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Elastic pp scattering and TOTEM experiment Jan Kas par TOTEM - - PowerPoint PPT Presentation
Elastic pp scattering and TOTEM experiment Jan Kas par TOTEM - - PowerPoint PPT Presentation
Elastic pp scattering and TOTEM experiment Jan Kas par TOTEM experiment (Total Cross Section, Elastic Scattering and Diffraction Dissociation at the LHC) The smallest, but independent experiment at the LHC at CERN. Its physical program [1] 1)
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There is no satisfactory theory of diffraction so far. On the other hand there is a variety of phenomenological models and the TOTEM experiment will provide a tool to sort out the models that are based on true physical nature.
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TOTEM detectors
It will share location with CMS. The forward inelastic detectors will be assembled in CMS. Elastic detectors will be placed far from IP in Roman pots
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Roman pots
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TOTEM/CMS acceptance
Angle corresponding to pseudorapidity η = 10 is ϑ ≈ 2e−η ≈ 9 · 10−5.
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TOTEM elastic scattering aims
During one run, dσ/dt can be measured in a relatively narrow t
- region. Finally, these pieces of data will have to be fitted together.
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What can one find in this thesis?
1) Elastic pp scattering analysis a) Islam’s model b) Impact parameter point of view c) Coulomb interference 2) Calibration of detector alignment – my summer project in CERN
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The model
- Interesting is small t region, where coupling constant is too
large to use perturbative approach within QCD ⇒ phenomeno- logical models are employed.
- Model of Islam et al. ([2], [3], [4]) was used. It is based on idea
- f proton structure. They start from SU(3)L × SU(3)R gauged
nonlinear sigma model.
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They distinguish 3 mechanisms (diffraction, core scattering and quark scattering) T(s, t) = TD(s, t) + TC(s, t) + TQ(s, t) dσ dt = π sp2 |T(s, t)|2 Diffraction is a consequence of interaction of the outer clouds. The corresponding amplitude is parametrised TD(s, t) ∼
∞
- bdb J0(b
√ −t)
- 1
1 + e
b−R a
+ 1 1 + e
−b−R a
− 1
- Scattering one core off the other gives birth to the core scattering.
It should be go via ω exchange, therefore TC(s, t) ∼ s F 2(t) m2
ω − t
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Quark amplitude reflects transition to perturbative regime of
- QCD. They take into account ”distribution functions”, elastic qq
- amplitude. The latter is obtained with BFKL theory. The former
leads to appearance of form factor F(q⊥). The quark amplitude reads TQ(s, t) ∼ s F(q⊥) |t| + r−2
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Alltogether, there are 17 (13) energy–independent parameters. Authors determined these parameters by fitting data on σtot(s), ρ(s) and dσ/dt for ¯ pp at energies √s = 541 GeV, 630 GeV and 1.8 TeV.
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Impact parameter representation
- Basic idea is to replace T(s, t) by A(s, b), where b is impact
parameter.
- It naturally follows from eikonal approximation in QM.
- There is precise formulation based on Fourier–Bessel transform
[5] U(q) = c
∞
- b db J0(bq) A(b)
A(b) = 1 c
∞
- q dq J0(bq) U(q)
- However, it requires both A(b) and U(q) to be defined on (0, ∞).
We want to identify U(q) with amplitude T(s, t = −q2), but it is constrained to tmin < t < 0 ⇒ we introduce arbitrary function ˜ T U(q) = T(s, t = −q2) for tmin < t < 0 ˜ T(s, t = −q2) for t < tmin
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- ˜
T represents ambiguity in i.p. amplitude ⇒ it is necessary to introduce a physical requirement for ˜ T, that would remove the ambiguity.
- A(s, b) = a(s, b) + ˜
a(s, b)
∞
- b db b2n ˜
a(s, b) = 0
∞
- b db ˜
a∗(s, b) a(s, b) = 0
- σtot(s) = 4π
∞
- d2b ℑA(s, b) ,
σel(s) = 4π
∞
- d2b |a(s, b)|2 ⇒
seems natural to interpret 4π ℑA(s, b) resp. 4π |a(s, b)|2 as total
- resp. elastic collisions density ̺el, tot(s, b). To do so, ℑA(s, b)
must be non–negative ⇒ physical constraint on ˜ a (or ˜ T).
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- If we adopt the density interpretation we can define mean
value of b2 (independent on ˜ T choice) b2(s)el, tot =
- d2b b2 ̺el, tot(s, b)
- d2b ̺el, tot(s, b)
−1 Islam model is central, with weak t dependence of amplitude phase ⇒ btot > bel. It is in contradiction with diffraction, which is described by peripheral models.
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Coulomb–Hadron interference
- Two fundamental forces are essential for pp scattering. Namely,
electromagnetic and strong. We know scattering amplitudes for each interaction separately. The problem, how to determine amplitude for both interactions acting simultaneously.
- One problem, two (approximate) solutions:
a) eikonal additivity way b) summing some relevant Feynman diagrams Eikonal approximation in QM gives following prescription A(s, b) = e2δ(s,b) − 1 2i δ(s, b) = − 1 2p
∞
- dz V (
- b2 + z2)
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Potential is additive ⇒ eikonal δ is additive. For V = V1 + V2, the total amplitude A(s, b) = A1(s, b) + A2(s, b) + 2iA1(s, b)A2(s, b) Jump to the final result [6], [7] T(s, t) = ∓αs t F 2(t) + TH(s, t) 1 ± iα
- tmin
dt′ ( lnt′ t dF 2(t′) dt − − TH(s, t′) TH(s, t) − 1
- I(t, t′)
I(t, t′) = 1 2π
2π
- dϕF 2(t′′)
t′′ t′′ = t + t′ + 2 √ tt′ cos ϕ – Two ways of possible use: T ⇒ TH and T ⇐ TH – One formula for whole t region.
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Feynman diagrams technique (West–Yennie formula) [8], [9] – Uses scalar field for protons. – Needs to know off shell hadron amplitude. – A ”model independent contribution” is introduced. It de- pends on amplitude on the mass shell only and does not suffer from IR divergence. – Final formula is equivalent with simplified formula derived within the previous approach. Conventional analysis – simplified WY formula assumes – Purely exponential decay of amplitude in whole t region. – Constant phase in whole t region. Relative importance of interference f(s, t) = |T(s, t)|2 − |TH(s, t)|2 − |TC(s, t)|2 |TH(s, t)|2
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For pp at energy √s = 53 GeV Prediction for LHC has the same character, with 2 quantitative
- differences. The blue and the red line are closer (diff. cross section
is higher) and the peak in f(t) is at t ≈ 5 · 10−1 GeV2 (position of diffraction dip).
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My summer project in CERN
calibration of detector alignment
- In every roman pot, there will be a bunch of 10 planar strip
detectors (with strip pitch 66µm). One half having the strips perpendicular to the other half.
- The detectors will be adapted for TOTEM special needs, they
will be edgeless.
- This new technology need to be tested. For this pourpose it is
crucial to know precise position of every detector.
- Estimated maximum error of mechanical placement is 20 µm.
This induces maximum possible slant 10−3.
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Test detector geometry
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The method
- We know distribution of tracks from beam source.
a = 0, σa = 3 · 10−3, b = 0 mm, σb ≈ 12 mm
- Difference in x positions measured in i-th and j-th detector
Dij = x′
i − x′ j = ax(zi − zj) − (∆xi − ∆xj) − ay(zisi − zjsj)−
− by(si − sj) + (∆yisi − ∆yjsj)
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Shifts
- Event-averaged value of Dij
¯ Dij = 1 N
N
- n=0
Dn
ij = − (∆xi − ∆xj) + ¯
ax(zi − zj)− − ¯ ay(zisi − zjsj) − ¯ by(si − sj)
- Deviation of arithmetic mean
σ¯
ax = σax
√ N
- ¯
Dij ≈ −(∆xi − ∆xj) with error 0.7 µm for 5 · 104 events.
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Slants
- Linear regression applied on data Dn
ij versus bn y
Tij = −(si − sj) + F
- It can be shown F = 0 m−1 and σF ∼ 1/
√ N
- Tij ≈ −(si − sj) with error 2.2 · 10−4 for 5 · 104 events.
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Real situation
h(a, b) = g(b)
bmax
- bmin
g(b) f(a)
amax(b)
- amin(b)
f(a)
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That’s all, thank you for attention :-)
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