antiprotons are accelerated Tevatron at
Fermilab with a kinetic energy of 1 TeV or
1000 GeV or 1 million MeV or T=1×1012 eV
How fast is a 1 TeV proton ?
We go to the limit where T>>mc 2
2
⎛ mc ⎞
v
1 ⎛ mc ⎞
= 1− ⎜
⎟ ≈ 1− ⎜
2 ⎟
+
c
2
T
mc
E
⎝
⎠
⎝
⎠
2
2
2
v
1 ⎛ mc 2 ⎞
1 ⎛ 940 ⎞
1
≈ 1− ⎜
=
−
⎟
⎜
⎟
2⎝ T ⎠
2 ⎝ 1×106 ⎠
c
v
≈ 1 − 4.4 ×10−7 Very close to c!
c
2
2
Another important particle is the photon which
is the carrier of light waves. Our formula shows
that the photon must be massless in order to
travel at the speed of light with finite energy.
Since the mass and velocity of the photon is
set, something else must control the photon’s
energy … this is the frequency of the photon.
The energy of a photon is E=hω where
h is Planck's constant and ω is the it's
angular frequency. Since ω = 2π f = 2π c / λ ,
E = ( 2π hc ) / λ = 1240 eV • nm / λ . A visible
light photon with λ = 600 nm has E γ ≈ 2 eV.
A photon is very small -- a 60 Watt bulb puts out
60 J • s -1 /(2 × 1.6 × 10-19 J/γ ) ≈ 2 × 1020 γ /s
Since the photon is massless E 2 / c 2 − p 2 = 0
and cp =E , and p% γ = hω / c hω / ckˆ where kˆ
(
)
is photon's direction. Boosting to a different
frame changes E and thus changes ω. This is
another way of looking at the Doppler shift.
6
The Doppler shift from 4 momenta
( ct
A%
(
r
0 0
r
r)
% ⇒ Θ
% = η%
η% is ideal choice for Θ
Recall
% • A%
Θ
(Θ )
ct =
% •Θ
%
Θ
)
%
Θ
η% •η% = γ 2 (c 2 − v 2 ) = c γ 2 (1 − β 2 ) = c
In our first lecture on Special Relativity, we
obtained the above formula for transforming
the time ( or A0 component) of interval A
into the Θ reference frame. The derivation of
this formula only depended on dot product
invariance so we can use it for any 4-vectors
Here is a transcription for photons.
p% γ
(
%
Θ
r
0 0
r ⎛ hω
p) = ⎜
⎝ c
(E / c
)
E
(Θ )
η% • p% γ
E (η ) η% • p% γ
=
=
where
c
c
η% •η%
⎛ hω
p% γ = ⎜
⎝ c
r
hω ˆ ⎞
k ⎟ and η% = γ ( c v )
c ⎠
hω ˆ ⎞
k⎟
c ⎠
% • A%
cΘ
=
% •Θ
%
Θ
7
Check of general Doppler formula
r
r
v⎞
⎛ hω hω ˆ ⎞
⎛
%
p
=
k⎟
η% = γ ( c v ) =cγ ⎜1
;
⎟ γ ⎜
c ⎠
⎝ c
⎝ c⎠
r
hω (η ) η% • p% γ
hω v ˆ
⎛ hω ⎞
=
=γ ⎜
•k
⎟ −γ
c
c
c c
⎝ c ⎠
r
hω (η )
hω ⎛ v • kˆ ⎞
→
=γ
⎜⎜1 −
⎟
c
c ⎝
c ⎟⎠
r⎞
ˆ
⎛
•
k
v
⇒ ω (η ) = γω ⎜1 −
⎟⎟
⎜
c
⎝
⎠
v
kˆ
x̂
Θ
r
kˆ = xˆ , v = −v xˆ
r⎞
ˆ
⎛
k
v
•
ω (rec) = ω (send)γ ⎜⎜1 −
⎟⎟
c
⎝
⎠
1+ β
⎛ v⎞
ω (rec) = ω (send)γ ⎜1 + ⎟ = ω (send)
1− β
⎝ c⎠
v<<c → 1 + β ω (send) 1st order
ω (rec) ⎯⎯⎯
(
)
r
r
If v ⊥ kˆ , kˆ • v = 0 → ω (rec) = γω (send)
2
⎛
⎞ (send)
β
v
c
<<
(rec)
2nd order
ω ⎯⎯⎯→ ⎜1 + ⎟ ω
2 ⎠
⎝
v Θ
kˆ x̂
8
Momentum Conservation
iˆ
π
0
We are use to the idea of conservation of
γ
γ2
1
momentum in all collision problems and the
( hω1 −hω1 ) / c ( mc 0 ) ( hω2 hω2 ) / c
conservation of energy in elastic collision
problems. In the subatomic world, all collisions
−hω1 + hω2
r r r
p
= 0 → ω1 = ω2
conserve energy since there are no internal
1 + p1 = 0 →
c
degrees of freedom which can absorb energy in
2
2
mc
mc
E
+
E
=
→
ω
+
ω
=
h
h
1
2
1
2
the form of heat or friction. Thus in a
subatomic collision, all 4 components of 4
→ 2hω1 = mc 2 ∴ hω1 = hω2 = ( mc 2 ) / 2
momenta are conserved.
A% + B% = C% + D%
A%
B%
C%
D%
In subatomic physics, particles often change
identities during an interaction and “matter”
can turn to “energy”.
An example is the decay π 0 → γγ where
the π 0 has a rest energy of mπ c 2 = 135 MeV
and a mean lifetime of 8.4 × 10-17 s. Lets find
the photon energies in the π 0 rest frame.
We thus get two back-to-back 62.5 MeV gamma
rays. Matter into energy? What if the π0 is moving to
the right in the lab with a velocity v? Solve it with
Doppler shifts. γ1 is red shifted; γ2 is blue shifted.
π0
γ1
mc 2 1 + β
hω2 =
2 1− β
mc 2
2 1− β
γ2
mc 2 1 − β
hω1 =
2 1+ β
mc 2 ⎛ 1 − β
1+ β
hω1 + hω2 =
+
⎜
2 ⎝⎜ 1 + β
1− β
hω1 + hω2 =
v
2
⎞
⎟⎟ =
⎠
( (1 − β ) + (1 + β ) )
9
= γ mc = Eπ 0 ∴ The γ energies sum to lab Eπ 0
2
Reconstructing the parent from the daughters
The π0 lives such a short time, it hardly exists
at all and must be deduced by measuring the
direction and energy of its daughter photons.
We can easily do this using conservation and
the 4 vector length.
p% π 0 = p%1(γ ) + p% 2(γ )
( m c ) = p% • p% =
( p% + p% ) • ( p% + p% )
r
r
p% + p% = ( E / c p ) + ( E / c p )
r
r
= (E / c + E / c p + p )
(m c) =
( E + E ) / c − ( pr + pr ) • ( pr + pr )
2
πo
π0
(γ )
1
π0
(γ )
2
(γ )
1
(γ )
2
(γ )
1
(γ )
1
(γ )
2
(γ )
1
(γ )
2
(γ )
1
(γ )
1
(γ )
2
(γ )
2
(γ )
2
2
πo
(γ )
1
(γ ) 2
2
2
(γ )
1
(γ )
2
(γ )
1
(γ )
2
x̂
γ1
Eγ
(2)
mc 2 1 + β
=
2 1− β
Eγ
(1)
γ2
mc 2 1 − β
=
2 1+ β
(2)
Eγ(1) ⎞
r (2) r (1) ⎛ Eγ
pγ + pγ = ⎜
xˆ −
xˆ ⎟
⎜ c
c ⎟⎠
⎝
r
r
mc ⎛ 1 + β
1− β ⎞
ˆ
ˆ
pγ(2) + pγ(1) =
x
x
−
⎜
⎟
2 ⎜⎝ 1 − β
1 + β ⎟⎠
1− β ⎞
mcxˆ ⎛ 1 + β
⎜
⎟ = βγ mcxˆ
=
−
2 ⎟
2 ⎜⎝ 1 − β 2
1− β ⎠
2
⎛ Eγ(1) + Eγ(2) ⎞
r
r
r
r
− ( pγ(1) + pγ(2) ) • ( pγ(1) + pγ( 2 ) )
P% • P% = ⎜
⎟
⎜
⎟
c
⎝
⎠
= ( γ mc ) − ( βγ mc ) = ( mc ) γ 2 (1 − β 2 )
2
Lets illustrate this technique by reconstructing
the mass of a πο from the energies and
directions of two photons which decay along
the direction of the πο velocity.
π
v
0
2
2
= ( mc ) It works!
2
10
Compton Scattering
)
Metal electron –photon scattering 4 vectors: photons k% and k% ' electrons p% and p '
k%
k% '
p%
p% '
⎛E
p% = ( mc 0 ) ; k% = ⎜
⎝c
p% • p% = ( mc )
2
E ˆ ⎞ % ⎛ E' E' ˆ ⎞
k⎟ ; k' =⎜
k '⎟
c ⎠
⎝c c ⎠
= p% '• p% ' ; k% • k% = k% '• k% ' = 0
Eliminate p% ' = p% + k% − k% ' since don't measure e'
(
)
(
) (
)
(
)
2
2
2
2
%
%
%
%
%
%
⎡
⎤
%
%
%
%
=
=
+
−
=
+
•
−
+
−
p
'
mc
p
k
k
'
p
2
p
k
k
'
k
k
'
; 2 p% • k% − k% ' = 2m ( E − E' )
( ) ( ) ⎣
(
)
⎦
2
EE'
EE'
k% − k% ' = k% • k% − 2k% • k% '+ k% '• k% ' = −2k% • k% ' = −2 2 1 − kˆ • kˆ ' = −2 2 (1 − cosθ )
c
c
(
2
(
)
∴ ( mc ) = ( mc ) + 2m ( E − E' ) − 2
2
2
Now some QM: E=hω = h
2π c
λ
&
)
EE'
E − E' 1 − cos θ
1
−
cos
θ
→
=
(
)
c2
EE'
mc 2
E − E' ⎡ 1 1 ⎤ ⎛ λ '− λ ⎞ 1 − cosθ
= ⎢ − ⎥ ⇒⎜
=
⎟
EE'
mc2
⎣ E' E ⎦ ⎝ 2π hc ⎠
2π hc
→ λ '− λ =
1 − cosθ ) = λCompton (1 − cosθ ) ;
2 (
mc
1240ev • nm
1
nm (x-ray)
hc = 197ev • nm → λCompton =
=
6
0.511×10 ev 412
λ
Metal rod
θ
e'
λ'
11
A%
Center of momentum
in
a
collision
%
B%
X
C
D%
A famous example is the creation of
antiprotons via p p → p p p p. This is the
simplest reaction which conserves baryon
number and charge. Both particles have the
same rest energy. It is easy to show that
2
that E (cm)
min = 4mc . Use this to find the threshold
kinetic energy required for an incident p to collide
with a stationary p to create the final p p p p .
There exists a reference frame where the vector
sum of the 3-momenta of incoming and
outgoing particles is zero. We will call this the
center of momentum (cm) frame. It is useful to
measure the total energy in the cm frame which
4
is easy to do using dot product invariants. The
(cm)
Emin = ∑
square of E(cm) is often called s. s is the same
i =1
for the incident and final state.
In any frame we have:
r ⎞
⎛
p% X = p% A + p% B = ⎜ ∑ Ei / c ∑ pi ⎟
i
⎝ i
⎠
( cm )
⎛
⎞
r ( cm ) r
E
( cm )
i
%
In cm: ∑ pi = 0 p X = ⎜ ∑
0⎟
c
i
⎝ i
⎠
2
⎛
⎞
2
s ≡ c 2 p% X • p% X = ⎜ ∑ Ei( cm ) ⎟ Ecm
is called s
⎝ i
⎠
We can use the cm to compute the minimum
(threshold) energy required to create particles.
( mc ) + ( cp ) > ∑ mc >4mc
( p% + p% ) • ( p% + p% ) > ( 4mc )
2 2
(cm) 2
i
4
2
2
i =1
2
2
Thus s = c 2 A
B
A
B
We can simplify this a bit by expanding:
( p% A + p% B ) • ( p% A + p% B ) = p% A • p% A + 2 p% A • p% B + p% B • p% B
= ( mc ) + 2 p% A • p% B + ( mc ) > 16 ( mc ) or
2
2
2
p% A • p% B > 7 ( mc ) Let B be the projectile proton
⎛ T+mc 2 r ⎞
%
with PB = ⎜
p ⎟ and A be at rest with
c
⎠
⎝
r
2
P% = mc 0 ;c 2 p% • p% = ( T+mc 2 ) mc 2 > 7 ( mc 2 )
2
A
(
)
A
B
or T > 6mc 2 > 6(940) > 5640 MeV
The antiproton was discovered once
12
such a powerful accelerator existed.
More on the CM frame
It is easy to find a boost into the cm frame.
% ΛB% =
We want a frame where ΛA+
r
s /c 0 .
(
C%
A%
)
%
%
B
r r ⎞
D
E
E
+
⎛
A
B
% ΛB% = Λ A+
% B% = Λ
ΛA+
PA + PB ⎟
⎜
c
⎝
⎠ In the cm frame, the above collision would look
r
r
= s / c 0 . We boost along the direction of like this:
C r
The two entering momenta r
B
the lab momentum sum. r
are equal and opposite. The A
r
x'
two exiting momenta are
PA + PB
β
r
equal and opposite.
r
D
(
(
)
(
)
)
PB
r
PA
Entering and exiting |P| are often unequal
⎛ E A + EB ⎞
− βγ ⎞ ⎜
⎟
c
γ ⎠⎟ ⎜⎜ r r ⎟⎟
P +P
⎝ A B ⎠
r r
E A + EB )
(
+ γ PA + PB
0 = − βγ
c
r r
r c PA + PB
β=
( E A + EB )
⎛ s⎞
⎜
⎟=⎛ γ
⎜ c ⎟ ⎝⎜ − βγ
⎜
⎟
⎝ 0 ⎠
(
(
(
)
)
)
It is easy to get the energies and momenta of
the particles in the cm frame using our old trick.
A% + B% • A%
A% + B% • A%
EA(cm)
=
=
c
%A + B% • A% + B%
s /c
r
(cm)
%
%
Since A + B
= s /c 0 ,
(
(
(
(
)
) (
(
)
)
(
)
)
(cm)
%A + B% • A% + B% = s / c 2 and A% (cm) = ⎛ EA
⎜
⎝ c
) (
r ⎞
PA ⎟
⎠
)
13
E
(cm)
A
=
(
)
c 2 A% + B% • A%
s
mc )
(
=
CM description
2 2
+ B% • A%
A
s
It is easy to get B% • A% from the masses
(
) (
)
s
= A% + B% • A% + B% = A% • A% + 2 B% • A% + B% • B%
2
c
c 2 B% • A% =
s − ( mc 2 ) − ( mc 2 )
A
2
And EB(cm) =
s + ( mc
B
→ EA(cm) =
) − ( mc )
2 2
2 2
B
A
2 s
r
B
r
A
s + ( mc 2 ) − ( mc 2 )
2
2
A
B
2 s
r
r
A= B =p
s = EA(cm) + EB(cm) =
+
( pc )
2
+ ( mc 2 )
( pc ) + ( mc 2 )
2
(E ) − (m c )
cpB(cm) =
(E ) − (m c )
(cm) 2
A
(cm) 2
B
B
We could solve for p using a quadratic
equation.
EA(cm) + EB(cm) = s
2 2
A
2 2
B
Analogous expressions exist for C and D
EC(cm) =
s + ( mc 2 ) − ( mc 2 )
2
2
C
D
2 s
ED(cm) =
s + ( mc 2 ) − ( mc 2 )
2
2
D
C
2 s
This problem can also be solved without
invariant methods by realizing that two
incoming momenta have equal magnitude
A
2
We can easily find momenta magnitudes using
cp A(cm) =
2
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