11
From the first part of the question, to show explicitly that p µ pµ for a system of
+
2.3
Show
the process
can not
occur intothe
vacuum.
particles
is that
Lorentz
invariant!iteise only
necessary
show
that Ei0 E 0j p0zi p0z j =
Underlying
Ei E j pzi pz j , the proof of which
is almostConcepts
identical to the first part of the question.
This is most easily demonstrated by considering the reaction in the rest frame of
2.6 e+ eForpair,
the decay
a!
+ 2, show
that the
of the particle
can In
bethis
expressed
the
namely
the1frame
in which
themass
total momentum
is azero.
frame
b)
pas=Here
0 and E = Ee + Ee+ > 2me . Since energy and momentum are conserved this
X ⇣ m2 = m2Particle
2 ⇣ Physics – Homework 1
⌘ mX
2E>1 E2m
✓) , consistent⌘ with E = p
2 (1e , which
1 2 cos
2a p 2=
would imply
0 and
ispnot
1 ++
2 +E
p µ the
pµ ⌘photonEhas
p
E
E
p
p
pzi pz j .
i
j
xi
x
j
yi py j
i
i
for the massless photon.
i
i, j
where 1 and 2 are the velocities of the daughter particles ( i = vi /c) and ✓ is the
angle between
From
first them.
part
of the
question,
to show
explicitly
p µ pµ with
for amomentum
system of
2.4
Athe
particle
of mass
3 GeV
is travelling
in the
positive that
z-direction
0 E0
particles
is
Lorentz
invariant
it
is
only
necessary
to
show
that
E
p0zi p0z j =
4 GeV, what
are2 its energy
and velocity?
i j
2
2
Since
pa andofenergy
momentum
are to
conserved
in the
Ei E j mpazi=pzE
whichand
is almost
identical
the first part
of decay
the question.
j , athe proof
Here the key equations
m2a =are(E(in
E2 )2 units):
(p1 + p2 )2
1 +natural
2.6
as
For the decay a ! 1 + 2, show that the mass of the particle a can be expressed
= E 2 + E 2 + 2E1 E2 p2 p22 22p1 · 2p2
E = 2m 1, 2p 2= 2 m and 1 E 2 =
p +m .
mam=2 m
2 cos✓✓) ,
1 +2m2 + 2E 1 E 2 (12p p1 cos
=
1 2
1 + m2 + 2E 1 E 2
2 the particle’s
From
p2 +2 mare
E = 51GeV
and
whereE 21 =and
the
the1 Eis
daughter
particles
( i = the
vi /cabove
) and expres✓ is the
= velocities
m21 + m22energy
+of2E
✓) ,from
2 (1
2 cos
sions
E andthem.
p,
angle for
between
where the last step follows from p = E.
= p/E = 0.8 .
2 and energy and momentum are conserved in the decay
Since
m2aa=collider
Ea2 pexperiment,
a
2.7 In
⇤ baryons can be identified from the decay ⇤ ! ⇡ p
that gives rise to a displaced
vertex2 in a tracking2 detector. In a particular decay, the
2
E2 ) (p
p20.75
) GeV
2.5
In the
frame,
⌃,to
particle
travelling
in the
z-direction
has
1 +denoted
1a+be
a =
momenta
of laboratory
the ⇡+ m
and
p(E
are
measured
and 4.25
GeV
respectively,
momentum
p
=
p
ẑ
and
energy
E
.
2
2
2
2
z
and the opening angle =
between
tracks is 9p. The
E1 + Ethe
p2masses
2p1 · of
p2 the pion and proton
2 + 2E 1 E 2
1
areUse
139.6
andtransformation
938.3 MeV. to find expressions for the momentum p0z and energy
a)
theMeV
Lorentz
= m2 0 +
m22 + is
2Emoving
2p
1 E2
2 cos ✓ v = +vẑ relative to ⌃, and
Ea)0 of
the particle
in a frame
, which
in 1apvelocity
Calculate
the 2mass
of the⌃1⇤
baryon.
2
0 2
0 2
show that E
pz = (E )= m(p
2 z) . 2
+ menergy
(1
cos
✓) , at a distance of 0.35 m
1 E 2 observed
1 2 to
2 + 2Eare
b) On average, ⇤ baryons of1 this
decay
b) For a system of particles, prove that the total four momentum squared,
from the
production.
Calculate
of the ⇤.
where
thepoint
lastofstep
follows from
p 0= the
E. 1lifetime
0
12
2
X
X
B
C
B
CC
BBB
CCC BBB
µ
p p⇤µ baryons
⌘ B@
Ei can
pi CCCA , from the decay ⇤ ! ⇡ p
CA be
B@ identified
2
2 experiment,
2
2.7 In a collider
a) From E = p + m the energies of the two decay products are
that gives rise to a displaced vertex ini a trackingi detector. In a particular decay, the
+
momenta
the ⇡Lorentz
and
are measured
to beE0.75
GeV and
E⇡ =ptransformations.
0.763
GeV and
GeV4.25
. GeV respectively,
p = 4.352
is
invariantof
under
and the opening angle between the tracks is 9 . The masses of the pion and proton
are
MeV and 938.3
MeV.( = p/E) are
The139.6
corresponding
velocities
2
2
a)
that of2 the
= 1/(1
a) Remembering
Calculate the mass
⇤ baryon.) or equivalently (1
= 0.983transformations
and
explicit energy-momentum⇡Lorentz
p = 0.976 .
2 ) = 1, and using the
b) On average, ⇤ baryons of this energy are observed to decay at a distance of 0.35 m
0 result
0
Usingthe
the
of thepzprevious
from
of the ⇤. 0
Epoint
= of(Eproduction.
) , Calculate
p0x =question,
p x , theplifetime
E) ,
y = py and pz = (pz
2
m2⇤2be
= m221 + m22 + 2E1 E2 (1
1 2 cos ✓) = 1.244 GeV .
then
E 0 Ep20 =
can
a) From
p + written:
m the energies of the two
decay products are
0 mass
2
2
HenceEthe
the ⇤ obtained
the measurements
give is m⇤ = 1.115 GeV.
p0 = of2 (E
p2xfrom
p2yand
(p
E)2 GeV
z)
E⇡ =p0.763
GeV
Ezp = 4.352
.
2 2
2 2
2
2
2 2
= (E
2 E p + pz ) arep x py
(pz 2 E pz + 2 E 2 )
The corresponding velocities (z = p/E)
=
2
(1
2
2
2 2
)E 2 p2x p2y
(1
)pz
⇡ = 0.983 and
p = 0.976 .
= E 2 p2 .
Using the result of the previous question,
m2⇤ = m21 + m22 + 2E1 E2 (1
2
1 2 cos ✓) = 1.244 GeV .
Hence the mass of the ⇤ obtained from the measurements give is m⇤ = 1.115 GeV.