Lecture 1: Addendumvirgilio.mib.infn.it/moroni/lectures/Luigi Lecture 1...5/6/2013 Experimental...
Transcript of Lecture 1: Addendumvirgilio.mib.infn.it/moroni/lectures/Luigi Lecture 1...5/6/2013 Experimental...
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Lecture 1: Addendum
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Cross Section
• Let’s consider a reaction of the form 𝑎 + 𝑏 𝑐 + 𝑑
• with 2 particles in the initial state • If we regard 𝑎 as the projectile/beam particle and 𝑏 the target
particle, – the Cross Section for the above reaction is defined as the
transition rate 𝑊 per unit incident flux per target particle• Let’s now explicitly calculate what this definition means for an
infinitesimal target area element 𝑑𝑥 𝑑𝑦 transverse to the beam • Calling 𝐹 = 𝐹 𝑥, 𝑦, 𝑡 the flux of beam particles per 𝑐𝑚2 per
𝑠𝑒𝑐 at 𝑥, 𝑦, 𝑡 and 𝐷 = 𝐷 𝑥, 𝑦, 𝑡 the target density i.e. the number of target particles per 𝑐𝑚2 at 𝑥, 𝑦, 𝑡
• we obtain
𝜎 = 𝑊 =𝑑𝑇
𝐷𝑑𝑥𝑑𝑦 𝐹• where 𝑑𝑇 is the transition rate due to the 𝑑𝑥 𝑑𝑦 portion of the
target
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Reversing the argument: from 𝜎 to 𝑑𝑇
• Last equation can be inverted to get the transition rate from the Cross Section, i.e.
𝑑𝑇 = 𝜎 𝐷 𝑥, 𝑦, 𝑡 𝐹 𝑥, 𝑦, 𝑡 𝑑𝑥 𝑑𝑦• and therefore integrating over the beam flux and
the target density, we can derive the total rate at instant 𝑡
𝑇 𝑡 = 𝜎 𝐷 𝑥, 𝑦, 𝑡 𝐹 𝑥, 𝑦, 𝑡 𝑑𝑥 𝑑𝑦• This is exactly what we use to compute the collider
luminosity• If 𝐷 𝑥, 𝑦, 𝑡 is constant for any 𝑥, 𝑦, 𝑡 where 𝐹 𝑥, 𝑦, 𝑡 ≠ 0, as for a typical fixed target experiment, we obtain
𝑇 𝑡 = 𝜎 𝐷 𝐹 𝑥, 𝑦, 𝑡 𝑑𝑥 𝑑𝑦
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1/14/2021 Experimental Methods in High Energy Physics - An Introductory Course (L. Moroni) 4
Understanding Synchrotron Radiation Shapes
• Actually, it’s pretty immediate.• In an inertial frame, where the particle is at rest at the time t=ti, we can legitimately calculate the
radiated field due to an instantaneous acceleration using the non-relativistic Hertz dipole formula • We can then boost it to the lab frame to obtain our result• Well, the Hertz dipole radiated field features surfaces of equal field intensity, both E and H, which are
toric and result from the rotation around the vector of a circle passing through the position of the particle at the time ti and tangent to
• Since and are normal to the vector and relatively normal too, the energy flux density, i.e. the
Poynting vector , will be directed as and have the same modulus on each point of the toric
surface.
• Therefore, the radiated energy per unit time and solid angle element about will just scale as or equivalently as
• Indeed,
)( itv
r
e
)( itv
)( itv
E
H
HE
r
d
dkdrHEdPrad 22sin
2r
2sin
)( itv
r
r
Z
X
Y
)( itv
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1/14/2021 Experimental Methods in High Energy Physics - An Introductory Course (L. Moroni) 5
A Further Step
• Now, boosting in a direction, we can understand how the radiated field looks for a relativistically moving particle
• To this extent, we would rather use a more intuitive argument based on the boost of the radiated photons instead of the rigorous and more complex one, which would directly transform the fields (F)
• To do this, we have simply to recall that a Lorentz boost will modify the direction of a photon independently of its energy
• In other words, if you pick a very collimated jet of photons radiated around a certain direction in the space, no matter of the energies of the single photons, it will remain a jet also after the boost: the jet can open up or shrink down by a certain amount, which does not depend from the energies of the single photons
• Even more, also the energy of each photons in the jet will change by a factor which does not depend from their energies
• This means that the radiated energy in one direction, which consists of photons of different energies, will remain energy radiated in one direction also in the new frame: the energy will be multiplied by a factor depending on its original direction only, and its direction transformed into the new one.
• Let’s justify these assertions
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1/14/2021 Experimental Methods in High Energy Physics - An Introductory Course (L. Moroni) 6
Boosting Photons
• Let’s consider a photon of energy E’=p’c in an inertial frame X’Y’Z’• We choose as Z’-axis that directed as , where is the boost
vector, and as Y’Z’ plane that containing the photon• ’ is the angle between the photon and the Z’-axis
X
Y
Z
zp
yp
c
c
• Let’s now boost this photon to the XYZ frame moving with speed in X’Y’Z’
cos
sin
0
00
0100
0010
00
p
p
p
p
p
p
p
z
y
x
c
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A Few Mathematics Yet
• Let’s see how the photon get transformed in the new frame and it compares with the original one
dd
d
p
p
E
E
pp
p
pp
p
p
p
p
p
p
p
z
y
z
y
x
cos1
1
cos
sin
cos1
coscos
cos1
sinsin
cos
sincos1
cos
sin
0
cos
cos
sin
0
00
0100
0010
00
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1/14/2021 Experimental Methods in High Energy Physics - An Introductory Course (L. Moroni) 8
The Results
• For large all the photons, but the small fraction of them with , get boosted in the forward direction with a much higher energy ( ) and at an angle of the order of
• Now we can easily reproduce the shapes we start with• Have just to boost each ray/photon of the toric distribution in the inertial
frame to the lab frame moving with -c relative velocity • A little caveat yet: since the radiated energy flux are per unit solid angle and
time, we have to put all the transformed radiated photons in the corresponding solid angle and time
• i.e. gets shrank by whereas E amplified by • Therefore a total factor• Putting everything together we obtain
cos
1EE
tdd
Ed
dtd
Ed
232
2
cos1
2cos1cos1cos1
sinsin
tddtdd
ddtdddtd
tdd 2cos1
32 cos1
cos1
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Acceleration parallel to the boost
• The boost is along the axis of the torus, therefore
• and the Y’Z’ section is symmetric • In this case we have
2
2232
2
sincos1
tdd
Ed
dtd
Ed
Z
r
e
Y
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Calculating Equal Field Intensity Lines
• Since
• and is by definition constant on the Equal Field Intensity Lines (EFIL) , we can infer that
• This means that the EFIL are described by the vector having polar angle and azimuthal angle equal to that of the particular section of the torus we start with
• Thus, calculating and we can easily draw the lines
HE
2
22322
2
sincos1
tdd
EdrHE
dtd
Ed
2322sincos1r
r
r
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EFIL for acceleration parallel to the boost
• symmetric• Radiation will be in the forward direction, no
matter about the sign of the acceleration
• Maximum radiated power at about =1/• EFIL scales longitudinally with -boost
– This is a common feature for all the EFIL
0
0,2
0,4
0,6
0,8
1
1,2
-20 0 20 40 60 80 100 120 140 160 180
boost = 10
boost =100
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1/14/2021 Experimental Methods in High Energy Physics - An Introductory Course (L. Moroni) 12
Acceleration normal to the boost
• Here we do not have any more the azimuthal symmetry, thus have to distinguish between the two extreme cases for =/2 and =0
• Thus, for =/2, we have
• and for =0
Z
r
e
Y
Z
r
e
X =/2 =0
2
232
2
cos1
tdd
Ed
dtd
Ed
2
2232
2
coscos1
tdd
Ed
dtd
Ed
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EFIL for acceleration normal to the boost
• For =/2
• For =0
0
0,2
0,4
0,6
0,8
1
1,2
-50 0 50 100 150 200 250 300
boost = 100
0
0,2
0,4
0,6
0,8
-50 0 50 100 150 200 250 300
bost = 100
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and what about the field intensity on these EFIL?
• All the previous EFIL are normalized to maximum value of the Hertz dipole radiation, i.e.
• Therefore, the fields on them are for the same or modulus of the force in the primed reference frame where the charged particle is instantaneously at rest
• This is why the field intensity on these lines resultsabout of the same order, , and also the total radiated energy
• If we instead require and be equal in
the lab, the picture dramatically changes!
116
2
32
00
2
2
td
pd
cm
eHE
d
Pd
2
td
pd
2
dt
pd T
2
dt
pd L
HE
2
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Let’s see how much it changes
• The ratio of force or to acceleration is when the force is normal to
the acceleration and in case they are parallel
• Therefore, , and for we get
• Transforming the acceleration to the particle rest frame, we obtain
and
• Since the ratio , and thus the radiated field for
acceleration normal to the boost acquire an extra factor 2
• In conclusion, for the same force acting on a charged particle moving with
velocity c in the lab, the radiated energy is a factor 2 larger when the
force is normal to the velocity
dt
pdm
m3
(see for instance Landau,The Classical Theory of Fields, paragraph 1-7 and 2-3)
cT ma
dt
pd L
L madt
pd 3dt
pd
dt
pd LT 2c
L
aa
cc aca 22c
cL ac
aca 2
2
23
L
c
a
a
td
pd
td
pd LT
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Space Charge Effect in the BeamSP
AC
E C
HA
RG
E, b
y K
. Sch
ind
l, C
ERN
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SPACE CHARGE, by K. Schindl, CERN
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SPACE CHARGE, by K. Schindl, CERN
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SPACE CHARGE, by K. Schindl, CERN
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SPACE CHARGE, by K. Schindl, CERN
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SPACE CHARGE, by K. Schindl, CERN
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The frequent student’s question
• How could it possibly be that in the beam particle CoM frame still the particles repel each other because of the Coulomb force?
• The trick is due to the fact that the time intervals get boosted by , which approachingthe speed of light diverges
• Thus, if in t sec a particle moves away from another one by 1 cm in the CoM transversal plane, the same 1 cm movement will be seen in the lab to happen in t sec– Just for 1 TeV proton, 1 sec will be boosted to 103
sec, about 16 minutes.