P780.02 Spring 2003 L2
Relativistic Kinematics
Richard Kass
In HEP the particles (e.g. protons, pions, electrons) we are concerned
with are usually moving at speeds close to the speed of light.
The classical relationship for the kinetic energy of the particle in terms
of its mass and velocity is not valid:
Read:
Appendix A of M&S
Kinetic Energy ≠ 1/2 mv2
Thus we must use special relativity to describe the energies and
momentum of the particles.
The total energy (E=rest +kinetic) of a particle with rest mass, mo, is:
E=m0
c2+T
2
E  mc 
moc 2
1 v / c2
  moc
2
Here v is its speed, c = speed of light, and m is sometimes called the
relativistic mass. The total momentum, p, of a particle with rest mass,
mov
mo, is:
p  mv 
 mo v
2
1 v / c 
We can also relate the total energy, E, to a particle's total momentum, p:
2
2
2 2
E  ( pc)  (moc )
P780.02 Spring 2003 L2
Richard Kass
Relativistic Kinematics 4-vectors
It is sometimes convenient to describe a particle (or a collection of
particles) by a 4-vector. The components of the momentum and
energy 4-vector, p, are given by:

p  ( E , p x , p y , p z ) or p  ( E , p ) or p  ( E , p) with c  1
The length of the 4-vector is given by:
2
2
2
2
2
2
mo2  E  p  E  px  py  pz
A particle is said to be
“on the mass shell” if
m0=rest mass
This relationship is true in ALL reference frames
(lab, center of mass,…) because it is a Lorentz invariant.
A 4-vector with length L2 is classified as follows:
Time like if L2 >0
Space like if L2<0
Light like if L2=0 (think photon!)
P780.02 Spring 2003 L2
Richard Kass
Lorentz Invariant Vs. Conserved quantity
With a Lorentz Invariant you get the same number in two different
reference systems (it is a scalar).
Let EL and pL be energy and momentum measured in LAB frame
Let Ecm and pcm be energy and momentum measured in center of mass frame
Then: E2cm-p2cm= E2L-p2L
Since (E, p) is a Lorentz invariant (as long as both are measured in same system)
With a conserved quantity you get the same number in the same
reference system but at a different time.
Let piL=initial momentum in lab (before collision)
Let pfL=final momentum in lab (after collision)
Let picm=initial momentum in CM (before collision)
Let pfcm=final momentum in CM (after collision)
Momentum conservation says:
piL= pfL AND picm= pfcm BUT NOT piL= pfcm
Relativistic Kinematics 4-vectorsRichard Kass
P780.02 Spring 2003 L2
We can also manipulate 4-vectors using contravariant/covariant
(up/down) notation: m02=pupu=guvpvpu
In this notation guv is a metric and is given (e.g.) by:
1 0 0 0 
0 1 0 0 
g  
0 0 1 0 


0 0 0 1
Thus the (scalar) product of two 4-vectors (a, b) is given by:

3
3
ab  a b    g a  b  g a  b  aob o  a1b1  a 2b 2  a3b3
 0  0
The sum of two 4-vectors is also a 4-vector.
p1  p2  (E1  E2 , p1  p2 )
The length of the sum of the 4-vectors of two particles (1,2) is:
(p1  p2 ) 2  (E1  E2 )2 ( p1  p2 )2
(p1  p2 ) 2  E12  E22  2E1E2  p12  p22  2 p1  p2
2
(p1  p2 ) 2  m12  m22  2(E1 E2 | p1|| p2 |cos  )  m12
=angle between particles, m1, m2 are rest masses
m12 is called the
invariant mass
or effective mass
P780.02 Spring 2003 L2
Richard Kass
Relativistic Kinematics 4-vectors
Example: Consider a proton at rest in the lab frame and an antiproton
with 10GeV/c of momentum also in the lab frame.
What is the energy of the antiproton in the lab frame?
Since the rest energy of a particle is a Lorentz invariant we can make us of:
2
2
c=1
mo2  E  p
For an antiproton the rest mass, m0, = 938 MeV/c2. We can re-write the above as:
2
2
E  p  mo  102  0.9382  10.044 GeV
Thus at high energies (E>>m0c2) E  |p|.
How fast is the anti-proton moving in the lab frame ?
We need to remember the energy/momentum relationship between the rest frame
(particle at rest) of the anti-proton and the lab frame:
m0 c 2
m0 v
1
Elab 
 m0 c 2 plab 
 m0 c
  v / c,  
1  (v / c) 2
1  (v / c) 2
1  2
plab / Elab  m0 c / m0 c 2    10 / 10.044  0.996
Thus v = 0.996c (fast!)
P780.02 Spring 2003 L2
More Relativistic Kinematics
Richard Kass
Colliding beam vs fixed target Collisions
As we will see later in the course cross sections and the energy available for new
particle production depend on the total energy in the center of mass (CM) frame. We
define the CM where the total vector momentum of the collision is zero.
By definition then, in the CM frame we have for two 4-vectors (p1, p2):
(p1+p2)= (E1+E2, p1+p2) = (E1+E2, 0)
If the masses of the two particles are equal as in the case of proton anti-proton
collisions then the above reduces to:
(p1+p2)= (E1+ E2, p1+ p2) = (E1+ E2, 0)= (2E, 0) (twice energy of either particle)
NOTE: the square of the energy in the CM is often called s. s= magnitude of (p1+p2)
How much energy is available in the CM from a 10 GeV/c anti-proton
colliding with a proton at rest?
Since (p1+p2) is a Lorentz invariant we evaluate in any frame we please!
We are given values in the lab frame: (p1+ p2) =(E1+mp, p1+0)
The magnitude of this 4-vector is: s = (E1+mp)2- p12= (10.044+0.938)2-102= 20.6GeV2
Thus the total energy in the CM is 4.54GeV
We could have gotten the same CM energy with two beams = 2.27 GeV !
In general the energy available for new particle production increases as:
(2mtargetEbeam)1/2 for fixed target experiments
colliding beams are
more efficient for
2Ebeam for colliding beam experiments
heavy particle production
P780.02 Spring 2003 L2
Richard Kass
Even More Relativistic Kinematics
Our final example for this section concerns the discovery of the anti-proton. In the
early 1950’s many labs were trying to find evidence of the anti-proton. At Berkeley
a new proton accelerator (BEVATRON) was being designed for this purpose.
Assuming fixed target proton-proton collisions would be used to create the
antiproton what energy proton beam (Eb) is necessary ?
The simplest reaction that conserves all the necessary quantities (energy, momentum,
electric charge, baryon number) is:
p p  p p p p w ith p = anti-proton
The total energy in the CM is given by: (pb+pt)2 the sum of the 4-vectors of
the
beam and target proton (assumed to be at rest):
(pb+pt)2 =(Eb+mp, pb)2= mb2+mt2+2mtEb=2mp2+2mpEb
The trick now is to remember that (pb+pt)2 is Lorentz invariant and can be
evaluated in any frame we choose. The most convenient frame is the one
where all the final state particles are produced at rest. Here we have:
(pb+pt)2inirtial =(Eb+mp, pb)2= (pb+pt)2final = (4mp)2
2mp2+2mpEb = (4mp)2
Eb = 7mp= 6.6 GeV
The anti-proton was discovered at Berkeley in 1955 (Nobel Prize 1959)
P780.02 Spring 2003 L2
Richard Kass
Time Dilation
Most of the particles we are concerned with in HEP are not stable, i.e. they
spontaneously decay into other particles after a certain amount of time. For example
consider the lepton family:
Lepton
Mean Lifetime (s)
electron
stable
muon ()
2x10-6
tau (t)
3x10-13
The above table gives the average lifetime of the leptons in their rest frame. However,
often we need to know how long a particle will live (on average) in a frame where the
particle is moving close to the speed of light (c). We need to use special relativity!
Consider a particle moving with speed v as pictured below:
v
cm
The particle is moving in the lab frame y
particl e at
with speed v along the x-axis. The
rest i n cm
relationship between time and distance
measured in the lab and CM frame is:
In CM: x1cm=x2cm
tlab= (tcm+ /c xcm) with = v/c.
x
xlab= (xcm+ c tcm)
l ab
In the lab frame the time between life and death of the particle is:
tlab= t2lab- t1lab= (t2cm + /c x2cm)- (t1cm + /c x1cm) = (t2cm-t1cm)=t
Note: t is called the proper lifetime and t tlab
P780.02 Spring 2003 L2
More Time Dilation
Richard Kass
Consider a muon (m0=0.106 GeV/c2) with 1 GeV energy in the lab frame.
On average how long does this particle live in the LAB?
In the muon’s rest frame it only lives (on average) t = 2 sec.
But in the lab frame it lives (on average):
tlab=t, and =E/(m0c2) = 1/0.106  10
tlab=t =(10)(2 sec) = 20 sec
On average how far does this particle travel in the LAB before decaying ?
Dxlab= ct with = p/(m0c) 1/0.106  10
Dxlab= ct = 10ct= (10)(3x108 m/s)(2x10-6 s) = 6x103 m
Big increase due to Special Relativity