Lecture 4 - Lorentz Invariants

E. Daw

March 2, 2012

1 Review of Lecture 3

Last time we studied the scattering of photons off particles. Defining

Eγi as the energy of the γ–ray before scattering, Eγf as its energy after
scattering, and θ as the scattering angle, we used conservation of energy
and the three components of momentum from before the scattering to
after it to show that
(Eγi − Eγf ) 1 − cos θ
= . (1)
Eγi Eγf m 0 c2

Defining ∆ to be the energy lost by the γ–ray, equal to Eγi − Eγf , we

were also able to show that the maximum energy imparted to a particle
of rest mass m is
(Eγi )2
∆= 2 . (2)
Eγi + mc2
This lost energy is imparted to the massive particle. If that particle
is an electron, it may interact in detectors designed to measure the
γ–rays. Figure 1 shows a spectrum taken with a γ–ray spectrometer.
Notice that there are two examples of narrow peaks at high energies,
then a gap, then the edge of a broader, flattish spectrum of lower en-
ergy particles. These are Compton scattered electrons, and the highest
energy they can have is δ given by the formula above. So given the
energy of the top of the Compton spectrum, you can relate this energy
∆ to the energy of the narrow peak, which is Eγi , using Equation 2.

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2 Transforming between different inertial
observers

So far we have emphasised conservation of energy and momentum, with

the caveat that these quantities are only conserved if the same observer
measures them before and after, this observer defining a single inertial
(non–accelerating) frame of reference in which all measurements are
made. However, we know how to transform between different inertial
observers. It’s time to review what we know about this and see how it
can be extended, again with a view to making calculations easier.

First, we know how to transform the space–time coordinates of an event

between their values measured by different observers having a relative
velocity v = βc between them aligned with the x–axis. Suppose the
coordinates of an event for observer O are: (ct, x, y, z), and those for a
second observer O0 are (ct0 , x0 , y 0 , z 0 ). If O0 is moving to the right along
the x–axis with respect to O at velocity v = βc, then the coordinates
are related by the Lorentz transformations, which are (of course),

ct0 = γ(ct − β x)
x0 = γ(x − β ct)
(3)
y0 = y
z0 = z

We also learned that another set of four quantities have exactly the
same coordinate transformations between their values for observers O
and O0 . These four quantities are (E/c, px , py , pz ), the energy divided
by c, and the three components of the momentum, of a particle. So we
can write the Lorentz transformations on energy and momentum like
this:

E 0 /c = γ(E/c − β px )
0
px = γ(px − β E/c)
0 (4)
py = py
0
pz = pz

It turns out that the Lorentz transformations for energy and momen-
tum are in many ways more useful than the Lorentz transformations
for time and position, especially when you are doing particle physics
problems. So let’s stop and do an example to see how we can use these
transformations.

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An electron and a positron are approaching each other with equal and
opposite momentum in the lab. The total energy of each particle in the
lab is 45 GeV. Find the total energy of one proton in the rest frame of
the other one.

To solve this one, pick a particle whose rest frame we are going to
transform to. Say the electron is moving in the +x direction and the
positron is moving in the −x direction. Let’s transform the positrons
energy into the lab frame of the electron. First, we write down the
transformation equation for the energy. Let E be the energy of the
positron in the lab, and E 0 be the energy of the positron in the rest
frame of the electron. Next, identify the regime the electron and
positron are in. They are highly relativistic because γ = E/(mc2 ),
which is 45GeV/0.511MeV=88, 000, which is much bigger than 7, the
lower limit for the highly relativistic regime. Next, we write down the
Lorentz transformation for energy from Equation 4. We note that the
primed frame is the rest frame of the positron, which is moving in the
+x direction with respect to the lab frame, hence the sign of β is correct
in Equations 4.
E 0 /c = γ(E/c − β px ). (5)
. But because we’re in the highly relativistic regime, we set β = 1 and
obtain
E 0 /c = γ(E/c − px ). (6)
Now for the electron, because again it’s highly relativistic, E = cp,
where p is the magnitude of the momentum of the electron in the lab
frame. The sign of the electron momentum is negative, so the momen-
tum is px = −E/c. The only other thing we don’t know is γ, which is
the gamma factor for the positron in the lab frame, which is 88, 000 as
calculated above. Putting numbers into Equation 6, we get

E 0 /c = γ(45[GeV]/c − (−45[GeV]/c)) (7)

Multiplying all terms by c and cancelling the double minus sign to add
up the two 45GeV terms to 90GeV, we obtain

E 0 = γ × 90[GeV] = 88, 000 × 90[GeV] = 7.9 × 106 [GeV]. (8)

Again, if we use energy units and are careful to make simplifying ap-
proximations, these calculations really aren’t difficult. The main hazard
with this one is to be clear about the sign of the momentum. Because
we chose to transform to the frame of the right–going particle, there
was already a − sign by the β, but this was cancelled by the negative
sign in the momentum of the positron, travelling to the left. Now,
suppose we had chosen to transform to the rest frame of the positron
from the lab frame. Then the Lorentz transformations would have had

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a + sign by the β, but the momentum of the electron in the lab is also
positive. In both cases, you end up with 90GeV in the bracket, in one
case because the two − signs cancel, and in the other because there are
no minus signs. So, everything works out.

3 Four vectors

Because (ct, x, y, z) and (E/c, px , py , pz ) have the same transformations

under changes of coordinate, we call them both 4–vectors. The vec-
tors with which you are familiar can be defined as objects possessing
a magnitude and a direction, or objects that transform in a well de-
fined way under rotations of coordinate system. 4–vectors are defined
as objects that transform under the Lorentz transformations when con-
verting between the measurements made by two different inertial (non–
accelerating) observers.

3.1 Four-displacement

An event at coordinates (ct, x, y, z) is represented by the 4–displacement

xν , where ν is a number that can take values 0,1,2 or 3. The four
components xν are x0 = ct, x1 = x, x2 = y, x3 = z. So far we
have considered displacements from the origin only. What about small
displacements between two neighbouring points. These too are four
displacements, dxν . You can see this by writing out two copies of the
Lorentz transform for neighbouring events, and subtracting them. So
suppose we have two events at (ct1 , x1 , y1 , z1 ) and (ct2 , x2 , y2 , z2 ), such
that ct2 − ct1 = c dt, x2 − x1 = dx, y2 − y1 = dy, and z2 − z1 = dz.
Then we could write for the time component of the Lorentz transform,

ct02 = γ(ct2 − β x2 )
(9)
ct01 = γ(ct1 − β x1 )

and therefore, subtracting the two equations

c dt0 = γ(c dt − β dx). (10)

And we see that small displacements are also four–vectors, because the
have the same transformation rules as the coordinates of events referred
to the origin.

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3.2 Four–momentum

Next, we define the four momentum of a particle as pµ , where p0 = E/c,

p1 = px , p2 = py , p3 = pz . From Equation 4, pµ is a four–vector.

4 Lorentz Invariants

We have seen that a class of objects called four–vectors transform

in a known way, by the Lorentz transformations, between observers.
Are there other objects that have known transformations between ob-
servers? Yes there most certainly are! Let’s see if we’ve already met
any. Recall the time dilation formula, relating the time interval dτ be-
tween two events in their rest frame, the frame where the two events
occur in the same place, and the time interval dt0 between the same two
events in an inertial (non–accelerating) frame where the two events oc-
cur in different places. So, for example, dτ could be the time between
two ticks on a clock at rest in the lab, hence occurring at the same
place in the lab frame, and dt0 could be the time interval between the
same two ticks in a frame in which the clock is moving at a constant
velocity. They are related by
dτ
dt0 = γ dτ = p . (11)
1 − β2

Because γ is always greater than 1, the events always have a greater

time interval between them in a frame in which the events do not occur
in the same place. Hence, moving clocks run slow. Now, any observer
is free to measure the speed of the moving clock and use this speed to
calculate γ between the clocks rest frame and his rest frame, then use
Equation 11 to calculate dτ . If multiple, different, inertial observers all
carry out this procedure, they will all get different answers for γ, but
they will all get the same answer for dτ . Therefore dτ is an invariant
quantity, a quantity that is the same when calculated by all inertial
observers. It is a an example of a Lorentz invariant.

Can we find some more Lorentz invariants? Yes we can. Recall the
relationship between energy, momentum and rest mass of a particle

E 2 = p2 c2 + m20 c4 . (12)

Now, rearrange this formula, and we obtain

p
m0 c2 = E 2 − p2 c2 (13)

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Once again, some inertial observer sees a particle, and measures its
total energy E and momentum magnitude p. This observer substitutes
these numbers into Equation 13, and calculates the rest energy of the
particle, m0 c2 . A different inertial observer gets different values E 0 and
p0 for energy and momentum, but the formula will give them the same
value for the rest energy of the particle. Therefore the rest energy of
the particle, m0 c2 is a Lorentz invariant quantity. Any inertial observer
calculating the rest energy will get the same answer.

5 Lorentz invariants from 4–vectors

There is a close relationship between Lorentz invariants and four vec-

tors. It turns out that one can always calculate a Lorentz invariant from
a four–vector, using the same procedure every time. The procedure is:
suppose we have a four vector Aµ . This is any quantity that transforms
under Lorentz transformations parallel to the x axis like dxµ and pµ :
0
A0 = γA0 − βγA1
0
A1 = −βγA0 + γA1
0 (14)
A2 = A2
0
A3 = A3 .

To construct an invariant quantity out of the components of Aµ , we

simply write down,

|A|2 = −(A0 )2 + (A1 )2 + (A2 )2 + (A3 )2 . (15)

It turns out, this combination is always Lorentz invariant, for any four–
vector. You can prove it easily if you like, by substitution, but I’m not
going to do it because I want to show you how to use it instead. Let’s
figure out the Lorentz invariants corresponding to dxµ and pα . For dxµ
we get

−(dx0 )2 +(dx1 )2 +(dx2 )2 +(dx3 )2 = −c2 dt2 +dx2 +dy 2 +dz 2 = ds2 (16)

This is the so–called Lorentz invariant interval between two events. The
fact that it is invariant follows directly from the time dilation formula
of Equation 11, if you set v dt0 = dx0 , where dx0 is the distance the clock
moves in time dt0 . Again, though, I’m not going to prove it now. For
pα we get
E
− (p0 )2 + (p1 )2 + (p2 )2 + (p3 )2 = −( )2 + (px )2 + (py )2 + (pz )2 . (17)
c

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Comparing with Equation 13 we can see that this sum is equal to
−m20 c2 .
E
− (p0 )2 + (p1 )2 + (p2 )2 + (p3 )2 = −( )2 + (px )2 + (py )2 + (pz )2 = −m20 c2 .
c
(18)
Because the result is equal to a product of the rest mass squared and
−c2 , and because both the rest mass and c are observer independent,
we see that we have again calculated an invariant quantity.

6 The centre of mass energy

We are now going to study another Lorentz invariant quantity, the

centre of mass energy available in a collision. Let us consider HERA,
an accelerator in which protons are accelerated up to a total energy of
920 GeV, and collided with electrons which have been accelerated up to
a total energy of 27.6 GeV. The question is, what is the highest mass
particle that could be created in such a collision? To find this out, we
need to calculate the centre of mass energy of the accelerator. This
is the total energy in the frame where the sum of the momenta of the
particles is zero. Let’s do this calculation using 4–vectors. Consider the
incident particles in the lab frame. Both are in the highly relativistic
regime. We have already written down the energies. The momentum
of each proton, taking them to be moving in the +x direction at the
collision point, is +920 GeV/c. The momentum of each electron, taking
them to be moving in the −x direction, is −27.6 GeV/c Hence the sum
of the energies is 947.6 GeV and the sum of the momenta is 892.4 GeV/c.
Hence, Now we calculate the Lorentz invariant quantity,
E2 2
− +p = −947.62 [GeV2 ]/c2 +892.42 [(GeV/c)2 ] = −101568[GeV2 /c2 ].
c2
(19)
Now, this quantity is equal to −M02 c2 . What is that? What is M0 ?
Remember that the whole quantity is Lorentz invariant, which means
that it’s the same in all frames. So, pick the centre of mass frame!
In this frame, there is no momentum, only energy. Any energy in this
frame is available to make new particles. And, the heaviest particle you
could create is a single particle whose rest energy is equal to M0 c2 , as
this would use up all the energy available, wasting nothing on kinetic
energy. Consider, you can only do this in the centre of mass frame,
because in any other frame there is momentum, and hence you have
kinetic energy. Therefore,
E2
− + p2 = −M02 c2 = −101568[GeV2 /c2 ]. (20)
c2
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and
M0 c2 = 318.7 [GeV]. (21)
2
M0 c√ is called the centre of mass energy, Sometimes it’s given the sym-
bol s. So the heaviest particle you could possibly create in this ac-
celerator would have an mass of 318.7 GeV/c2 . Note that in reality
this would never happen - the protons are not of course really point
particles, so you could never really have a point-like collision between
an electron and a whole proton; instead you would get a deep inelastic
scattering event where the electron strikes a constituent quark or gluon
of the proton. Indeed, this is how quarks were first discovered!

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 Eγi

∆
EγI

∆ Eγi ∆ Eγi

Figure 1: The energy spectrum of gamma rays from an

AmBe source, as measured by a gamma ray spectrometer.
Taken from http://upload.wikimedia.org/wikipedia/commons/f/
f2/Am-Be-SourceSpectrum.jpg