## 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）

1. Consider an infinite parallel-plate capacitor, with the lower plate (at $z=-d / 2$ ) carrying the charge density $-\sigma$, and the upper plate (at $z=+d / 2$ ) carrying the charge density $+\sigma$.
a. Determine all nine elements of the stress tensor, in the region between the plates
b. Determine the force per unit area on the top plate.

a. The stress tensor for an electrostatic field in vacuum is given by: \sigma_{ij}=-\epsilon_0\left(E_iE_j-\frac{1}{2}\delta_{ij}E^2\right)σij​=−ϵ0​(Ei​Ej​−21​δij​E2) where $\sigma_{ij}$ is the stress tensor, $E_i$ is the electric field, $\delta_{ij}$ is the Kronecker delta symbol, and $\epsilon_0$ is the permittivity of free space.

For an infinite parallel-plate capacitor with the lower plate carrying the charge density $-\sigma$ and the upper plate carrying the charge density $+\sigma$, the electric field is constant in the region between the plates and zero outside the plates. Therefore, the stress tensor reduces to:

\sigma_{ij}=-\epsilon_0\left(E_iE_j-\frac{1}{2}\delta_{ij}E^2\right)=-\epsilon_0\frac{1}{2}\delta_{ij}E^2σij​=−ϵ0​(Ei​Ej​−21​δij​E2)=−ϵ0​21​δij​E2

Since the electric field is pointing in the $z$-direction, the only non-zero elements of the stress tensor are:

\sigma_{xx}=\sigma_{yy}=-\frac{1}{2}\epsilon_0E^2σxx​=σyy​=−21​ϵ0​E2 \sigma_{zz}=\epsilon_0E^2σzz​=ϵ0​E2

The off-diagonal elements are zero since there is no shear stress.

b. The force per unit area on the top plate is given by the integral of the normal component of the stress tensor over the surface of the plate:

\mathbf{F}=\int_S \mathbf{\sigma} \cdot \mathbf{\hat{n}} dSF=∫S​σ⋅n^dS

where $\mathbf{\hat{n}}$ is the unit normal vector to the surface $S$.

For the top plate at $z=+d/2$, the unit normal vector is $\mathbf{\hat{n}}=-\mathbf{\hat{z}}$, and the surface element is $dS=dxdy$. Therefore, the force per unit area on the top plate is:

\frac{F}{A}=\frac{\int_{-L/2}^{L/2}\int_{-W/2}^{W/2}\sigma_{zz}dxdy}{LW}=\frac{\sigma_{zz}WL}{LW}=\sigma_{zz}=\epsilon_0E^2AF​=LW∫−L/2L/2​∫−W/2W/2​σzz​dxdy​=LWσzz​WL​=σzz​=ϵ0​E2

where $L$ and $W$ are the dimensions of the plates. Thus, the force per unit area on the top plate is equal to the normal stress on the plate, which is $\epsilon_0E^2$.

1. A piece of wire bent into a loop as shown figure below, carries a current that increases linearly with time: $I(t)=k t$ Calculate the retarded vector potential $\mathbf{A}$ at center. Find the electric field at center. Why does this wire produce an electric field? Or why can’t you determine the magnetic field from this expression for $\mathbf{A}$

To find the vector potential $\mathbf{A}$ at the center of the loop, we can use the Biot-Savart law, which relates the magnetic field $\mathbf{B}$ produced by a current to the vector potential $\mathbf{A}$:

\mathbf{B}(\mathbf{r})=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{r’})\times(\mathbf{r}-\mathbf{r’})}{|\mathbf{r}-\mathbf{r’}|^3}d^3r’B(r)=4πμ0​​∫∣rr′∣3J(r′)×(rr′)​d3r′

where $\mathbf{J}$ is the current density, $\mu_0$ is the permeability of free space, and the integral is taken over the volume of the wire.

For a circular loop of radius $R$ carrying a current $I(t)=kt$, the current density is given by $\mathbf{J}=\frac{I}{\pi R^2}\mathbf{\hat{\phi}}$, where $\mathbf{\hat{\phi}}$ is the unit vector in the azimuthal direction. Thus, the magnetic field at the center of the loop is:

\mathbf{B}(\mathbf{0})=\frac{\mu_0 k}{4\pi R^2}\int_0^{2\pi}\frac{\mathbf{\hat{\phi}}\times(\mathbf{\hat{z}}R)}{R^2}Rd\phi=\frac{\mu_0 k}{2R}\mathbf{\hat{z}}B(0)=4πR2μ0​k​∫02π​R2ϕ^​×(z^R)​Rdϕ=2Rμ0​k​z^

where we have used the fact that $\mathbf{r}-\mathbf{r’}=R\mathbf{\hat{\phi}}$ and $|\mathbf{r}-\mathbf{r’}|=R$ at the center of the loop. Note that the magnetic field is directed in the $z$-direction.

The retarded vector potential $\mathbf{A}$ is related to the magnetic field by:

\mathbf{B}=\nabla\times\mathbf{A}B=∇×A

Since the magnetic field has only a $z$-component, the vector potential must have only an $x$- and $y$-component. Using the expression for the magnetic field, we can solve for the $x$- and $y$-components of the vector potential by taking the curl:

\nabla\times\mathbf{B}=\nabla\times(\nabla\times\mathbf{A})=\nabla(\nabla\cdot\mathbf{A})-\nabla^2\mathbf{A}=\mu_0\frac{\partial\mathbf{B}}{\partial t}=0∇×B=∇×(∇×A)=∇(∇⋅A)−∇2A=μ0​∂t∂B​=0

Since $\nabla\cdot\mathbf{A}=0$ (the vector potential is divergence-free), we have:

\nabla^2\mathbf{A}=-\mu_0\frac{\partial\mathbf{B}}{\partial t}=0∇2A=−μ0​∂t∂B​=0

Thus, the vector potential must satisfy Laplace’s equation. Since the magnetic field is only in the $z$-direction, we can choose the vector potential to have the form:

\mathbf{A}(x,y,0)=\frac{\mu_0 k}{4\pi}\frac{1}{\sqrt{x^2+y^2}}(-y\mathbf{\hat{x}}+x\mathbf{\hat{y}})A(x,y,0)=4πμ0​k​x2+y2​1​(−yx^+xy^​)

where we have used the fact that the current is increasing linearly with time to justify the use of the retarded time

## 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.

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。

## 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）

Coaxial waveguides are often used in particle accelerators. There are several reasons for this;

• The waveguide can receive, transport and transmit electromagnetic waves.
• There is no leakage of waves.
• The fundamental mode is the TEM mode, which propagates for all frequencies.
• Both the phase speed and group speed of the fundamental mode are equal to the speed of light for all frequencies.

The losses in coaxial waveguides comes from the currents running along the outer and inner conductor. The losses decrease with increasing radii of the conductors. If the radii increase they eventually reach values where higher order modes start to propagate.

a) Consider a coaxial waveguide with vacuum between the conductors. Let the inner conductor have radius $a=1 \mathrm{~cm}$ and the outer conductor an inner radius of $b=2.3 \mathrm{~cm}$. Find the frequency where the second mode can start to propagate.
b) Determine if the second mode is a TE- or TM-mode.
Hints: 2D. Electromagnetic waves. Eigenfrequency. Booleans and Partitions. Difference. The fundamental mode does not show up as a resonance in Comsol so you should look for the lowest resonance frequency that is larger than zero.

a) To find the frequency where the second mode can start to propagate, we need to determine the cutoff frequency for the second mode. The cutoff frequency is the frequency below which a mode cannot propagate.

In a coaxial waveguide with vacuum between the conductors, the cutoff frequency for the $n$-th mode is given by:

f_{c,n}=\frac{1}{2\pi\sqrt{\epsilon_r}}\frac{c}{\sqrt{\left(\frac{m\pi}{b}\right)^2+\left(\frac{n\pi}{a}\right)^2}}fc,n​=2πϵr​​1​(bmπ​)2+(anπ​)2​c​

where $\epsilon_r$ is the relative permittivity of the dielectric material between the conductors (in this case, vacuum), $c$ is the speed of light, $a$ is the radius of the inner conductor, $b$ is the inner radius of the outer conductor, $m$ and $n$ are integers that determine the mode.

For the second mode, we have $m=1$ and $n=2$. Plugging in the values, we get:

f_{c,2}=\frac{1}{2\pi\sqrt{1}}\frac{c}{\sqrt{\left(\frac{\pi}{2.3 \mathrm{~cm}}\right)^2+\left(\frac{2\pi}{1 \mathrm{~cm}}\right)^2}}\approx 1.04 \mathrm{~GHz}fc,2​=2π1​1​(2.3 cmπ​)2+(1 cm2π​)2​c​≈1.04 GHz

Therefore, the frequency where the second mode can start to propagate is approximately $1.04 \mathrm{~GHz}$.

b) To determine whether the second mode is a TE- or TM-mode, we need to look at the electric and magnetic fields of the mode.

Consider that you like to use the coaxial waveguide in problem 1 for transmitting signals with frequencies between $100 \mathrm{MHz}$ and $2.2 \mathrm{GHz}$. Close to the transmitter there is a source that generates a signal with frequency $1.6 \mathrm{GHz}$. Also this signal is transmitted and you have to get rid of it. To do this you build a bandstop filter. You cut the coaxial waveguide in two pieces and put a short coaxial waveguide between the two pieces, as in the figure. The extra waveguide also has the radius $1 \mathrm{~cm}$ of the inner conductor. To find the proper radius $c$ of the outer conductor and the height $h$ you use Comsol. The flat surfaces between the outer conductors, marked in the figure, are also made of metal.

Determine $a$ and $h$ such that all frequencies except $1.6 \mathrm{GHz}$ pass. The width of the stop band should be less than $50 \mathrm{MHz}$ at the $3 \mathrm{~dB}$ level.

Hints: Axial symmetry. Type of port: Coaxial. S-parameters. What affects the frequency and what affects the bandwidth?

To design the bandstop filter for the coaxial waveguide, we need to determine the appropriate dimensions of the added short coaxial waveguide. We can use simulation software like Comsol to model the waveguide structure and simulate its response to different frequencies.

To start, we can define the geometry of the coaxial waveguide and the added short waveguide in Comsol, with axial symmetry as indicated in the problem statement. We can then define the material properties of the waveguide, such as the conductivity and permittivity of the metal, and set up a port boundary condition at the input and output of the waveguide.

We can then use the S-parameters of the waveguide to analyze its frequency response. The S-parameters are a set of parameters that describe the relationship between the incident and reflected signals at the input and output of the waveguide. By analyzing the S-parameters, we can determine the frequency and bandwidth of the filter.

To design a bandstop filter, we need to find the dimensions of the added short waveguide that will create a resonant structure at the frequency of the unwanted signal (1.6 GHz), which will cause the signal to be reflected and attenuated. The dimensions of the added waveguide will affect the frequency of the reson

## 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.

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。

## 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）

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

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.

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。

## 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）

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

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.

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。