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物理代写|Physics6561 Electrodynamics

Physics6561课程简介

As an undergraduate you learn the language of physics, the laws of physics, and even some physics trivia. And thanks to the skill of your instructors, the subject hums like a well oiled machine. If there’s anything that’s daunting, occasionally, it’s some of the math that comes with the territory. But that’s incidental since you know that, fundamentally, physics is simple and elegant.

But by the time you’re a graduate student you realize that the world doesn’t always seem to play by the rules of the physics laid out in textbooks. There are subtleties, apparent contradictions, paradoxes. You find out that eventually these things got resolved, but when they first appeared on the radar they were called effects. Coming to terms with the effects of physics is a good way to think of the next phase of your education.

Prerequisites

No doubt some effects, such as Aharonov-Bohm, came up in undergraduate physics. The best effects, such as this one, increase your understanding in a deep way, like the vector potential is a more fundamental description of the electromagnetic field. While not all effects are this famous, every one of them provided an important lesson in physics at some point in time. In fact, the name of the effect often becomes a proxy for a general concept, as for example Hanbury-Brown and Twiss effect might be used to describe an experiment involving electrons when in fact the original effect arose in stellar interferometry.

Physics6561 Electrodynamics HELP（EXAM HELP， ONLINE TUTOR）

Because any plane obeys $A x+B y+C z+D$, it is easy to show that the planar surface of the tetrahedon obeys $x+y+z=1$. It is easiest to do the z-integral last, so doing the $\mathrm{x}$-integral we get,
$$\int_0^1 z^2 d z \int_0^{1-z} d y \int_0^{1-y-z} d x=\int_0^1 z^2 d z \int_0^{1-z} d y(1-y-z)$$
Now doing the $\mathrm{y}$ integral and then the z-integral we find,
$$\left.\int_0^1 d z z^2\left((1-z) y-y^2 / 2\right)\right|_0 ^{1-z}=\frac{1}{2} \int_0^1 d z z^2(1-z)^2=\frac{1}{6}-\frac{1}{4}+\frac{1}{10}=\frac{1}{60}$$

We want to demonstrate the divergence theorem $\oint \vec{v} \cdot d \vec{a}=\int(\vec{\nabla} \cdot \vec{v}) d \tau$ for the particular case $\vec{v}=x y \hat{x}+2 y z \hat{y}+3 z x \hat{z}$. First the volume integral where $\vec{\nabla} \cdot \vec{v}=y+2 z+3 x$, so the integral becomes,
$$\int_0^2 d x \int_0^2 d y \int_0^2 d z(y+2 z+3 x)=4\left(\int_0^2 y d y+2 \int_0^2 z d z+3 \int_0^2 x d x\right)=48$$
The surface integral must consider all 6 surfaces of the cube. The surface normals point out of the cube and are $\hat{x},-\hat{x}, \hat{y},-\hat{y}, \hat{z},-\hat{z}$. We also have: $\vec{v} \cdot \hat{x}=x y, \vec{v} \cdot \hat{y}=2 y z, \vec{v} \cdot \hat{z}=3 z x$. So the surface integrals to be considered are, for the surfaces with normals in the $\hat{x}$ and $-\hat{x}$ directions,
$$\left.\int_0^2 d y \int_0^2 d z x y\right|{x=2}=8, \quad \text { and } \quad-\left.\int_0^2 d y \int_0^2 d z x y\right|{x=0}=0$$
and for the surfaces in the $\hat{y}$ and $-\hat{y}$ directions,
$$\left.\int_0^2 d x \int_0^2 d z 2 y z\right|{y=2}=16, \quad \text { and } \quad-\left.\int_0^2 d x \int_0^2 d z 2 y z\right|{y=0}=0,$$
and for the surfaces with normals in the $\hat{z}$ and $-\hat{z}$ directions
$$\left.\int_0^2 d y \int_0^2 d z 3 x z\right|{z=2}=24, \quad \text { and } \quad-\left.\int_0^2 d y \int_0^2 d y 3 x z\right|{z=0}=0,$$
The sum of the surface integrals is 48 as required to satisfy the divergence theorem.

The easiest way to do this problem is to draw the unit vectors and to use geometric reasoning, to find,
\begin{aligned} & \hat{r}=\sin (\theta) \cos (\phi) \hat{x}+\sin (\theta) \sin (\phi) \hat{y}+\cos (\theta) \hat{z}, \ & \hat{\theta}=\cos (\theta) \cos (\phi) \hat{x}+\cos (\theta) \sin (\phi) \hat{y}-\sin (\theta) \hat{z}, \end{aligned}
and
$$\hat{\phi}=-\sin (\phi) \hat{x}+\cos (\phi) \hat{y},$$
It can also be done algebraically, but it does not provide much illumination.

We want to verify the divergence theorem, $\oint \vec{v} \cdot d \vec{a}=\int(\vec{\nabla} \cdot \vec{v}) d \tau$ for a hemisphere in the upper half plane, centered at the origin and of radius $\mathrm{R}$. The function is, $\vec{v}=r \cos (\theta) \hat{r}+r \sin (\theta) \hat{\theta}+r \sin (\theta) \cos (\phi) \hat{\phi}$. First the volume integral where we first evaluate the divergence in spherical polar co-ordinates,
$$\vec{\nabla} \cdot \vec{v}=\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2(r \cos (\theta))\right)+\frac{1}{r \sin (\theta)} \frac{\partial}{\partial \theta}(\sin (\theta)(r \sin (\theta)))+\frac{1}{r \sin (\theta)} \frac{\partial}{\partial \phi}(r \sin (\theta) \cos (\phi))=5 \cos (\theta)-\sin (\phi)$$
The volume integral in spherical polar co-ordinates is then,
$$\int_0^R r^2 d r \int_0^{\pi / 2} \sin (\theta) d \theta \int_0^{2 \pi} d \phi(5 \cos (\theta)-\sin (\phi))$$

The second term is zero as the integral of $\sin (\phi)$ over the interval $[0,2 \pi]$ is zero. The $\phi$ integral gives $2 \pi$ for the first term. Also doing the $R$ integral and using a half angle formula for the $\theta$ term yields,
$$\frac{2 \pi R^3}{3} \frac{5}{2} \int_0^{\pi / 2} d \theta \sin (2 \theta)=\frac{5 \pi R^3}{3}$$
The surface normals are $\hat{r}$ for the spherical part and $-\hat{z}$ for the flat part of the hemisphere. The surface integral for the spherical part is then,
$$\int_0^{2 \pi} d \phi \int_0^{\pi / 2} d \theta R^2 \sin (\theta)(R \cos (\theta))=\pi R^3 .$$
Using $\hat{z}=\hat{r} \cos (\theta)-\hat{\theta} \sin (\theta)$, The integral over the flat part is,
$$\int_0^R \int_0^{2 \pi} r d r d \phi\left(-r \cos ^2(\theta)+r \sin ^2(\theta)\right)=\frac{2 \pi R^3}{3}$$
The sum of the surface contributions gives the volume integral, as required.

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