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Vector Analysis 2

Thu, 2021-09-16
Vector Analysis 2
本作品采用知识共享署名-非商业性使用-相同方式共享 4.0 国际许可协议进行许可。

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):


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: \(\int (\nabla \cdot \bold{A}) dV = \int_{\partial V} \bold{A} \cdot d\bold{S}\)

From the \(\int \nabla \cdot \bold{A} dV = \int \bold{A} \cdot d\bold{S}\), we get \(\nabla \cdot \bold{A} = \frac{1}{dV}(\bold{A} \cdot d\bold{S})\)

  1. \[ \nabla \cdot \bold{A} dV \equiv \nabla \cdot \bold{A} h_1 h_2 h_3 dr d\theta d\phi \]
  2. \[ \bold{A} \cdot d\bold{S} = \Delta_r(A_r dl_\theta dl_\phi) + \Delta_\theta(A_\theta dl_r dl_\phi) + \Delta_\phi (A_\phi dl_\theta dl_r) \]

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:


  1. \[ \nabla \times A \cdot d\bold{S}_r = \nabla \times A \ h_2h_3 d\theta d\phi \ \hat{\theta} \times \hat{\phi} =(\nabla \times A)_r \ h_2h_3 d\theta d\phi \]

  2. \[ \bold{A}_r \cdot d\bold{l}_r = - \Delta_\phi(A_\theta h_2d\theta) + \Delta_\theta(A_\phi h_3 d\phi) \]

\[ (\nabla \times \bold{A})_r = \frac{\bold{A}\cdot d\bold{l}}{dS}=- \frac{\Delta_\phi(A_\theta h_2) }{h_2h_3d\phi} + \frac{\Delta_\theta(A_\phi h_3)}{h_2h_3d\theta} \]

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): \[ \begin{aligned} \nabla \cdot \bold{v} &= \frac{\partial }{ \partial x} \left(\frac{x}{r^3}\right) + \frac{\partial }{ \partial y} \left(\frac{y}{r^3}\right) + \frac{\partial }{ \partial z} \left(\frac{z}{r^3}\right)\\ &=\frac{\partial }{ \partial x} \left[x(x^2+y^2+z^2)^{-\frac{3}{2}}\right] +\frac{\partial }{ \partial y} \left[y(x^2+y^2+z^2)^{-\frac{3}{2}}\right] +\frac{\partial }{ \partial z} \left[z(x^2+y^2+z^2)^{-\frac{3}{2}}\right]\\ &=()^{-\frac{3}{2}} + x(-\frac{3}{2})()^{-\frac{5}{2}}2x +()^{-\frac{3}{2}} + y(-\frac{3}{2})()^{-\frac{5}{2}}2y +()^{-\frac{3}{2}} + z(-\frac{3}{2})()^{-\frac{5}{2}}2z\\ &=3r^{-3}-3r^{-5}(x^2+y^2+z^2)=3r^{-3}-3r^{-3}=0 \end{aligned} \]


\[ \nabla \cdot \bold{v} = \frac{1}{r^2} \frac{\partial }{ \partial r} \left(r^2 \frac{1}{r^2}\right) = 0 \]


\[ \oint \bold{v} \cdot d\bold{S} = \int_{0}^{R} \left(\frac{1}{r^2}\hat{r}\right) \cdot (R^2 \sin \theta d \theta d\phi \hat{r}) = 4\pi \] 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 \(\frac{0}{0}\)).

\[ d\Omega \equiv \sin\theta d \theta d\phi \]

Define Dirac function


\[ \delta (x-a) = \begin{cases} 0, &\text{ if } x\neq a \\ \infty &\text{ if }x = a \\ \end{cases} \text{ and} \int_{-\infty}^{+\infty} \delta(x-a)dx=1 \]


\[ \nabla \cdot \left(\frac{1}{r^2}\hat{\bold{r}}\right) \equiv \frac{1}{4\pi}\delta(r) \]


\[ \int_{-\infty}^{+\infty} f(x) \delta(x-a) dx =f(a) \] \[ \delta(kx) = \frac{1}{|k|} \delta(x) \]

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:\(\bold{F}=\bold{0}\) \(F=yz \hat{\bold{x}} + zx \hat{\bold{y}} + xy \hat{\bold{z}}\) \(F=\sin x\cosh y \hat{\bold{x}} - \cos x \sinh \hat{\bold{y}}\)


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(\(V\)):

\[ \nabla \times \bold{F} = 0\Leftrightarrow \bold{F} = - \nabla V. \]

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): \[ \nabla \cdot \bold{F} =0 \Leftrightarrow \bold{F} = \nabla \times \bold{A} \]

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

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: \[ \nabla \cdot r =\nabla \cdot r = 3 \]

\[ \nabla(r \cdot r) = 2x\hat{x} +2y \hat{y} +2z \hat{z} = 2 r \]

\[ \nabla p \cdot r = \nabla (p_1x + p_2y+p_3z) = p_1 \]