Q1: Derive the formula for the volume of a sphere
\[ \begin{aligned} V &=\int dV \\ &= \int r^2\sin\theta dr d\theta d\phi\\ &=\frac{1}{3} r^3|_{0}^{r} \times (-\cos\theta)|_{0}^{\pi}\times 2\pi\\ &=\frac{4}{3}r^3 \end{aligned} \]
\(r^2\sin\theta\): it can also be derivative form Jacobi formula.
Curvilinear Coordinates
Spherical polar coordinates (SPC):
review
- 4 types integrate
- Curvilinear Coordinates
- \(dl\), \(dl_\theta\), \(dl_\phi\) 对球坐标和柱坐标
Derivatives in SPC
1. Gradient in the SPC
Definition of the gradient:
\[ df \equiv \frac{\partial f}{ \partial r} dr+ \frac{\partial f}{ \partial \theta} d\theta + \frac{\partial f}{ \partial \phi } d\phi \]
We define \(\nabla f\) as
\[ df = \nabla f \cdot d\bold{l} = \nabla f \cdot ( dl_r \hat{r} + dl_\theta \hat{\theta} + dl_\phi \hat{\phi}) \]
because \(dl\) defined as \[ d\bold{l} \equiv dl_r \hat{r} + dl_\theta \hat{\theta} + dl_\phi \hat{\phi} = \hat{r} dr + \hat{\theta}\sin\theta d\theta + \hat{\phi} r\sin\theta d\phi \]
To meet the definition of \(df\),
\[ \nabla f = \frac{\partial f}{ \partial r} \hat{r}+ \frac{1}{r}\frac{\partial f}{ \partial \theta} \hat{\theta} + \frac{1}{r\sin\theta} \frac{\partial f}{ \partial \phi} \hat{\phi}\\ = \frac{1}{h_1} dr + \frac{1}{h_2}d\theta + \frac{1}{h_3}d\phi \]
2. Divergence in the SPC
To get \[ \nabla \cdot \bold{A} \]
We need start from: \(\int (\nabla \cdot \bold{A}) dV = \int_{\partial V} \bold{A} \cdot d\bold{S}\)
\(dS_z=\hat{x}dx \times \hat{y} dy=dxdy\hat{z}\)
So for a small cubic: \(dS=dl_\phi dl_\theta\hat{r}+dl_r dl_\phi \hat{\theta} + dl_r dl_\theta\phi\)
From the \(\int \nabla \cdot \bold{A} dV = \int \bold{A} \cdot d\bold{S}\), we get \(\nabla \cdot \bold{A} = \frac{1}{dV}(\bold{A} \cdot d\bold{S})\)
- \[ \nabla \cdot \bold{A} dV \equiv \nabla \cdot \bold{A} h_1 h_2 h_3 dr d\theta d\phi \]
- \[ \bold{A} \cdot d\bold{S} = \Delta_r(A_r dl_\theta dl_\phi) + \Delta_\theta(A_\theta dl_r dl_\phi) + \Delta_\phi (A_\phi dl_\theta dl_r) \]
- 第一项是穿过\(\theta \phi\) 面的通量
Then \[ \nabla \cdot \bold{A} = \frac{1}{h_1h_2h_3} \left[\frac{\partial (A_r h_2 h_3)}{ \partial r} +\frac{\partial (A_\theta h_1 h_3)}{ \partial \theta} + \frac{\partial (A_\phi h_1h_2)}{ \partial \phi} \right] \]
Eg. \[ \frac{1}{h_1h_2h_3} \frac{\partial A_rh_2h_3}{ \partial r} = \frac{1}{r^2\sin\theta} \frac{\partial \frac{1}{r^2} r^2\sin\theta}{ \partial r} =\frac{1}{r^2} \frac{\partial A_r r^2}{ \partial r} \]
3. Curl in the SPC
矢量场中一个方向的分量和另外两个坐标有关，和标量场不同。比方说，沿着\(y\) 轴向上，只有\(y\) 在变，而此时，\(x\) 方向的电场会变小。
To get \(\nabla \times \bold{A}\), we start from
\[ \int (\nabla \times \bold{A}) \cdot d\bold{S} = \oint_{\partial S} \bold{A} \cdot d\bold{l} \]
我们先只写 \(r\) 方向的分量：
\[ \nabla \times A \cdot d\bold{S}_r = \nabla \times A \ h_2h_3 d\theta d\phi \ \hat{\theta} \times \hat{\phi} =(\nabla \times A)_r \ h_2h_3 d\theta d\phi \]
\[ \bold{A}_r \cdot d\bold{l}_r = - \Delta_\phi(A_\theta h_2d\theta) + \Delta_\theta(A_\phi h_3 d\phi) \]
\[ (\nabla \times \bold{A})_r = \frac{\bold{A}\cdot d\bold{l}}{dS}=- \frac{\Delta_\phi(A_\theta h_2) }{h_2h_3d\phi} + \frac{\Delta_\theta(A_\phi h_3)}{h_2h_3d\theta} \]
- here the \(dS=h_2h_3 d\theta d\phi\), which is a scalar.
So, curl in SPC
\[ \nabla \times \bold{A} = \frac{1}{h_1h_2h_3} \begin{pmatrix} h_1\hat{r} & h_2\hat{\theta} & h_3\hat{\phi}\\ \partial_r & \partial_\theta & \partial_\phi\\ h_1A_r & h_2A_\theta & h_3A_\phi \end{pmatrix} \]
Q2: Find the formula of curl in the spherical polar coordinate \[ \nabla \times A = \]
Dirac Delta Function
Consider the vector function \(v=\frac{1}{r^2} \hat{r}\), from the very begin (problem 1.16): \[ \begin{aligned} \nabla \cdot \bold{v} &= \frac{\partial }{ \partial x} \left(\frac{x}{r^3}\right) + \frac{\partial }{ \partial y} \left(\frac{y}{r^3}\right) + \frac{\partial }{ \partial z} \left(\frac{z}{r^3}\right)\\ &=\frac{\partial }{ \partial x} \left[x(x^2+y^2+z^2)^{-\frac{3}{2}}\right] +\frac{\partial }{ \partial y} \left[y(x^2+y^2+z^2)^{-\frac{3}{2}}\right] +\frac{\partial }{ \partial z} \left[z(x^2+y^2+z^2)^{-\frac{3}{2}}\right]\\ &=()^{-\frac{3}{2}} + x(-\frac{3}{2})()^{-\frac{5}{2}}2x +()^{-\frac{3}{2}} + y(-\frac{3}{2})()^{-\frac{5}{2}}2y +()^{-\frac{3}{2}} + z(-\frac{3}{2})()^{-\frac{5}{2}}2z\\ &=3r^{-3}-3r^{-5}(x^2+y^2+z^2)=3r^{-3}-3r^{-3}=0 \end{aligned} \]
or
\[ \nabla \cdot \bold{v} = \frac{1}{r^2} \frac{\partial }{ \partial r} \left(r^2 \frac{1}{r^2}\right) = 0 \]
But
\[ \oint \bold{v} \cdot d\bold{S} = \int_{0}^{R} \left(\frac{1}{r^2}\hat{r}\right) \cdot (R^2 \sin \theta d \theta d\phi \hat{r}) = 4\pi \] where, \[ d\bold{S}=dl_\theta \hat{\theta} \times dl_\phi \hat{\phi} = R d\theta R \sin \theta d\phi \hat{r} \]
The contradiction comes from the point \(r = 0\), where \(\frac{1}{r^2}\) blows up. So the point \(r = 0\) contributes to the integration, giving the \(4\pi\) result.
The Answer is that \(\nabla \cdot \bold{v} =0\) everywhere except at the origin. The calculation of \(\frac{\partial }{ \partial x} \left(\frac{x}{r^2}\right)\) at origin doesn’t work. (or \(\frac{\partial }{ \partial r} \left(r^2 \frac{1}{r^2}\right)\) not work at origin \(\frac{0}{0}\)).
- 空间角
\[ d\Omega \equiv \sin\theta d \theta d\phi \]
- 来源：当r取R时，上面那个式子没有\(r\)了，就变成\(\int d\Omega\)了，或者说这个式子本身和\(r\) 无关，或者说可以用在求flux上。
Define Dirac function
1D:
\[ \delta (x-a) = \begin{cases} 0, &\text{ if } x\neq a \\ \infty &\text{ if }x = a \\ \end{cases} \text{ and} \int_{-\infty}^{+\infty} \delta(x-a)dx=1 \]
3D
\[ \nabla \cdot \left(\frac{1}{r^2}\hat{\bold{r}}\right) \equiv \frac{1}{4\pi}\delta(r) \]
性质：
\[ \int_{-\infty}^{+\infty} f(x) \delta(x-a) dx =f(a) \] \[ \delta(kx) = \frac{1}{|k|} \delta(x) \]
LS变瘦了,原来的\(1\)，现在\(\frac{1}{k}\) 就够了
\[ \int_{a}^{b} f(x) \delta[ g(x) - g(x_0) ] dx = f(x_j) / \left|\frac{dg}{dx}\right||_{x=x_j} \]
The Theory of Vector Fields
Helmholtz’s Theorem:
Maxwell reduce the \(E\) and \(B\) into four equations, and this four equations are all about the divergence and the curl of \(\bold{E}\) and \(\bold{B}\). So here raise an important mathematical question: To what extent is a vector function determined by its divergence and curl?(一个向量函数在多大程度上是由它的散度和卷度决定的), 即给一个向量函数的散度和旋度，是否能得到这个向量函数。
事实上，是不行的，需要一个边界条件，Helmholtz theorem 告诉说给一些边界条件之后可以唯一确定这个函数。
数学的表示： the divergence of \(\bold{F}\) (which stands for \(\bold{E}\) or \(\bold{B}\)) is a specified (scalar) function D, \[ \nabla \cdot \bold{F} = D \]
and the curl of \(\bold{F}\) is a specified (vector) function \(\bold{C}\),
\[ \nabla \times \bold{F}=\bold{C} \]
当然 \(\bold{C}\) must be divergenceless, \(\nabla \cdot \bold{C}=0\)，旋度的散度为\(0\)
Eg: divergence =0 , curl = 0, 的解：\(\bold{F}=\bold{0}\) \(F=yz \hat{\bold{x}} + zx \hat{\bold{y}} + xy \hat{\bold{z}}\) \(F=\sin x\cosh y \hat{\bold{x}} - \cos x \sinh \hat{\bold{y}}\)
Potentials
Curl-less field
If the curl of a vector field(\(\bold{F}\)) vanishes(everywhere), then \(\bold{F}\) can be written as gradient of a Scalar potential(\(V\)):
\[ \nabla \times \bold{F} = 0\Leftrightarrow \bold{F} = - \nabla V. \]
- Theorem 1 Curl-less
(“irrotational”) fields:(保守场)
- \(\nabla \times \bold{F} = 0\) everywhere
- \(\int_a^{b} \bold{F} \cdot d\bold{l}\) is independent of path, for any given end points.
- \(\oint\bold{F} \cdot d\bold{l} =0\) for any closed loop
- \(\bold{F}\) is gradient of some scalar, \(\bold{F} = -\nabla V\)(+ any constant)
Divergence-less field
If the divergence of a vector field (\(\bold{F}\)) vanishes(everywhere), then \(\bold{F}\) can be expressed as the curl of Vector potential(A): \[ \nabla \cdot \bold{F} =0 \Leftrightarrow \bold{F} = \nabla \times \bold{A} \]
- Theorem 2 Divergence-less (or
“solenoidal”) fields:
- \(\nabla \cdot \bold{F} = 0\) everywhere
- \(\int \bold{F} \cdot d\bold{a}\) is independent of surface, for any given bound.
- \(\oint\bold{F} \cdot d\bold{a} =0\) for any closed surface
- \(\bold{F}\) is curl of some scalar, \(\bold{F} = -\nabla \times \bold{A}\)(+gradient of any scalar function)
Any vector field \(\bold{F}\) can be written as gradient of a scalar of a scalar plus the curl of a vector: \[ \bold{F} = -\nabla V + \nabla \times A \]
Any vector field resolved into curl-free vector field and divergence-free vector field
- curl free: \(\nabla \times (\nabla \phi) = 0\)
- divergence free: \(\nabla \cdot (\nabla \times \vec{A}) = 0\)
- \(\nabla \cdot F= D\) and \(\nabla \times F = C\)
- physical: \(\nabla \cdot (\nabla \phi) =\nabla \cdot E\) and \(\nabla \times A = \vec{B}\)
Q3: In the following, \(p\) is a constant vector, and \(r=x\hat{x}+y\hat{y}+z\hat{z}\) \[ \begin{aligned} \nabla \cdot r\\ \nabla(r \cdot r)\\ \end{aligned} \] and \[ \begin{aligned} \nabla p \cdot r\\ \end{aligned} \]
Solve: \[ \nabla \cdot r =\nabla \cdot r = 3 \]
\[ \nabla(r \cdot r) = 2x\hat{x} +2y \hat{y} +2z \hat{z} = 2 r \]
\[ \nabla p \cdot r = \nabla (p_1x + p_2y+p_3z) = p_1 \]