# 热力学代写 Thermodynamics代考2023

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## 热力学代写Thermodynamics

### 统计力学Statistical mechanics代写

• Chemical thermodynamics化学热力学
• Equilibrium thermodynamics平衡热力学

## 热力学的历史

Thermodynamics is a branch of classical physics and chemistry that studies and describes the thermodynamic transformations induced from heat to work in a thermodynamic system in processes involving changes in the state variables temperature and energy.

Classical thermodynamics is based on the concept of a macroscopic system, i.e. a part of the mass physically or conceptually separated from the external environment, which for convenience is usually assumed not to be disturbed by the exchange of energy with the system (isolated system): the state of a macroscopic system in equilibrium is specified by quantities called thermodynamic variables or state functions, such as temperature, pressure, volume and chemical composition. The main notations of chemical thermodynamics have been established by IUPAC.

## 热力学相关课后作业代写

A state function for a Van der Waals gas is given by an equation between thermodynamic variables that depend on model parameters $A, B$, and a physical constant $R$ :
$$\left(P+\frac{A N^2}{V^2}\right)(V-N B)=N R T$$
where $A N^2 / V^2$ is referred to as the internal pressure due to the attraction between molecules and $N B$ is an extra volume, sometimes associated with the the volume per molecule.

Write out a differential expression for $d N$ in terms of differentials of the thermodynamic variables.

A state function for a Van der Waals gas is given by an equation between thermodynamic variables that depend on model parameters $A, B$, and a physical constant $R$ :
$$\left(P+\frac{A N^2}{V^2}\right)(V-N B)=N R T$$
where $A N^2 / V^2$ is referred to as the internal pressure due to the attraction between molecules and $N B$ is an extra volume, sometimes associated with the volume per molecule.

Write out a differential expression for $d N$ in terms of differentials of the thermodynamic variables.

The solution is pretty straightforward. One way is to differentiate the entire expression and group the terms corresponding to $d N, d P, d T$, and $d V$. Another way to do is by implicit differentiation. The real gas equation can be rewritten such that,
\begin{aligned} N & =N(T, V, P) \ d N & =\left(\frac{\partial N}{\partial P}\right){T, V} d P+\left(\frac{\partial N}{\partial V}\right){T, P} d V+\left(\frac{\partial N}{\partial T}\right){P, V} d T \end{aligned} In an equivalent way, you could have written the function $P=P(V, T, N)$ and extract $d N$ from the following. $$d P=\left(\frac{\partial P}{\partial N}\right){T, V} d N+\left(\frac{\partial P}{\partial V}\right){T, N} d V+\left(\frac{\partial P}{\partial T}\right){N, V} d T$$
For instance, the first term $\left(\frac{\partial P}{\partial N}\right){T, V}$ can be evaluated as $$\left(\frac{\partial P}{\partial T}\right){P, V}=\frac{N R}{V-N B}$$
Using any one of the methods you would get
$$d N=\frac{(V-B N) d P+\left(P-\left(A N^2 / V^2\right)+\left(2 A B N^3 / V^3\right)\right) d V-R N d T}{B P+R T+\left(3 A B N^2 / V^2\right)-(2 A N / V)}$$

# 光学代写 optics代考2023

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## 光学代写optics

### 量子光学Quantum optics代写

• 反射（物理学）Reflection (physics)
• 折射Refraction

## 光学的历史

The history of optics is a part of the history of science. The term optics comes from the ancient Greek τα ὀπτικά. It is originally the science of everything related to the eye. The Greeks distinguish optics from dioptrics and catoptrics. We would probably call the first science of vision, the second science of lenses and the third science of mirrors. The great names of Greek optics are Euclid, Heron of Alexandria and Ptolemy.

Since antiquity, optics has undergone many developments. The very meaning of the word has varied and from the study of vision, it has passed in several stages to the study of light, before being incorporated recently into a broader body of physics.

## 光学相关课后作业代写

A glass plate is sprayed with uniform opaque particles. When a distant point source of light is observed looking through the plate, a diffuse halo is seen whose angular width is about $2^{\circ}$. Estimate the size of the particles. (Hint: consider Fraunhoffer diffraction through random gratings, and use Babinet’s principle)

The diffraction pattern of an opaque circular particle is complementary to that due to circular apertures of the same size in an otherwise opaque screen.
Under the Fraunhofer condition $\left(\frac{k\left(x^{\prime 2}+y^{\prime 2}\right)}{2 z} \ll 1, \frac{k\left(x^2+y^2\right)}{2 z} \ll 1\right)$
\begin{aligned} & E\left(x^{\prime}, y^{\prime}\right) \approx \frac{1}{z} \iint \exp \left(-i k\left(\theta_{x^{\prime}} x+\theta_{y^{\prime}} y\right)\right) t(x, y) E(x, y) d x d y \ & \text { Where } \theta_{x^{\prime}} \approx \frac{x \prime}{z}, \theta_{y^{\prime}} \approx \frac{y \prime}{z} \ & \end{aligned}
For the given problem, we may further assume $\mathrm{E}(\mathrm{x}, \mathrm{y})$ is a plane wave at normal incidence, and the transmission function $t(x, y)$ for a single can be expressed as:
$$t(x, y)=1-\operatorname{circ}\left(\frac{\sqrt{x^2+y^2}}{R}\right)$$
Where $R$ is the radius of the opaque particles.
$$\begin{gathered} E\left(x^{\prime}, y^{\prime}\right) \approx \frac{1}{z} \iint \exp \left(-i k\left(\theta_{x^{\prime}} x+\theta_{y^{\prime}} y\right)\right)\left[1-\operatorname{circ}\left(\frac{\sqrt{x^2+y^2}}{R}\right)\right] d x d y \ E\left(x^{\prime}, y^{\prime}\right) \approx \frac{1}{z} \mathcal{F}\left[1-\operatorname{circ}\left(\frac{\sqrt{x^2+y^2}}{R}\right)\right] \ \text { With } x^{\prime}=\frac{z}{k} k_x, y^{\prime}=\frac{z}{k} k_y \ E\left(k_x, k_y\right) \approx \frac{1}{z}\left[\delta\left(\sqrt{k_x{ }^2+k_y{ }^2}\right)-|R|^2 \frac{2 \pi J_1\left(R \sqrt{k_x^2+k_y{ }^2}\right)}{R \sqrt{k_x{ }^2+k_y{ }^2}}\right] \end{gathered}$$

Where $\gamma=R \sqrt{k_x^2+k_y^2}=\frac{2 \pi}{\lambda} R \theta$
From the above table,
$$\frac{2 \pi}{\lambda} R \Delta \theta=7.106-3.832=3.274$$
Taking central wavelength at visible frequency, $\lambda=500 \mathrm{~nm}$ and given $\Delta \theta=2^{\circ}$, we find the radius of the particle:
$$R=\lambda \frac{3.274}{(2 \pi)^2\left(\frac{\Delta \theta}{360}\right)}=500 \mathrm{~nm} \times\left(\frac{3.274}{(2 \pi)^2 \frac{2}{360}}\right)=7463 \mathrm{~nm}=7.4 \mu \mathrm{m}$$

# 电磁学代写 Electromagnetism代考2023

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## 电磁学代写Electromagnetism

### 经典电磁学Classical electrodynamics代写

1600年，威廉-吉尔伯特在他的De Magnete中提出，电和磁虽然都能引起物体的吸引和排斥，但却是不同的效果。海员们注意到，雷击有能力干扰罗盘针。直到本杰明-富兰克林在1752年提出的实验，法国的托马斯-弗朗索瓦-达利巴德在1752年5月10日用一根40英尺高（12米）的铁棒代替风筝进行了实验，他成功地从云中提取了电火花，这才证实了闪电和电力之间的联系。

• Nonlinear system非线性系统
• Magnetohydrodynamics磁流体力学

## 电磁学的历史

The earliest study of this phenomenon probably goes back to the Greek philosopher Thales (600 BC), who studied the electrical properties of amber, a fossil resin that attracts other substances when rubbed: its Greek name is elektron (ἤλεκτρον), from which the word ‘electricity’ is derived. The ancient Greeks realised that amber could attract light objects, such as hair, and that repeated rubbing of the amber itself could even produce sparks.

## 电磁学相关课后作业代写

Show that $S^4$ has no symplectic structure. Show that $S^2 \times S^4$ has no symplectic structure.

To show that $S^4$ has no symplectic structure, we will use the following fact from symplectic geometry: a compact symplectic manifold has even dimension.

Suppose that $S^4$ has a symplectic structure. Then $S^4$ is a compact symplectic manifold, so its dimension must be even. However, the dimension of $S^4$ is $4$, which is not even. Therefore, $S^4$ cannot have a symplectic structure.

To show that $S^2 \times S^4$ has no symplectic structure, we will use the following fact: the product of two symplectic manifolds is symplectic if and only if both factors have even dimension.

Suppose that $S^2 \times S^4$ has a symplectic structure. Then both $S^2$ and $S^4$ are symplectic manifolds, so their dimensions must both be even. However, the dimension of $S^2$ is $2$, which is not even. Therefore, $S^2 \times S^4$ cannot have a symplectic structure.

# 几何学代写 Geometry代考2023

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## 几何学代写Geometry

### 点（几何学）Point (geometry)代写

• Line (geometry)线（几何学）
• Plane (geometry)平（几何学）

## 几何学的历史

The birth of geometry can be traced back to the time of the ancient Egyptians. Herodotus recounts that due to the erosion and deposition of the Nile floods, the extent of the Egyptians’ land changed from year to year, making it necessary to recalculate taxes. Therefore, it was necessary to invent land surveying techniques (geometry, in the original meaning of the term).

The development of practical geometry is very ancient, and since it could have many applications and was developed for those applications, in ancient times, practical geometry was sometimes reserved for a class of wise men with priestly attributes. In ancient Greece, mainly due to the influence of the Athenian philosopher Plato, and even before him, Anaximander of Miletus [citation needed], the use of rules and compasses became common (although these tools seem to have been invented elsewhere), and above all, new ideas using presentation techniques were born. Greek geometry became the basis for the development of geography, astronomy, optics, mechanics and other sciences, as well as various technologies (such as navigation technology).

In Greek civilization, in addition to Euclidean geometry and conic theory, which were still studied in schools, spherical geometry and trigonometry (plane and sphere) were born.

## 几何学相关课后作业代写

Show that $S^4$ has no symplectic structure. Show that $S^2 \times S^4$ has no symplectic structure.

To show that $S^4$ has no symplectic structure, we will use the following fact from symplectic geometry: a compact symplectic manifold has even dimension.

Suppose that $S^4$ has a symplectic structure. Then $S^4$ is a compact symplectic manifold, so its dimension must be even. However, the dimension of $S^4$ is $4$, which is not even. Therefore, $S^4$ cannot have a symplectic structure.

To show that $S^2 \times S^4$ has no symplectic structure, we will use the following fact: the product of two symplectic manifolds is symplectic if and only if both factors have even dimension.

Suppose that $S^2 \times S^4$ has a symplectic structure. Then both $S^2$ and $S^4$ are symplectic manifolds, so their dimensions must both be even. However, the dimension of $S^2$ is $2$, which is not even. Therefore, $S^2 \times S^4$ cannot have a symplectic structure.

# 力学代写 mechanics代考2023

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## 力学代写mechanics

### 统计力学Statistical mechanics代写

• Theory of relativity相对论
• Quantum mechanics量子力学

## 力学的相关

The history and development of differential geometry as a discipline can be traced back at least to the ancient classics. It is closely related to the development of geometry, the concepts of space and form, and the study of topology, particularly manifolds. In this section we focus on the history of the application of infinitesimal methods to geometry, and then on the idea of tangent spaces, and finally on the development of the modern formalism of the discipline in terms of tensors and tensor fields.

## 力学相关课后作业代写

Hooke’s law, a constitutive equation for a linear, elastic material, can be written in general form as:
$$\sigma_{i j}=\lambda \varepsilon_{k k} \delta_{i j}+2 \mu \varepsilon_{i j} \text { where } \lambda \text { and } \mu \text { are Làme constants. }$$
a) Expand Hooke’s Law. How many independent equations are there?

a) Hooke’s law
$$\sigma_{i j}=\lambda \varepsilon_{k k} \delta_{i j}+2 \mu \varepsilon_{i j}$$
where
$$\delta_{i j}=\left{\begin{array}{l} 1, \mathrm{i}=\mathrm{j} \ 0, \mathrm{i} \neq \mathrm{j} \end{array}\right.$$
For $i=1$
\begin{aligned} & \sigma_{11}=\lambda\left(\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}\right)+2 \mu \varepsilon_{11} \ & \sigma_{12}=2 \mu \varepsilon_{12} \ & \sigma_{13}=2 \mu \varepsilon_{13} \end{aligned}

For $\mathrm{i}=2$
\begin{aligned} & \sigma_{21}=2 \mu \varepsilon_{21} \ & \sigma_{22}=\lambda\left(\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}\right)+2 \mu \varepsilon_{22} \ & \sigma_{23}=2 \mu \varepsilon_{23} \end{aligned}
For $\mathrm{i}=3$
\begin{aligned} & \sigma_{31}=2 \mu \varepsilon_{31} \ & \sigma_{32}=2 \mu \varepsilon_{32} \ & \sigma_{33}=\lambda\left(\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}\right)+2 \mu \varepsilon_{33} \end{aligned}

# 微分几何代写 Differential Geometry代考2023

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## 微分几何代写Differential Geometry

### 伪黎曼尼几何学Pseudo-Riemannian geometry代写

• Finsler manifold芬斯勒流形
• Symplectic geometry辛几何

## 微分几何的历史

The history and development of differential geometry as a discipline can be traced back at least to the ancient classics. It is closely related to the development of geometry, the concepts of space and form, and the study of topology, particularly manifolds. In this section we focus on the history of the application of infinitesimal methods to geometry, and then on the idea of tangent spaces, and finally on the development of the modern formalism of the discipline in terms of tensors and tensor fields.

## 微分几何相关课后作业代写

Let $c$ be a regular curve such that $|c(s)| \leq 1$ for all $s$. Suppose that there is a point $t$ where $|c(t)|=1$. Prove that the curvature at that point satisfies $|\kappa(t)| \geq 1$.

Let $c$ be a regular curve with arc length parameterization $s$. By definition, the curvature $\kappa(s)$ of the curve $c$ at the point $c(s)$ is given by $\kappa(s) = |\boldsymbol{c}”(s)|$, where $\boldsymbol{c}”(s)$ denotes the second derivative of $\boldsymbol{c}(s)$ with respect to $s$.

Since $|\boldsymbol{c}(s)| \leq 1$ for all $s$, we have $|\boldsymbol{c}'(s)| = 1$ for all $s$. Moreover, since $|\boldsymbol{c}(t)| = 1$, we have $|\boldsymbol{c}'(t)| = 0$, which implies that $\boldsymbol{c}”(t)$ is perpendicular to $\boldsymbol{c}'(t)$. Therefore, we have

$\left|\boldsymbol{c}^{\prime \prime}(t)\right|=\left|\boldsymbol{c}^{\prime \prime}(t) \cdot \frac{c^{\prime}(t)}{\left|c^{\prime}(t)\right|}\right|=\left|c^{\prime \prime}(t) \cdot \frac{c^{\prime}(t)}{\left|c^{\prime}(t)\right|}\right|$.

Now, we observe that $\boldsymbol{c}”(t) \cdot \boldsymbol{c}'(t)$ is the tangential component of $\boldsymbol{c}”(t)$ at the point $c(t)$, which is given by $\boldsymbol{c}”(t) \cdot \boldsymbol{c}'(t) = \frac{d}{ds}\left(|\boldsymbol{c}'(s)|^2\right)\bigg|{s=t} = 2|\boldsymbol{c}'(t)|\boldsymbol{c}”(t) \cdot \boldsymbol{c}'(t) = 0$. Thus, we can write $\boldsymbol{c}”(t) = \boldsymbol{c}”(t){\perp} + \boldsymbol{c}”(t){\parallel}$, where $\boldsymbol{c}”(t){\perp}$ is the perpendicular component of $\boldsymbol{c}”(t)$ to $\boldsymbol{c}'(t)$ and $\boldsymbol{c}”(t){\parallel}$ is the tangential component of $\boldsymbol{c}”(t)$ to $\boldsymbol{c}'(t)$. Since $|\boldsymbol{c}'(t)| = 0$, we have $\boldsymbol{c}”(t){\parallel} = 0$. Therefore, we have

$\left|\boldsymbol{c}^{\prime \prime}(t)\right|=\left|\boldsymbol{c}^{\prime \prime}(t){\perp}\right| \leq\left|\boldsymbol{c}^{\prime \prime}(t)\right|{\max }$

where $|\boldsymbol{c}”(t)|_{\max}$ denotes the maximum value of $|\boldsymbol{c}”(t)|$ over all vectors $\boldsymbol{v}$ perpendicular to $\boldsymbol{c}'(t)$ with $|\boldsymbol{v}| = 1$.