分类: PHYS322 electrodynamics

统计代写|PHYS322 electrodynamics

Statistics-lab™可以为您提供washington.edu PHYS322 electrodynamics电动力学的代写代考辅导服务!

统计代写|PHYS322 electrodynamics

PHYS322 electrodynamics课程简介

This course provides a deeper look into the theory of electricity and magnetism.

We shall study:

– magnetostatics

– magnetic fields in matter

– Maxwell’s equations

– electromagnetic waves

PREREQUISITES 

Upon successful completion of this course you will be able to:

  • Understand the implications of Ampère’s and Biot-Savart’s laws
  • Understand the nature of the vector potential
  • Understand the behaviour of magnetic fields in medium
  • Be perfectly familiar with Maxwell’s equations
  • Know the nature of momentum and angular moment in Electrodynamics
  • Understand the details of propagation of electromagnetic waves

PHYS322 electrodynamics HELP(EXAM HELP, ONLINE TUTOR)

问题 1.

1.2.3. Energy-momentum tensor for a massive scalar field Consider the massive scalar field from ch. $0 \S 2.5$ :
$$
\mathcal{L}=\frac{1}{2}\left(\partial_\mu \varphi\right)\left(\partial^\mu \varphi\right)-\frac{m^2}{2} \varphi^2
$$
and the tensor field $H_\mu{ }^\nu$ defined analogously to Problem 1.2.1:
$$
H_\mu{ }^\nu=\left(\partial_\mu \varphi\right) \frac{\partial \mathcal{L}}{\partial\left(\partial_\nu \varphi\right)}-\delta_\mu{ }^\nu \mathcal{L}
$$
Determine $H_\mu^\nu$ explicitly and show that
$$
\partial_\nu H_\mu^\nu=0
$$
hint: Use the Euler-Lagrange equation determined in ch. $0 \S 2.5$.

We have the Lagrangian density for the massive scalar field: $$\mathcal{L}=\frac{1}{2}(\partial_\mu \varphi)(\partial^\mu \varphi)-\frac{m^2}{2} \varphi^2.$$ The corresponding energy-momentum tensor is given by: $$T_{\mu\nu}=\frac{\partial\mathcal{L}}{\partial(\partial^\mu\varphi)}\partial_\nu\varphi+\frac{\partial\mathcal{L}}{\partial(\partial^\mu\partial^\rho\varphi)}\partial_\nu\partial_\rho\varphi-\delta_{\mu\nu}\mathcal{L}.$$ For the scalar field Lagrangian, we have: $$\frac{\partial\mathcal{L}}{\partial(\partial^\mu\varphi)}=\partial^\mu\varphi,\qquad\frac{\partial\mathcal{L}}{\partial(\partial^\mu\partial^\rho\varphi)}=\frac{1}{2}(\delta^\mu_\rho\partial^\sigma\varphi+\delta^\sigma_\rho\partial^\mu\varphi),\qquad\mathcal{L}=\frac{1}{2}(\partial_\mu \varphi)(\partial^\mu \varphi)-\frac{m^2}{2} \varphi^2.$$ Thus, we can compute the components of $T_{\mu\nu}$: \begin{align} T_{00}&=\frac{1}{2}(\partial_0\varphi)(\partial^0\varphi)+\frac{1}{2}(\partial_k\varphi)(\partial^k\varphi)+\frac{m^2}{2} \varphi^2,\ T_{0i}&=\frac{1}{2}(\partial_0\varphi)(\partial^i\varphi)+\frac{1}{2}(\partial_k\varphi)(\partial^i\varphi),\ T_{i0}&=\frac{1}{2}(\partial_i\varphi)(\partial^0\varphi)+\frac{1}{2}(\partial_i\varphi)(\partial^k\varphi),\ T_{ij}&=\frac{1}{2}(\partial_i\varphi)(\partial^j\varphi)+\frac{1}{2}(\partial_j\varphi)(\partial^i\varphi)-\delta_{ij}\left(\frac{1}{2}(\partial_k\varphi)(\partial^k\varphi)+\frac{m^2}{2} \varphi^2\right). \end{align} To verify that $\partial_\nu H_\mu^\nu=0$, we first compute $H_\mu^\nu$: \begin{align} H_\mu^\nu&=\left(\partial_\mu \varphi\right) \frac{\partial \mathcal{L}}{\partial(\partial_\nu \varphi)}-\delta_\mu{ }^\nu \mathcal{L}\ &=\left(\partial_\mu \varphi\right) \partial^\nu \varphi – \delta_\mu{ }^\nu \left[\frac{1}{2}(\partial_\rho \varphi)(\partial^\rho \varphi)-\frac{m^2}{2} \varphi^2\right]\ &=\left(\partial_\mu \varphi\right) \partial^\nu \varphi

问题 2.

Consider the 4-vector potential $A^\mu(x)=(\varphi(x), A(x))$. Show that one can always find a gauge transformation such that
$$
\nabla \cdot A(x)=0
$$
This choice is called Coulomb gauge.

In general, a gauge transformation of a vector potential $A^\mu(x)$ is given by

A^\mu(x) \rightarrow A^{\prime\mu}(x) = A^\mu(x) + \partial^\mu \Lambda(x),Aμ(x)→A′μ(x)=Aμ(x)+∂μΛ(x),

where $\Lambda(x)$ is an arbitrary scalar function of spacetime. In terms of the new potential, the electric and magnetic fields are given by \begin{align*} \mathbf{E} &= -\nabla \varphi – \frac{\partial \mathbf{A}}{\partial t}, \ \mathbf{B} &= \nabla \times \mathbf{A}. \end{align*} We want to find a gauge transformation that sets $\nabla \cdot \mathbf{A}=0$, i.e., $\partial_i A^i = 0$.

We can start by choosing $\Lambda(x)$ such that

\partial_i A^{\prime i}(x) = 0.∂iAi(x)=0.

Substituting the gauge transformation formula, we have \begin{align*} \partial_i A^{\prime i}(x) &= \partial_i A^i(x) + \partial_i \partial^i \Lambda(x) \ &= \partial_i A^i(x) + \nabla^2 \Lambda(x) = 0. \end{align*} Solving for $\Lambda(x)$, we get

\Lambda(x) = -\frac{1}{\nabla^2} \partial_i A^i(x),Λ(x)=−∇21​∂iAi(x),

provided that $\nabla^2$ is invertible. If $\nabla^2$ has zero modes, then there may be gauge transformations that cannot be obtained in this way.

Thus, we have found a gauge transformation that sets $\partial_i A^{\prime i}(x) = 0$. This is the Coulomb gauge. Note that this choice of gauge does not uniquely determine the potential $A^\mu(x)$, since we can still perform gauge transformations that satisfy $\partial_\mu \Lambda(x)=0$. However, it simplifies the equations of motion and makes the electric potential $\varphi(x)$ behave like the Coulomb potential in electrostatics.

Textbooks


• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

此图像的alt属性为空;文件名为%E7%B2%89%E7%AC%94%E5%AD%97%E6%B5%B7%E6%8A%A5-1024x575-10.png
PHYS322 electrodynamics

Statistics-lab™可以为您提供washington.edu PHYS322 electrodynamics电动力学的代写代考辅导服务! 请认准Statistics-lab™. Statistics-lab™为您的留学生涯保驾护航。

统计代写|PHYS322 electrodynamics

Statistics-lab™可以为您提供washington.edu PHYS322 electrodynamics电动力学的代写代考辅导服务!

统计代写|PHYS322 electrodynamics

PHYS322 electrodynamics课程简介

This course provides a deeper look into the theory of electricity and magnetism.

We shall study:

– magnetostatics

– magnetic fields in matter

– Maxwell’s equations

– electromagnetic waves

PREREQUISITES 

Upon successful completion of this course you will be able to:

  • Understand the implications of Ampère’s and Biot-Savart’s laws
  • Understand the nature of the vector potential
  • Understand the behaviour of magnetic fields in medium
  • Be perfectly familiar with Maxwell’s equations
  • Know the nature of momentum and angular moment in Electrodynamics
  • Understand the details of propagation of electromagnetic waves

PHYS322 electrodynamics HELP(EXAM HELP, ONLINE TUTOR)

问题 1.

1.2.1. Energy-momentum tensor
Consider the electromagnetic field in the absence of matter.
a) Show that the tensor field
$$
H_\mu{ }^\nu(x)=\left(\partial_\mu A_\alpha(x)\right) \frac{\partial \mathcal{L}}{\partial\left(\partial_\nu A_\alpha(x)\right)}-\delta_\mu^\nu \mathcal{L}
$$
obeys the continuity equation
$$
\partial_\nu H_\mu^\nu(x)=0 \quad() $$ note: Notice that $H_\mu{ }^\nu(x)$ is a generalization of Jacobi’s integral in Classical Mechanics. b) Show that () also holds for
$$
\tilde{T}\mu^\nu=H\mu^\nu+\partial_\alpha \psi_\mu^{\nu \alpha}
$$
where $\psi_\mu^{\nu \alpha}$ is any tensor field that is antisymmetric in the second and third indices, $\psi_\mu^{\nu \alpha}(x)=$ $-\psi_\mu^{\alpha \nu}(x)$.
c) Show that $\psi_\mu{ }^{\nu \alpha}$ can be chosen such that $\tilde{T}\mu{ }^\nu(x)=T\mu{ }^\nu(x)$, which provides an alternative proof that $T_\mu{ }^\nu(x)$ obeys $\left({ }^*\right)$.

a) To show that $H_\mu{ }^\nu(x)$ obeys the continuity equation, we start by taking the divergence of $H_\mu{ }^\nu(x)$ with respect to $\nu$: \begin{align*} \partial_\nu H_\mu{ }^\nu(x) &= \partial_\nu \left[\left(\partial_\mu A_\alpha(x)\right) \frac{\partial \mathcal{L}}{\partial\left(\partial_\nu A_\alpha(x)\right)}-\delta_\mu^\nu \mathcal{L}\right]\ &=\left(\partial_\mu \partial_\nu A_\alpha(x)\right) \frac{\partial \mathcal{L}}{\partial\left(\partial_\nu A_\alpha(x)\right)} + \left(\partial_\mu A_\alpha(x)\right) \frac{\partial^2 \mathcal{L}}{\partial\left(\partial_\nu A_\alpha(x)\right)^2} – \partial_\nu(\delta_\mu^\nu\mathcal{L})\ &= \left(\partial_\nu \partial_\mu A_\alpha(x)\right) \frac{\partial \mathcal{L}}{\partial\left(\partial_\nu A_\alpha(x)\right)} + \left(\partial_\mu A_\alpha(x)\right) \frac{\partial^2 \mathcal{L}}{\partial\left(\partial_\nu A_\alpha(x)\right)^2} – \delta_\mu^\nu\partial_\nu \mathcal{L} \ &= \partial_\nu \left(\partial_\mu A_\alpha(x) \frac{\partial \mathcal{L}}{\partial\left(\partial_\nu A_\alpha(x)\right)}\right) – \delta_\mu^\nu \partial_\nu \mathcal{L} \ &= \partial_\nu F_{\mu\nu}(x) – \delta_\mu^\nu \partial_\nu \mathcal{L}, \end{align*} where we have used the definition of the electromagnetic field tensor, $F_{\mu\nu}(x) = \partial_\mu A_\nu(x) – \partial_\nu A_\mu(x)$, in the third line. Note that the last term in the above equation can be simplified using the Euler-Lagrange equation: \begin{align*} \delta_\mu^\nu \partial_\nu \mathcal{L} &= \partial_\nu \left(\delta_\mu^\nu \mathcal{L}\right) – \partial_\mu \mathcal{L} \ &= \partial_\mu \left(\partial_\alpha A_\beta(x) \frac{\partial \mathcal{L}}{\partial\left(\partial_\mu A_\beta(x)\right)}\right) – \partial_\mu \mathcal{L} \ &= \partial_\alpha \left(\partial_\mu A_\beta(x) \frac{\partial \mathcal{L}}{\partial\left(\partial_\mu A_\beta(x)\right)}\right) \ &= \partial_\alpha \left(\frac{\partial \mathcal{L}}{\partial\left(\partial_\mu A_\beta(x)\right)}\partial_\mu A_\beta(x)\right) \ &= \partial_\alpha \left(T_\mu{ }^\beta(x)\partial_\mu A_\beta(x)\right) \ &= \partial_\alpha (\partial_\mu A_\beta(x))T_\mu{ }^\beta(x) \ &= \partial_\mu (\partial

问题 2.

1.2.2. Energy-momentum conservation in the presence of matter
Prove the corollary of ch. $1 \S 2.3$ : In the presence of matter, the energy-momentum tensor obeys the continuity equation
$$
\partial_\nu T_\mu{ }^\nu(x)=\frac{-1}{c} F_\mu{ }^\nu(x) J_\nu(x)
$$

In the presence of matter, the total Lagrangian density $\mathcal{L}_\text{tot}$ is given by

\mathcal{L}_\text{tot} = \mathcal{L}_\text{EM} + \mathcal{L}_\text{matter},Ltot​=LEM​+Lmatter​,

where $\mathcal{L}\text{EM}$ is the Lagrangian density of the electromagnetic field and $\mathcal{L}\text{matter}$ is the Lagrangian density of matter. The energy-momentum tensor for the electromagnetic field and matter is given by

T_\mu{ }^\nu = \frac{\partial \mathcal{L}_\text{tot}}{\partial(\partial_\nu \phi)}\partial_\mu \phi – \delta_\mu^\nu \mathcal{L}_\text{tot},Tμ​ν=∂(∂ν​ϕ)∂Ltot​​∂μ​ϕ−δμν​Ltot​,

where $\phi$ denotes either the electromagnetic field $A_\mu$ or the matter fields $\psi_i$.

Taking the divergence of $T_\mu{ }^\nu$, we have \begin{align*} \partial_\nu T_\mu{ }^\nu &= \frac{\partial}{\partial x^\nu} \left(\frac{\partial \mathcal{L}\text{tot}}{\partial(\partial\nu \phi)}\partial_\mu \phi\right) – \partial_\nu (\delta_\mu^\nu \mathcal{L}\text{tot}) \ &= \frac{\partial}{\partial x^\nu} \left(\frac{\partial \mathcal{L}\text{tot}}{\partial(\partial_\nu \phi)}\right) \partial_\mu \phi + \frac{\partial \mathcal{L}\text{tot}}{\partial(\partial\nu \phi)} \frac{\partial}{\partial x^\nu}(\partial_\mu \phi) – \partial_\mu \mathcal{L}\text{tot} \ &= \frac{\partial}{\partial x^\nu} \left(\frac{\partial \mathcal{L}\text{tot}}{\partial(\partial_\nu \phi)}\right) \partial_\mu \phi – \partial_\mu \mathcal{L}\text{tot} \ &= \partial\nu \left(\frac{\partial \mathcal{L}\text{tot}}{\partial(\partial\nu \phi)}\partial_\mu \phi\right) – \frac{\partial^2 \mathcal{L}\text{tot}}{\partial x^\mu \partial(\partial\nu \phi)} \partial_\nu \phi – \frac{\partial \mathcal{L}\text{tot}}{\partial \phi} \partial\mu \phi \ &= \partial_\nu \left(\frac{\partial \mathcal{L}\text{tot}}{\partial(\partial\nu \phi)}\partial_\mu \phi\right) – \frac{\partial}{\partial x^\mu} \left(\frac{\partial \mathcal{L}\text{tot}}{\partial(\partial\nu \phi)}\partial_\nu \phi\right) – \frac{\partial \mathcal{L}\text{tot}}{\partial \phi} \partial\mu \phi \ &= \partial_\nu \left(\frac{\partial \mathcal{L}\text{tot}}{\partial(\partial\nu \phi)}\partial_\mu \phi\right) – \partial_\mu \left(\frac{\partial \mathcal{L}_\text{

Textbooks


• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

此图像的alt属性为空;文件名为%E7%B2%89%E7%AC%94%E5%AD%97%E6%B5%B7%E6%8A%A5-1024x575-10.png
PHYS322 electrodynamics

Statistics-lab™可以为您提供washington.edu PHYS322 electrodynamics电动力学的代写代考辅导服务! 请认准Statistics-lab™. Statistics-lab™为您的留学生涯保驾护航。