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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

So far, we have assumed that the charge distribution is spherically symmetric. If that were the case we would have $$\left\langle x^2\right\rangle=\left\langle y^2\right\rangle=\left\langle z^2\right\rangle=\frac{1}{3}\left\langle r^2\right\rangle \text {, }$$
where
$$=\frac{1}{Z e} \int x^2 \rho(\boldsymbol{r}) d^3 \boldsymbol{r},$$
etc.
However, for many nuclei this is not the case. They usually still have an axis of symmetry, which we set to be the $z$-axis, so that they are symmetric in the $x-y$ plane but asymmetric in the $x-z$ or $y-z$ planes. In polar coordinates $(r, \vartheta, \phi)$, this means that the charge distribution is a function of the polar angle, $\vartheta$, but for nuclei which still maintain one axis of symmetry (the $z$-axis), the charge distribution is independent of the azimuthal angle, $\phi$.
We can still determine a charge radius
$$R=\int r \rho(\boldsymbol{r}) d^3 \boldsymbol{r},$$
even if $\rho(\boldsymbol{r})$ is not a function of the radial component, $r$, alone. However, if we look at the expectation values of the squares of individual component of $\boldsymbol{r}$
$$\left\langle x^2\right\rangle=\int x^2 \rho(\boldsymbol{r}) d^3 \boldsymbol{r},\left\langle y^2\right\rangle=\int y^2 \rho(\boldsymbol{r}) d^3 \boldsymbol{r},\left\langle z^2\right\rangle=\int z^2 \rho(\boldsymbol{r}) d^3 \boldsymbol{r},$$
we find that these are not equal. Nuclei which nevertheless have an axis of symmetry have the same values of $\left\langle x^2\right\rangle$ and $\left\langle y^2\right\rangle$, but a different value for $\left\langle z^2\right\rangle$.

Nuclei with a charge distribution that is not spherically symmetric possess an “electric quadrupole moment” defined (with respect to the $z$-axis) as
$$\mathcal{Q}=3\left\langle z^2\right\rangle-\left\langle r^2\right\rangle=\int\left(3 z^2-r^2\right) \rho(\boldsymbol{r}) d^3 \boldsymbol{r},$$
where
$$\left\langle r^2\right\rangle=\left\langle x^2\right\rangle+\left\langle y^2\right\rangle+\left\langle z^2\right\rangle .$$

## 物理代写|粒子物理代写particle physics代考|Strong Force Distribution

The protons and neutrons inside a nucleus are held together by a strong nuclear force. This has to be strong enough to overcome the Coulomb repulsion between the protons, but unlike the Coulomb force, it extends only over a short range of a few fm.

Electrons are used to probe the charge distribution of the target nuclei, because they interact with the electric field, but not with the strong forces. Likewise, scattering of neutrons from a nucleus can be used to probe the strong force distribution, but not the electric charge distribution since neutrons are uncharged but interact strongly.

As in the case of electron scattering, the cross section of neutron scattering displays a diffraction pattern if the de Broglie wavelength of the neutrons is of the order of the nuclear size. In such a case the wave from different parts of the nucleus interfere to produce diffraction maxima and minima at different scattering angles. This can be seen in Fig. $2.9$ in which neutrons with kinetic energy $14 \mathrm{MeV}$ are scattered from a Ni (nickel) nucleus. The de Broglie wavelength of neutrons with kinetic energy $14 \mathrm{MeV}$ is approximately $1.2 \mathrm{fm}$ so that the first minimum of the differential cross section, at a scattering angle of $42^{\circ}$, implies an effective radius of the strong force distribution of a few fm. This is similar to the charge radius of the nuclens. We would expect the total nucleon distrihution to have the same range as the proton (charge) distribution. However, whereas the Coulomb potential from the nucleus is long-range, being attenuated as the inverse of the distance from the centre of the nucleus, the strong force is rapidly attenuated and becomes negligible after a few fim from the nucleus.

The differential cross section can be expressed in terms of a “scattering amplitude”, $f(\theta, \phi)$ (usually a function of the scattering angle, $\theta$, only, but could in some cases depend on the azimuthal angle, $\phi$, of the outgoing particle), via the relation
$$\frac{d \sigma}{d \Omega}=|f(\theta, \phi)|^2$$

# 粒子物理代考

$$\left\langle x^2\right\rangle=\left\langle y^2\right\rangle=\left\langle z^2\right\rangle=\frac{1}{3}\left\langle r^2\right\rangle,$$

$$=\frac{1}{Z e} \int x^2 \rho(\boldsymbol{r}) d^3 \boldsymbol{r},$$

$$R=\int r \rho(\boldsymbol{r}) d^3 \boldsymbol{r},$$

$$\left\langle x^2\right\rangle=\int x^2 \rho(\boldsymbol{r}) d^3 \boldsymbol{r},\left\langle y^2\right\rangle=\int y^2 \rho(\boldsymbol{r}) d^3 \boldsymbol{r},\left\langle z^2\right\rangle=\int z^2 \rho(\boldsymbol{r}) d^3 \boldsymbol{r},$$

$$\mathcal{Q}=3\left\langle z^2\right\rangle-\left\langle r^2\right\rangle=\int\left(3 z^2-r^2\right) \rho(\boldsymbol{r}) d^3 \boldsymbol{r},$$

$$\left\langle r^2\right\rangle=\left\langle x^2\right\rangle+\left\langle y^2\right\rangle+\left\langle z^2\right\rangle .$$

## 物理代写|粒子物理代写particle physics代考|Strong Force Distribution

$$\frac{d \sigma}{d \Omega}=|f(\theta, \phi)|^2$$

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师