牛顿力学代写|NEWTONIAN MECHANICS MATH122 University of Liverpool Assignment

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Newtonian mechanics, also known as classical mechanics, is a branch of physics that deals with the study of motion and its causes. It was founded by Sir Isaac Newton in the 17th century and is based on three laws known as Newton’s laws of motion.

Newton’s first law of motion states that an object at rest will remain at rest, and an object in motion will continue in motion with a constant velocity unless acted upon by an external force. This law is also known as the law of inertia.

Newton’s second law of motion states that the force acting on an object is equal to the mass of the object times its acceleration. Mathematically, F=ma, where F is the force, m is the mass, and a is the acceleration.

Newton’s third law of motion states that for every action, there is an equal and opposite reaction. This law implies that forces always occur in pairs.

In addition to these laws, Newtonian mechanics also includes the concepts of energy, work, and momentum. The laws of conservation of energy, work-energy theorem, and conservation of momentum are also important principles in this branch of physics.

Newtonian mechanics has been instrumental in the development of modern physics, and its principles are still widely used in many fields of science and engineering.

牛顿力学代写|NEWTONIAN MECHANICS MATH122 University of Liverpool Assignment

问题 1.

A rocket in zero gravitational field has a mass of $m_{r, i}=2.81 \times 10^7 \mathrm{~kg}$, which is the sum of the mass of the fuel $m_{f, i}=2.46 \times 10^7 \mathrm{~kg}$ and the dry mass of the rocket (empty of fuel) $m_{r, d} \equiv m_{r, i}-m_{f, i}=0.35 \times 10^7 \mathrm{~kg}$. The fuel is ejected at a speed $u=3000$ $\mathrm{m} / \mathrm{s}$ relative to the rocket. The total burn time is $510 \mathrm{~s}$ and the fuel is burned at a constant rate. (a) What is the final speed $v_f$ of the rocket in meters/second after all the fuel is burned assuming it starts from rest?

证明 .

The total mass of the rocket decreases during the burn due to the ejection of fuel. The change in velocity of the rocket is given by the rocket equation:

$\Delta v=v_{f}-v_{i}=u \ln \left(\frac{m_{r, i}}{m_{r, f}}\right)$

where $v_i = 0$ is the initial velocity of the rocket, $v_f$ is the final velocity of the rocket, $m_{r,i}$ is the initial mass of the rocket, $m_{r,f}$ is the final mass of the rocket, and $u$ is the relative velocity of the ejected fuel. At the end of the burn, all the fuel has been ejected, so the final mass of the rocket is the dry mass $m_{r,d}$:

$m_{r, f}=m_{r, d}=0.35 \times 10^7 \mathrm{~kg}$

Thus,

$\Delta v=u \ln \left(\frac{m_{r, i}}{m_{r, d}}\right)=3000 \mathrm{~m} / \mathrm{s} \ln \left(\frac{2.81 \times 10^7 \mathrm{~kg}}{0.35 \times 10^7 \mathrm{~kg}}\right) \approx 7974.4 \mathrm{~m} / \mathrm{s}$

Therefore, the final speed of the rocket is $v_f = \Delta v = 7974.4$ m/s.

问题 2.

(b) Now suppose that the same rocket burns the fuel in two stages, expelling the fuel in each stage at the same relative speed $u=3000 \mathrm{~m} / \mathrm{s}$. In stage one, the available fuel to burn is $m_{f, 1, i}=2.03 \times 10^7 \mathrm{~kg}$ with burn time $150 \mathrm{~s}$. The total mass of the rocket after all the fuel in stage 1 is burned is $m_{r, 1, d}=m_{r, i}-m_{f, 1, i}=0.78 \times 10^7$ kg. What is the change in speed after stage one is complete?

证明 .

In stage one, the initial mass of the rocket is $m_{r,1,i}=m_{r,i}=2.81 \times 10^7 \mathrm{~kg}$ and the initial mass of fuel is $m_{f,1,i}=2.03 \times 10^7 \mathrm{~kg}$. The mass of the rocket after all the fuel in stage 1 is burned is $m_{r,1,f}=m_{r,1,i}-m_{f,1,i}=0.78 \times 10^7$ kg. By the principle of conservation of momentum, the change in velocity of the rocket after stage one is complete is given by:

$\Delta v_1=u\ln \left(\frac{m_{r,1,i}}{m_{r,1,f}}\right)$

where $u=3000 \mathrm{~m}/\mathrm{s}$ is the relative speed of the ejected fuel and $m_{r,1,i}$ and $m_{r,1,f}$ are the initial and final masses of the rocket in stage one, respectively. Substituting the values:

$\Delta v_1=3000 \mathrm{~m}/\mathrm{s} \ln \left(\frac{2.81 \times 10^7 \mathrm{~kg}}{0.78 \times 10^7 \mathrm{~kg}}\right) \approx 3066 \mathrm{~m}/\mathrm{s}$

Therefore, the change in speed after stage one is complete is approximately $3066 \mathrm{~m}/\mathrm{s}$.

问题 3.

(c) Next, the empty fuel tank and accessories from stage one are disconnected from the rest of the rocket. These disconnected parts had a mass $m=1.4 \times 10^6 \mathrm{~kg}$, hence the remaining dry mass of the rocket is $m_{r, 2, d}=2.1 \times 10^6 \mathrm{~kg}$. All the remaining fuel with mass $m_{f, 2, i}=4.3 \times 10^6 \mathrm{~kg}$ is burned during stage 2 with burn time of $360 \mathrm{~s}$. What is the change in speed in meters/second after stage two is complete?

证明 .

To calculate the change in speed after stage two, we can use the rocket equation, which relates the change in velocity of a rocket to the mass ratio of the rocket before and after the burn, and the effective exhaust velocity of the propellant.

The mass ratio of the rocket before and after stage two is:

$$\frac{m_{i}}{m_{f, 2}}=\frac{m_{r, i}+m_{f, i}+m}{m_{r, 2, d}}=\frac{2.81 \times 10^7 \mathrm{~kg}+2.46 \times 10^7 \mathrm{~kg}+1.4 \times 10^6 \mathrm{~kg}}{2.1 \times 10^6 \mathrm{~kg}}=22.95$$

where $m_{i}$ is the initial mass of the rocket (including all the fuel and the dry mass) and $m_{f,2}$ is the final mass of the rocket after stage two (including the dry mass and all the fuel burned in stage two).

The effective exhaust velocity $v_{e}$ is given by:

$$v_{e}=u \ln\frac{m_{i}}{m_{f,2}}=3000 \mathrm{~m/s} \ln\frac{2.81 \times 10^7 \mathrm{~kg}}{2.1 \times 10^6 \mathrm{~kg}+4.3 \times 10^6 \mathrm{~kg}}=2445.5 \mathrm{~m/s}$$

where $u$ is the speed of the ejected fuel relative to the rocket.

Now we can use the rocket equation:

$$\Delta v=v_{e} \ln\frac{m_{i}}{m_{f,2}}=2445.5 \mathrm{~m/s} \ln\frac{2.81 \times 10^7 \mathrm{~kg}}{2.1 \times 10^6 \mathrm{~kg}+4.3 \times 10^6 \mathrm{~kg}}=2434.4 \mathrm{~m/s}$$

Therefore, the change in speed after stage two is $\Delta v = 2434.4$ $\mathrm{m/s}$.

这是一份2023年的利物浦大学University of Liverpool NEWTONIAN MECHANICS MATH122代写的成功案例

计算物理学代写|INTRODUCTION TO COMPUTATIONAL PHYSICS PHYS105 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool INTRODUCTION TO COMPUTATIONAL PHYSICS PHYS105计算物理学代写代考辅导服务!

Instructions:

Computational physics is a subfield of physics that involves the use of computers to solve problems in physics that are too complex to be solved analytically. The use of computational techniques has become increasingly important in physics research and has led to many breakthroughs in our understanding of the physical world.

To get started with computational physics, it’s important to have a strong foundation in mathematics and programming. In terms of programming languages, Python is a popular choice for computational physics because of its ease of use and large user community.

There are many resources available online to help you learn computational physics. Some good places to start include:

  • The Open Source Physics Project: This project provides a collection of open-source software tools and simulations for teaching and researching physics. They have a range of resources and materials available for learning computational physics, including tutorials and sample code.
  • Coursera: Coursera offers a range of online courses in computational physics, including courses on Python programming and numerical methods for physics.
  • GitHub: GitHub is a platform for collaborative software development, and there are many open-source repositories available on GitHub that contain code for solving physics problems. You can use these repositories to learn from existing code and contribute to the development of new code.
  • Textbooks: There are many textbooks available on computational physics, including “Computational Physics” by Mark Newman and “An Introduction to Computational Physics” by Tao Pang.

I hope this information is helpful! Let me know if you have any other questions.

计算物理学代写|INTRODUCTION TO COMPUTATIONAL PHYSICS PHYS105 University of Liverpool Assignment

问题 1.

(a) Write a procedure that solves quadratic equations using the quadratic formula: It should take as arguments three numbers a, b, and c. It should print error messages if a is zero, or if the roots are complex. Otherwise it should print the two roots.

证明 .

# a.) code for roots function
def roots(a, b, c):
    #form of equation is a*x**2 + b*x + c = 0
    #quadratic formula is   x = (-b + sqrt(b**2 - 4 * a * c))/(2 * a)
    #                   or  x = (-b - sqrt(b**2 - 4 * a * c))/(2 * a)
    #roots are complex when the discriminant (b**2 - 4*a*c) is negative
    discriminant = b**2 - (4 * a * c)
    if discriminant < 0:
        return "Roots are complex"
    return "x = "+str((-b + math.sqrt(discriminant)) / (2 * a))+" or x = "+\
        str((-b - math.sqrt(discriminant)) / (2 * a))

# Test Cases (assertions are just to automate)
##print '\nTesting roots'
##
##print roots(1, 2, 1) #(x+1)^2: double root at x = -1
##assert roots(1, 2, 1) == 'x = -1.0 or x = -1.0'
##
##print roots(1, -2, -3) #(x+1)(x-3): roots at x = 3 or x = -1
##assert roots(1, -2, -3) == 'x = 3.0 or x = -1.0'
##
##print roots(2, 2, 2) #2x^2 + 2x + 2: complex roots
##assert roots(2, 2, 2) == 'Roots are complex'

问题 2.

(b) Modify your procedure to handle the case of complex roots.

证明 .

# b.) same code, modified to handle complex roots
def roots(a, b, c):
    discriminant = b**2 - (4 * a * c)
    if discriminant < 0:
        discriminant = discriminant + 0j
    return "x = "+str((-b + discriminant**0.5) / (2 * a))+ " or x = " +\
        str((-b - discriminant**0.5) / (2 * a)) # math.sqrt does not work

# Test Cases (assertions are just to automate)
##print '\nTesting roots'
##
##print roots(1, 2, 1) #(x+1)^2: double root at x = -1
##assert roots(1, 2, 1) == 'x = -1.0 or x = -1.0'
##
##print roots(1, -2, -3) #(x+1)(x-3): roots at x = 3 or x = -1
##assert roots(1, -2, -3) == 'x = 3.0 or x = -1.0'
##
##print roots(2, 2, 2) #2x^2 + 2x + 2: complex roots
##assert roots(2, 2, 2) == 'x = (-0.5+0.866025403784j) or x = (-0.5-0.866025403784j)'

###########################################
## 2.) procedure for evaluating polynomials
###########################################
def eval_poly(x, coeffs):
    total=0 #will keep a running total of the sum
    coeffs.reverse() # to put the low order coeffs first
    for i in range(len(coeffs)):
        total+=coeffs[i]*(x**i) #add the curent term to the total
    return total

# Test cases
##print '/nTesting Polynomial'
##
##print eval_poly(1,[1,2,3])
##assert eval_poly(1,[1,2,3]) == 6
##
##print eval_poly(2,[1,2,3,4])
##assert eval_poly(2,[1,2,3,4]) == 26

问题 3.

Write a procedure that evaluates polynomials. It should take two arguments. The first is a number $x$. The second is a list of of coefficients ordered from highest to lowest:
$$
a_n, a_{n-1}, \ldots, a_2, a_1, a_0
$$
Your procedure should return the value of the polynomial evaluated at $x$ :
$$
a_n x^n+a_{n-1} x^{n-1}+\ldots+a_2 x+a_1 x+a_0
$$

证明 .

) procedure for evaluating polynomials
###########################################
def eval_poly(x, coeffs):
    total=0 #will keep a running total of the sum
    coeffs.reverse() # to put the low order coeffs first
    for i in range(len(coeffs)):
        total+=coeffs[i]*(x**i) #add the curent term to the total
    return total

# Test cases
##print '/nTesting Polynomial'
##
##print eval_poly(1,[1,2,3])
##assert eval_poly(1,[1,2,3]) == 6
##
##print eval_poly(2,[1,2,3,4])
##assert eval_poly(2,[1,2,3,4]) == 26

###########################################
## 3.) procedure to make change
###########################################
def make_change(cost, paid):
    change=paid-cost #calculate the change due
    bills={20:0,10:0,5:0,2:0,1:0} #dictionary that maps each bill to the amount of it
    bill_list=bills.keys() #bill_list is a list of the available bills
    bill_list.sort() #sort it in ascending order
    bill_list.reverse() #now reverse it to be in descending order
                        #this is to make sure you get the smallest number of bills
    for bill in bill_list: 
        while change >=bill: 
            bills[bill]+=1 #increment the amount of that bill
            change-=bill #decrease the change by that bill
            
    print "Change is:"
    for bill in bill_list:
        if bills[bill] == 1:
            print bills[bill], bill, "dollar bill"
        elif bills[bill] > 1:
            print bills[bill], bill, "dollar bills"

    #return bills #return the dictionary with amounts of each bill needed

make_change(1, 6)
make_change(4, 109)

这是一份2023年的利物浦大学University of Liverpool INTRODUCTION TO COMPUTATIONAL PHYSICS PHYS105代写的成功案例

天体物理学代写|ASTRONOMY 1 ASTRO1001 University of Glasgow AssignmentONLINE TUTOR

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Assignment-daixieTM为您提供格拉斯哥大学University of Glasgow ASTRONOMY 1 ASTRO1001天体物理学代写代考辅导服务!

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Astronomy is the scientific study of celestial objects and phenomena beyond the Earth’s atmosphere, including planets, stars, galaxies, and other celestial bodies. It is a fascinating field that has captured the attention and imagination of people for thousands of years.

There are several subfields within astronomy, including solar system physics, positional astronomy, and dynamical astronomy.

Solar system physics is the study of the physical properties of the planets, moons, asteroids, and comets in our solar system. This includes their composition, structure, atmospheres, and orbits. Solar system physics also investigates the formation and evolution of the solar system, as well as the interactions between the various bodies.

Positional astronomy is concerned with measuring the positions, distances, and motions of celestial objects. This involves the use of telescopes, cameras, and other instruments to gather data on the positions of stars, galaxies, and other objects. Positional astronomy also includes the study of celestial coordinate systems, which are used to locate objects in the sky.

Dynamical astronomy is the study of the motions of celestial objects, including their orbits and gravitational interactions. This field includes the study of celestial mechanics, which deals with the mathematical description of the motions of objects in space. Dynamical astronomy also investigates the formation and evolution of galaxies, and the interactions between galaxies and other celestial bodies.

In conclusion, astronomy is a vast and complex field that encompasses many different areas of study. Solar system physics, positional astronomy, and dynamical astronomy are just a few of the many subfields within this fascinating discipline. The study of astronomy has helped us to understand the universe around us and has inspired us to ask even more questions about the nature of the cosmos.

天体物理学代写|ASTRONOMY 1 ASTRO1001 University of Glasgow Assignment

问题 1.

a. A globular cluster has $10^6$ stars each of apparent magnitude +8 . What is the combined apparent magnitude of the entire cluster?

证明 .

$\begin{gathered}+8=-2.5 \log \left(F / F_0\right) \ F=6.3 \times 10^{-4} F_0 \ F_{\text {cluster }}=10^6 \times 6.3 \times 10^{-4} F_0=630 F_0 \ m_{\text {cluster }}=-2.5 \log (630)=-7\end{gathered}$

问题 2.

b. Find the distance modulus to the Andromeda galaxy (M31). Take the distance to Andromeda to be $750 \mathrm{kpc}$.

证明 .

$\mathrm{DM}=5 \log \left(\frac{d}{10 \mathrm{pc}}\right)=5 \log (75,000)=24.4$

问题 3.

c. An eclipsing binary consists of two stars of different radii and effective temperatures. Star 1 has radius $R_1$ and $T_1$, and Star 2 has $R_2=0.5 R_1$ and $T_2=2 T_1$. Find the change in bolometric magnitude of the binary, $\Delta m_{\text {bol }}$, when the smaller star is behind the larger star. (Consider only bolometric magnitudes so you don’t have to worry about color differences.)

证明 .

$$
\begin{gathered}
\mathcal{F}{1 \& 2}=4 \pi \sigma\left(T_1^4 R_1^2+T_2^4 R_2^2\right) \ \mathcal{F}{\text {eclipse }}=4 \pi \sigma T_1^4 R_1^2 \
\Delta m=-2.5 \log \left(\frac{\mathcal{F}{1 \& 2}}{\mathcal{F}{\text {eclipse }}}\right) \
\Delta m=-2.5 \log \left(1+\frac{T_2^4 R_2^2}{T_1^4 R_1^2}\right) \
\Delta m=-2.5 \log \left(1+\frac{16}{4}\right)=-1.75
\end{gathered}
$$
So, the binary is 1.75 magnitudes brighter out of eclipse than when star 2 is behind star 1 .

这是一份2023年的格拉斯哥大学University of Glasgow天体物理学ASTRO1001代写的成功案例

物理代写|PHYSICS 1 PHYS1001 University of Glasgow Assignment

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Assignment-daixieTM为您提供格拉斯哥大学University of Glasgow PHYSICS 1 PHYS1001物理代写代考辅导服务!

Instructions:

Physics is a branch of science that deals with the study of matter, energy, and their interactions. It seeks to understand the fundamental laws of nature and the behavior of the universe at the most basic level. Physicists study everything from the tiniest subatomic particles to the largest structures in the universe, such as galaxies and black holes. They use mathematical models and experimental methods to develop theories and test them against observations. Physics plays a critical role in many areas of modern technology, including electronics, communications, and medicine. It has also contributed significantly to our understanding of the natural world and the universe we live in.

物理代写|PHYSICS 1 PHYS1001 Cardiff University Assignment

问题 1.

The line integral of a scalar function $f(x, y, z)$ along a path $C$ is defined as
$$
\int_C f(x, y, z) d s=\lim {\substack{N \rightarrow \infty \ \Delta s_i \rightarrow 0}} \sum{i=1}^N f\left(x_i, y_i, z_i\right) \Delta s_i
$$
where $C$ has been subdivided into $N$ segments, each with a length $\Delta s_i$. To evaluate the line integral, it is convenient to parameterize $C$ in terms of the arc length parameter $s$. With $x=x(s)$, $y=y(s)$ and $z=z(s)$, the above line integral can be rewritten as an ordinary definite integral:
$$
\int_C f(x, y, z) d s=\int_{s_1}^{s_2} f[x(s), y(s), z(s)] d s
$$

证明 .


Yes, that is correct.

When we parameterize the curve $C$ using the arc length parameter $s$, we can express the coordinates of any point on the curve as functions of $s$ as $x(s)$, $y(s)$, and $z(s)$.

Using this parameterization, we can express the line integral as an ordinary definite integral over $s$ as shown above. The integrand $f[x(s), y(s), z(s)]$ represents the value of the scalar function $f$ evaluated at the point on the curve corresponding to the arc length parameter $s$.

The limit as $N\to\infty$ and $\Delta s_i\to 0$ in the definition of the line integral is necessary because we want to calculate the exact value of the integral, which involves summing up infinitely many infinitesimal contributions along the path $C$.

问题 2.

A solid cylinder of length $L$ and radius $R$, with $L \gg R$, is uniformly filled with a total charge $Q$. a. What is the volume charge density $\rho$ ? Check units!

证明 .

a. The volume charge density $\rho$ is defined as the charge $Q$ per unit volume. In this case, the total charge $Q$ is uniformly distributed throughout the volume of the cylinder. The volume of a cylinder is given by $V=\pi R^2L$. Therefore, the volume charge density is:

$$\rho = \frac{Q}{V} = \frac{Q}{\pi R^2 L}$$

Checking units: $[Q]=C$, $[R]=m$, and $[L]=m$, so $[\rho]=\frac{C}{m^3}$.

问题 3.

b. Suppose you go very far away from the cylinder to a distance much greater than $R$. The cylinder now looks like a line of charge. What is the linear charge density $\lambda$ of that apparent line of charge? Check units!

证明 .

b. When we are far away from the cylinder, it appears as a line of charge with linear charge density $\lambda$. The linear charge density is defined as the charge per unit length. We can find $\lambda$ by considering an element of length $dl$ on the cylinder. The charge on this element is $dq=\rho Adl$, where $A$ is the cross-sectional area of the cylinder (which is just a circle of radius $R$). The element of length $dl$ subtends an angle of $d\theta=\frac{dl}{R}$ at the observation point, and the distance from the element to the observation point is $r\gg R$. Therefore, the contribution to the electric field from this element is:

$$dE=\frac{1}{4\pi\epsilon_0}\frac{dq}{r^2}=\frac{1}{4\pi\epsilon_0}\frac{\rho Adl}{r^2}$$

The total electric field at the observation point is obtained by integrating over the entire length of the cylinder:

$$E=\int_{-L/2}^{L/2} dE = \int_{-L/2}^{L/2} \frac{1}{4\pi\epsilon_0}\frac{\rho A}{r^2}dl=\frac{\rho A}{4\pi\epsilon_0 r^2}\int_{-L/2}^{L/2}dl$$

The integral gives the total length of the cylinder, which is just $L$. Therefore, we have:

$$E=\frac{\rho AL}{4\pi\epsilon_0 r^2}$$

On the other hand, we know that the electric field due to a line of charge with linear charge density $\lambda$ is given by:

$$E=\frac{\lambda}{2\pi\epsilon_0 r}$$

Comparing the two expressions for $E$, we get:

$$\frac{\lambda}{2\pi\epsilon_0 r}=\frac{\rho AL}{4\pi\epsilon_0 r^2}$$

Solving for $\lambda$, we obtain:

$$\lambda=\frac{\rho AL}{2r}=\frac{Q}{2\pi R L}\qquad\text{(since }A=\pi R^2\text{)}$$

Checking units: $[Q]=C$, $[R]=m$, and $[L]=m$, so $[\lambda]=\frac{C}{m}$.

这是一份2023年的格拉斯哥大学University of Glasgow物理PHYS1001代写的成功案例

机械学和物质代写|Mechanics and Matter PX1121 Cardiff University Assignment

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Assignment-daixieTM为您提供卡迪夫大学Cardiff University PX1121 Mechanics and Matter代写代考辅导服务!

Instructions:

Mechanics is the branch of physics that deals with the motion of objects under the influence of forces. It is concerned with describing and predicting the behavior of physical systems, including both macroscopic objects like cars and airplanes, as well as microscopic particles like atoms and molecules.

Matter, on the other hand, refers to anything that has mass and takes up space. It is the basic building block of the universe, and includes everything from tiny subatomic particles to massive planets and stars.

Mechanics and matter are intimately connected, as the behavior of matter is governed by the laws of mechanics. For example, the motion of a ball thrown into the air can be described using the principles of mechanics, as can the behavior of atoms and molecules in a gas or liquid.

Understanding the relationship between mechanics and matter is crucial for many areas of science and technology, including engineering, materials science, and many others.

机械学和物质代写|Mechanics and Matter PX1121 Cardiff University Assignment

问题 1.

Surface tension: Thermodynamic properties of the interface between two phases are described by a state function called the surface tension $\mathcal{S}$. It is defined in terms of the work required to increase the surface area by an amount $d A$ through $d W=\mathcal{S} d A$.
(a) By considering the work done against surface tension in an infinitesimal change in radius, show that the pressure inside a spherical drop of water of radius $R$ is larger than outside pressure by $2 \mathcal{S} / R$. What is the air pressure inside a soap bubble of radius $R$ ?

证明 .

The work done by a water droplet on the outside world, needed to increase the radius from $R$ to $R+\Delta R$ is
$$
\Delta W=\left(P-P_o\right) \cdot 4 \pi R^2 \cdot \Delta R
$$
where $P$ is the pressure inside the drop and $P_o$ is the atmospheric pressure. In equilibrium, this should be equal to the increase in the surface energy $\mathcal{S} \Delta A=\mathcal{S} \cdot 8 \pi R \cdot \Delta R$, where $\mathcal{S}$ is the surface tension, and
$$
\Delta W_{\text {total }}=0, \Longrightarrow \Delta W_{\text {pressure }}=-\Delta W_{\text {surface }},
$$
resulting in
$$
\left(P-P_o\right) \cdot 4 \pi R^2 \cdot \Delta R=\mathcal{S} \cdot 8 \pi R \cdot \Delta R, \quad \Longrightarrow \quad\left(P-P_o\right)=\frac{2 \mathcal{S}}{R}
$$
In a soap bubble, there are two air-soap surfaces with almost equal radii of curvatures, and
$$
P_{\text {film }}-P_o=P_{\text {interior }}-P_{\text {film }}=\frac{2 \mathcal{S}}{R}
$$
leading to
$$
P_{\text {interior }}-P_o=\frac{4 \mathcal{S}}{R}
$$
Hence, the air pressure inside the bubble is larger than atmospheric pressure by $4 \mathcal{S} / R$.

问题 2.

(b) A water droplet condenses on a solid surface. There are three surface tensions involved $\mathcal{S}{a w}, \mathcal{S}{s w}$, and $\mathcal{S}_{s a}$, where $a, s$, and $w$ refer to air, solid and water respectively. Calculate the angle of contact, and find the condition for the appearance of a water film (complete wetting).

证明 .

When steam condenses on a solid surface, water either forms a droplet, or spreads on the surface. There are two ways to consider this problem:
Method 1: Energy associated with the interfaces
In equilibrium, the total energy associated with the three interfaces should be minimum, and therefore
$$
d E=S_{a w} d A_{a w}+S_{a s} d A_{a s}+S_{w s} d A_{w s}=0 .
$$

Since the total surface area of the solid is constant,
$$
d A_{a s}+d A_{w s}=0 .
$$
From geometrical considerations (see proof below), we obtain
$$
d A_{w s} \cos \theta=d A_{a w} .
$$
From these equations, we obtain
$$
d E=\left(S_{a w} \cos \theta-S_{a s}+S_{w s}\right) d A_{w s}=0, \quad \Longrightarrow \quad \cos \theta=\frac{S_{a s}-S_{w s}}{S_{a w}} .
$$
Proof of $d A_{w s} \cos \theta=d A_{a w}$ : Consider a droplet which is part of a sphere of radius $R$, which is cut by the substrate at an angle $\theta$. The areas of the involved surfaces are
$$
A_{w s}=\pi(R \sin \theta)^2, \quad \text { and } \quad A_{a w}=2 \pi R^2(1-\cos \theta) .
$$
Let us consider a small change in shape, accompanied by changes in $R$ and $\theta$. These variations should preserve the volume of water, i.e. constrained by
$$
V=\frac{\pi R^3}{3}\left(\cos ^3 \theta-3 \cos \theta+2\right)
$$
Introducing $x=\cos \theta$, we can re-write the above results as
$$
\left{\begin{aligned}
A_{w s} & =\pi R^2\left(1-x^2\right), \
A_{a w} & =2 \pi R^2(1-x), \
V & =\frac{\pi R^3}{3}\left(x^3-3 x+2\right) .
\end{aligned}\right.
$$

The variations of these quantities are then obtained from
$$
\left{\begin{aligned}
d A_{w s} & =2 \pi R\left[\frac{d R}{d x}\left(1-x^2\right)-R x\right] d x \
d A_{a w} & =2 \pi R\left[2 \frac{d R}{d x}(1-x)-R\right] d x \
d V & =\pi R^2\left[\frac{d R}{d x}\left(x^3-3 x+2\right)+R\left(x^2-x\right)\right] d x=0 .
\end{aligned}\right.
$$
From the last equation, we conclude
$$
\frac{1}{R} \frac{d R}{d x}=-\frac{x^2-x}{x^3-3 x+2}=-\frac{x+1}{(x-1)(x+2)}
$$

Substituting for $d R / d x$ gives,
$$
d A_{w s}=2 \pi R^2 \frac{d x}{x+2}, \quad \text { and } \quad d A_{a w}=2 \pi R^2 \frac{x \cdot d x}{x+2},
$$
resulting in the required result of
$$
d A_{a w}=x \cdot d A_{w s}=d A_{w s} \cos \theta .
$$
Method 2: Balancing forces on the contact line
Another way to interpret the result is to consider the force balance of the equilibrium surface tension on the contact line. There are four forces acting on the line: (1) the surface tension at the water-gas interface, (2) the surface tension at the solid-water interface, (3) the surface tension at the gas-solid interface, and (4) the force downward by solid-contact line interaction. The last force ensures that the contact line stays on the solid surface, and is downward since the contact line is allowed to move only horizontally without friction. These forces should cancel along both the $y$-direction $x$-directions. The latter gives the condition for the contact angle known as Young’s equation,
$$
\mathcal{S}{a s}=\mathcal{S}{a w} \cdot \cos \theta+\mathcal{S}{w s}, \Longrightarrow \cos \theta=\frac{\mathcal{S}{a s}-\mathcal{S}{w s}}{\mathcal{S}{a w}} .
$$
The critical condition for the complete wetting occurs when $\theta=0$, or $\cos \theta=1$, i.e. for
$$
\cos \theta_C=\frac{\mathcal{S}{a s}-\mathcal{S}{w s}}{\mathcal{S}{a w}}=1 $$ Complete wetting of the substrate thus occurs whenever $$ \mathcal{S}{a w} \leq \mathcal{S}{a s}-\mathcal{S}{w s} .
$$

问题 3.

(c) In the realm of “large” bodies gravity is the dominant force, while at “small” distances surface tension effects are all important. At room temperature, the surface tension of water is $\mathcal{S}_o \approx 7 \times 10^{-2} \mathrm{Nm}^{-1}$. Estimate the typical length-scale that separates “large” and “small” behaviors. Give a couple of examples for where this length-scale is important.

证明 .

(c) In the realm of “large” bodies gravity is the dominant force, while at “small” distances surface tension effects are all important. At room temperature, the surface tension of water is $\mathcal{S}_o \approx 7 \times 10^{-2} \mathrm{Nm}^{-1}$. Estimate the typical length-scale that separates “large” and “small” behaviors. Give a couple of examples for where this length-scale is important.

  • Typical length scales at which the surface tension effects become significant are given by the condition that the forces exerted by surface tension and relevant pressures become comparable, or by the condition that the surface energy is comparable to the other energy changes of interest.

Example 1: Size of water drops not much deformed on a non-wetting surface. This is given by equalizing the surface energy and the gravitational energy,
$$
S \cdot 4 \pi R^2 \approx m g R=\rho V g R=\frac{4 \pi}{3} R^4 g
$$

leading to
$$
R \approx \sqrt{\frac{3 S}{\rho g}} \approx \sqrt{\frac{3 \cdot 7 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{10^3 \mathrm{~kg} / \mathrm{m}^3 \times 10 \mathrm{~m} / \mathrm{s}^2}} \approx 1.5 \times 10^{-3} \mathrm{~m}=1.5 \mathrm{~mm} .
$$
Example 2: Swelling of spherical gels in a saturated vapor: Osmotic pressure of the gel (about $1 \mathrm{~atm})=$ surface tension of water, gives
$$
\pi_{g e l} \approx \frac{N}{V} k_B T \approx \frac{2 S}{R}
$$
where $N$ is the number of counter ions within the gel. Thus,
$$
R \approx\left(\frac{2 \times 7 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{10^5 \mathrm{~N} / \mathrm{m}^2}\right) \approx 10^{-6} \mathrm{~m} .
$$

这是一份2023年的卡迪夫大学Cardiff University 机械学和物质PX1121代写的成功案例

电、磁和波代写|Electricity, Magnetism and Waves PX1221 Cardiff University Assignment

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Assignment-daixieTM为您提供卡迪夫大学Cardiff University PX1221 Electricity, Magnetism and Waves电、磁和波代写代考辅导服务!

Instructions:

Electricity, magnetism, and waves are three interconnected branches of physics that deal with the behavior and properties of electromagnetic fields and phenomena.

Electricity refers to the flow of electric charge, usually carried by electrons in a conductor, and is characterized by voltage, current, and resistance. It is essential to many aspects of modern life, from powering our homes and electronic devices to driving industry and transportation.

Magnetism refers to the behavior of magnets and magnetic fields, which are produced by moving electric charges or certain materials called ferromagnetic materials. Magnetic fields are important in many applications, including electric motors, generators, and medical imaging.

Waves refer to the propagation of energy through space, which can be in the form of electromagnetic waves or mechanical waves like sound waves. Electromagnetic waves, such as radio waves, microwaves, light waves, and X-rays, are characterized by their wavelength, frequency, and amplitude and play a vital role in communication, broadcasting, and imaging.

The study of electricity, magnetism, and waves is fundamental to understanding many aspects of the physical world, from the behavior of atoms and molecules to the functioning of modern technology.

电、磁和波代写|Electricity, Magnetism and Waves PX1221 Cardiff University Assignment

问题 1.

(a) Find the total charge $Q$ on the rectangular surface of length $a$ ( $x$ direction from $x=0$ to $x=a$ ) and width $b(y$ direction from $y=0$ to $y=b)$, if the charge density is $\sigma(x, y)=k x y$, where $k$ is a constant.

证明 .

To find the total charge $Q$ on the rectangular surface, we need to integrate the charge density $\sigma(x, y)$ over the surface area:

$$Q = \int_{0}^{b} \int_{0}^{a} \sigma(x, y) dx dy$$

Substituting $\sigma(x,y)=kxy$ into the above expression and performing the integrals yields:

\begin{align*} Q &= \int_{0}^{b} \int_{0}^{a} k x y dx dy \ &= k \int_{0}^{b} \left[\frac{x^2}{2} y \right]{x=0}^{x=a} dy \ &= k \int{0}^{b} \frac{a^2}{2} y dy \ &= \frac{k a^2}{2} \left[\frac{y^2}{2} \right]_{y=0}^{y=b} \ &= \frac{k a^2 b^2}{4} \end{align*}

Therefore, the total charge on the rectangular surface is $\boxed{\frac{k a^2 b^2}{4}}$.

问题 2.

(b) Find the total charge on a circular plate of radius $R$ if the charge distribution is $\sigma(r, \theta)=k r(1-\sin \theta)$.

证明 .

To find the total charge on the circular plate of radius $R$ with charge distribution $\sigma(r,\theta) = kr(1-\sin\theta)$, we need to integrate the charge density over the entire area of the plate.

The charge density $\sigma$ is a function of $r$ and $\theta$, where $r$ is the radial distance from the center of the plate and $\theta$ is the angle measured from some reference direction. Since the plate is circular, we can integrate over $r$ and $\theta$ as follows:

\begin{align*} Q &= \int_{0}^{2\pi}\int_{0}^{R} \sigma(r,\theta)r,dr,d\theta \ &= \int_{0}^{2\pi}\int_{0}^{R} k r^2 (1-\sin\theta),dr,d\theta\ &= k \int_{0}^{2\pi}\left[\frac{r^3}{3} – r^2\cos\theta\right]{r=0}^{r=R},d\theta\ &= k \int{0}^{2\pi}\left(\frac{R^3}{3} – R^2\cos\theta\right),d\theta\ &= k\left[\frac{R^3}{3}\theta – R^2\sin\theta\right]_{\theta=0}^{\theta=2\pi}\ &= k\left(\frac{2}{3}\pi R^3 – 0\right)\ &= \frac{2}{3}\pi k R^3 \end{align*}

Therefore, the total charge on the circular plate is $\frac{2}{3}\pi k R^3$.


问题 3.

Use Gauss’s Law to find the direction and magnitude of the electric field in the between the inner and outer cylinders $(a<r<b)$. Express your answer in terms of the total charge $Q$ on the inner cylinder cylinder, the radii $a$ and $b$, the height $l$, and any other constants which you may find necessary.

证明 .

To apply Gauss’s Law, we need to choose a closed surface that encloses the region of interest, which in this case is the region between the inner and outer cylinders. A convenient choice is a cylindrical Gaussian surface of radius $r$ and height $l$, centered on the axis of the cylinders. The electric field will be perpendicular to the surface at every point, so the flux of the electric field through the surface is simply the product of the magnitude of the field and the area of the curved surface:

$$\Phi_E = \oint_S \vec{E} \cdot d\vec{A} = E(r) 2\pi rl$$

where $E(r)$ is the magnitude of the electric field at radius $r$ and $d\vec{A}$ is a differential area element of the surface.

By Gauss’s Law, the flux through the surface is equal to the enclosed charge divided by the permittivity of free space:

$$\Phi_E = \frac{Q_{enc}}{\epsilon_0}$$

where $Q_{enc}$ is the charge enclosed by the Gaussian surface. Since the inner cylinder is the only source of charge inside the surface, the enclosed charge is simply $Q_{enc}=Q$.

Setting these two equations equal to each other, we have

$$E(r) 2\pi rl = \frac{Q}{\epsilon_0}$$

Solving for $E(r)$, we obtain:

$$E(r) = \frac{Q}{2\pi\epsilon_0 rl}$$

The direction of the electric field is radial, i.e., pointing outward from the axis of the cylinders if $Q$ is positive and inward if $Q$ is negative.

Therefore, the magnitude of the electric field between the cylinders is given by:

$$\boxed{E(r) = \frac{Q}{2\pi\epsilon_0 rl}}$$

这是一份2023年的卡迪夫大学Cardiff University 电、磁和波PX1221代写的成功案例

动态环境代写|World of Dynamic Environments EA1300 Cardiff University Assignment

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Assignment-daixieTM为您提供卡迪夫大学Cardiff University EA1300 World of Dynamic Environments动态环境代写代考辅导服务!

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Dynamic environments refer to environments that are constantly changing and evolving over time. These changes can be driven by various factors, such as the behavior of agents within the environment, external events, or the passage of time.

Examples of dynamic environments include weather systems, traffic patterns, and financial markets. In these environments, the state of the system can change rapidly and unpredictably, requiring agents to adapt and adjust their strategies in real-time.

Dynamic environments are often modeled using dynamic systems theory, which provides a mathematical framework for understanding how systems change over time. This theory can be used to develop models and algorithms that can help agents navigate dynamic environments and make decisions in the face of uncertainty.

Machine learning techniques such as reinforcement learning, which involves training agents to learn optimal actions through trial and error, can be particularly effective in dynamic environments. By continually updating their policies based on new information, these agents can learn to adapt to changing conditions and optimize their behavior over time

动态环境代写|World of Dynamic Environments EA1300 Cardiff University Assignment

问题 1.

Show that
$$
\rho \frac{D}{D t}\left(\frac{q_i q_i}{2}\right)=\rho f_i q_i+\frac{\partial\left(\sigma_{i j} q_i\right)}{\partial x_j}-\sigma_{i j} \frac{\partial q_i}{\partial x_j}
$$
where $\sigma_{i j}$ is the viscous stress tensor.

证明 .

Starting from the left-hand side of the equation, we have:

$\rho \frac{D}{D t}\left(\frac{q_i q_i}{2}\right)=\rho \frac{d}{d t}\left(\frac{q_i q_i}{2}\right)+\rho \frac{\partial}{\partial x_j}\left(u_j \frac{q_i q_i}{2}\right)$

where we have used the material derivative and the fact that $\rho u_i$ is the momentum density, which is related to the velocity by $u_i=q_i/\rho$. Now we can expand the first term on the right-hand side using the chain rule:

$\begin{aligned} \rho \frac{d}{d t}\left(\frac{q_i q_i}{2}\right) & =\rho \frac{\partial}{\partial t}\left(\frac{q_i q_i}{2}\right)+\rho u_j \frac{\partial}{\partial x_j}\left(\frac{q_i q_i}{2}\right) \ & =\rho \frac{\partial}{\partial t}\left(\frac{q_i q_i}{2}\right)+\rho q_i \frac{\partial u_j}{\partial x_j}+\rho u_j q_i \frac{\partial}{\partial x_j}\left(q_i\right) \ & =\rho \frac{\partial}{\partial t}\left(\frac{q_i q_i}{2}\right)+\rho q_i f_i+\rho u_j q_i \frac{\partial}{\partial x_j}\left(q_i\right)\end{aligned}$

where we have used the continuity equation $\partial \rho/\partial t + \partial(\rho u_j)/\partial x_j = 0$ and the equation of motion $\rho D u_i/Dt = \rho f_i – \partial \sigma_{ij}/\partial x_j$ to simplify the second term. Using the identity $\partial (q_i^2)/\partial x_j=2q_i\partial q_i/\partial x_j$ and the Einstein summation convention, we can rewrite the last term as $\partial(q_i q_j)/\partial x_j=q_i f_i+\partial(\sigma_{ij} q_i)/\partial x_j-\sigma_{ij}\partial q_i/\partial x_j$. Putting everything together, we get:

$\rho \frac{D}{D t}\left(\frac{q_i q_i}{2}\right)=\rho f_i q_i+\frac{\partial\left(\sigma_{i j} q_i\right)}{\partial x_j}-\sigma_{i j} \frac{\partial q_i}{\partial x_j}$

as desired.

问题 2.

Derive the explicit expression for the last term
$$
\Phi=\sigma_{i j} \frac{\partial q_i}{\partial x_j}
$$
for a two dimensionnal flow in term of $u, v$ and $x, y$ and comment on the sign of $\Phi$.

证明 .

In two-dimensional flow, the stress tensor $\sigma_{ij}$ has only three non-zero components: $\sigma_{xx}$, $\sigma_{yy}$, and $\sigma_{xy}=\sigma_{yx}$. Therefore, we can write

$\Phi=\sigma_{x x} \frac{\partial q}{\partial x}+\sigma_{y y} \frac{\partial p}{\partial y}+2 \sigma_{x y} \frac{\partial q}{\partial y}$

where $q$ and $p$ are the velocity potential and stream function, respectively. In two-dimensional flow, the velocity components can be written as $u=\frac{\partial \phi}{\partial y}$ and $v=-\frac{\partial \phi}{\partial x}$, where $\phi=q+ip$ is the complex potential.

Using the Cauchy-Riemann equations, we can write $\frac{\partial q}{\partial x}=\frac{\partial p}{\partial y}$ and $\frac{\partial q}{\partial y}=-\frac{\partial p}{\partial x}$. Substituting these relations and using the fact that $\sigma_{xx}=2\mu \frac{\partial u}{\partial x}$, $\sigma_{yy}=2\mu \frac{\partial v}{\partial y}$, and $\sigma_{xy}=\sigma_{yx}= \mu \left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)$ (where $\mu$ is the dynamic viscosity), we get

$\Phi=2 \mu\left(\frac{\partial u}{\partial x} \frac{\partial p}{\partial y}+\frac{\partial v}{\partial y} \frac{\partial p}{\partial x}\right)+2 \mu\left(\frac{\partial u}{\partial y} \frac{\partial q}{\partial y}-\frac{\partial v}{\partial x} \frac{\partial q}{\partial y}\right)$

Using the definition of the complex potential, we can write

$\frac{\partial u}{\partial x}=\frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial y}\right)=\frac{\partial^2 \phi}{\partial x \partial y}, \quad \frac{\partial u}{\partial y}=-\frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial x}\right)=-\frac{\partial^2 \phi}{\partial x \partial y}$

and

$\frac{\partial v}{\partial x}=-\frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial x}\right)=-\frac{\partial^2 \phi}{\partial x^2}, \quad \frac{\partial v}{\partial y}=\frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial y}\right)=\frac{\partial^2 \phi}{\partial y^2}$

问题 3.

For long-scale slow motion show that mass conservation requires
$$
\frac{\partial h}{\partial t}+\frac{\partial q}{\partial x}=0
$$
where $q$ is the discharge rate.
$$
q=\int_0^h u(x, y, t) d y
$$
Assume the profile obtained for uniform flow with $h$ and $h_o$ depending on $x, t$. Obtain a set of equations for $h(x, t)$ and $h_o(x, t)$.

证明 .

The control volume is defined by two cross-sectional areas, $A_1$ and $A_2$, and a length, $\Delta x$. The volume of fluid entering the control volume at $x$ per unit time is $q(x,t)$, and the volume leaving the control volume at $x+\Delta x$ per unit time is $q(x+\Delta x,t)$. The change in volume of the fluid within the control volume per unit time is $\frac{\partial h}{\partial t} \Delta x$, where $h(x,t)$ is the height of the fluid at $x$. By the principle of conservation of mass, we have:

$\frac{\partial h}{\partial t} \Delta x+q(x, t)-q(x+\Delta x, t)=0$

Dividing by $\Delta x$ and taking the limit as $\Delta x \rightarrow 0$, we obtain:

$\frac{\partial h}{\partial t}+\frac{\partial q}{\partial x}=0$

where $q(x,t)$ is the discharge rate at $x$ and time $t$, defined as:

$q(x, t)=\int_0^{h(x, t)} u(x, y, t) d y$

where $u(x,y,t)$ is the velocity profile of the fluid.

For the case of uniform flow, the velocity profile is constant in the $y$ direction, and we have:

$u(x, y, t)=\frac{Q}{h(x, t)}$

where $Q$ is the discharge per unit width of the channel. Substituting this into the definition of $q(x,t)$, we have:

$q(x, t)=\int_0^{h(x, t)} \frac{Q}{h(x, t)} d y=Q$

Therefore, for uniform flow, the discharge rate is constant along the channel. Substituting this into the mass conservation equation, we obtain:

$\frac{\partial h}{\partial t}+\frac{\partial Q}{\partial x}=0$

To obtain a set of equations for $h(x,t)$ and $h_o(x,t)$, we need an additional equation that relates the depth of the fluid to the width of the channel. For a rectangular channel of width $b$, the area of the cross-section is $A(x,t) = b h(x,t)$, and the wetted perimeter is $P(x,t) = b + 2h(x,t)$. Using these relations, we can derive the equation of motion for $h(x,t)$ by applying the principle of conservation of momentum to the control volume shown in the figure above.

Assuming the flow is steady and inviscid, the momentum equation reduces to:

$\frac{\partial}{\partial x}\left(\frac{h^2}{2}+g h+\frac{Q^2}{2 g b^2}\right)=0$

where $g$ is the acceleration due to gravity. This equation can be integrated to obtain:

$\frac{h^2}{2}+g h+\frac{Q^2}{2 g b^2}=C(x)$

where $C(x)$ is the integration constant. The value of $C(x)$ can be determined from the boundary conditions. For example, if the depth of the fluid is $

这是一份2023年的卡迪夫大学Cardiff University EA1300动态环境代写的成功案例




















电磁学和电波代写ELECTROMAGNETISM AND WAVES|PHY-4005A University of East Anglia Assignment

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Assignment-daixieTM为您提供东英吉利大学University of East Anglia  PHY-4005A APPLIED ELECTROMAGNETISM AND WAVES电磁学和电波代写代考辅导服务!

Instructions:

Electricity and Magnetism: Electricity and magnetism are two closely related phenomena that together form the foundation of the study of electromagnetism. The theory of electricity and magnetism deals with the behavior of charged particles and their interactions with electric and magnetic fields.

Electric charge is a fundamental property of matter, and there are two types of charge, positive and negative. When like charges are brought close together, they repel each other, while opposite charges attract each other. The force between two charged particles is known as the Coulomb force and is proportional to the product of the charges and inversely proportional to the distance between them.

When charges are in motion, they create a magnetic field, and the motion of charges in a magnetic field creates a force on the charges. This is known as the Lorentz force and is proportional to the velocity of the charges and the strength of the magnetic field.

Electric and magnetic fields are related and can create each other. When an electric charge is moving, it creates a magnetic field, and when a magnetic field changes, it creates an electric field. This is known as electromagnetic induction and is the basis of many electrical devices, including generators, transformers, and motors.

电磁学和电波代写ELECTROMAGNETISM AND WAVES|PHY-4005A University of East Anglia Assignment

问题 1.

In this problem, we consider several equivalent situations for a plane wave propagation in the $\hat{z}$ direction. Let the electric field be $\hat{x}$ directed. $$ \bar{E}=\hat{x} E_0 e^{i k z}, \quad \bar{H}=\hat{y} \frac{1}{\eta} E_0 e^{i k z} $$ and the region of interest be $z>0$. (a) Put an electric current sheet with $\bar{J}_s=A \hat{x}$. What is the value of $A$ so that the same field is preserved in the region of interest?

证明 .

(a) The electric current sheet produces a magnetic field $\bar{H}_s = \hat{y} \frac{A}{\epsilon_0} \delta(z)$. To preserve the same field in the region of interest, this magnetic field must cancel the magnetic field produced by the plane wave. Since $\bar{H}_s$ only exists at $z=0$, we need to consider the $z>0$ region where the plane wave is propagating. Using the relationship $\bar{E} = \eta \bar{H}$, we have $\bar{H} = \frac{1}{\eta} \hat{y} E_0 e^{ikz}$ for $z>0$. The magnetic field produced by the electric current sheet at $z=0$ is $\bar{H}_s = \hat{y} \frac{A}{\eta}$. Therefore, we need $A = E_0$ for the same field to be preserved in the region of interest.

问题 2.

(b) Put a magnetic current sheet with $\bar{M}_s=B \hat{y}$. What is the value of $B$ so that the same field is preserved in the region of interest?

证明 .

(b) The magnetic current sheet produces an electric field $\bar{E}_s = \hat{x} \frac{B}{\mu_0} \delta(z)$. To preserve the same field in the region of interest, this electric field must cancel the electric field produced by the plane wave. Since $\bar{H} = \frac{1}{\eta} \bar{E}$, we have $\bar{E} = \eta \hat{x} E_0 e^{ikz}$ for $z>0$. The electric field produced by the magnetic current sheet at $z=0$ is $\bar{E}_s = \hat{x} \frac{B}{\mu_0}$. Therefore, we need $B = \frac{\mu_0}{\eta} E_0$ for the same field to be preserved in the region of interest.

问题 3.

(c) Replace the region $z<0$ with a perfect conductor. Place in front of the conductor an electric sheet with $\bar{J}_s=C \hat{x}$ and a magnetic current sheet with $\bar{M}_s=D \hat{y}$. What is the value of $C$ and $D$ so that the same field is preserved in the region of interest?

证明 .

(c) The perfect conductor reflects the incident wave, so the reflected wave will interfere with the incident wave. To preserve the same field in the region of interest, we need to ensure that the interference results in no net field. The electric current sheet produces a reflected magnetic field $\bar{H}{sr} = -\hat{y} \frac{C}{\epsilon_0} \delta(z)$. The magnetic current sheet produces a reflected electric field $\bar{E}{sr} = \hat{x} \frac{D}{\mu_0} \delta(z)$. The total magnetic field at $z>0$ is the sum of the incident and reflected fields:

$\bar{H}_t=\hat{y} \frac{1}{\eta} E_0 e^{i k z}-\hat{y} \frac{C}{\eta \epsilon_0} e^{-i k z}$

The total electric field at $z>0$ is the sum of the incident and reflected fields:

$\bar{E}_t=\hat{x} \eta E_0 e^{i k z}+\hat{x} \frac{D}{\mu_0} e^{-i k z}$

For the same field to be preserved, we need $\bar{H}_t = 0$ and $\bar{E}_t = 0$ for $z>0$. Solving for $C$ and $D$, we get:

$C=E_0, \quad D=-\frac{\mu_0}{\eta} E_0$

Therefore, to preserve the same field in the region of interest, we need to place an electric current sheet with $A=E_0$

这是一份2023年的东英吉利大学University of East Anglia  PHY-4005A电磁学和电波代写的成功案例




















热、原子和固体代写HEAT, ATOMS AND SOLIDS|PHY-4005A University of East Anglia Assignment

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Assignment-daixieTM为您提供东英吉利大学University of East Anglia  PHY-4005A HEAT, ATOMS AND SOLIDS热原子和固体代写代考辅导服务!

Instructions:

Thermodynamics is the study of energy and its transformation. It deals with macroscopic systems and how they respond to changes in energy. Condensed matter physics is the study of materials in which the atoms are close together, and they interact strongly with each other. This includes solids, liquids, and some gases.

Electronic Structure: The Sommerfeld and Band Theories

In the free-electron Sommerfeld theory, electrons are treated as free particles moving in a periodic potential created by the ions. This model explains some properties of metals, such as their electrical conductivity and heat capacity. However, it fails to account for the diversity of properties observed in different materials.

The band theory of solids describes the electronic structure of materials in terms of energy bands. In a periodic crystal lattice, the wave functions of the electrons overlap, forming bands of allowed energies. The electrons can occupy these bands, up to a maximum called the Fermi energy. This theory explains the diversity of electronic properties of materials, including their conductivity, magnetism, and optical properties.

Phonons and Heat Capacity

Phonons are quantized vibrations of the atoms in a crystal lattice. They contribute to the heat capacity of a solid by carrying energy and increasing the number of ways in which energy can be distributed among the atoms.

Structure, Bonding, and Properties of Solids

The properties of solids depend on their atomic structure and bonding. The types of bonding include ionic, covalent, and metallic bonding. The electronic structure and bonding determine the electrical conductivity, magnetism, and other properties of the material.

热、原子和固体代写HEAT, ATOMS AND SOLIDS|PHY-4005A University of East Anglia Assignment

问题 1.

(a) Prove the finite temperature version of the fluctuation dissipation theorem
$$
\chi^{\prime \prime}(q, \omega)=\frac{1}{2}\left(e^{-\beta \omega}-1\right) S(q, \omega),
$$
and
$$
S(q, \omega)=-2\left(n_B(\omega)+1\right) \chi^{\prime \prime}(q, \omega)
$$
where $S(q, \omega)=\int d \vec{x} d t e^{-i \vec{q}-\vec{x}} e^{i \omega x}\langle\rho(\vec{x}, t) \rho(0,)\rangle_T$ and $n_B(\omega)=\left(e^{\beta \omega}-1\right)^{-1}$ is the Bose occupation factor.

证明 .

The fluctuation-dissipation theorem relates the correlation function of a fluctuating variable to its response to an external perturbation. Specifically, it relates the imaginary part of the susceptibility, $\chi^{\prime \prime}(q,\omega)$, to the power spectrum of the fluctuations, $S(q,\omega)$.

Consider a system described by a Hamiltonian $H$ at temperature $T$. Let $\rho(\vec{x},t)$ be an operator that measures the local density fluctuations of the system. Then, the susceptibility $\chi(q,\omega)$ is defined as the linear response of the density fluctuations to a spatially varying potential $V(\vec{x})$ of wavevector $\vec{q}$ and frequency $\omega$:

$$\delta \langle\rho(\vec{q},\omega)\rangle_T = \chi(q,\omega) \delta \langle V(\vec{q},\omega)\rangle_T$$

where $\delta$ denotes the fluctuation about the thermal equilibrium value and $\langle \rangle_T$ denotes the thermal average.

The imaginary part of the susceptibility is related to the power spectrum of the density fluctuations as follows:

$$\chi^{\prime\prime}(q,\omega) = \frac{1}{2i} \left[\chi(q,\omega) – \chi^*(q,\omega)\right]$$

where $\chi^*(q,\omega)$ is the complex conjugate of $\chi(q,\omega)$.

Using the Kubo formula, the susceptibility can be expressed in terms of the time-dependent density correlation function:

$$\chi(q,\omega) = \frac{1}{i\omega} \int_0^\infty dt, e^{i\omega t} \langle[\rho(\vec{q},t),\rho(-\vec{q},0)]\rangle_T$$

where $[\cdot,\cdot]$ denotes the commutator.

The power spectrum of the density fluctuations is defined as:

$$S(q,\omega) = \int dt \int d\vec{x}, e^{-i\vec{q}\cdot\vec{x}} e^{i\omega t} \langle\rho(\vec{x},t)\rho(0,0)\rangle_T$$

Using these definitions, we can now derive the finite temperature version of the fluctuation-dissipation theorem.

First, we express the susceptibility in terms of the density correlation function using the Kubo formula:

$$\chi(q,\omega) = \frac{1}{i\omega} \int_0^\infty dt, e^{i\omega t} \langle[\rho(\vec{q},t),\rho(-\vec{q},0)]\rangle_T$$

Taking the imaginary part, we have:

$$\chi^{\prime\prime}(q,\omega) = -\frac{1}{2}\int_0^\infty dt, e^{i\omega t} \langle[\rho(\vec{q},t),\rho(-\vec{q},0)]\rangle_T + \text{c.c.}$$

where $\text{c.c.}$ denotes the complex conjugate of the first term.

问题 2.

(b) Show that $\chi^{\prime \prime}(q, \omega)=-\chi^{\prime \prime}(-q,-\omega)$ and $S(-q,-\omega)=e^{-\beta \omega} S(q, \omega)$. In terms of the scattering probability, show that this is consistent with detailed balance.

证明 .

To show that $\chi^{\prime \prime}(q, \omega)=-\chi^{\prime \prime}(-q,-\omega)$, we start with the definition of the dynamic structure factor:

$\chi^{\prime \prime}(q, \omega)=-\frac{1}{\pi} \operatorname{Im} \int_{-\infty}^{\infty} d t e^{i \omega t} \int d^3 r e^{-i q \cdot r}\langle\hat{\rho}(\mathbf{r}, t) \hat{\rho}(0,0)\rangle$

where $\hat{\rho}(\mathbf{r}, t)$ is the density operator at position $\mathbf{r}$ and time $t$, and $\langle \rangle$ denotes the statistical average over an ensemble of systems in thermal equilibrium. Using the fact that the correlation function $\left\langle\hat{\rho}(\mathbf{r}, t) \hat{\rho}(0,0)\right\rangle$ is real, we can rewrite the integral as

$\chi^{\prime \prime}(q, \omega)=\frac{1}{\pi} \int_{-\infty}^{\infty} d t \sin (\omega t) \int d^3 r e^{-i q \cdot r}\langle\hat{\rho}(\mathbf{r}, t) \hat{\rho}(0,0)\rangle$

Changing variables $t\rightarrow -t$ and $r\rightarrow -r$ gives

$\begin{aligned} \chi^{\prime \prime}(-q,-\omega) & =\frac{1}{\pi} \int_{-\infty}^{\infty} d t \sin (-\omega t) \int d^3 r e^{i(-q) \cdot r}\langle\hat{\rho}(-\mathbf{r},-t) \hat{\rho}(0,0)\rangle \ & =\frac{1}{\pi} \int_{-\infty}^{\infty} d t \sin (\omega t) \int d^3 r e^{-i q \cdot r}\langle\hat{\rho}(\mathbf{r}, t) \hat{\rho}(0,0)\rangle \ & =-\chi^{\prime \prime}(q, \omega)\end{aligned}$

Thus, we have shown that $\chi^{\prime \prime}(q, \omega)=-\chi^{\prime \prime}(-q,-\omega)$.

Next, we show that $S(-q,-\omega)=e^{-\beta \omega} S(q, \omega)$. Starting with the definition of the dynamic structure factor,

$S(q, \omega)=\frac{1}{\pi} \frac{\operatorname{Im} \chi(q, \omega)}{1-e^{-\beta \omega}}$

we can rewrite $\chi(q,\omega)$ using the fluctuation-dissipation theorem as

$\chi(q, \omega)=\frac{1}{\beta}\left[\chi^{\prime \prime}(q, \omega)+i \frac{\chi^{\prime}(q, \omega)}{\omega}\right]$

问题 3.

(a) Using linear response theory, derive the following expression for the magnetic susceptibility $\chi_{|}=\partial M_z / \partial H_z$.
$$
\chi_{|}=\lim _{q \rightarrow 0} \int \frac{d \omega}{2 \pi}\left\langle S_z(q, \omega) S_z(-q,-\omega)\right\rangle \frac{\left(1-e^{-\frac{\hbar \omega}{k T}}\right)}{\omega}
$$

证明 .

Linear response theory is a framework for understanding the response of a physical system to small perturbations. In this case, we want to calculate the magnetic susceptibility $\chi_{|}$, which is the response of the magnetization $M_z$ to a small magnetic field $H_z$. We can use the Kubo formula to relate the susceptibility to the spin-spin correlation function:

$$\chi_{|}=\frac{1}{k_BT}\lim_{\omega\rightarrow 0}\frac{1}{\omega}\int_{0}^{\infty}dt\int d\mathbf{r} e^{i\omega t}\left\langle\left[\hat{S}_z(\mathbf{r},t),\hat{S}z(\mathbf{0},0)\right]\right\rangle{H_z=0}$$

where $\hat{S}z(\mathbf{r},t)$ is the $z$-component of the spin density operator at position $\mathbf{r}$ and time $t$, and $\left\langle\cdots\right\rangle{H_z=0}$ denotes the thermal average in the absence of an external magnetic field. The Fourier transform of the spin density operator is defined by

$$\hat{S}_z(\mathbf{q},\omega)=\int dt\int d\mathbf{r} e^{i(\mathbf{q}\cdot\mathbf{r}-\omega t)}\hat{S}_z(\mathbf{r},t)$$

Using this definition, we can rewrite the susceptibility as

$$\chi_{|}=\frac{1}{k_BT}\lim_{\omega\rightarrow 0}\frac{1}{\omega}\int_{0}^{\infty}dt\int d\mathbf{r} e^{i\omega t}\int\frac{d\mathbf{q}}{(2\pi)^3}e^{-i\mathbf{q}\cdot\mathbf{r}}\left\langle\left[\hat{S}_z(\mathbf{q},\omega),\hat{S}z(-\mathbf{q},-\omega)\right]\right\rangle{H_z=0}$$

Now, we use the fact that the commutator is related to the imaginary part of the retarded correlation function:

$$\left[\hat{S}_z(\mathbf{q},\omega),\hat{S}z(-\mathbf{q},-\omega)\right]=2i\operatorname{Im}\left{\hat{\chi}{zz}(\mathbf{q},\omega)\right}$$

where $\hat{\chi}_{zz}(\mathbf{q},\omega)$ is the Fourier transform of the spin-spin correlation function $\left\langle\hat{S}z(\mathbf{r},t)\hat{S}z(\mathbf{0},0)\right\rangle{H_z=0}$. Substituting this into the expression for $\chi{|}$ and taking the limit $\omega\rightarrow 0$, we obtain

$$\chi_{|}=\frac{2}{k_BT}\int_{0}^{\infty}dt\int d\mathbf{r}\int\frac{d\mathbf{q}}{(2\pi)^3}e^{-i\mathbf{q}\cdot\mathbf{r}}\operatorname{Im}\left{\hat{\chi}_{zz}(\mathbf{q},\omega=0)\right}$$

这是一份2023年的东英吉利大学University of East Anglia  PHY-5004B热原子和固体代写的成功案例




















波与量子代写Waves and Quanta|PA1140 University of Leicester Assignment

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Waves and quanta are two fundamental concepts in physics that are closely related but also distinct.

A wave is a disturbance that travels through space and time, carrying energy without the transfer of matter. Waves can take many forms, including electromagnetic waves (such as light and radio waves), sound waves, and water waves. Waves are characterized by their wavelength, frequency, and amplitude, which determine their properties such as speed and energy.

On the other hand, a quantum is a discrete packet of energy that behaves both as a particle and as a wave. In the quantum world, particles can exhibit wave-like behavior, and waves can behave like particles. This duality is known as wave-particle duality. The behavior of quanta is described by quantum mechanics, a branch of physics that deals with the behavior of matter and energy at the atomic and subatomic level.

The relationship between waves and quanta is that waves can be thought of as a collection of quanta. For example, electromagnetic waves consist of a stream of photons, which are quanta of electromagnetic radiation. Similarly, matter waves, which describe the behavior of particles such as electrons, can be thought of as a collection of quanta.

Overall, waves and quanta are both fundamental concepts in physics that are used to explain and understand the behavior of matter and energy in our universe.

波与量子代写Waves and Quanta|PA1140 University of Leicester Assignment

问题 1.

A long, uniform taut string (mass per unit length $\rho$, tension $T$ ) along $-\infty<x<\infty$ is supported on an elastic foundation of stiffness $\alpha$, and a point mass $M$ is attached at $x=0$.
Suppose that a time-harmonic vertical force
$$
F \cos \Omega t
$$
is applied to the mass at $x=0$.
Determine the steady-state displacement response of the string for $-\infty<x<\infty$.

证明 .

The equation of motion for the string is given by the wave equation:

$\rho \frac{\partial^2 u}{\partial t^2}=T \frac{\partial^2 u}{\partial x^2}-\alpha u$

where $u(x,t)$ is the displacement of the string from its equilibrium position at point $x$ and time $t$. The first term on the right-hand side represents the tension in the string, and the second term represents the elastic foundation. We can assume a solution of the form:

$u(x, t)=U(x) e^{i \omega t}$

where $\omega = \Omega + i\beta$ is the complex frequency and $U(x)$ is the complex amplitude of the displacement.

Substituting this solution into the wave equation and simplifying, we get:

$\frac{d^2 U}{d x^2}+k^2 U=0$

where $k^2 = \frac{\omega^2 \rho}{T} – i\frac{\alpha}{T}$. This is a second-order homogeneous differential equation with constant coefficients, which has solutions of the form:

$U(x)=A e^{i k x}+B e^{-i k x}$

where $A$ and $B$ are complex constants to be determined by the boundary conditions.

At $x=0$, the displacement of the string must be equal to the displacement of the mass, which is given by:

$u(0, t)=U(0) e^{i \omega t}=\frac{F}{M} e^{i \omega t}$

Therefore, $U(0) = \frac{F}{M}$.

At $x=\pm \infty$, the string must be taut and have no displacement. Therefore, we can assume that $U(\pm \infty) = 0$.

Using these boundary conditions, we can solve for $A$ and $B$:

$A=-\frac{F}{2 M} e^{-i k x} \quad$ and $\quad B=\frac{F}{2 M} e^{i k x}$

Therefore, the displacement of the string is given by:

$u(x, t)=\frac{F}{2 M}\left(e^{i(k x-\omega t)}-e^{i(-k x-\omega t)}\right)$

Simplifying, we get:

$u(x, t)=\frac{F}{M} \sin (k x) \sin (\omega t)$

where $k = \sqrt{\frac{\omega^2 \rho}{T} – i\frac{\alpha}{T}}$. This is the steady-state displacement response of the string for $-\infty < x < \infty$.

问题 2.

The propagation of free uni-directional surface waves of small amplitude on moderately shallow water is governed by the equation
$$
\frac{\partial \eta}{\partial t}+c_0 \frac{\partial \eta}{\partial x}+\beta \frac{\partial^3 \eta}{\partial x^3}=0
$$
where $\eta(x, t)$ is the free-surface elevation and $c_0$ and $\beta$ are constants.
(a) Suppose that an external localized pressure disturbance traveling with constant speed $V$ acts on the free surface. Determine the wavenumber(s) of the excited steadystate radiating wave(s), depending on the forcing speed $V$. Sketch the position of these waves relative to the forcing. (Take $c_0>0$ and consider $\beta>0$ and $\beta<0$ as well as $V>0$ and $V<0$.)

证明 .

To determine the wavenumber(s) of the excited steady-state radiating wave(s), we start by assuming a solution of the form:

$\eta(x, t)=A e^{i(k x-\omega t)}$

where $A$ is the amplitude of the wave, $k$ is the wavenumber, and $\omega$ is the frequency. Substituting this solution into the given equation yields:

$-i \omega A+i c_0 k A-i \beta k^3 A=0$

Dividing through by $-iA$ and solving for $k$, we get:

$k^2=\frac{\omega}{c_0} \pm \sqrt{\left(\frac{\omega}{c_0}\right)^2-\frac{\beta}{c_0}}$

Now suppose that an external localized pressure disturbance traveling with constant speed $V$ acts on the free surface. This means that the phase velocity of the wave excited by the disturbance will be $V$, i.e. $\omega/k=V$. Substituting this into the above expression for $k^2$ yields:

$k^2=\frac{V^2}{c_0^2} \pm \sqrt{\left(\frac{V^2}{c_0^2}\right)^2-\frac{\beta}{c_0^3}}$

Depending on the sign of $\beta$, we have two cases to consider:

Case 1: $\beta>0$

In this case, there are two possible values of $k$ given by:

$k_1=\frac{V}{c_0}+\sqrt{\left(\frac{V}{c_0}\right)^2-\frac{\beta}{c_0^3}}, \quad k_2=\frac{V}{c_0}-\sqrt{\left(\frac{V}{c_0}\right)^2-\frac{\beta}{c_0^3}}$

The two corresponding frequencies are:

$\omega_1=c k_1=c_0\left(\frac{V}{c_0}+\sqrt{\left(\frac{V}{c_0}\right)^2-\frac{\beta}{c_0^3}}\right), \quad \omega_2=c_0 k_2=c_0\left(\frac{V}{c_0}-\sqrt{\left(\frac{V}{c_0}\right)^2-\frac{\beta}{c_0^3}}\right)$

where $c=\sqrt{c_0^2+\beta k^2}$ is the phase velocity of the wave.

Note that $k_1$ and $k_2$ have opposite signs, which means that the two waves have opposite directions of propagation. Furthermore, the two waves have different wavelengths and frequencies. The wave with wavenumber $k_1$ has a longer wavelength and a lower frequency than the wave with wavenumber $k_2$.

To sketch the position of these waves relative to the forcing, we note that the wave with wavenumber $k_1$ travels in the same direction as the external disturbance, while the wave with wavenumber $k_2$ travels in the opposite direction. The amplitude of the waves decreases with distance from the source, so the wave with wavenumber $k_1$ will be closer to the source than the wave with wavenumber $k_2$.

Case 2: $\beta<0$

In this case, there is only one possible value of $k$ given

问题 3.

(b) Suppose at $t=0$ a localized initial wave disturbance is introduced in the vicinity of $x=0$. Sketch qualitatively the time history of the response for $t>0$ at a fixed station $x=L>0$, far from the region of the initial disturbance. Sketch qualitatively a snapshot of the disturbance for $-\infty<x<\infty$ at time $t=T$, long after the initial excitation. Justify your answers. (Again, $c_0>0$ and consider $\beta>0$ and $\beta<0$.)

证明 .

To sketch qualitatively the time history of the response for $t>0$ at a fixed station $x=L>0$, we first need to find the solution to the given equation. We can use the method of characteristics to solve this equation.

Let $dx/dt=c_0$ and $d^2x/dt^2=0$, which gives $x=c_0t+x_0$ and $x_0=L$ for $t=0$. Then, we have

$\frac{d \eta}{d t}=-\beta \frac{\partial^3 \eta}{\partial x^3}$

We assume a solution of the form $\eta(x,t)=f(x-ct)$, where $c$ is the wave speed. Substituting this into the above equation and integrating three times with respect to $x$, we get

$\eta(x, t)=A e^{-k(x-c t)} \cos (k(x-c t)+\phi)$

where $k=\sqrt[3]{\beta/c_0}$, $A$ and $\phi$ are constants determined by the initial conditions.

Now, we can sketch the time history of the response for $t>0$ at a fixed station $x=L>0$. At $x=L$, the free surface elevation is given by

$\eta(L, t)=A e^{-k(L-c t)} \cos (k(L-c t)+\phi)$

As $t$ increases, the amplitude of the cosine function decays exponentially with a decay rate of $k(c t-L)$. Therefore, the free surface elevation decreases in amplitude and oscillates with a decreasing frequency as time progresses. The decay rate is faster for larger values of $k$ (i.e., for larger values of $\beta$ or smaller values of $c_0$). If $\beta<0$, then $k$ is imaginary, which means the wave amplitude decays exponentially without oscillation.

To sketch qualitatively a snapshot of the disturbance for $-\infty<x<\infty$ at time $t=T$, long after the initial excitation, we can use the same solution as above, but with $t=T$. Thus, we have

$\eta(x, T)=A e^{-k(x-c T)} \cos (k(x-c T)+\phi)$、

The cosine function represents a wave with wavelength $\lambda=2\pi/k$ and wave number $k$. The wave is propagating to the right if $c>c_0$, or to the left if $c<c_0$. The amplitude of the wave decreases exponentially as $x$ increases or decreases from the initial position of the disturbance. The decay rate is faster for larger values of $k$ (i.e., for larger values of $\beta$ or smaller values of $c_0$). If $\beta<0$, then $k$ is imaginary, which means the wave amplitude decays exponentially without oscillation. The phase velocity of the wave is given by $c/k$, which decreases as $k$ increases. If $\beta>0$, then the group velocity of the wave is given by $c-(2/3)k^2c_0$, which is smaller than the phase velocity. If $\beta<0$, then the group velocity is undefined because $k$ is imaginary.

这是一份2023年的莱斯特大学University of Leicester PA1140波与量子代写的成功案例