2. Relativistic kinematics
2.1 Lorentz transformations
We want to write down the Lorentz transformation of space-time coordinates from a system K
to a system K ′ , that is moving at constant velocity βc along the x-axis.
x = γ (x′ + βct′ )
y = y′
z = z′
ct = γ (βx′ + ct′ )
(2.1.1)
√
where γ = 1/ 1 − β 2 is the Lorentz factor. If the relative velocity between the two systems,
~βc, is not parallel to a coordinate axis, then
γ ~ ′
hβ · ~x i + ct′
1+γ
~x = ~x + ~βγ
′
ct = γ h~β · ~x′ i + ct′
!
(2.1.2)
Particle velocities can be transformed as ~v = d~x/dt. Let θ be the angle between ~v and ~β. One
then obtains using h~β · ~x′ i = βx′ cos θ′
v cos θ =
v sin θ =
v ′ cos θ′ + βc
′
1 + β vc cos θ′
v ′ sin θ′
′
γ 1 + β vc cos θ′
(2.1.3)
For photons and other massless particles we find using v = v ′ = c
cos θ =
cos θ′ + β
1 + β cos θ′
sin θ =
sin θ′
γ(1 + β cos θ′ )
(2.1.4)
2.2 Four-vectors
We define a generalized vector s = (τ, x, y, z) with associated distance ds = (dτ, dx, dy, dz) by
writing
ds2 = c2 dt2 − dx2 − dy 2 − dz 2 = −dx2 − dy 2 − dz 2 − dτ 2
(2.2.1)
where we have used the generalized time
τ = ıct
(2.2.2)
Inserting the Lorentz transformations (2.1.2) we find that s2 is invariant. Four-dimensional
vectors P that transform as s are called four-vectors. Consequently, ds2 , P 2 , P Q und (P +
Q)2 are Lorentz-invariant. We already know the space-time four-vector. Important for many
1
applications is the energy-momentum four vector ( 1c E,~p). It being a four-vector implies the
Lorentz transformations of total energy and momentum are
E = γ E ′ + ~p′ ~β c
~p = ~p′ +
′
~β γ 2
~β · ~p′ + ~β γ E
1+γ
c
(2.2.3)
The momentum transformation can be broken down into that for the momentum components
parallel and perpendicular to the transformation direction ~β.
p⊥ = p
′
⊥
pk = γ
β
p + E′
c
′
k
!
(2.2.4)
2.3 Center-of-momentum frame and threshold energy
The center-of-momentum frame (indicated with ∗ ) is of particular importance when studying
threshold energies for various processes. Consider two particles with masses m1 , m2 and energymomentum four-vectors P1 , P2 . The center-of-momentum frame is defined by ~p∗1 + ~p∗2 = 0. So
with e = 1c E we can write down the total four-vector
P ∗ = P1∗ + P2∗ = (e∗1 + e∗2 , 0)
⇒ in any frame (P1∗ + P2∗ )2 = (P1 + P2 )2 = M 2 c2
(2.3.1)
The center-of-momentum frame behaves like a particle with mass M and energy-momentum
four-vector P = P1 + P2 . If we consider only one particle, then it would be at rest in the
center-of-momentum frame, so that frame would be the same as the rest frame of the particle.
The center-of-momentum frame may be moving, and one finds its Lorentz factor as
γCM =
E
e1 + e2
=q
2
Mc
(e1 + e2 )2 − (~p1 + ~p2 )2
(2.3.2)
For the collision of an energetic proton with a proton at rest for example we find
γp + 1
=
γCM = q
(γp + 1)2 − (γp2 − 1)
s
1 + γp
2
(2.3.3)
Incidentally, this is also the Lorentz factor of the protons in the center-of-momentum frame.
If we considered the creation of particles with mass δm in the center-of-momentum frame of
two particles 1 and 2, then the total energy must be at least the threshold energy
∗
∗ 2
Eth
= Mth
c = m1 c2 + m2 c2 + δmc2
(2.3.4)
At the same time we have in the laboratory system
M ∗ 2 c2 = P 2 = (e1 + e2 ,~p1 + ~p2 )2 = m21 c2 + m22 c2 + 2e1 e2 − 2~p1~p2
2
(2.3.5)
Inserting the square of (2.3.4) into (2.3.5) one obtains
e1 e2 − ~p1~p2 = e1 e2 − p1 p2 cos θ = m1 m2 c2 + δmc2 (m1 + m2 + 0.5 δm)
(2.3.6)
If both particles have rest mass, then (2.3.6) can be written using Lorentz factors.
γ1 γ2 −
q
γ12 − 1
1
δm
1
γ22 − 1 cos θ = 1 + δm
+
+
m1 m2 2 m1 m2
q
!
where θ is the angle between the momentum vectors of the two colliding particles.
Example: Consider the reaction p + p → p + p + π 0 , where one proton is initially at rest,
implying ~p2 = 0. This could be the collision of a cosmic rays particle with a hydrogen nucleus
in the interstellar medium. Then
γ1 = 1 + mπ
mπ
2
+
mp 2m2p
!
≃ 1.2984
(2.3.7)
So about 280 MeV in kinetic energy is needed to produce the pion, although its restmass energy
is only 135 MeV.
If one of the collision partners is a photon, implying e2 = p2 c and m2 = 0, then the threshold
energy would be
!
q
δm
2
e2 (γ1 − γ1 − 1 cos θ) = δmc 1 +
(2.3.8)
2m1
Example: Consider the reaction p + γ → p + π 0 on photons of the microwave background
(e ≃ 7 · 10−4 eV/c). The threshold energy is given by the most favorable collision angle in
(2.3.8), namely cos θ = −1, a head-on collision. Then
γ2 ≃ 1011
Greisen − Zatsepin − Kuzmin Cut − off
(2.3.9)
Above this threshold cosmic ray protons have a strongly reduced lifetime in intergalactic space.
The AUGER experiment has just observed a turn-over in the cosmic-ray spectrum at the GZK
cut-off energy.
If both collision partners are photons, for example in pair creation, we find
e1 e2 = 0.25 δm2 c2
(2.3.10)
2.4 Energy distributions of decay products
In high-energy astrophysics we are often concerned with so-called secondary particles, that are
produced in inelastic particle collisions or by decay of unstable particles. We already mentioned
the production of pions in proton-proton collisions as example when we talked about threshold
energies. The pions themselves are unstable and decay in to e± , ν, and γ, depending on the
charge of the pion. All these particles would be secondaries, whereas primaries would have
been accelerated from low energy.
3
It is important to know the relation between the energy distributions of the primary and
secondary particles. Let us first consider the center-of-momentum frame (∗ ) of the one or many
primary particles. In this system the secondary particle is emitted with energy E ∗ (Lorentz
factor γ ∗ ) at an angle θ∗ relative to ~βCM . If the center-of-momentum frame moves with Lorentz
factor γCM in the laboratory frame, then the energy of the secondary particle would be measured
in the lab as
q
γ = γ ∗ γCM + (γ ∗ 2 − 1)(γCM 2 − 1)x∗
(2.4.1)
with x∗ = cos θ∗ or
E = E ∗ γCM +
q
γCM 2 − 1 x∗
(2.4.2)
in the case of massless particles, for which the concept of Lorentz factor doesn’t apply. Suppose
the secondary particles are produced with a distribution
f1 (γ ∗ , x∗ ; γCM ) =
dn
dγ ∗ dx∗
(2.4.3)
where I am allowing for a functional dependence on the center-of-momentum energy. Typically
we will find a large number of primary particles with various energies, corresponding to a
distribution of center-of-momentum energies or Lorentz factors,
f2 (γCM ) =
dn
dγCM
(2.4.4)
The final energy distribution in the laboratory system, f3 (γ) = dn/dγ, must obey particle
number conservation.
f3 (γ) dγ = f1 (γ ∗ , x∗ , γCM ) f2(γCM ) dγ ∗ dx∗ dγCM
(2.4.5)
Eq. (2.4.1) provides us with a functional relation between the differential quantity on the lefthand side and the three differential quantities on the right-hand side. If we change one of the
three differential quantities on the right-hand side to dγ, then dγ cancels and we have derived
the desired energy distribution in the laboratory frame as an integral over the two remaining
variables. Let us perform the transformation dx∗ → dγ. Rewrite (2.4.1) to isolate x∗ and
obtain
γ − γCM γ ∗
x∗ = q
(2.4.6)
(γ ∗ 2 − 1)(γCM 2 − 1)
or using (2.4.2)
x∗ =
The derivative is
E − γCM E ∗
√
E ∗ γCM 2 − 1
dx∗
1
=q
dγ
(γ ∗ 2 − 1)(γCM 2 − 1)
4
(2.4.7)
(2.4.8)
or, respectively,
1
dx∗
= ∗√ 2
dE
E γCM − 1
(2.4.9)
The variable x∗ is limited to the interval [−1, 1] and we have to preserve those limits. For a
given γ and γ ∗ this implies limits to γCM which we then choose for the inner integral and obtain
f3 (γ) =
Z
∞
dγ
∗
Z
γ+
γ−
1
dγCM
f1 (γ ∗ , x∗ (γ, γ ∗ , γCM ), γCM ) f2 (γCM )
q
(γ ∗ 2 − 1)(γCM 2 − 1)
(2.4.10)
with
γ ± = γγ ∗ ±
or, again in the case of massless particles,
f3 (E) =
Z
0
∞
dE ∗
Z
∞
γmin
dγCM
q
(γ ∗ 2 − 1)(γ 2 − 1)
f1 (E ∗ , x∗ (E, E ∗ , γCM ), γCM ) f2 (γCM )
√
E ∗ γCM 2 − 1
with
γmin
1 E
E∗
=
+
2 E∗
E
(2.4.11)
(2.4.12)
(2.4.13)
Example: The decay π 0 → 2γ.
In the rest frame of the decaying pion both photons must carry the same momentum and hence
the same energy, but have opposite directions. The angular distribution of the photons is in
good approximation isotropic, and we will happily ignore small spin-related correction to that.
For the distribution of the photons in the pion rest frame (=center-of-momentum frame in this
case) we therefore find
!
mπ c2
∗
∗
∗
f1 (E , x ) = δ E −
(2.4.14)
2
Inserting this into (2.4.12) we then obtain the desired photon spectral distribution as integral over the center-of-momentum Lorentz factors, which is identical to pion Lorentz-factor
distribution.
2 Z∞
f2 (γCM )
(2.4.15)
⇒ fγ (E) =
dγ
CM √
2
2
m
c
mπ c mπEc2 + 4πE
γCM 2 − 1
2.5 Transformation of differential quantities
How do we transform the differential cross section dσ/dE of a reaction from one frame into
another? What are the transformation rules for the power spectrum of a radiation source?
In both cases we are dealing with differential quantities, whose relation to Lorentz-invariant
functions are not a priori known. Generally, it is useful to know those invariants, for we can
derive the transformation rules by relating the function in question to invariants. Examples of
invariants are pure numbers of particles or the particle distribution function f (~x,~p).
5
As in the preceding subsection it is useful to write the differential quantities as integrals, for
example for a differential cross section
σ=
Z
dE
I
dΩ
dσ
.
dE dΩ
(2.5.1)
The total cross section in Lorentz invariant, because it only concerns the number of particles
that undergo a reaction. Mathematically we would transform the integral (2.5.1) with the
Jacobi-Determinant (a.k.a. the Jacobian).
Z
dE
I
dσ
dΩ
=
dE dΩ
Z
dE
′
I
dΩ′
∂(E, Ω)
dσ
dE dΩ ∂(E ′ , Ω′ )
(2.5.2)
with the Jacobi-Determinant
∂x1
∂y1
...
...
...
∂(x1 , . . . , xn )
= ...
∂x1
∂(y1 , . . . , yn)
∂yn
∂xn
∂y1
...
∂xn
∂yn
(2.5.3)
where care must be exercised in keeping the right variables constant in the derivatives, namely
all xj except the xi of which the partial derivative is taken.
It is more practical to split the transformation into small steps, during which only one variable
is transformed. In our example (2.5.2) that might be done as shown below.
∂(E, Ω) ∂(p, Ω) ∂(p1 , p2 , p3) ∂(p′1 , p′2 , p′3 ) ∂(p, Ω)
∂(E, Ω)
=
∂(E ′ , Ω′ )
∂(p, Ω) ∂(p1 , p2 , p3 ) ∂(p′1 , p′2 , p′3) ∂(p′ , Ω′ ) ∂(E ′ , Ω′ )
p
=
E
1
p2
E
∂p3
=
∂p′3
E′
!
p′
2
E′
p′
sin θ
=
=
p′
p
sin θ′
(2.5.4)
where we have used that the perpendicular momentum p⊥ = p sin θ is invariant. The identity
in brackets was calculated as
∂p3
=γ
∂p′3
β ∂E ′
1+
c ∂p′k
!
=γ
6
p′k
1+βc ′
E
!
=
E
E′
(2.5.5)