# 物理代写|粒子物理学作业代写Nuclear and Particle Physics代考|vectors and Invariants

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## 物理代写|粒子物理学作业代写Nuclear and Particle Physics代考|Four-vectors and Invariants

An extremely useful concept in particle physics is that of four-vectors. Recall from your special relativity lectures the time-space four-vector: $(c t, \vec{x})=(c t, x, y, z)$.

In particle physics we make extensive use of the energy-momentum four-vector (known as the fourmomentum $: P=(E, c \vec{p})=\left(E, c p_{x}, c p_{y}, c p_{z}\right)$. As you will have learnt in your relativity courses, under a change of reference frame these four-vectors undergo Lorentz transformations (see the lecture slides for the matrix form of the Lorentz transformations).

It turns out that the inner product of a pair of four-vectors is invariant under a change of reference frame (just like the inner product of two spatial vectors is invariant under coordinate system rotations). The inner product of a four-vector $A$ is defined as:
$$A^{2}=A \cdot A=A^{0} A^{0}-\vec{A} \cdot \vec{A}$$
For an energy-momentum four-vector we have:
$$P^{2}=P . P=E^{2}-\vec{p}^{2} c^{2} .$$
You can convince yourselves that this inner product will not change under a Lorentz transformation, by substituting the explicit form of the Lorentz transformation into the above. Since this is true in any frame,it can be conveniently evaluated in the rest frame where $\vec{p}=0$. This really simplifies things as
$$P^{2}=E^{2}$$
and we know that $E^{2}=m^{2} c^{4}$ when a particle is at rest, therefore:
$$P^{2}=E^{2}-\vec{p}^{2} c^{2}=m^{2} c^{4}$$
is true in any frame, which is just the Einstein energy-momentum relationship. This is encouraging, it tells us that the rest mass of a particle is independent of the frame it is in.

We can define the invariant mass, $W=P / c^{2}$. For a system of $N$ particles, the total four-momenta is the sum of the individual four-momenta, so we have:
$$W^{2} c^{4}=\left(\sum_{i}^{N} E_{i}\right)^{2}-\left(c \sum_{i}^{N} \vec{p}_{i}\right)^{2}$$
which is an invariant quantity (i.e. it is the same in any frame of reference). It is also a conserved quantity. For one particle the invariant mass is the same as the rest mass. A very important use of the invariant mass is in determining the mass of a short-lived particle. If a particle decays to a set of $N$ particles its mass can be determined from the invariant mass of the decay products, since it is conserved.

## 物理代写|粒子物理学作业代写Nuclear and Particle Physics代考|Centre-of-mass frame

For a system of particles the centre-of-mass (or centre-of-momentum) frame is the frame in which $\sum_{i} \vec{p}{i}$ is zero. In a collider experiment the energy in the centre-of-mass frame is a very useful quantity, as it gives the maximum energy available to create a massive particle: $E=m c^{2}$. In any other frame (including the laboratory frame) some of the energy is tied up in the conserving the momentum. The square of the centreof-mass energy, $E{C M}^{2}$, is just equal to the invariant mass squared, in the centre-of-mass frame. Since the invariant mass squared is frame invariant, it can be evaluated from beam energies given in the laboratory (or any other) frame, and this will give us the centre-of-mass energy squared. Below are a couple of worked examples for evaluating the centre-of-mass energy:

Example 1 Consider a simple case in which two beams are colliding with equal energy in opposite directions. The centre-of-mass frame is the same as the lab frame (i.e. the frame in which we observe the beams). In order to determine the centre-of-mass energy we just evaluate the invariant mass in the centre-of-mass frame (for convenience I am dropping factors of $c$ ), using Equation $1.2$ we have:
$$W^{2}=\left(2 \times E_{\text {beam }}\right)^{2}-(0)^{2}=4 E_{\text {beam }}^{2}$$
giving:
$$E_{\mathrm{CM}}=2 E_{\text {beam }}$$
where $E_{\text {beam }}$ is the beam energies in the lab frame and $E_{\mathrm{CM}}$ is the centre-of-mass energy.

Example 2 A slightly more complicated example is for a fixed-target experiment, where there is one moving beam of particles colliding with a stationary target. The mass of the particles in the target and the beam can usually be assumed to be much less than the energy of the beam. We evaluate the invariant mass in the lab-frame:
$$W^{2}=\left(E_{\text {beam }}+m_{\text {target }}\right)^{2}-\left(\vec{p}{\text {beam }}\right)^{2}=E{\text {beam }}^{2}+m_{\text {target }}^{2}+2 E_{\text {beam }} m_{\text {target }}-\vec{p}{\text {beam }}^{2}$$ substituting in $E{\text {beam }}^{2}=\vec{p}{\text {beam }}^{2}+m{\text {beam }}^{2}$ we obtain:
$$W^{2}=m_{\text {beam }}^{2}+m_{\text {target }}^{2}+2 E_{\text {beam }} m_{\text {target }}$$
we can neglect the first two terms in the limit that the beam energy is much larger than the masses of the particles in the beams, giving:
$$W^{2}=2 E_{\text {beam }} m_{\text {target }}$$
and therefore:
$$E_{\mathrm{CM}}=\sqrt{W^{2}}=\sqrt{2 E_{\text {beam }} m_{\text {target }}} .$$

## 物理代写|粒子物理学作业代写NUCLEAR AND PARTICLE PHYSICS代考|FOUR-VECTORS AND INVARIANTS

$$W^{2}=m_{\text {beam }}^{2}+m_{\text {target }}^{2}+2 E_{\text {beam }} m_{\text {target }}$$
we can neglect the first two terms in the limit that the beam energy is much larger than the masses of the particles in the beams, giving:
$$W^{2}=2 E_{\text {beam }} m_{\text {target }}$$
and therefore:
$$E_{\mathrm{CM}}=\sqrt{W^{2}}=\sqrt{2 E_{\text {beam }} m_{\text {target }}} .$$

## Matlab代写

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