统计代写|PHYS322 electrodynamics
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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.∂iA′i(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.

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