PHGN462: Advanced E&M
Lorentz 4-vector Exercise Sheet – October 12, 2009
Work in groups of two, but each student must turn in his/her own worksheet. In preparation for our treatment of electromagnetic potentials using Lorentz 4-vectors, it will be useful to review the standard results of special relativity and recast
them in terms of 4-vectors. We will only consider the case where the relative velocity, v, of two inertial frames of reference
is along the x-axis. To keep the units of each component of the four vector the same, we multiply the time by c. Start with
the Lorentz transformations:
x0 = γ(x − βct)
y0 = y
z0 = z
ct0 = γ(ct − βx)
p
where β = v/c and γ = 1/ 1 − β 2 . The idea of using 4-vectors is to give a matrix representation to the Lorentz transformations. Define the so-called contravariant and covariant 4-vectors:




x
x
 y 
 y 
~x
~x
=


xµ = 
=
,
x
=
(1)
µ
 z 
 z 
ct
−ct
ct
−ct
These provide a convenient way to construct the Lorentz invariant:
~x
xµ xµ → (xµ )T xµ = ~x ct
= ~x · ~x − (ct)2 .
−ct
(2)
where we have adopted the Einstein summation convention whereby any repeated index is implicitly summed over.
Exercise 1: Show that x0µ x0µ = xµ xµ .
In terms of 4-vectors the Lorentz transformation can be written:

x0µ = Λµν xν ,
γ
 0
µ

where Λν = 
0
−βγ
The derivative operations also make up a 4-vector: ∂µ =
∂
∂xµ
and ∂ µ =
Exercise 2: a. What is the invariant: ∂µ xµ ?
b. What is the invariant ∂ µ ∂µ ?
1
∂
∂xµ .

0 0 −βγ
1 0
0 
.
0 1
0 
0 0
γ
(Note the contra- and co-variant forms.)
(3)
The velocity addition formulas are given (without derivation) by:
u0x + v
1 + u0x v/c2
u0y
uy =
γ(1 + u0x v/c2 )
ux − v
1 − ux v/c2
uy
u0y =
γ(1 − ux v/c2 )
u0x =
ux =
(4)
The relativistic energy and momentum of a point particle are given by
E = γmc2
p~ = γm~u
(5)
p
where here γ = 1/ 1 − u2 /c2 and m is the rest mass.
Exercise 3: Derive the Einstein triangle relation: E 2 = (pc)2 + (mc2 )2 .
Exercise 4: For the special case where ~u0 = (0, u0y , 0, 0), use the velocity addition formulas to show that the energy and
momentum (times c) satisfy the same Lorentz transformation formulas as that of the space-time transformation Eq. [1],
namely,
p0x c = γ(px c − βE)
p0y c = py c
p0z c = pz c
E 0 = γ(E − βpx c)
2
This means that the energy-momentum 4-vector,


px c
 py c 

pµ = 
 pz c  ,
E
(6)
transforms exactly as xµ . Any vector for which this is true we call a Lorentz vector.
Exercise 5: What is the invariant pµ pµ ?
Exercise 6: Let’s conclude with a look at energy-momentum conservation in relativistic collisions. The Lorentz transformation is linear; so any sum of Lorentz vectors will still be a Lorentz vector. Specifically, the total 4-momentum of a
collision is a Lorentz vector, and, since the energy and momentum are conserved, the total Lorentz 4-momentum will be
conserved as well. As an example, consider the laboratory production of a pion by colliding a high energy proton into a
hydrogen (proton) target. Our goal is to find the threshold laboratory kinetic energy needed to produce a pion. In the center
of momentum frame this would be very easy, because at threshold after the collision all the particles would be at rest. Let
Ecm = mp c2 + Tcm be the initial energy of each of the protons where mp is the proton rest mass. The total energy is 2Ecm
(and in the cm frame, of course, the total momentum is zero.) Therefore, at threshold we would have 2Ecm = 2mp c2 + mπ c2 ;
so Tcm = 12 mπ c2 as we would have guessed. We could use the Lorentz transformations to find pL , but an easier way is to
construct Lorentz invariants using the square of the total 4-momentum. In the cm frame the total 4-momentum squared is
(2Ecm )2 = (2mp c2 + mπ c2 )2 . We now proceed to calculate the same quantity in the lab frame.
Suppressing the y- and z-directions, let
pL c
pµ1L =
EL
be the 4-momentum in the lab frame of particle 1 (projectile ) where pL and EL are the relativistic momentum and energy
of the projectile proton. The four momentum of the target proton (particle 2) in the lab frame is just,
0
pµ2L =
.
mp c2
Thus, the total 4-momentum in the lab frame is given by:
pµL = pµ2L + pµ2L =
pL c
EL + mp c2
Now its your turn. Square the total 4-momentum in the lab frame and, since it is a Lorentz invariant, we can set it equal to
the same quantity we already calculated in the cm frame. Use the Einstein triangle relation to solve for pL and from that
calculate TL = EL − mp c2 . Plug in the values mp c2 = 938 MeV and mπ c2 = 135 MeV, to find TL in MeV.
3