lecture 5 - Department of Physics, The Chinese University of Hong

Report
Topics in Contemporary Physics
Basic concepts 3
Luis Roberto Flores Castillo
Chinese University of Hong Kong
Hong Kong SAR
January 16, 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
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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
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V)
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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 16, 2015
2
… last time: Basic concepts 2
• Review of units & particle dynamics
– Feynman diagram examples
• Decays and conservation laws
• A word on Unification
L. R. Flores Castillo
CUHK
January 16, 2015
3
Reminder: interactions
QED:
QCD:
Weak:
W/Z:
L. R. Flores Castillo
W/Z/γ:
CUHK
January 16, 2015
4
Reminder: adding possible histories
+…
L. R. Flores Castillo
CUHK
January 16, 2015
5
A few key concepts
• Formally, the W boson can only link
‘up-type’ quarks (u,c,t) into the
corresponding ‘down-type’ (d,s,b).
• However, experimentally, some times it
mixes generations
• Solution: the weak force “sees” slightly
rotated versions of the down quarks:
Cabibbo-Kobayashi-Maskawa matrix
L. R. Flores Castillo
CUHK
January 16, 2015
6
About unification
•
•
•
•
Electricity + Magnetism
Glashow, Weinberg and Salam: EM + Weak = EW
Chromodynamics + EW ?
The “running” of the coupling constants hints at it
?
L. R. Flores Castillo
CUHK
January 16, 2015
7
Today’s outline
• Relativistic Kinematics
• Symmetries
L. R. Flores Castillo
CUHK
January 16, 2015
8
Relativistic Kinematics
9
Lorentz transformations
• S’ moves with velocity v wrt S
• Along the common x/x’ axis
• t = t’ = 0 when x = x’ = 0.
L. R. Flores Castillo
CUHK
January 16, 2015
10
Lorentz transformations
An event occurs at (x,y,z) and time t
in S; coordinates in S’:
x ' = g (x - vt)
x = g (x '+ vt)
y' = y
y = y'
z' = z
æ
v
t ' = g çt - 2
è c
z = z'
æ
v
t = g ç t '+ 2
è c
ö
x÷
ø
g=
L. R. Flores Castillo
CUHK
ö
x÷
ø
1
1- v 2 / c 2
Inverse transformation: v-v
January 16, 2015
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Consequences
• Relativity of simultaneity:
events A and B, at the same time in S, but in different
locations, do not occur at the same time in S’:
t ' A = t 'B +
gv
c
2
(xB - x A )
• Lorentz contraction:
a stick at rest in S’, from x’=0 to L’; size in S?
– its ends at the same time in S (t=0): x=0, x = L’/γ (from LT)
– Length = L’/ γ
– γ > 1, so moving objects are shortened by a factor γ
L. R. Flores Castillo
CUHK
January 16, 2015
12
Consequences
• Time dilation:
– Clock at x’=0 ticks at t’=0, t’=T’; interval in S?
from t=0 to (t’=T’, x’=0), so t=γT’
– T = γT’ ; clocks in S mark a longer interval
– “Moving clocks run slow”
– All processes are affected
– Including the “internal clock” of elementary particles
– Moving particles lasts longer by a factor of γ
– Examples:
• Muons from the upper atmosphere
• B mesons in our HEP detectors
L. R. Flores Castillo
CUHK
January 16, 2015
13
A few quick examples
v
130 km/h
beta
gamma
14,000 km/h
0.5 c
0.99 c
?
L. R. Flores Castillo
CUHK
6,500
January 16, 2015
14
Four-vector notation
Position-time four-vector:
x = ct, x = x, x = y, x = z
0
1
2
3
Lorentz transformations take a nicely symmetrical form:
æ
v ö
t ' = g çt - 2 x ÷
è c ø
x ' = g (x - vt)
y' = y
x 0 ' = g (x 0 - b x1 )
z' = z
x3 ' = x3
L. R. Flores Castillo
CUHK
x ' = g (x - b x )
1
1
0
x2 ' = x2
January 16, 2015
v
bº
c
15
Four-vector notation
This can be written as:
3
x ' = åL x
m
m n
n
(m = 0,1, 2, 3)
v=0
with
Einstein’s ‘summation convention’:
m
m n
n
x '=L x
L. R. Flores Castillo
CUHK
January 16, 2015
16
Four-vector notation
Very concise, and allows generalizations.
• It works also for transformations not along the x axis
From S to S’, coordinate values change, but the following
combination remains the same:
I º (x ) - (x ) - (x ) - (x )
0 2
1 2
2 2
3 2
= (x ') - (x ') - (x ') - (x ')
0
2
1
2
2
2
3
2
3
To express it as sum
m m
x
å x
m =0
we need to change the sign of the spatial components.
L. R. Flores Castillo
CUHK
January 16, 2015
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Four-vector notation
We can do it by introducing:
so that
We can also define
L. R. Flores Castillo
CUHK
January 16, 2015
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Four-vector notation
So that, now
m
m
I º xm x = x xm
This expresses the invariant in a very simple expression.
Too much trouble just to include those minus signs?
It generalizes to non-Cartesian coordinates and to GR.
L. R. Flores Castillo
CUHK
January 16, 2015
19
Four-vector notation
Other quantities also transform as the position-time fourvector xμ.
m
m n
n
a '=L a
“contravariant”
We define a four-vector as any object that transforms like
this when we go from one inertial system to another.
We can always define a corresponding vector with negative
signs in the spatial components:
n
am = gmn a
(which is easy to convert back:
“covariant”
m
mn
a = g an
)
[ formally, gμν are the elements of g-1, but it’s its own inverse ]
L. R. Flores Castillo
CUHK
January 16, 2015
20
Four-vector notation
For any two four-vectors, their scalar product is
m
m
a bm = am b = a b - a b - a b - a b
0 0
1 1
It is also invariant.
a × b = a 0b0 - a × b
a º a × a = (a ) - a
2
0 2
2 2
3 3
a × b º am b
m
2
Four vectors are called
timelike
if a2>0
spacelike if a2<0
lightlike
if a2=0
L. R. Flores Castillo
CUHK
January 16, 2015
21
Energy and momentum
When two reference frames are considered, the
proper time advances by a smaller amount
dt =
dt
g
This is also an invariant (all observers agree on the proper
time of a given particle, satellite or space traveler).
A related quantity, similar to velocity
is the “proper velocity”:
dx
h=
dt
dx
v=
dt
h = gv
Useful because, under a LT, only the numerator transforms
(and as a four-vector); this is indeed a four-vector too.
L. R. Flores Castillo
CUHK
January 16, 2015
22
Energy and momentum
Components:
h = g (c, vx , vy, vz )
m
In this case, the invariant ημημ is:
hmh m = g 2 (c2 - vx2 - vy2 - vz2 ) = g 2c2 (1- v 2 / c2 ) = c2
How should we define momentum?
mv or mη ?
Better use η, because then, if momentum is conserved in
one system, it will also be in all others.
m
m
p º mh
L. R. Flores Castillo
CUHK
January 16, 2015
23
Energy and momentum
Proper velocity is a four-momentum, so also is pμ
p º mh
m
m
p0 = g mc
p = g mv =
mv
1- v 2 / c 2
We define the relativistic energy as
E = g mc =
2
mc 2
1- v 2 / c 2
So, the 0-th component of the four-momentum is E/c.
æE
ö
p = ç , px , py , pz ÷
èc
ø
m
L. R. Flores Castillo
CUHK
January 16, 2015
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Energy and momentum
Computing the invariant pμpμ:
From the definition:
pm pm = (mhm )(mh m ) = m2hmh m = m2c 2
From its components:
E
pm p = 2 - p2
c
2
m
E2
2
2 2
-p = m c
2
c
E 2 = m2 c 4 + p2 c 2
L. R. Flores Castillo
CUHK
January 16, 2015
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Energy and momentum
Energy:
æ
1
2
2ç
E = g mc = mc
ç 1- v22
è
c
ö
÷ = 1+ 1 v22 + 3 v44 +...
2 c
8 c
÷
ø
= mc 2 + 12 mv 2 + 83 m vc4 +...
4
Rest energy
Relativistic kinetic energy
T º mc (g -1)
2
L. R. Flores Castillo
CUHK
January 16, 2015
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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 16, 2015
E = hv
27
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 16, 2015
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Collisions
• In general, mass is not conserved: π0  γ + γ
• Elastic: same particles come out as went in
L. R. Flores Castillo
CUHK
January 16, 2015
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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
2
Mc = 2Em =
= (2mc 2 )
1- (3 / 5)2 4
5 2
M = mc
2
Notice that M > m1 + m2
L. R. Flores Castillo
CUHK
January 16, 2015
30
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 2 = 2g mc 2 =
2m
1- v 2 / c 2
v = c 1- (2m / M )
L. R. Flores Castillo
CUHK
2
Only if M>2m
For M=2m, v=0
For M>>2m, vc
January 16, 2015
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