Mathematical Methods for Physics and Engineering
1. Some conclusions:
(1) Ratio theorem:
Line segment AB is divided into AP and PB. AP:PB = \lambda:\mu. Then \mathbf{OP}=\frac{\mu}{\lambda+\mu} \mathbf{OA}+\frac{\lambda}{\lambda+\mu} \mathbf{OB}.
(2) Centroid G of a triangle:
\mathbf{g}=\frac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c}).
(3) The cosine of the angle \theta between a and b:
\cos\theta=\frac{a_{x}}a\frac{b_{x}}b+\frac{a_{y}}a\frac{b_{y}}b+\frac{a_{z}}a\frac{b_{z}}b.
(4) Complex vector is possible:
\mathbf{a}\cdot\mathbf{b}=a_x^*b_x+a_y^*b_y+a_z^*b_z and \mathbf{a}\cdot\mathbf{b}=(\mathbf{b}\cdot\mathbf{a})^{*}.
(5) Vector product is non-associative:
(\mathbf{a}\times\mathbf{b})\times\mathbf{c}\neq\mathbf{a}\times(\mathbf{b}\times\mathbf{c}).
(6) Scalar triple product:
a. [\mathbf{a},\mathbf{b},\mathbf{c}]\equiv\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})\equiv(\mathbf{a}\times\mathbf{b})\cdot\mathbf{c}\equiv|\mathbf{a}||\mathbf{ b}||\mathbf{ c}| sin\theta cos\phi
b. Especially for real vector: \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})\equiv\mathbf{c}\cdot(\mathbf{a}\times\mathbf{b}).
(7) Vector triple product:
a. \mathbf{a\times(b\times c)=(a\cdot c)b-(a\cdot b)c}
b. \mathbf{(a\times b)\times c=(a\cdot c)b-(b\cdot c)a}
c. \mathbf{a\times(b\times c)+b\times(c\times a)+c\times(a\times b)=0}
(8) Lagrange's identity:
(\mathbf{a\times b})\cdot(\mathbf{c\times d})\equiv(\mathbf{a\cdot c})(\mathbf{b\cdot d})-(\mathbf{a\cdot d})(\mathbf{b\cdot c}).
2. Definition of basis vectors:
The basis vectors must satisfy two conditions:
(1) The number of basis vectors is the same as the dimensions.
(2) The basis vectors must be linear independent, namely:
c_1\mathbf{e}_1+c_2\mathbf{e}_2+\cdots+c_N\mathbf{e}_N\neq\mathbf{0} except c_1=c_2=\cdots=c_N=0.
3. Equations of lines, planes and spheres:
(1) Lines:
a. \mathbf{r=a+\lambda b}: \mathbf{a} is its start point. By manipulating \mathbf{\lambda}, it becomes a line having the same direction as \mathbf{b}.
b. \frac{x-a_{x}}{b_{x}}=\frac{y-a_{y}}{b_{y}}=\frac{z-a_{z}}{b_{z}}=\mathrm{\lambda}: This is deduced from a.
c. (\mathbf{r-a})\times\mathbf{b}=\mathbf{0}: This is deduced from a. \mathbf{a} is its start point, and it has the same direction as \mathbf{b}.
d. \mathbf{r}=\mathbf{a}+\lambda(\mathbf{c}-\mathbf{a}): This is deduced from a. It passes two fixed points \mathbf{a} and \mathbf{c}.
e. \mathbf{r}=\mu\mathbf{a}+\lambda\mathbf{c},\mu+\lambda=1: This is deduced from d. It passes two fixed points \mathbf{a} and \mathbf{c}.
(2) Planes:
a. (\mathbf{r-a})\cdot\mathbf{\hat{n}}=0: point \mathbf{a} and a direction vector \mathbf{\hat{n}} decided this plane.
b. lx+my+nz=d: This is deduced from a. \mathbf{r}=(x,y,z) and \mathbf{\hat{n}}:=(l,m,n). Be aware of the distance d between this plane and origin O.
c. \mathbf{r=a+\lambda(b-a)+\mu(c-a)}: This is deduced from a. It passes three fixed points \mathbf{a}, \mathbf{b} and \mathbf{c}.
d. \mathbf{r}=\alpha\mathbf{a}+\beta\mathbf{b}+\gamma\mathbf{c},\alpha+\beta+\gamma=1: This is deduced from c. It passes three fixed points \mathbf{a}, \mathbf{b} and \mathbf{c}.
(3) Spheres:
a. |\mathbf{r-c}|^2=a^2.
Example:
Find the radius \rho of the circle that is the intersection of the plane \mathbf{\hat{n}}\cdot\mathbf{r}=p and the sphere
of radius a centered on the point with position vector \mathbf{c}. Answer is on the page 229.
4. Distance:
(1) Distance from a point to a line:
d=|(\mathbf{p}-\mathbf{a})\times\mathbf{\hat{b}}| where \mathbf{\hat{b}} is an unit direction vector of this line and \mathbf{a} is a point on this line.
(2) Distance from a point to a plane:
d=(\mathbf{a}-\mathbf{p})\cdot\mathbf{\hat{n}} where \mathbf{\hat{n}} is an unit direction vector of this plane and \mathbf{a} is a point on this plane.
(3) Distance from a line to a line:
d=|(\mathbf{p}-\mathbf{q})\cdot\mathbf{\hat{n}}| where \mathbf{\hat{n}} is an unit direction vector(\hat{\mathbf{n}}=\mathbf{\hat{a}}\times\mathbf{\hat{b}}, \mathbf{\hat{a}}, \mathbf{\hat{b}} are unit direction vectors of two lines) and \mathbf{p}, \mathbf{q} is two points on two lines.
(4) Distance from a line to a plane:
d=|(\mathbf{a}-\mathbf{r})\cdot\mathbf{\hat{n}}| where \mathbf{a} is a point on this line and \mathbf{r} is a point on this plane. This is valid only when \mathbf{b}\cdot\mathbf{\hat{n}}=0(This line is parallel with the plane).
5. Reciprocal vectors:
Definition:
The two sets of vectors \mathbf{a},\mathbf{b},\mathbf{c} and\mathbf{a}^{\prime},\mathbf{b}^{\prime},\mathbf{c}^{\prime} are called reciprocal sets if \mathbf{a}\cdot\mathbf{a}^{\prime}=\mathbf{b}\cdot\mathbf{b}^{\prime}=\mathbf{c}\cdot\mathbf{c}^{\prime}=1
and \mathbf{a^{\prime}}\cdot\mathbf{b}=\mathbf{a^{\prime}}\cdot\mathbf{c}=\mathbf{b^{\prime}}\cdot\mathbf{a}=\mathbf{b^{\prime}}\cdot\mathbf{c}=\mathbf{c^{\prime}}\cdot\mathbf{a}=\mathbf{c^{\prime}}\cdot\mathbf{b}=0.
How to construct:
\mathbf{a}^{\prime}=\frac{\mathbf{b}\times\mathbf{c}}{\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})},\\ \mathbf{b}^{\prime}=\frac{\mathbf{c}\times\mathbf{a}}{\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})},\\ \mathbf{c}^{\prime}=\frac{\mathbf{a}\times\mathbf{b}}{\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})}
where \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})\neq0(not coplanar).
Rewrite a vector using reciprocal vectors:
\mathbf{a}=(\mathbf{a}\cdot\mathbf{e}_1')\mathbf{e}_1+(\mathbf{a}\cdot\mathbf{e}_2')\mathbf{e}_2+(\mathbf{a}\cdot\mathbf{e}_3')\mathbf{e}_3. Especially for Cartesian basis vectors \mathbf{i, j, k}: \mathbf{a}=(\mathbf{a}\cdot\mathbf{i})\mathbf{i}+(\mathbf{a}\cdot\mathbf{j})\mathbf{j}+(\mathbf{a}\cdot\mathbf{k})\mathbf{k}.
6. Vector analysis:
(1) Cylinder coordinate:
\begin{aligned} &\left\{\begin{array}{c}x=\rho cos\phi\\y=\rho sin\phi\\z=z\end{array}\right. \\ &\left.\left\{\begin{array}{c}\rho=\sqrt{x^2+y^2}\\\phi=arctan(\frac yx)\\z=z\end{array}\right.\right. \\ &dS=\rho d\phi dz \\ &dV=\rho d\rho d\phi dz \end{aligned}
(2) Circular coordinate:
\left\{\begin{array}{c}x=r sin\theta cos\phi\\y=r sin\theta sin\phi\\z=r cos\theta\end{array}\right.\\ \left\{\begin{array}{c}r=\sqrt{x^2+y^2+z^2}\\\theta=arctan(\frac{\sqrt{x^2+y^2}}z)\\\phi=arctan(\frac{\tilde{y}}x)\end{array}\right.\\ dS=r^2 sin\theta d\theta d\phi \\ dV=r^2 sin\theta dr d\theta d\phi
(3) Coordinate transition:
\begin{gathered} \begin{bmatrix}A_\rho\\A_\phi\\A_z\end{bmatrix}=\begin{bmatrix}cos\phi&sin\phi&0\\-sin\phi&cos\phi&0\\0&0&1\end{bmatrix}\begin{bmatrix}A_x\\A_y\\A_z\end{bmatrix} \\ \begin{bmatrix}A_r\\A_\theta\\A_\phi\end{bmatrix}=\begin{bmatrix}sin\theta cos\phi&sin\theta sin\phi&cos\theta\\cos\theta cos\phi&cos\theta sin\phi&-sin\theta\\-sin\phi&cos\phi&0\end{bmatrix}\begin{bmatrix}A_x\\A_y\\A_z\end{bmatrix} \\ \begin{bmatrix}A_r\\A_\theta\\A_\phi\end{bmatrix}=\begin{bmatrix}sin\theta&0&cos\theta\\cos\theta&0&-sin\phi\\0&1&0\end{bmatrix}\begin{bmatrix}A_\rho\\A_\phi\\A_z\end{bmatrix} \end{gathered}
(4) Gradient, curl, divergence:
\begin{gathered} \nabla f=\frac{\partial f}{\partial x}\vec{e_x}+\frac{\partial f}{\partial y}\vec{e_y}+\frac{\partial f}{\partial z}\vec{e_z} \\ =\frac{\partial f}{\partial\rho}\vec{e_\rho}+\frac1\rho\frac{\partial f}{\partial\phi}\vec{e_\phi}+\frac{\partial f}{\partial z}\vec{e_z} \\ =\frac{\partial f}{\partial r}\vec{e_r}+\frac1r\frac{\partial f}{\partial\theta}\vec{e_\theta}+\frac1{rsin\theta}\frac{\partial f}{\partial\phi}\vec{e_\phi} \end{gathered}
\nabla\times\vec{\boldsymbol{B}}=\begin{vmatrix}\vec{e_x}&\vec{e_y}&\vec{e_z}\\\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\\\vec{B_x}&\vec{B_y}&\vec{B_z}\end{vmatrix}\\ =(\frac{\partial\mathrm{B_z}}{\partial\mathrm{y}} - \frac{\partial\mathrm{B_y}}{\partial\mathrm{z}} )\overrightarrow{\mathrm{e_x}} +(\frac{\partial\mathrm{B_x}}{\partial\mathrm{z}} - \frac{\partial\mathrm{B_z}}{\partial\mathrm{x}} )\overrightarrow{\mathrm{e_y}} +(\frac{\partial\mathrm{B_y}}{\partial\mathrm{x}} - \frac{\partial\mathrm{B_x}}{\partial\mathrm{y}} )\overrightarrow{\mathrm{e_z}}\\ =\frac1\rho\begin{vmatrix}\vec{e_\rho}&\rho\vec{e_\phi}&\vec{e_z}\\\\\frac\partial{\partial\rho}&\frac\partial{\partial\phi}&\frac\partial{\partial z}\\\\\vec{B_\rho}&\rho\vec{B_\phi}&\vec{B_z}\end{vmatrix} =\frac1{r^2sin\theta}\begin{vmatrix}\vec{e_r}&r\vec{e_\theta}&rsin\theta\vec{e_\phi}\\\\\frac\partial{\partial r}&\frac\partial{\partial\theta}&\frac\partial{\partial\phi}\\\\\vec{B_r}&r\vec{B_\theta}&rsin\theta\vec{B_\phi}\end{vmatrix}
\begin{gathered}\nabla\cdot\vec{\boldsymbol{A}}=\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\\\\=\frac1\rho\frac{\partial(\rho A_\rho)}{\partial\rho}+\frac1\rho\frac{\partial(A_\phi)}{\partial\phi}+\frac{\partial(A_z)}{\partial z}\end{gathered}\\ =\frac{1}{\rho^{2}}\frac{\partial(\rho^{2}A_{\rho})}{\partial \rho}+\frac{1}{\rho\sin\theta}\frac{\partial(\sin\theta A_{\theta})}{\partial\theta}+\frac{1}{r\sin\theta}\frac{\partial A_{\phi}}{\partial\phi}
(5) Laplace operator:
\begin{gathered} \nabla^2f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2} \\ =\frac1\rho\frac\partial{\partial\rho}(\rho\frac{\partial f}{\partial\rho})+\frac1{\rho^2}\frac{\partial^2f}{\partial\phi^2}+\frac{\partial^2f}{\partial z^2} \end{gathered}\\ ={\frac{1}{\rho^{2}}}{\frac{\partial}{\partial \rho}}\left(\rho^{2}{\frac{\partial f}{\partial \rho}}\right)+{\frac{1}{\rho^{2}\sin\theta}}{\frac{\partial}{\partial\theta}}\left(\sin\theta{\frac{\partial f}{\partial\theta}}\right)+{\frac{1}{\rho^{2}\sin^{2}\theta}}{\frac{\partial^{2}f}{\partial\phi^{2}}}
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