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assignmentutor-lab™ 为您的留学生涯保驾护航 在代写宇宙学cosmology方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写宇宙学cosmology代写方面经验极为丰富，各种代写宇宙学cosmology相关的作业也就用不着说。

• 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

## 物理代写|宇宙学代写cosmology代考|Neutron abundance

We begin by solving for the neutron-proton ratio. Protons can be converted into neutrons via weak interactions, $p+e^{-} \rightarrow n+v_e$ for example. As we will see, reactions of this sort keep neutrons and protons in equilibrium until $T \sim \mathrm{MeV}$. Thereafter, one must solve Eq. (4.8) to track the neutron abundance.

From Eq. (4.5), the proton/neutron equilibrium ratio in the nonrelativistic limit (so that $\left.E_i(p)=m_i+p^2 / 2 m_i\right)$ is
$$\frac{n_p^{(0)}}{n_n^{(0)}}=\frac{e^{-m_p / T} \int d p p^2 e^{-p^2 / 2 m_p T}}{e^{-m_n / T} \int d p p^2 e^{-p^2 / 2 m_n T}}$$
The integrals here are proportional to $m^{3 / 2}$. The resulting ratio $\left(m_p / m_n\right)^{3 / 2}$ is sufficiently close to unity that we can neglect the mass difference. However, in the exponential the mass difference is very important, and we are left with
$$\frac{n_p^{(0)}}{n_n^{(0)}}=e^{\mathcal{Q} / T}$$
with $\mathcal{Q} \equiv m_n-m_p=1.293 \mathrm{MeV}$. Therefore, at high temperatures, there are as many neutrons as protons. As the temperature drops beneath $1 \mathrm{MeV}$, the neutron fraction goes down. If weak interactions operated efficiently enough to maintain equilibrium indefinitely, then it would drop to zero (even if free neutrons were stable). The main task of this section is to find what happens in the real world where weak interactions are not so efficient. It is convenient to define
$$X_n \equiv \frac{n_n}{n_n+n_p}$$

that is, $X_n$ is the ratio of neutrons to total nuclei. In equilibrium,
$$X_n \rightarrow X_{n, \mathrm{EQ}} \equiv \frac{1}{1+n_p^{(0)} / n_n^{(0)}}$$

## 物理代写|宇宙学代写cosmology代考|Light element abundances

A useful way to approximate light element production is that it occurs instantaneously at a temperature $T_{\text {nuc }}$ when the energetics compensates for the small baryon-to-photon ratio. Let us consider deuterium production as an example, with Eq. (4.16) as our guide. The equilibrium deuterium abundance is of order the baryon abundance (i.e. if the universe stayed in equilibrium, all neutrons and protons would form deuterium) when the righthand side of Eq. (4.16) is of order unity, or
$$\ln \left(\eta_{\mathrm{b}}\right)+\frac{3}{2} \ln \left(T_{\text {nuc }} / m_p\right) \sim-\frac{B_{\mathrm{D}}}{T_{\text {nuc }}} .$$
Eq. (4.29) suggests that deuterium production takes place at $T_{\text {nuc }} \sim 0.07 \mathrm{MeV}$.
Since the binding energy of helium is larger than that of deuterium, the exponential factor $e^{B / T}$ favors helium over deuterium. Indeed, Fig. $4.3$ illustrates that helium is produced almost immediately after deuterium. Virtually all remaining neutrons at $T \sim T_{\text {nuc }}$ then are processed into ${ }^4 \mathrm{He}$. Since two neutrons go into ${ }^4 \mathrm{He}$, the final ${ }^4 \mathrm{He}$ abundance is equal to half the neutron abundance at $T_{\text {nuc. }}$. Often, results are quoted in terms of mass fraction; then,
$$Y_P \equiv \frac{4 n\left({ }^4 \mathrm{He}\right)}{n_{\mathrm{b}}}=2 X_n\left(T_{\mathrm{nuc}}\right),$$
which yields a final helium mass fraction of $0.22$. This rough estimate, obtained by solving a single differential equation, is in remarkable agreement with a full numerical calculation, which can be fit via (Olive, 2000)
$$Y_P=0.2262+0.0135 \ln \left(\eta_{\mathrm{b}} / 10^{-10}\right)$$

# 宇宙学代考

## 物理代写|宇宙学代写cosmology代考|中子丰度

.

$$\frac{n_p^{(0)}}{n_n^{(0)}}=\frac{e^{-m_p / T} \int d p p^2 e^{-p^2 / 2 m_p T}}{e^{-m_n / T} \int d p p^2 e^{-p^2 / 2 m_n T}}$$

$$\frac{n_p^{(0)}}{n_n^{(0)}}=e^{\mathcal{Q} / T}$$

$$X_n \equiv \frac{n_n}{n_n+n_p}$$

，即$X_n$是中子与总核的比率。在平衡状态下，
$$X_n \rightarrow X_{n, \mathrm{EQ}} \equiv \frac{1}{1+n_p^{(0)} / n_n^{(0)}}$$

## 物理代写|宇宙学代写cosmology代考|轻元素丰度

.

$$\ln \left(\eta_{\mathrm{b}}\right)+\frac{3}{2} \ln \left(T_{\text {nuc }} / m_p\right) \sim-\frac{B_{\mathrm{D}}}{T_{\text {nuc }}} .$$

$$Y_P \equiv \frac{4 n\left({ }^4 \mathrm{He}\right)}{n_{\mathrm{b}}}=2 X_n\left(T_{\mathrm{nuc}}\right),$$
，它产生了$0.22$的最终氦质量分数。通过求解单个微分方程得到的这个粗略估计与完整的数值计算非常一致，可以通过(Olive, 2000)
$$Y_P=0.2262+0.0135 \ln \left(\eta_{\mathrm{b}} / 10^{-10}\right)$$ 进行拟合

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

assignmentutor™您的专属作业导师
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