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# 物理代写|高能物理代写High Energy Physics代考|PH4204 Secondary beams

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## 物理代写|高能物理代写High Energy Physics代考|Secondary beams

If one wants to investigate the interactions of neutral particles (e.g. $K^{0}$ ), unstable particles (e.g. $\pi^{\pm}, K^{\pm}$) or antiprotons with matter one has first to produce these particles. To this end the extracted proton beam hits a primary target. The particles produced in these collisions are selected and directed as secondary beams towards the target of the experiment. A beam transport system containing a set of dipole magnets and quadrupoles is used to transport charged particles in a certain angle and momentum interval.

To produce a photon beam, electrons are first directed onto a target of high atomic number $A$ and then deflected by a magnetic field. The energy spectrum of the photons is inversely proportional to the momentum of the photons, thus many low-energy photons are produced. By having lithium hydride in the beam the number of low-energy photons can be reduced. When these photons are directed towards another target they produce $e^{+} e^{-}$pairs. The positrons can then be selected by a magnetic field and a collimator.

## 物理代写|高能物理代写High Energy Physics代考|Energy balance in scattering experiments

In fixed-target experiments part of the energy of the incoming particles is not available for the interaction but wasted in boosting the particles in the forward direction. Storage rings or colliders are used to increase the energy available in an interaction. In a collider two beams from opposite directions interact with each other. The energy $W$ available in an interaction can be expressed in a Lorentz invariant form:
$$W^{2}=s=\left(p_{1}+p_{2}\right)^{2}$$
where $s$ is the square of the invariant energy, $p_{1}=\left(m_{1}, \mathbf{p}{1}\right)$ is the four vector of the incoming particle, $p{2}=\left(m_{2}, \mathbf{p}{2}\right)$ is the four vector of the target particle, $\left(p{2}=\left(m_{2}, 0\right)\right.$ for a fixed target $)$. When a proton of energy $E=400 \mathrm{GeV}$ hits a stationary hydrogen target (protons) the square of the invariant mass is
\begin{aligned} &s=m_{\mathrm{p}}^{2}+m_{\mathrm{p}}^{2}+2 E m_{\mathrm{p}}=754 \mathrm{GeV} \ &\left(m_{\mathrm{p}}=0.938 \mathrm{GeV}\right) \ &W=\sqrt{s}=27.5 \mathrm{GeV} \end{aligned}
Subtracting twice the proton mass for the surviving protons (baryon number conservation) leads to the energy available in the reaction
$$M_{x}=W-2 m_{\mathrm{p}}=25.6 \mathrm{GeV}=2 \times 12.8 \mathrm{GeV}$$
which means that two colliding protons of $12.8 \mathrm{GeV}$ each release as much energy as a proton of $400 \mathrm{GeV}$ hitting a proton at rest. The equivalent beam energy for a fixed target machine for a given storage ring beam energy $E$ of a storage ring and a particle mass $m$ is then:
$$E_{\text {fixed target }}=2 E_{\text {storage ring }}^{2} / m$$
Three different types of collider can be distinguished: hadron-hadron machines, lepton-lepton machines and hadron-lepton machines. Several of these machines are listed in Table $1.1$.

## 物理代写|高能物理代写High Energy Physics代考|Energy balance in scattering experiments

$$W^{2}=s=\left(p_{1}+p_{2}\right)^{2}$$

$$s=m_{\mathrm{p}}^{2}+m_{\mathrm{p}}^{2}+2 E m_{\mathrm{p}}=754 \mathrm{GeV} \quad\left(m_{\mathrm{p}}=0.938 \mathrm{GeV}\right) W=\sqrt{s}=27.5 \mathrm{GeV}$$

$$M_{x}=W-2 m_{\mathrm{p}}=25.6 \mathrm{GeV}=2 \times 12.8 \mathrm{GeV}$$

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