MY-ASSIGNMENTEXPERT™可以为您提供cornell.edu Physics6561 Electrodynamics电动力学课程的代写代考和辅导服务！
这是康乃尔大学电动力学课程的代写成功案例。
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）
For convenience we place the cube with one corner at the origin and the cube in the positive quadrant. Then the vectors of the two body diagonals are $\vec{b}_1=(1,1,1)$ and $\vec{b}_2=(-1,1,1)$. The angle between the two vectors is found from the definition of the dot product,
$$
\vec{b}_1 \cdot \vec{b}_2=\left|\vec{b}_1\right|\left|\overrightarrow{b_2}\right| \cos (\theta) \text { so that } 1=3 \cos (\theta)
$$
The two vectors $\vec{v}_1=(0,2,0)-(1,0,0)=(-1,2,0)$ and $\vec{v}_2=(0,0,3)-(1,0,0)=(-1,0,3)$ define the plane of the surface of interest. The cross product of these two vectors, $\vec{v}_1 \wedge \vec{v}_2=(6,-3,2)$, is normal to the plane of interest. The unit normal of interest is then $(6 / 7,-3 / 7,2 / 7)$.
In these calculations we can usually just look at one component and then note that the other components have a similar behavior, with the appropriate change of variables. a) The gradient operator is $\vec{\nabla}=$ $\hat{x} \partial_x+\hat{y} \partial_y+\hat{z} \partial_z$ and $r^2=x^2+y^2+z^2$. Since $\partial_x\left(r^2\right)=2 x$, with similar expressions for the other components, it is clear that $\vec{\nabla}\left(r^2\right)=2 \vec{r}=2 r \hat{r}$.
b) We have,
$$
\vec{\nabla}\left(\frac{1}{r}\right)=\left(\hat{x} \partial_x+\hat{y} \partial_y+\hat{z} \partial_z\right)\left(\frac{1}{\left(x^2+y^2+z^2\right)^{1 / 2}}\right)
$$
Since $\partial_x(1 / r)=-x / r^3$, it is clear that $\vec{\nabla}\left(\frac{1}{r}\right)=-\vec{r} / r^3=-\hat{r} / r^2$. c) Using a similar procedure to that used in b) it can be shown that $\vec{\nabla} r^n=n r^{n-1} \hat{r}$
In these proofs we can do the calculation for one component and then extend to the other components. For the first quotient formula we have,
$$
\vec{\nabla}\left(\frac{f}{g}\right)=\hat{x} \partial_x\left(\frac{f}{g}\right)+\hat{y} \partial_y\left(\frac{f}{g}\right)+\hat{z} \partial_z\left(\frac{f}{g}\right)
$$
For the first component we have,
$$
\frac{\partial}{\partial x}\left(\frac{f}{g}\right)=\frac{g \frac{\partial f}{\partial x}-f \frac{\partial g}{\partial x}}{g^2}=\left(\frac{1}{g} \vec{\nabla} f-\frac{f}{g^2} \vec{\nabla} g\right)_x
$$
The other components proceed in the same way, completing the proof. The second quotient formula is
$$
\vec{\nabla} \cdot\left(\frac{\vec{A}}{g}\right)=\partial_x\left(\frac{A_x}{g}\right)+\partial_y\left(\frac{A_y}{g}\right)+\partial_z\left(\frac{A_z}{g}\right)
$$
The first term gives,
$$
\frac{\partial}{\partial x}\left(\frac{A_x}{g}\right)=\frac{g \frac{\partial A_x}{\partial x}-A_x \frac{\partial g}{\partial x}}{g^2}
$$
Similar terms occur for the second term, using $x \rightarrow y$, and third term, using $x \rightarrow z$. It is then evident that this is equivalent to $\vec{\nabla} \cdot \vec{A} / g-\vec{A} \cdot(\vec{\nabla} g) / g^2$ as required. The third quotient rule considers,
$$
\vec{\nabla} \wedge\left(\frac{\vec{A}}{g}\right)=\left|\begin{array}{ccc}
\hat{x} & \hat{y} & \hat{z} \
A_x & A_y & A_z \
B_x & B_y & B_z
\end{array}\right|=\hat{x}\left(\partial_y\left(\frac{A_z}{g}\right)-\partial_z\left(\frac{A_y}{g}\right)\right)-\hat{y}\left(\partial_x\left(\frac{A_z}{g}\right)-\partial_z\left(\frac{A_x}{g}\right)\right)+\hat{z}\left(\partial_x\left(\frac{A_y}{g}\right)-\partial_y\left(\frac{A_x}{g}\right)\right)
$$
The $\mathrm{x}$-component of this expression gives,
$$
\frac{g \frac{\partial A_z}{\partial y}-A_z \frac{\partial g}{\partial y}}{g^2}-\frac{g \frac{\partial A_y}{\partial z}-A_y \frac{\partial g}{\partial z}}{g^2}=\left(\frac{1}{g}(\vec{\nabla} \wedge \vec{A})+\frac{1}{g^2} \vec{A} \wedge(\vec{\nabla} g)\right)_x
$$
The other components follow in the same way.
MY-ASSIGNMENTEXPERT™可以为您提供CORNELL.EDU PHYSICS6561 ELECTRODYNAMICS电动力学课程的代写代考和辅导服务！
