lecture 6

Report
Topics in Contemporary Physics
Basic concepts 4
Luis Roberto Flores Castillo
Chinese University of Hong Kong
Hong Kong SAR
January 23, 2015
PART 1
• Brief history
• Basic concepts
• Colliders & detectors
5σ
• From Collisions to
papers
S
ATLA
(*)
15
GeV
d
Selecte
s=7
2000
1800
TeV,
s=8
ò
TeV,
ò Ldt =
-1
5.9 fb
1600
1400
1200
1000
800
600
10
ATLAS
400
5
150
0
GeV)
sample
2
126.5
and 201 fit (m H =
2011
usive
Data
Bkg incl
-1
Sig +
nomial
4.8 fb
er poly
Ldt =
4th ord
n
diphoto
2400
2200
/
Events
-1
fb
t = 4.8
V: òLd
-1
5.8 fb
Ldt =
TeV: ò
Te
s=7
s=8
®4l
100
nary
Prelimi
200
250
0
]
[GeV
200 m100
4l
- Bkg
Even
• The Higgs discovery
c.
Un
Syst.
20
H®ZZ
Data
V
ts/5 Ge
(*)
Data
ZZ
round
s, tt
Backg
Z+jet
round
V)
Backg
25 Ge
l (m H=1
Signa
25
150
140
-100
100
160
V]
mg g [Ge
130
0
120
110
• BSM
• MVA Techniques
• The future
L. R. Flores Castillo
CUHK
January 23, 2015
2
… last time: Basic concepts 3
• Relativistic kinematics
– Lorentz transformations
– Consequences
– A few examples
– Four-vector notation
• Four-vectors
• Lorentz transformation in matrix form
• The metric, the scalar product, invariants
• Energy and momentum
L. R. Flores Castillo
CUHK
January 23, 2015
3
Reminder: interactions
QED:
QCD:
Weak:
W/Z:
L. R. Flores Castillo
W/Z/γ:
CUHK
January 23, 2015
4
Reminder: the CKM matrix
• The W boson can ‘jump across
generations’
• In the SM, the weak force “sees”
slightly rotated versions of the down
quarks:
Cabibbo-Kobayashi-Maskawa matrix
L. R. Flores Castillo
CUHK
January 23, 2015
5
Reminder: Relativistic kinematics
Maxwell
equations
æ
v ö
t ' = g çt - 2 x ÷
è c ø
x ' = g (x - vt)
y' = y
z' = z
Lorentz
transformations
c for all observers
x 0 ' = g (x 0 - b x1 )
Four-vector
x m ' = Lnm xn
x1 ' = g (x1 - b x 0 )
x2 ' = x2
x 0 = ct,
x = x,
1
x3 ' = x3
x 2 = y,
x =z
3
bº
time-position:
proper velocity:
energy-momentum:
v
c
E = g mc2 = mc 2 + 12 mv2 + 83 m vc4 +...
4
contravariant
I º (x 0 )2 - (x1 )2 - (x 2 )2 - (x3 )2
I = xm x
is Lorentz-invariant
Energy-momentum
a × b º am bm
Useful:
p = g mv
pm pm = (mhm )(mh m ) = m2 c2
E2
pm p = 2 - p2
c
m
L. R. Flores Castillo
Scalar product:
m
covariant
CUHK
E = m c +p c
2
xμ = (ct, x, y, z)
ημ=dxμ/dτ = γ(c, vx, vy, vz)
pμ = mημ = (E/c, px, py, pz)
2 4
2 2
E = g mc
January 23, 2015
2
p / E = v / c2
v = pc 2 / E
6
Relativistic Kinematics II
7
Energy and momentum
What if m=0?
Would the energy/momentum necessarily be 0?
mv
p=
1- v / c
2
E=
2
mc
2
1- v / c
2
2
not if v=c !!
The equation
can still hold.
Hence:
E =m c +p c
2
v = c,
2 4
2 2
E = |p| c
And
L. R. Flores Castillo
CUHK
January 23, 2015
E = hv
8
Collisions
• Classical
– Mass is conserved: mA+mB=mC+mD
– Momentum is conserved: pA+pB=pC+pD
– Kinetic energy may or may not be conserved:
• Sticky (kinetic energy decreases):
TA+TB > TC+TD
• Explosive (kinetic energy increases):
TA+TB < TC+TD
• Elastic (kinetic energy decreases):
TA+TB = TC+TD
• Relativistic
– Energy is conserved: EA+EB=EC+ED
m
m
m
m
p
+
p
=
p
+
p
A
B
C
D
– Momentum is conserved: pA+pB=pC+pD
– Kinetic energy may or may not be conserved
• Sticky (T decreases):
rest energy and mass increase
• Explosive (T increases): rest energy and mass decrease
• Elastic (T is conserved): rest energy and mass are conserved
L. R. Flores Castillo
CUHK
January 23, 2015
9
Collisions
• In general, mass is not conserved: π0  γ + γ
• Elastic: same particles come out as went in
• Consistency?
– Rest mass vs internal energy
– Hadron mass
L. R. Flores Castillo
CUHK
January 23, 2015
10
Example 1
Two lumps of clay, each of mass m, collide head-on at 3c/5; they
stick together. What is the mass M of the final composite lump?
Conservation of momentum: trivial. p1 + p2 = pM = 0
Conservation of Energy:
2mc 2
5
Mc = 2Em =
= (2mc 2 )
1- (3 / 5)2 4
2
5
M= m
2
Notice that M > m1 + m2
L. R. Flores Castillo
CUHK
January 23, 2015
11
Example 2
A particle of mass M, initially at rest, decays into two
pieces, each of mass m. What is the speed of each piece
as it flies off?
Conservation of momentum: equal and opposite speeds
Conservation of energy:
Mc = 2g mc =
2
2
2mc 2
1- v 2 / c 2
v = c 1- (2m / M )
L. R. Flores Castillo
CUHK
2
Only makes sense if M>2m
For M=2m, v=0
For M>>2m, vc
β depends only on m/M
January 23, 2015
12
Example 3
A pion at rest decays into a
muon plus a neutrino. What is
the speed of the muon?
(mμ=105.66 MeV/c2; mπ=139.57 MeV/c2; mν=0)
Ep = Em + En
pp = pm + pn
pp = 0 ® pm = -pn
mp c 2 = pm c + c mm2c 2 + pm2
pm =
mp2 - mm2
2mp
c
Subst. in E2=m2c4+p2c2:
E 2 = m2 c 4 + p2 c 2 :
Ep = mp c 2
En = pn c = p m c
Em = c mm2 c 2 + p m2
L. R. Flores Castillo
CUHK
Em =
mp2 + mm2
2mp
c2
Subst. in v = p c2 / E :
mp2 - mm2
vm = 2
c = 0.27c
2
mp + mm
January 23, 2015
13
Example 3. Second solution.
A pion at rest decays into a
muon plus a neutrino. What is
the speed of the muon?
(mμ=105.66 MeV/c2; mπ=139.57 MeV/c2; mν=0)
pp = pm + pn
Scalar product w/itself:
In all frames:
pn = pp - pm
pn2 = pp2 + pm2 - 2 pp × pm
pn2 = 0; pp2 = mp2c2 ; pm2 = mm2 c2
With the pion at rest: pp × pm =
Similarly, from
:
2 2
Em =
Ep Em Ep
= 2 Em = mp Em
c c c
mp2 + mm2
2mp
c2
pm = pp - pn
mm c = mp2c 2 - 2mp En
Similarly, from En = pn c = pm c and v = p c2 / E:
L. R. Flores Castillo
0 = mp2c 2 + mm2 c2 - 2mp Em
CUHK
mp2 - mm2
vm = 2
c = 0.27c
2
mp + mm
January 23, 2015
14
Example 4
The Bevatron at Berkeley aimed at producing antiprotons via
p+ p ® p+ p+ p+ p
What is the minimum energy of the incident proton that allows it?
Lab frame:
CM frame:
L. R. Flores Castillo
CUHK
January 23, 2015
15
Example 4
The Bevatron at Berkeley aimed at producing antiprotons via
p+ p ® p+ p+ p+ p
What is the minimum energy of the incident proton that allows it?
Minimum = all particles at
rest in the CM frame.
m
æ E + mc 2
ö
=ç
, p , 0, 0 ÷
è c
ø
Before the collision:
pTOT
After the collision:
m
pTOT
' = ( 4mc, 0, 0, 0)
m
m
pTOT
pm TOT = pTOT
' pm TOT '
æ E + mc 2 ö
2
2
ç
÷ - p = (4mc)
è c ø
2
Eliminating p w/
E2=m2c4+p2c2:
E = 7mc
2
i.e., the incident proton should have a kinetic energy above 6TeV
(and, indeed, the antiproton was discovered with E ~ 6000MeV)
L. R. Flores Castillo
CUHK
January 23, 2015
16
Example 5
Two identical particles, each with mass m and kinetic energy T,
collide head-on. What is their relative kinetic energy, T’?
(the kinetic energy of one on the rest system of the other)
Fixed target
Colliding beams
m
pTOT
( p ) = ( p ')
m
2
TOT
æ E '+ mc 2 ö
pTOT ' = ç
, p'÷
è c
ø
æ 2E ö
= ç , 0÷
è c ø
m
TOT
m
2
æ 2E ö æ E '+ mc 2 ö
2
ç ÷ =ç
÷ - p'
è c ø è c ø
2
2
:
(
Eliminating p w/ E2=m2c4+p2c2: 2E 2 = mc 2 E '+ mc 2
In terms of
L. R. Flores Castillo
T=E-mc2
and
CUHK
T’=E’-mc2:
)
æ
T ö
T ' = 4T ç1+
÷
è 2mc 2 ø
January 23, 2015
Classical result: only T’=4T
Reduces to it when T<<mc2
In B’s frame, A has twice the
speed, hence 4 times the E.
17

similar documents