Selected Topics in Physics
a lecture course for 1st year students
by W.B. von Schlippe
Spring Semester 2007
Lecture 7
1. Relativistic Mechanics
Charged particle in magnetic field
2. Relativistic Kinematics of
Particle Collisions
1
Particle of mass m and charge q in a magnetic field B:
The equation of motion is
dp
= qv × B
dt
where v is the particle’s velocity,
p = γ mv is its momentum
and
F = qv × B
For simplicity of notation write
Thus
is the Lorentz force
B for qB
and restore
q
at the end.
dp
1
p× B
=v×B =
dt
γm
hence
dp
p⋅
= 0,
dt
(
)
∵ p⋅ p× B = 0
2
or
dp 2
=0
dt
i.e.
p 2 = const.
and hence
E 2 = p 2 c 2 + m 2 c 4 = const.
and
p
v = = const.
E
Consider a solenoidal field in
z direction:
B = B ( 0, 0,1) ≡ Bzˆ
3
and then consider two cases: (i) p ║ B, and (ii) p ┴ B.
Case (i): p ║ B:
then
hence
dp
=0
dt
p = const.; v = const.
i.e. the particle continues to travel with its initial velocity.
Case (ii): p ┴ B:
p × B = B ( p y , − px , 0 )
hence
and if initially
dpz
= 0,
dt
pz = const.
pz = 0, then the particle continues to move
in the (x,y) plane
4
Thus
p = p ( cos ϕ ,sin ϕ , 0 )
p = pϕ ( − sin ϕ , cos ϕ , 0 )
and since
we get
1
pB
p=
p× B =
( cos ϕ ,sin ϕ , 0 ) × ( 0, 0,1)
γm
E
pB
=−
( − sin ϕ , cos ϕ , 0 )
E
ϕ =−
hence
w.l.o.g.
B
E
B
ϕ = ϕ 0 − ( t − t0 )
E
ϕ0 = 0,
and
t0 = 0
5
hence
B
ϕ =− t
E
Now since the vectors v and p are parallel, we have:
v = r = v ( cos ϕ ,sin ϕ , 0 )
hence
t
r = r0 + v ∫ ( cos ϕ ,sin ϕ , 0 ) dt = r0 + v
0
E
( sin ϕ , − cos ϕ , 0 )
B
and finally
2
⎛ p ⎞
2
r − r0 = ⎜
⎟ ,
⎝ qB ⎠
this is the equation of a circle of radius
had absorbed the charge
q
in
∵v = p E
R = p/qB, remembering that we
B.
6
Example of an application of our result: mass spectrometer.
Two spectrometers to measure
the momenta of electrons, ejected
from a metal foil illuminated by
x-rays.
a)
The magnetic field is perpendicular
to the plane of the drawing.
a) narrow source, wide slit: the
image of the source lies on the
photographic emulsion.
b)
c)
x rays
From R.W. Pohl, Optik und Atomphysik
b) broad source, narrow slit: the
image of the slit lies on the photographic emulsion.
c) Spectrum of electrons ejected
from silver atoms by x-rays of
60 keV. The different lines
correspond to different atomic
energy levels from which the
electrons are ejected.
7
The next three pictures are photographs of bubble chamber events.
A bubble chamber is a volume of superheated liquid, placed in a strong
magnetic field. A charged particle passing through the liquid causes
ionization along its track. The ions serve as seeds to begin boiling.
When the bubbles have grown to a visible size, but before the entire
volume of the chamber is filled with bubbles, a light flash is triggered
and a photograph is taken
From the measured curvature of the tracks the momenta of the particles
are calculated. To do this we write the formula we have derived in the form of
p = qBR
The sense of the curvature gives the sign of the charge of the particle.
The reconstruction of bubble chamber pictures has enabled particle
physicists to discover many unknown particles, including neutral particles
which leave no tracks but decay into pairs of charged particles.
8
e+e- pair
recoil proton
track of incident pion
9
10
The discovery of the
Ω- meson gave strong
support to Gell-Mann’s
quark model.
11
The advantage of bubble chambers is the high quality of their pictures.
The disadvantage is the slow rate at which pictures can be taken: there is a
time of several seconds before the bubble chamber is ready for the next
event. Therefore bubble chambers are today replaced by detectors consisting
of an inner gas-filled tracking chamber surrounded by a calorimeter.
The particle tracks are reconstructed by dedicated computers which determine
the momenta of the charged particles from the curvature of the tracks.
Calorimeters are devices which measure the energies of particles.
Combining the information on energies and momenta one can calculate
the particle masses.
The following slides show an example of such a detector, the H1 detector
at the DESY laboratory in Hamburg, Germany.
12
13
superconducting coil
1.2 Tesla; diameter 5.2 m
Central tracking chamber; length 2.6 m; diameter 2 m
14
Cross section of the H1 detector
at the HERA electron-proton
collider in DESY.
superconducting coil
1.2 Tesla; diameter 5.2 m
15
Computer reconstruction of a collision event in the
H1 detector. Momenta of particles are found from
the curvature of the tracks; their energies are
measured in the calorimeters.
calorimeters
16
17
CMS (Compact Muon Solenoid) under construction at CERN
the superconducting coil gives a magnetic field of 4 Tesla.
18
January 2006: First cosmic rays seen in a
complete sector of CMS muon chambers.
The CMS solenoid at 100 Kelvin
19
14000 tons is an awful lot of weight, about the weight of a good-sized
battle cruiser including 1000 sailors. And all this is hard stuff. Mostly iron.
So you may be surprised that a particle, like a muon for instance, can pass
through. But the even more surprising thing is that this huge bulk is mostly
empty space to the muon.
Let’s assume for simplicity that the CMS detector is made purely of iron.
Iron has atomic number A = 56, so one atom of iron weighs about 56 times
the weight of a proton (or neutron, which is practically the same).
But one proton weighs roughly 1.7 times 10-27 kg, so one iron atom
weighs about 95 times 10-27 kg or, between friends, 10-25 kg
Therefore the number of iron atoms in that CMS detector is
14000 tons
32
=
×
N=
1.4
10
10−25 kg
20
Now the volume of a nucleus is
4π 3
V=
R
3
and the nuclear radii are to a good approximation given by
R = A1/ 3 R0 ; R0 ≈ 1.4 × 10−15 m
so the volume occupied by a single iron nucleus is
V ≈ 4 AR03 = 4 × 56 × 1.43 × 10− 45 m3 ≈ 0.6 × 10-42 m3
and hence the volume taken up by all iron nuclei of the CMS detector is
Viron / CMS = N × V ≈ 1.4 × 1032 × 0.6 × 10−42 m3
≈ 10−10 m 3 = 0.1 mm 3
the rest of the bulk of the detector is empty space.
21
A sector of the cross section of the CMS detector. The field of the SC solenoid is 4 T;
outside the field is about 2 T in the return yoke.
Simulated tracks of some charged particles are shown.
22
Particle collisions at relativistic energies
Collision processes are of two kinds:
(i) elastic collisions,
(ii) inelastic collisions
In relativistic kinematics such processes are studied within the framework
of conservation laws: conservation of energy and of momentum.
Relativistic kinematics is not concerned with the forces acting between
the colliding particles.
A typical problem of relativistic kinematics is the following:
Consider two particles,
a and b. Let b
be initially at rest and
with 4-momentum pa. After their collision particle
momentum and particle
b
a
a moving
has a different
has also acquired a momentum. Then one
wants to know what the relation is between the momenta of
a and b.
23
24
We begin solving the problem by writing down the equation of
4-momentum conservation:
p a + pb = p′a + p′b
To these we can add the four equations for the invariant squares of the
4-momenta:
p a2 = ( Ea c ) − pa2 = ( ma c )
2
2
p′a2 = ( ma c ) , pb2 = p′b2 = ( mb c )
2
2
We can simplify our notation if we set c = 1.
There is an additional advantage in doing that: now all equations
must be homogenous in masses, energies and momenta, so it is
easy to keep track of dimensionally correct calculations!
Thus our invariant squares of 4-momenta will be of the general form of
p2 = E 2 − p 2 = m2
25
and the 4-momenta of particles
scattering are
a
and
b
in the case of elastic
p a = ( Ea , 0, 0, pa ) , p′a = ( Ea′ , pa′ x , pa′ y , pa′ z ) ,
pb = ( mb , 0, 0, 0 ) ,
p′b = ( Eb′ , pb′ x , pb′ y , pb′ z )
Given the momentum of the
incident particle and the scattering
angle we can now for example find
the momentum of the scattered
particle, and then also the recoil
momentum and recoil angle.
To do this we write down the
energy conservation equation:
W ≡ Ea + mb = Ea′ + Eb′
26
then express the energies in terms of the corresponding 3-momenta,
hence
W = ma2 + pa′2 + mb2 + pb′2
and use momentum conservation to express the recoil momentum
in terms of the momenta of the incident and the scattered particles:
pa = pa′ + pb′
hence
W = ma2 + pa′2 + mb2 + ( pa′ − pa )
2
= ma2 + pa′2 + mb2 + pa2 + pa′2 − 2 pa pa′ cos θ
and solving for the momentum of the scattered particle we get
pa′ ( ± ) = pa
(s + m
2
a
− mb2 ) cos θ ± 2W mb2 − ma2 sin 2 θ
2 ( s + p sin θ )
2
a
2
(A)
27
where
s = ( p a + pb )
2
This is an invariant, and we shall make use of its invariance to express
s
in different inertial frames. But at the moment we shall write it explicitly
for the frame we have specified (i.e. with the target particle at rest):
s = ma2 + mb2 + 2mb Ea
There are two features in our result (A) that need discussing:
(i) the expression under the square root
(ii) the ± in front of the square root.
(i) the expression under the square root (“radicand”) is
mb2 − ma2 sin 2 θ
and this must be greater than or equal to zero for the
momentum of the scattered particle to be real:
28
thus the reality condition demands
mb2 − ma2 sin 2 θ ≥ 0
This will be satisfied for all scattering angles θ if the mass
target particle is greater than the mass
ma
However, if the mass ma is greater than
restricts the scattering angles:
mb
of the
of the incident particle.
mb, then the reality condition
sin θ ≤ mb ma < 1
(ii) The ± sign of the square root:
if ma < mb, then the lower sign gives a negative momentum
and must therefore be rejected;
if ma > mb, then both signs give a positive momentum, there are
therefore two values of the momentum of the scattered particle
for every value of θ
29
ma < mb
all scattering angles θ give
a real momentum; the lower sign
must be rejected.
ma > mb
sin θ max = mb ma < 1
both signs of the square root
must be kept.
30
Having found the momentum of the scattered particle we can get the
recoil momentum and recoil angle from momentum conservation:
pa = pa′ + pb′
hence
p′b2 = ( pa − pa′ ) = p 2a + pa′ 2 − 2 pa pa′ cos θ
2
pa′ sin θ
tan ϕ =
pa − pa′ cos θ
These results can be applied for instance in the design of collision
experiments.
Assume that we want to study elastic collisions at an energy at which
some of the collisions are inelastic. Then we can select among all collisions
the elastic ones by placing detectors at the appropriate angles.
31
32
Compton scattering
Consider the particular case when the incident particle is a photon,
i.e. ma = 0. Then we have
scattered photon is
pa′ = pa
Ea = pa etc. and the momentum of the
2mb pa cos θ + 2 ( pa + mb ) mb
2 ( mb2 + 2mb pa + pa2 sin 2 θ )
= mb pa
pa cos θ + ( pa + mb )
( pa + mb ) − pa2 cos 2 θ
2
where we have kept only the “+” sign because the incident particle
(photon) has a smaller mass than the target particle (electron).
Dividing by mbpa and taking the reciprocal we get
33
2
2
mb pa ( pa + mb ) − pa cos θ
=
= ( pa + mb ) − pa cos θ
pa′
pa cos θ + ( pa + mb )
2
= mb + pa (1 − cos θ )
hence
1
1
1
−
=
(1 − cos θ )
′
pa pa mb
or using the de Broglie relation
and reinstating
p=h λ
c we get
λ′ − λ =
h
(1 − cos θ )
mb c
i.e. we have recovered the famous Compton effect formula.
34
Of course, we could have found the Compton effect formula much more
easily in a few lines if we had started with the assumption of a zero-mass
incident particle.
Beginning with 4-momentum
conservation,
k + p = k ′ + p′
we have
p′2 = ( k − k ′ + p )
2
= p 2 + ( k − k ′ ) + 2p ⋅ ( k − k ′ )
2
or
m 2 = m 2 + ( k − k ′ ) + 2m ( k − k ′ )
2
hence
m ( k − k ′ ) = kk ′ (1 − cos θ )
and finally
λ′ − λ =
1
(1 − cos θ )
m
35
Laboratory and centre-of-mass frames
The reference frame that has a target particle initially at rest is called
laboratory frame (LAB).
For many theoretical investigations it is more convenient to work in the
centre-of-mass frame (CMS) which is defined by the momenta of the
initial particles being equal in magnitude and opposite in direction.
36
I will label all CMS variables by an asterisk.
Thus the 4-momenta for the case of elastic scattering
in the CMS and LAB:
CMS
a + b → a + b are
LAB
p*a = ( Ea* , 0, 0, p* ) ,
p a = ( Ea , 0, 0, pa ) ,
pb = ( mb , 0, 0, 0 ) ,
p*b = ( Eb* , 0, 0, − p* ) ,
p′a = ( Ea′ , pa′ x , pa′ y , pa′ z ) ,
p′a* = ( Ea′* , p′* ) ,
p′b = ( Eb′ , pb′ x , pb′ y , pb′ z ) .
p′b* = ( Eb′* , − p′* ) .
I have previously defined the invariant
s:
s = ( p a + pb )
2
and we can see that expressed in LAB variables it is given by
s = ma2 + mb2 + 2mb Ea
37
and in the CMS it is
s = (E + E
*
a
)
* 2
b
i.e. s is the square of the total CMS energy.
Now since
Ea* = ma2 + p*2 , Eb* = mb2 + p*2
we can solve for
p =
p*:
1
*
2
⎡ s − ( m − m ) ⎤ ⎡ s − ( m + m ) ⎤}
{
⎦⎣
⎦
s ⎣
2
a
b
2
a
12
b
The velocity of the CMS relative to the LAB (“boost velocity”) is found by
writing down the LT formula for the target particle and noting that by
definition its LAB momentum is zero:
pbz = γ ( pbz* + VEb* ) = 0,
38
hence
pbz*
paz*
V =− * = *
Eb
Eb
where in the last step I have used the definition of the CMS:
paz* = − pbz*
The scattering angles in the LAB and CMS are related by
sin θ *
sin θ
*
tan θ =
;
tan
θ
=
γ ( cos θ − V v )
γ ( cos θ * + V v* )
where
v* = p3* E3*
and v = p3 E3
are the velocities of the scattered particle in the LAB and CMS,
respectively.
39