RELATIVISTIC KINEMATICS
JUNGIL LEE
DEPARTMENT OF PHYSICS
KOREA UNIVERSITY
HEP Summer School
- Physics at Hadron Colliders -
CHEP, June 16-17, 2006
COPYRIGHT c© 2006, BY JUNGIL LEE
2
0.1 Special Theory of Relativity
0.1.1 Proper time
1. Pi = (ti, xi, yi, zi) is a particular event in an inertial reference frame K and P ′i =
(t′i, x′i, y
′i, z
′i) is the same event observed in another inertial frame K ′.
2. Spatial distance between two events i and j in the frame K is
dij =√
(xi − xj)2 + (yi − yj)2 + (zi − zj)2. (1)
3. If a signal is propagating with the speed of light,
(xi − xj)2 + (yi − yj)
2 + (zi − zj)2 = c2(ti − tj)
2, (2a)
(x′i − x′j)2 + (y′i − y′j)
2 + (z′i − z′j)2 = c2(t′i − t′j)
2, (2b)
where the speed of light c is invariant in any inertial reference frame.
4. In the limit xj = xi + dx→ xi and so forth,
(cdτ)2 ≡ (cdt)2 − (dx)2 − (dy)2 − (dz)2 (3)
= (cdt′)2 − (dx′)2 − (dy′)2 − (dz′)2 = 0 = constant. (4)
0.1.2 Time Dilation
5. A particle at rest in an inertial reference frame K ′ was created at (t′1,0) and then decayedat (t′2,0). In another inertial frame K both events were observed as (t1, x1) and (t2, x2).The relative velocity between the two frames is v = βc.
(a) Show that the particle is moving with the velocity v in frame K.
v =x2 − x1
t2 − t1. (5)
(b) Proper time interval τ is measured at the frame K ′ where the particle is at rest.
(cτ)2 = (c∆t)2 − (∆x)2 = (c∆t)2(1− β2
), (6a)
τ = t′2 − t′1 =t2 − t1
γ, where (6b)
γ =1√
1− β2. (6c)
(c) If K ′ is not an inertial frame, the particle does not move with a constant velocity inan inertial frame K. In this case show that
τ = t′2 − t′1 =
∫ t2
t1
dt
γ(t). (7)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
0.1. SPECIAL THEORY OF RELATIVITY 3
The lifetime of a particle represents that in its rest frame.
6. π± and π0 are the lightest hadrons in nature. mπ0 ≈ 135 MeV and mπ± ≈ 140 MeV.The mean life time of π± is (2.6033± 0.0005)× 10−8s. If a charged pion is observed at alaboratory with v = 0.999c, how long does track of the pion?
τ = (2.6033± 0.0005)× 10−8s, (8a)
γ =1√
1− 0.9992= 22.3662720421 · · · , (8b)
∆t = γτ, (8c)
∆x = v ×∆t = 0.9× 299792458m/s× 22.3662720421× (2.6033± 0.0005)× 10−8s
= (157.10± 0.03)m. (8d)
7. Speed of light c is an invariant quantity. As we have learned, life time of a particle variesdepending on the frame. Minimum value of the life time can be measured if the particleis at rest. The life time of a particle usually means the proper life time.
8. A collection of particles with identical mean life time τ is at rest. The number of particlesN(t) decreases as time passes by.
N(t) = N(t = 0)e−Γt. (9)
Decay rate is inverse of proper life time
Γ ≡ 1
τ. (10)
The ratio of the number of particles to the initial number measured at proper life time is
N(τ)
N(0)=
1
e= 0.367879441171 · · · . (11)
9. Decay rate is not Lorentz invariant quantity. Decay rate of a particle approaches themaximum value at the rest frame.
10. The most important assumption of the special theory of relativity is that the speed oflight is invariant in any inertial frame. Explain why there is no inertial frame where lightis at rest.
0.1.3 Length Contraction
11. Consider a cube with volume L30 placed at an inertial frame K ′ with a corner matching
to the spatial axes. The frame K ′ is moving with velocity vx with respect to the frameK.
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
4
12. At t′ = 0 the origins of the two frames meet each other. At t′, the origin of K passes theend of the side. Show that the length of the side along the x′−axis in the K ′ is
L0 = vt′. (12)
13. Show that the two events in the K ′ frame are
(0, 0, 0, 0) and (t′, L0, 0, 0) =
(L0
v, L0, 0, 0
). (13)
14. The origin of K is at rest in the frame K. Show that the above two events measured atthe frame K are
(0, 0, 0, 0) and (t, 0, 0, 0) =
(L
v, 0, 0, 0
). (14)
15. The length L of the side along the x−axis measured in the frame K is the speed of thecube v times
L = v × t. (15)
Using Lorentz invariance, show that
(ct)2 = (ct′)2 − L20 → L2
β2=
L20
β2− L2
0
→ L =√
1− β2L0 =L0
γ. (16a)
16. Show that the volume of the cube measured in the frame K is
V =V0
γ. (17)
17. There is a box with rest volume V containing N particles. If the box is at rest in theframe K ′ and the frame is moving with speed v with respect to another inertial frame K,the density of the particle in the box is depending on the frame.
ρ[K ′] =N
V, (18a)
ρ[K] =N
V/γ= γρ[K ′]. (18b)
Therefore, the particle density increases with the multiplicative factor γ compared to thatmeasured in the rest frame.
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
0.2. LORENTZ TRANSFORMATION 5
0.2 Lorentz Transformation
1. We would like to use the same dimensions for both time and length. Therefore, wemultiply the speed of light to time so that ct has the dimensions of length. And we definethe four vector xµ as
xµ = (x0, x1, x2, x3) ≡ (ct, x, y, z). (19)
We call this four-vector by contravariant vector.
2. We assumed(cτ)2 = (ct)2 − x2 − y2 − z2 (20)
is invariant for any inertial frame. Especially, the proper time dτ for the light is vanishingin any case so that the light can travel with the speed of light. Lorentz transformation isthe relation between the two four vectors satisfying the invariance condition (20).
3. We user2 = r2 = x2 + y2 + z2, (21)
for three-vectors.
4. For four-vectors, we have to define such inner product in a different way because of thesign difference between space and time. A convenient way of writing inner product isdefining the four-squared length as
x2 ≡ (ct)2 − x2 − y2 − z2. (22)
If we define covariant vector xµ as
xµ = (x0, x1, x2, x3) ≡ (ct,−x,−y,−z), (23)
we can writex2 ≡ xµx
µ. (24)
Note that xµxµ and xµxµ are not Lorentz invariant.
xµxµ = xµxµ = (ct)2 + x2 + y2 + z2.← wrong! (25)
5. Lorentz transformation is thenx′µ = Λµ
νxν , (26)
keeping the lengthx′2 = x2. (27)
6. For a while, we neglect the transverse directions with respect to the relative motion.We consider the case in which the relative motion is along the x−axis. By explicitevaluation to see if c2t2 − x2 = c2t′2 − x′2, show that the following transform is a Lorentztransformation (
ct′
x′
)=
(cosh α − sinh α− sinh α cosh α
)(ctx
), (28)
where α is a real number. Use cosh2 α− sinh2 α = 1.
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
6
7. Show that the inverse transform is(ctx
)=
(cosh α sinh αsinh α cosh α
)(ct′
x′
). (29)
8. Show that the Lorentz transform (28) can be written as(ct′
x′
)=
(γ −γβ−γβ γ
)(ctx
)(30)(
ctx
)=
(γ γβγβ γ
)(ct′
x′
)(31)
where α = cosh−1 γ = tanh−1 β
9. Show that K ′ is moving with a constant velocity v = βc along the positive x-axis in K.
10. K ′ is moving with the velocity βc in K. x = x‖ + x⊥, β · x⊥ = 0, and β · x‖ = β · x.Show that
x′ = x′‖ + x′
⊥ (32a)
x′‖ = β
β · x′
β2(32b)
x′⊥ = x′ − β
β · x′
β2(32c)
ct′ = γ(ct− β · x‖
)= γ (ct− β · x) (32d)
x′‖ = γ
(−βct + x‖
)= βγ
(−ct +
β · xβ2
)(32e)
x′⊥ = x⊥ = x− β
β · xβ2
(32f)
11. Arranging previous results, show that
ct′ = γ (ct− β · x) (33)
x′ = x + β
(−γct +
γ − 1
β2β · x
)(34)
= x + βγ
(−ct +
γ
1 + γβ · x
)(35)
12. From Eq. (35), read the matrix representation of the transform matrix Λµν .
x′µ = Λµνx
ν (36a)
Λµν =
γ −γβ1 −γβ2 −γβ3
−γβ1 1 + (γ−1)(β1)2
β2
(γ−1)β1β2
β2
(γ−1)β1β3
β2
−γβ2 (γ−1)β2β1
β2 1 + (γ−1)(β2)2
β2
(γ−1)β2β3
β2
−γβ3 (γ−1)β3β1
β2
(γ−1)β3β2
β2 1 + (γ−1)(β3)2
β2
µν
(36b)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
0.3. FOUR-VECTORS AND INVARIANTS 7
From Eq. (35), read the matrix representation of the inverse transform matrix.
x′µ = Λµνx
ν (37a)
xµ = (Λ−1)µνx
′ν (37b)
(Λ−1)µν = Λµ
ν(β → −β) (37c)
=
γ γβ1 γβ2 γβ3
γβ1 1 + (γ−1)(β1)2
β2
(γ−1)β1β2
β2
(γ−1)β1β3
β2
γβ2 (γ−1)β2β1
β2 1 + (γ−1)(β2)2
β2
(γ−1)β2β3
β2
γβ3 (γ−1)β3β1
β2
(γ−1)β3β2
β2 1 + (γ−1)(β3)2
β2
µν
(37d)
Show that
Λ−1(β) = Λ(−β) = gT Λ(β)g. (38)
13. Taking the non-relativistic limit, show that in the limit β → 0
t′ = t− v · xc2
, (39a)
x′ = x− vt. (39b)
0.3 Four-vectors and Invariants
0.3.1 Four-vector Notation and Metric Tensor
1. If a vector A satisfies the same transformation condition as that of x, A is a four-vector.
x′µ = Λµνx
ν (40)
A′µ = ΛµνA
ν (41)
where any repeated indices are summed over 0,1,2, and 3.
2. If both Aµ and Bµ are four-vectors, their inner product is Lorentz invariant.
A′ ·B′ = A ·B : A′µB′µ = AµBµ. (42)
3. Recall covariant vector Aµ and contravariant vector Aµ:
Aµ ≡ (A0,−A), Aµ ≡ (A0, +A) (43)
and check the following.
A ·B ≡ A0B0 −A ·B = AµBµ = AµBµ. (44)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
8
4. Metric tensor gµν = gµν is useful in the transformation between covariant and contravari-ant vectors.
gµν = gµν ≡
1 0 0 00 −1 0 00 0 −1 00 0 0 −1
µν
. (45)
Show thatAµ = gµνAν , Aµ = gµνA
ν , A ·B = gµνAµBν = gµνAµBν (46)
One of any pair of repeated indices is contravariant. And the other is covariant.
5. Show that
gµν = gµαgαν = gµ
ν = gµαgαν =
1 0 0 00 1 0 00 0 1 00 0 0 1
µν
(47)
6. Show that the proof A′ ·B′ = A ·B is equivalent to prove
ΛµαΛµ
β = gαβ =
1 0 0 00 1 0 00 0 1 00 0 0 1
αβ
. (48)
0.3.2 Four-Derivative
7. Show that∂xµ
∂xν=
1 if µ = ν,0 if µ 6= ν.
(49)
8. Show that∂xµ
∂xµ= 4 and
∂xµ
∂xµ= 4. (50)
Both ∂xµ
∂xµ and ∂xµ
∂xµare found to be Lorentz invariant. Therefore,
∂
∂xµ
= derivative”µ”(contravariant) and∂
∂xµ= derivative ”µ”(covariant) (51)
9. If you do not trust above argument, you can check it through brutal-force calculation.
(a) From the definition of Lorentz transformation, show that
Λµν =
∂x′µ
∂xν. (52)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
0.3. FOUR-VECTORS AND INVARIANTS 9
(b) From the identity (52), show that
∂µ ≡ ∂
∂xµ
(53)
is a contravariant vector∂′µ = Λµ
ν∂ν . (54)
(c) With the same method, show that ∂µ ≡ ∂∂xµ is a covariant vector
∂′µ = Λµν∂ν . (55)
10. Show that four-divergence is Lorentz invariant.
∂ · J = ∂µJµ = ∂µJµ =
∂ρ
∂t+ ∇ · J , (56)
where Jµ = (ρ, J).
11. Show that four-dimensional Laplacian of a Lorentz invariant function is also Lorentzinvariant.
∂2 ≡ ∂ · ∂ =∂2
∂t2−∇2 =
∂2
∂t′2−∇′2 = ∂′ · ∂′. (57)
0.3.3 Derivation of Useful Four-vectors
In this subsection we derive useful four-vectors using known Lorentz covariant four-vectorsand Lorentz invariant quantities.
12. We know xµ is a four-vector.x′µ = Λµ
νxν . (58)
13.
14. If Aµ is a four-vector and a is a scalar(invariant), aAµ is a four-vector.
15. Consider a particle at rest in K ′
ct = γ(cτ + vx′/c) = γcτ (59a)
x = γ(x′ + vτ) = γβcτ (59b)
dxµ
dt= (c, βc) = (c, v) → dxµ
dt
dxµ
dt= c2(1− β2) (59c)
dx′µ
dt′=
d
dτ(cτ,0) = (c,0) → dx′µ
dt′dx′µdt′
= c2 (59d)
dxµ
dtis NOT a four-vector because t is not a scalar.
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
10
16. Four-velocityxµ is a four-vector and proper time τ is a scalar. Show that uµ
uµ ≡ dxµ
dτ= γ(c, v) → u2 = c2 (60)
is a four-vector. uµ is the four-velocity of a particle. At rest the four-velocity reduces into
uµ = (c,0). (61)
17. Four-momentumuµ is a four-vector and the rest mass m0 is a scalar. Four momentum is a four-vector.
pµ ≡ m0uµ = m0γ(c, v), (62a)
p2 = m20c
2. (62b)
At rest the four-momentum becomes
pµ = (m0c,0). (63)
18. Show that
pµ =
(E
c, p
)→ E = mc2 = m0γc2 (64a)
E =√
(m0c2)2 + p2c2 (64b)
γ =E
m0c2(64c)
β =pc
E(64d)
19. Show that the Lorentz transform from a frame where a particle is at rest to the framewhere the particle is moving with momentum pµ = (E/c, p) is(
γ γβγβ γ
)=
1
mc2
(E pcpc E
), (65)
where m is the rest mass of the particle.
20. Velocity additionFrom the Lorentz transformation
t′ = γ
(t− β · x
c
), (66a)
x′ = x + βγ
(γβ · x1 + γ
− ct
), (66b)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
0.4. 2-BODY KINEMATICS 11
derive the formula for velocity addition
u′ =dx′
dt′=
u + vγ[
γu·vc2(1+γ)
− 1]
γ(1− u·v
c2
) , (67)
where u = dxdt
and v = βc.
21. From the velocity addition formula, derive the relation for the angle of the trajectory.
tan θ′ =u′2
u′1=
u sin θ
γ(u cos θ − v), (68)
where u = (u1, u2, 0), u′ = (u′1, u′2, 0), β = β(1, 0, 0), tan θ = u2/u1
0.4 2-body Kinematics
0.4.1 Center-of-Momentum Frame
1. Let us consider a collision of two particles. In general their momenta are written as
p1 = (E1, p1), p21 = m2
1, (69a)
p2 = (E2, p2), p22 = m2
2. (69b)
2. Invariant mass of the two particles is defined by
M2 = (p1 + p2)2 = (E1 + E2)
2 − (p1 + p2)2. (70)
Evidently, the invariant mass is Lorentz invariant.
3. In a center of momentum frame, expressions are greatly simplified.
p1 = (E∗1 , p
∗1), (71a)
p2 = (E∗2 , p
∗2), p∗
1 + p∗2 = 0, (71b)
M2 = (p1 + p2)2 = (E∗
1 + E∗2)
2. (71c)
Note that the center-of-momentum energy E∗1+E∗
2 , total energy in the center-of-momentumframe, is expressed in terms of Lorentz invariant quantity (p1 +p2)
2. If we know the com-plete components of the four-momenta for the two particle in any inertial frame, we cancalculate the center-of-momentum energy
√(p1 + p2)2 because the expression is Lorentz
invariant.
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
12
0.4.2 Fixed-Target or CM
In a fixed target frame, a particle hits the other particle at rest. In the center-of-momentum (CM) frame, the sum of two three-momenta of the colliding particles is van-ishing.
4. Let us consider a collision of two particles in any frame K.
(a) Sum of three-momenta for the two particles are
P = p1 + p2. (72)
If we are in the CM frame, the P vanishes.
(b) Sum of energies of the two particles are given by
E = E1 + E2. (73)
(c) Invariant mass is
M =√
(p1 + p2)2 = E∗1 + E∗
2 . (74)
(d) The velocity and γ factor of the system of particle P µ = (E, P ) are expressed asfollows.
β =P
E=
p1 + p2
E1 + E2
, (75a)
γ =E
M=
E√E2 − P 2
=E1 + E2√
(E1 + E2)2 − (p1 + p2)2. (75b)
5. Let us define θ by the scattering angle at the fixed-target frame.
(a) Show that cos θ can be expressed in terms of Lorentz invariant quantity.
cos θ =E1E2 − p1 · p2√
(E21 −m2
1)(E22 −m2
2). (76)
(b) Show thatM2 = m2
1 + m22 + 2(E1E2 − p1 · p2), (77)
where M2 = (p1 + p2)2.
6. Let us consider the case p1 is at rest; p1 = (m1,0) and p2 = (E2, p2),
(a) Express the energy and momentum of the incoming particle (p2) in terms of Lorentzinvariant quantities.
E2 =p1 · p2
m1
, (78a)
|p2| =
√(p1 · p2)2 −m2
1m22
m1
. (78b)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
0.4. 2-BODY KINEMATICS 13
(b) Express the γ factor and velocity β of the incoming particle in terms of Lorentzinvariant quantities.
γ =E2
m2
=p1 · p2
m1m2
(79a)
β =|p2|E2
=
√(p1 · p2)2 −m2
1m22
p1 · p2
. (79b)
7. Energy and momentum of each particle in two-body collision can be expressed in termsof Lorentz invariants. Let us first choose the center-of-momentum frame. The four-momenta for the two particles and their sum are p1 = (E∗
1 , p∗), p2 = (E∗
2 ,−p∗), andP = p1 + p2 = (M,0). Show that
M = E∗1 + E∗
2 , (80a)
E∗1 =
P · p1
M=
M2 + m21 −m2
2
2M, (80b)
E∗2 =
P · p2
M=
M2 + m22 −m2
1
2M. (80c)
8. The magnitude of the three-momentum of the two particles are same in the center-of-momentum frame. Derive the following formula for the magnitude of the three-momentumin the center-of-momentum frame.
|p∗| =λ1/2(M2, m2
1, m22)
2M(81a)
=
√[M2 − (m1 + m2)2] [M2 − (m1 −m2)2]
2M(81b)
λ(a, b, c) = a2 + b2 + c2 − 2(ab + bc + ca) (81c)
=
[a−
(√b +√
c)2] [
a−(√
b−√
c)2]
. (81d)
0.4.3 Lorentz transformation to the rest frame of an arbitraryparticle
9. pµ1 is a four-vector. Show that the components of this vector in the frame where P µ =
(E, P ) is at rest becomes
E ′1 = γ (E1 − β · p1) (82a)
p′1 = p1 + βγ
(γβ · p1
1 + γ− E1
)(82b)
β =P
E(82c)
γ =E
M, P 2 = M2 (82d)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
14
10. If p1 = P , show that
E ′1 = M (83a)
p′1 = 0 (83b)
11. If P = p1 + p2, show that p′1 and p′2 in the rest frame of the P become
E ′1 =
1
M(EE1 − P · p1) =
P · p1
M(84a)
E ′2 =
1
M(EE2 − P · p2) =
P · p2
M(84b)
p′1 = p1 +
P
M
(P · p1
E + M− E1
)(84c)
p′2 = p2 +
P
M
(P · p2
E + M− E2
)(84d)
E ′1 + E ′
2 = M (84e)
p′1 + p′
2 = 0 (84f)
0.5 Lorentz Transformation of integrals
1. There is a function f(x), x = (x1, · · · , xn) and we want to change the variables into
yk = yk(x1, · · · , xn), k = 1, 2, · · · , n (85)
Show that ∫Rx
f(x)dnx =
∫Ry
g(y)dny (86a)
g(y) =∂(x1, · · · , xn)
∂(y1, · · · , yn)f [x(y)], (86b)
where Rx and Ry are the region of integration with respect to xk or yk, respectively.
2. The Jacobian is defined by
∂(x1, · · · , xn)
∂(y1, · · · , yn)=
∣∣∣∣∣∣∣∣∣∣
∂x1
∂y1
∂x2
∂y1· · · ∂xn
∂y1
.
.
.∂x1
∂yn
∂x2
∂yn· · · ∂xn
∂yn
∣∣∣∣∣∣∣∣∣∣. (87)
3. If β = (β, 0, 0), show that p′µ = Λµνp
ν
Λµν =
γ −γβ 0 0−γβ γ 0 0
0 0 1 00 0 0 1
µν
. (88)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
0.6. KINEMATICS OF 2-2 SCATTERING 15
4. Show that∂(p0p1p2p3)
∂(p′0p′1p′2p′3)= DetΛ = 1. (89)
5. Using the general Lorentz Boost matrix, prove
d4p = d4p′ ↔ dp0dp1dp2dp3 = dp′0dp′1dp′2dp′3. (90)
This is equivalent to∂(p0p1p2p3)
∂(p′0p′1p′2p′3)= DetΛ = 1. (91)
Therefore, d4p is Lorentz invariant.
6. Show that
δ[f(x)] =∑ δ(x− xi)∣∣ df
dx(xi)
∣∣ , (92a)
where f(x) has a simple pole at x = xi.
7. Show that
δ(x2 − a2) =δ(x + |a|) + δ(x− |a|)
2|a|. (93)
8. Show that δ(p2 −m2)d4p is Lorentz invariant. E =√
m2 + p2
δ(p2 −m2)d4p = δ((p0 − E)(p0 + E)
)dp0dp1dp2dp3
=δ(p0 − E) + δ(p0 + E)
2Edp0dp1dp2dp3. (94a)
9. Show that
θ(p0)δ(p2 −m2)d4p =d3p
2E. (95)
is Lorentz invariant.
10. Explain the Lorentz invariance of d4p and d3p/(2E) using only time dilation and lengthcontraction.
0.6 Kinematics of 2-2 Scattering
0.6.1 Independent Kinematic Variables
1. Transition amplitude of four particles with momenta p1, p2, p3, and p4 is a function ofthe four four-momenta
Tfi = T (p1, p2, p3, p4), (96)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
16
where we ignore any other variables like spin, isospin, so on. The function is a Lorentzscalar. We want to know how many independent Lorentz scalars made of the scalarproducts of the four four-momenta exist.
2. Show that there are 10 scalar products made of the four four-vectors.
4(from p2i ) +
4 · 32
(from pi · pj, i 6= j) = 10. (97)
3. Among the 10 scalar products
p21 = m2
1, p22 = m2
2, p23 = m2
3, p24 = m2
4 (98)
are known from on-shell conditions so that we have at most 10−4 = 6 independent scalarproducts.
4. In addition, there are four more conditions from energy-momentum conservation
p1 + p2 + p3 + p4 = 0. (99)
Finally we end up with only two independent scalar products.
0.6.2 Lab(fixed-target) system
5.
p1(m1) + p2(m2) → p3(m3) + p4(m4) (100a)
p1 = (E1, p1), p2 = (m2,0) (100b)
p3 = (E3, p3), p4 = (E4, p4) (100c)
6. Energy-momentum conservation requires
E1 + m2 = E3 + E4, p1 = p3 + p4. (101)
7. For i = 1, 3, 4, show that
Ei =p2 · pi
m2
(102a)
|pi| =
√(p2 · pi)2 −m2
2m2i
m2
(102b)
8. If the scattering angle θi(i = 3, 4) is the angle between p1 and pi, show that
cos θi =p1 · p2pi · p2 − p1 · pim
22√
[(p1 · p2)2 −m21m
22] [(pi · p2)2 −m2
i m22]
. (103a)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
0.6. KINEMATICS OF 2-2 SCATTERING 17
0.6.3 Center-of-Momentum(CM) system
9.
p1(m1) + p2(m2) → p3(m3) + p4(m4), (104a)
p1 = (E1, p1), p2 = (E2, p2), (104b)
p3 = (E3, p3), p4 = (E4, p4), (104c)
P = p1 + p2 = p3 + p4 = (E, P ). (104d)
In the CM system, energy-momentum conservation requires
E = M = E1 + E2 = E3 + E4, P = p1 + p2 = p3 + p4 = 0. (105)
10. Show that√
s is the CM energy E;
s ≡ (p1 + p2)2 = (p3 + p4)
2 = E2, (106a)
p1 · p2 =1
2
(s−m2
1 −m22
), p3 · p4 =
1
2
(s−m2
3 −m24
), (106b)
11. Show that
p1 · p3 = −1
2
(t−m2
1 −m23
), p2 · p4 = −1
2
(t−m2
2 −m24
), (107a)
p1 · p4 = −1
2
(t−m2
1 −m24
), p2 · p3 = −1
2
(t−m2
2 −m23
), (107b)
t ≡ (p1 − p3)2 = (p2 − p4)
2, u ≡ (p1 − p4)2 = (p2 − p3)
2, (107c)
s + t + u = m21 + m2
2 + m23 + m2
4. (107d)
12. Show that in the CM system
E1 =p1 · PM
=s + m2
1 −m22
2√
s, E2 =
p2 · PM
=s + m2
2 −m21
2√
s(108a)
E3 =p3 · PM
=s + m2
3 −m24
2√
s, E4 =
p4 · PM
=s + m2
4 −m23
2√
s. (108b)
13. Show that in the CM system
|p1| = |p2| =√
(p1 · P )2 −m21s√
s=
√(p2 · P )2 −m2
2s√s
=λ1/2(s, m2
1, m22)
2√
s=
√[s− (m1 + m2)2] [s− (m1 −m2)2]
2√
s(109a)
|p3| = |p4| =√
(p3 · P )2 −m23s√
s=
√(p4 · P )2 −m2
4s√s
=λ1/2(s, m2
3, m24)
2√
s=
√[s− (m3 + m4)2] [s− (m3 −m4)2]
2√
s(109b)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
18
14. Show that
1
2(1− cos θ) = sin2 θ
2, (110a)
1
2(1 + cos θ) = cos2 θ
2. (110b)
15. If the scattering angle θ is the angle between p1 and p3 in the CM frame, show that
t = (p1 − p3)2 = (E1 − E3)
2 − (p1 − p3)2 (111a)
= (E1 − E3)2 − (|p1| − |p3|)2 − 4|p1||p3| sin2 θ
2, (111b)
tmin < t < tmax, (111c)
tmax = (E1 − E3)2 − (|p1| − |p3|)2, (111d)
tmin = (E1 − E3)2 − (|p1|+ |p3|)2, (111e)
E1 − E3 =m2
1 −m22 −m2
3 + m24
2√
s. (111f)
16. If the scattering angle θ is the angle between p1 and p3, show that
cos θ =p1 · Pp3 · P − p1 · p3s√
[(p1 · P )2 −m21s] [(p3 · P )2 −m2
3s]. (112)
0.7 Phase Space
1. Show that ∫eipxdx = 2πδ(p) (113)∫
eip·xd4x = (2π)4δ4(p) = (2π)4δ(p0)δ(p1)δ(p2)δ(p3) (114)
2. Consider a scattering
k1 + k2 → p1(m1) + p2(m2) + · · ·+ pn(mn). (115)
Show that the energy-momentum-conservation delta function∫ei(p1+p2+···+pn)·xe−i(k1+k2)·xd4x = (2π)4δ4
[n∑
i=1
pi − (k1 + k2)
](116)
is Lorentz invariant.
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
0.7. PHASE SPACE 19
3. Show that the phase-space element
δ4
[n∑
i=1
pi − (k1 + k2)
]n∏
i=1
δ(p2i −m2
i )d4pi (117)
is Lorentz invariant.
4. Show that
δ4
[n∑
i=1
pi − (k1 + k2)
]n∏
i=1
δ(p2i −m2
i )d4pi (118a)
= δ
(n∑
i=1
Ei − E1(k1)− E2(k2)
)δ3
[n∑
i=1
pi − k1 − k2
]n∏
i=1
d3pi
2Ei
(118b)
is Lorentz invariant, where p0i = Ei =
√m2
i + p2i and Ei(ki) =
√M2
i + k2i .
5. Show that
d3p = |p|2d|p|dΩ =1
2|p|dp2dΩ, (119a)
dΩ = sin θdθdφ, 0 < θ < π, 0 < φ < 2π (119b)
= d cos θdφ, − 1 < cos θ < 1. (119c)
6. If the angular dependence of the integrand is independent of the particular direction ofp1, show that
d3p1d3p2 = 4π|p1|2d|p1||p2|2d|p2|d cos θ12dφ12, (120a)
where Ω12(θ12, φ12) is the angle between p1 and p2.
7. n-body phase space element
dΦn = (2π)4δ4
[n∑
i=1
pi − (k1 + k2)
]n∏
i=1
d3pi
2Ei(2π)3(121)
8. 2-body phase space element
dΦ2 = (2π)4δ4(p1 + p2 − k1 − k2)d3p1d
3p2
4E1E2(2π)6(122a)
= (2π)−2δ(E1 + E2 −√
s)d3p1
4E1E2
(122b)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
20
9. Show that in the CM frame
E1 =√
m21 + p2, E2 =
√m2
2 + p2 (123a)√
s = E1 + E2 (123b)
p1 = −p2 = p, |p| = λ1/2(s, m21, m
22)
2√
s(123c)
∂E1
∂|p|=|p|E1
,∂E2
∂|p|=|p|E2
, (123d)
δ(E1 + E2 −√
s) =δ(|p1| − |p|)|p|E1
+|p|E2
(123e)
=E1E2
|p|√
sδ(|p1| − |p|) (123f)
10. Therefore
dΦ2 =1
(2π)2· |p|4√
sdΩ (124a)
|p| =λ1/2(s, p2
1, p22)
2√
s(124b)
this is valid in any inertial frame because of the Lorentz invariance.In case there is no angular dependence in the integrand(matrix element),∫
dΦ2 =|p|
4π√
s(125)
If p21 = p2
2 = 0, ∫dΦ2 =
1
8π(126)
11. 3-body phase space element Show that
dΦ3 = (2π)4δ4(p1 + p2 + p3 − k1 − k2)d3p1d
3p2d3p3
8E1E2E3(2π)9(127a)
= (2π)−5δ(E1 + E2 + E3 −
√s) d3p1d
3p2
8E1E2E3
(127b)
In the k1 + k2 CM frame show that
E1 =√
m21 + p2
1, E2 =√
m21 + p2
2 (128a)
E3 =√
m23 + p2
1 + p22 + 2|p1||p2| cos θ12 (128b)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
0.7. PHASE SPACE 21
12. Show that
∂E1
∂|p1|=|p1|E1
,∂E2
∂|p2|=|p2|E2
(129a)
∂E3
∂ cos θ12
=|p1||p2|
E3
(129b)
d|p1|d|p1|d cos θ12 =E1E2E3
p21p
22
dE1dE2dE3 (129c)
dΦ3 = (2π)−5δ(E1 + E2 + E3 −
√s)× 4πd cos θ12dφ12 (130a)
× p21d|p1|p2
2d|p2|8E1E2E3
(130b)
= (2π)−5dE1dE2πdφ12
2. (130c)
Relativistic Kinematics, COPYRIGHT c© 2006, BY JUNGIL LEE May 18, 2006
