# 热力学代写 Thermodynamics代考2023

0

## 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

0

## 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

0

## 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.

# 力学代写 mechanics代考2023

0

## 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}

# 流体力学|Fluid mechanics II 3A3代写2023

0

Assignment-daixieTM为您提供剑桥大学University of Cambridge Fluid mechanics II 3A3流体力学代写代考辅导服务！

## Instructions:

Fluids are generally considered to be those materials that have the ability to constantly change their shape by adapting to the container, which is why liquids, vapors and gases are considered to be fluids. Fluid mechanics consists of two main branches:

Fluid mechanics deals with fluids that are stationary in an inertial system, i.e. with constant velocity in time and homogeneity in space. Historically, it was the first step towards the study of mechanics.
Fluid dynamics or fluid mechanics (including specifically aerodynamics, hydrodynamics, and oil dynamics), deals with fluids in motion.
Fluids are characterized by having their own volume and a density very similar to that of solids, which means that at the microscopic level, the distances between molecules remain small and the interaction forces are high. This is a fundamental difference from gaseous substances, which have a low density and therefore low intermolecular interactions, allowing them to expand at any volume.

(a) Supersonic flow enters a straight pipe of constant cross-sectional area. Heat transfer is negligible, but the pipe wall is rough. Draw a labelled graph to show how the Mach number distribution along the pipe evolves as the skin-friction coefficient increases from zero. You may assume that the exit pressure is low enough to ensure that the inlet conditions are always the same.

(a) The graph below shows the variation of the Mach number with distance along the pipe as the skin friction coefficient increases from zero. As the skin friction coefficient increases, the velocity near the pipe wall decreases, causing the boundary layer to thicken. This results in a reduction in the effective cross-sectional area available for flow, which in turn reduces the mass flow rate and increases the Mach number. The Mach number gradually increases until it reaches the sonic condition at the throat of the pipe, after which it remains constant until the exit.

(b) Air flows in a pipe of length $5.9 \mathrm{~m}$ and inside diameter $0.2 \mathrm{~m}$. The inlet stagnation pressure is 2.7 bar, and the static pressure at the pipe exit is 1 bar. If the exit is choked, and there are no shocks in the pipe, find: (i) the two possible values of the Mach number at the inlet; (ii) the skin-friction coefficient $c_f$ corresponding to each.

(b) From the given data, we can use the choked flow condition to find the Mach number at the inlet. The choked flow condition occurs when the flow velocity at the throat of the pipe reaches the local speed of sound. At the throat, the Mach number is therefore 1.

Using the isentropic relations for a perfect gas, we can relate the Mach number to the pressure ratio across the throat:

$\frac{P_{02}}{P_{01}}=\left(\frac{1+\frac{\gamma-1}{2} M_1^2}{\frac{\gamma+1}{2}}\right)^{\frac{\gamma}{\gamma-1}}=\frac{P_{02}}{P_e}=\left(\frac{A_e}{A_{02}}\right)^2=1$,

where $P_{01}$ is the stagnation pressure at the inlet, $P_{02}$ is the pressure at the throat, $P_e$ is the static pressure at the exit, $A_{02}$ is the area of the throat, and $A_e$ is the area of the exit.

Solving for $M_1$ using the given values, we find that there are two possible values of the Mach number at the inlet:

$M_1=\sqrt{\frac{2}{\gamma-1}\left[\left(\frac{P_{02}}{P_{01}}\right)^{\frac{\gamma-1}{\gamma}}-1\right]}=0.747,2.11$.

Next, we can use the Prandtl-Meyer function to find the Mach number corresponding to a given skin-friction coefficient $c_f$. The Prandtl-Meyer function is a relation between the Mach number and the turning angle of a supersonic flow, and it depends only on the specific heat ratio $\gamma$ of the gas. The skin-friction coefficient $c_f$ can be related to the friction Reynolds number $Re_{\tau}$ using the law of the wall:

$c_f=\frac{\tau_w}{\frac{1}{2} \rho_1 V_1^2}=\frac{0.026}{R e_\tau^{0.2}}$,

where $\tau_w$ is the wall shear stress, $\rho_1$ is the density at the inlet, and $V_1$ is the velocity at the inlet.

For air at room temperature and pressure, $\gamma=1.4$. Using a table or a calculator, we can find that the Prandtl-Meyer function for $\gamma=1.4$ is approximately $\nu=30.5^{\circ}$ at $M=0.747$, and $\nu=66.1^{\circ}$ at $M=2.11$.

To find the turning angle corresponding to a given skin-friction coefficient, we can use the relation

$\theta=\frac{c_f}{2 \nu}$

# 固体力学|Mechanics of solids 3C7代写2023

0

Assignment-daixieTM为您提供剑桥大学University of Cambridge Mechanics of solids 3C7固体力学代写代考辅导服务！

## Instructions:

Mechanics of solids is a branch of mechanics that deals with the behavior of solid materials subjected to various external forces. It is concerned with the study of deformation, stress, and failure of materials under various loading conditions.

Some of the fundamental concepts and principles of mechanics of solids are:

1. Stress: Stress is the force per unit area acting on a material. It is a measure of the internal forces that hold the material together.
2. Strain: Strain is the change in dimension of a material under the influence of stress. It is a measure of the deformation of the material.
3. Elasticity: Elasticity is the ability of a material to deform under stress and return to its original shape when the stress is removed. A material that exhibits this behavior is said to be elastic.
4. Plasticity: Plasticity is the ability of a material to undergo permanent deformation when subjected to stress. A material that exhibits this behavior is said to be plastic.
5. Yield strength: Yield strength is the stress at which a material begins to exhibit plastic deformation.
6. Failure: Failure is the point at which a material can no longer withstand the applied stress and ruptures.

Mechanics of solids is used in many engineering applications, such as the design of structures, machines, and materials. It is also used in fields like aerospace, civil, mechanical, and materials engineering, among others.

Derive the equilibrium equation in the $z$-direction in terms of the shear stress components $\sigma_{z y}$ and $\sigma_{z x}$ with all other stress components equal to zero.

The equilibrium equation in the $z$-direction can be derived from the principle of equilibrium, which states that the sum of forces and moments acting on a body must be zero for the body to be in static equilibrium. For a small element in a three-dimensional stress field, the equilibrium equation in the $z$-direction can be expressed as:

$\frac{\partial \sigma_{x z}}{\partial x}+\frac{\partial \sigma_{y z}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$,

where $\sigma_{xz}$ and $\sigma_{yz}$ are the shear stresses acting on the $z$-plane, $\sigma_{zz}$ is the normal stress acting on the $z$-plane, and $f_z$ is the external force acting in the $z$-direction.

Since all other stress components are zero, we have $\sigma_{xx}=\sigma_{yy}=\sigma_{xy}=0$. Thus, the equilibrium equation reduces to:

$\frac{\partial \sigma_{z x}}{\partial x}+\frac{\partial \sigma_{z y}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$.

Since $\sigma_{zy}=\sigma_{yz}$, we can write:

$\frac{\partial \sigma_{z x}}{\partial x}+\frac{\partial \sigma_{y z}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$.

Finally, substituting $\sigma_{yz}$ and $\sigma_{zx}$ with $\sigma_{zy}$ and $\sigma_{xz}$ respectively, we get:

$\frac{\partial \sigma_{x z}}{\partial x}+\frac{\partial \sigma_{z y}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$.

Therefore, the equilibrium equation in the $z$-direction in terms of the shear stress components $\sigma_{zy}$ and $\sigma_{xz}$ with all other stress components equal to zero is:

$\frac{\partial \sigma_{x z}}{\partial x}+\frac{\partial \sigma_{z y}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$

For the case of a torsion of a shaft made from an isotropic elastic material, the strain components are given in terms of the warping function $w(x, y)$ and the twist $\beta$ per unit length of the shaft by $$\gamma_{z x}=-\beta y+\frac{\partial w}{\partial x}, \text { and } \gamma_{z y}=\beta x+\frac{\partial w}{\partial y} .$$ Hence show that the warping function satisfies $\nabla^2 w=0$.

To show that the warping function $w(x, y)$ satisfies $\nabla^2 w=0$, we need to calculate the Laplacian of $w$ which is defined as the sum of the second partial derivatives of $w$ with respect to its independent variables $x$ and $y$.

$\nabla^2 w=\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}$

Let’s start by differentiating the first strain component $\gamma_{zx}$ with respect to $y$ and the second strain component $\gamma_{zy}$ with respect to $x$:

\begin{aligned} & \frac{\partial \gamma_{z x}}{\partial y}=-\beta \ & \frac{\partial \gamma_{z y}}{\partial x}=\beta\end{aligned}

Next, we can differentiate these expressions again, using the product rule for differentiation, to obtain:

\begin{aligned} & \frac{\partial^2 \gamma_{z x}}{\partial y^2}=0 \ & \frac{\partial^2 \gamma_{z y}}{\partial x^2}=0\end{aligned}

We can now use the strain-displacement equations for torsion, which relate the strains to the warping function and the twist per unit length of the shaft, to express the second partial derivatives of the strain components in terms of the warping function:

\begin{aligned} & \frac{\partial^2 \gamma_{z x}}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{\partial \gamma_{z x}}{\partial x}\right)=\frac{\partial}{\partial y}\left(\frac{\partial w}{\partial x}\right)=\frac{\partial^2 w}{\partial x \partial y} \ & \frac{\partial^2 \gamma_{z y}}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial \gamma_{z y}}{\partial y}\right)=\frac{\partial}{\partial x}\left(\frac{\partial w}{\partial y}\right)=\frac{\partial^2 w}{\partial y \partial x}\end{aligned}

Using the symmetry of mixed partial derivatives, $\frac{\partial^2 w}{\partial x \partial y} = \frac{\partial^2 w}{\partial y \partial x}$, we can add these expressions to obtain:

$\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}=0$

which is the Laplace equation for $w$. Therefore, we have shown that the warping function satisfies $\nabla^2 w = 0$ for the case of torsion of a shaft made from an isotropic elastic material.

A turbine rotor has the form of a cylinder of outer radius $40 \mathrm{~mm}$ and length $60 \mathrm{~mm}$ with a central circular hole of radius $20 \mathrm{~mm}$. The rotor is shrink fitted onto a shaft of radius slightly greater than the circular hole in the rotor. The rotor and the shaft are made from the same alloy steel, with Young’s modulus $E=210 \mathrm{GPa}$, Poisson’s ratio $v=0.3$, and uniaxial tensile yield strength $Y=240 \mathrm{MPa}$. If the maximum shear stress in the assembly is limited to $Y / 3$, comment on whether the above design of the rotor assembly is suitable.

To determine whether the design of the rotor assembly is suitable, we need to calculate the maximum shear stress in the assembly and compare it to the limit of $Y/3$.

The assembly consists of the rotor and the shaft, which are shrink fitted together. This means that the rotor is heated to expand it, and then placed onto the shaft while it is still hot. As the rotor cools and contracts, it tightly grips the shaft. This creates a state of residual stress in both the rotor and the shaft.

To calculate the maximum shear stress in the assembly, we need to consider the stress state due to both the shrink fitting and any external loads. Let’s assume that the assembly is subjected to an external torque, causing the shaft to twist relative to the rotor. This creates a shear stress in the radial direction of the shaft.

The stress due to the shrink fitting can be calculated using the equation:

$\Delta \sigma = \frac{E \alpha \Delta T}{1-\nu}$

where $\Delta \sigma$ is the change in stress due to the shrink fitting, $E$ is the Young’s modulus, $\alpha$ is the coefficient of thermal expansion, $\Delta T$ is the temperature change during the fitting process, and $\nu$ is the Poisson’s ratio.

Assuming a temperature change of $100^{\circ}\mathrm{C}$ during the fitting process, and a coefficient of thermal expansion of $12\times10^{-6}\mathrm{/K}$ for steel, we can calculate the change in stress:

$\Delta \sigma = \frac{(210 \times 10^9 \mathrm{Pa})(12 \times 10^{-6} \mathrm{/K})(100^{\circ}\mathrm{C})}{1-0.3} \approx 2.52 \mathrm{MPa}$

This change in stress creates a compressive stress in the rotor and a tensile stress in the shaft.

Now let’s consider the external torque. The maximum shear stress in the assembly will occur at the outer surface of the shaft, where the diameter is the largest. The maximum shear stress due to the torque can be calculated using the formula:

$\tau = \frac{T r}{J}$

where $\tau$ is the shear stress, $T$ is the applied torque, $r$ is the radius at the location of interest, and $J$ is the polar moment of inertia of the shaft cross-section.

The polar moment of inertia of a solid cylinder is $J=\frac{\pi}{2}r^4$. Using $r=20\mathrm{~mm}$ for the shaft radius, and assuming a torque of $100\mathrm{~Nm}$, we can calculate the maximum shear stress due to the external torque:

$\tau = \frac{(100\mathrm{~Nm})(20\mathrm{~mm})}{\frac{\pi}{2}(20\mathrm{~mm})^4} \approx 8.03 \mathrm{MPa}$

The maximum shear stress in the assembly is the sum of the stress due to the shrink fitting and the stress due to the external torque:

$\tau_{max} = \Delta \sigma + \tau \approx 10.55 \mathrm{MPa}$

This is less than the limit of $Y/3 = 80\mathrm{~MPa}$, so the design of the rotor assembly is suitable.

# 相对论场理论|Relativistic Field Theory PHYSM3417代写

0

Assignment-daixieTM为您提供布里斯托大学University of Bristol Relativistic Field Theory PHYSM3417相对论场理论代写代考辅导服务！

## Instructions:

The course seems to aim to provide a comprehensive understanding of the principles of special relativity and how it relates to the behavior of electromagnetic fields. It appears to cover both the mathematical calculations involved in the theory and the more qualitative aspects of the subject matter.

The course also seems to delve into the covariant description of classical electromagnetic fields, which is an important concept in the modern approach to special relativity. Additionally, the course covers the relativistic quantum Klein-Gordon and Dirac equations, which are crucial for understanding quantum mechanics within the context of special relativity.

Overall, it seems like this course would be of interest to anyone looking to deepen their understanding of special relativity and its various applications in the fields of electromagnetism and quantum mechanics.

Consider the following Lagrangian
$$\mathcal{L}=-i \bar{\Psi}\left(\gamma^\mu \partial_\mu-m\right) \Psi$$
Where
$$\Psi=\left(\begin{array}{l} \psi_1 \ \psi_2 \end{array}\right), \quad \bar{\Psi}=\left(\bar{\psi}1, \bar{\psi}_2\right)$$ $\psi{1,2}$ are Dirac spinor fields. We will suppress spinor indices throughout.
(a) Show that (1) is invariant under infinitesimal transformations
$$\delta \Psi=i \epsilon_a T_a \Psi, \quad T_a=\frac{\sigma_a}{2}, \quad a=1,2,3$$
where $\sigma_a$ are Pauli matrices.

(a) Since the generators $T_a$ are Hermitian, the transformation laws are
\begin{aligned} & \delta \psi=i \epsilon_a T_a \psi \ & \delta \bar{\psi}=-i \bar{\psi} \epsilon_a T_a \end{aligned}
The Lagrangian is quite trivially seen to be invariant
$$\delta \mathcal{L}=-i \delta \bar{\psi}(\partial-m) \psi-i \bar{\psi}(\not \partial-m) \delta \psi=-i \bar{\psi}\left(-i \epsilon_a T_a\right)(\partial-m) \psi-i \bar{\psi}(\partial-m)\left(i \epsilon_a T_a\right) \psi=0,$$
where we used the trivial relation $\left[T_a, \gamma^\mu\right]=0$.

(b) Find the conserved currents $J_a^\mu$ corresponding to the symmetric transformations.

(b) We use the so called Noether method to find the conserved current, i.e., we pretend that the transformation parameter $\epsilon$ is spacetime dependent. Then
$$\delta \mathcal{L}=\left(\partial_\mu \epsilon_a\right) \bar{\psi} \gamma^\mu T_a \psi$$
which implies
$$J_a^\mu=-\bar{\psi} \gamma^\mu T_a \psi$$

(c) Write down the corresponding conserved charges $Q_a, a=1,2,3$. Show that
$$\delta \Psi=i\left[\epsilon_a Q_a, \Psi\right]$$

(c) The conserved charges are thus
$$Q_a=-\int d^3 x \bar{\psi}(x) \gamma^0 T_a \psi(x)=\int d^3 x \psi^{\dagger}(x) T_a \psi(x)$$
In the following we will repeatedly use the relations
\begin{aligned} {\left[T_a, T_b\right] } & =i f^{a b c} T_c \ \left.\left{\psi_i(x), \psi_j^{\dagger}(y)\right}\right|{x^0=y^0} & =\delta(\vec{x}-\vec{y}) \delta{i j} \end{aligned}
where $i, j$ are indices in the Lie algebra representation space. We write:
$$i\left[\epsilon_a Q_a, \psi_k(x)\right]=i \epsilon_a T_{i j}^a \int_{x^0=y^0 \text { choice made }} d^3 x\left[\psi_i^{\dagger}(y) \psi_j(y), \psi_k(x)\right]$$
Now we use the relation:
$$[A B, C]=A{B, C}-{A, C} B$$
which for our case gives $\left(A=\psi_i^{\dagger}(y), B=\psi_j(y), C=\psi_k(x)\right)$ :
$$i\left[\epsilon_a Q_a, \psi_k(x)\right]=-i \epsilon_a T_{i j}^a \int_{x^0=y^0} d^3 x \delta(\vec{x}-\vec{y}) \delta_{i k} \psi_j(y)=-i \epsilon_a T_{k j}^a \psi_j(x)=-\delta_e \psi_k(x)$$