# 矢量微积分在流体力学中的应用 Vector Calculus With Applications in Fluid Mechanics MATH225

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$$\frac{\partial}{\partial x_{k}}\left(\frac{\partial f_{i}}{\partial x_{j}}\right)(\boldsymbol{x})$$
(which we call second-order derivatives of $f_{i}$ ) all exist and are continuous on $U$ for each $i=1, \ldots, m$ and $j, k=1, \ldots, n$. We denote the above second-order derivative by
$$\frac{\partial^{2} f_{i}}{\partial x_{k} \partial x_{j}}(\boldsymbol{x})$$
One can clearly iterate this process in order to define a function of class $C^{p}$ on $U$. If $f$ is of class $C^{p}$ for every $p \in \mathbb{N}$, then we say that it is of class $C^{\infty}$ on $U$.

## MATH225 COURSE NOTES ：

$$g: U \subset \mathbb{R}^{n} \rightarrow \mathbb{R}$$
be a function of class $C^{1}$ on the open set $U$. Then
$$\mathbf{F}:=\nabla g: U \rightarrow \mathbb{R}^{n}, \mathbf{F}(\boldsymbol{x})=\nabla g(\boldsymbol{x}):=\left(\frac{\partial g}{\partial x_{1}}(\boldsymbol{x}), \ldots, \frac{\partial g}{\partial x_{n}}(\boldsymbol{x})\right)$$
is a vector field and is called the gradient field of $g$.

# 量子和原子物理学 Quantum and Atomic Physics PHYS203

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The integration over $t$ can be performed by means of the following evident relation:
$$\frac{\mathrm{d}}{\mathrm{d} t}\left(\mathrm{e}^{\mathrm{i} H t} \mathrm{e}^{-\mathrm{i} H_{0} t}\right)=\mathrm{ie}^{\mathrm{i} H t}\left(H-H_{0}\right) \mathrm{e}^{-\mathrm{i} H_{0} t}=\mathrm{ie}^{\mathrm{i} H t} V \mathrm{e}^{-\mathrm{i} H_{0} t}$$
which finally yieldswhere $U_{0}(t)$ and $U(t)$ are the evolution operators:
$$\begin{gathered} U(t)=\mathrm{e}^{-i H t} \quad(-\infty<t<+\infty) \ U_{0}(t)=\mathrm{e}^{-i H_{0} t} \quad(-\infty<t<+\infty) \end{gathered}$$
The domain of the definition of the operators $U(t)$ and $U_{0}(t)$ is the entire Hilbert space $\mathcal{H}$. The family of the operators $U(t)$ satisfies the relation
$$U(t) U\left(t^{\prime}\right)=U\left(t+t^{\prime}\right)=U\left(t^{\prime}\right) U(t)$$

## PHYS203 COURSE NOTES ：

$$H_{0}=S^{\dagger} H_{0} S$$
and, therefore, according to equation (7.32b):
$$\bar{E}{f}=\bar{E}{i} \quad \text { (QED). }$$
Here it remains only to provide an explanation for the claim that the quantities $\left\langle\Psi_{i}\left|H_{0}\right| \Psi_{i}\right\rangle$ and $\left\langle\Psi_{f}\left|H_{0}\right| \Psi_{f}\right\rangle$ indeed represent the initial and final energy of the total system. Namely, it is anticipated that the actual total energy $E$ for a certain state $\Psi$ is, in fact, the expected value of the Hamiltonian $H$ :
$$E=\langle\Psi|H| \Psi\rangle .$$
However, since we have $\Psi=\Omega \Psi_{0}$, where $\Psi_{0}$ is the eigenfunction of the operator $H_{0}$, it will be
$$E=\langle\Psi|H| \Psi\rangle=\left\langle\Psi_{0} \mid \Omega^{\dagger} H \Omega \Psi_{0}\right\rangle=\left\langle\Psi_{0}\left|H_{0}\right| \Psi_{0}\right\rangle$$

# 核与粒子物理学 Nuclear and Particle Physics PHYS204

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$$V(r)=-\frac{Z e^{2}}{\rho}\left(\frac{3}{2}-\frac{r^{2}}{2 \rho^{2}}\right)=-\frac{3 Z e^{2}}{2 \rho}+\frac{1}{2}\left(\frac{Z e^{2}}{\rho^{3}}\right) r^{2}$$
The dependence of the potential on $r$ suggests that the $\mu^{-}$may be treated as an isotropic harmonic oscillator of eigenfrequency $\omega=\sqrt{\frac{Z_{e^{2}}}{m_{\mu} \rho^{3}}}$. The energy levels are therefore
$$E_{n}=\hbar \omega\left(n+\frac{3}{2}\right)-\frac{3 Z e^{2}}{2 \rho}$$
where $n=0,1,2, \ldots, \rho \approx 1.2 \times 10^{-13} A^{1 / 3} \mathrm{~cm}$.

## PHYS204COURSE NOTES ：

$$\frac{d E}{d t}=-P$$
i.e.,
$$\frac{e^{2}}{2 r^{2}} \frac{d r}{d t}=-\frac{2 e^{2}}{3 c^{3}} \cdot \frac{e^{4}}{r^{4} M^{2}}$$
Integration gives
$$R^{3}-r^{3}=\frac{4}{c^{3}} \cdot \frac{e^{4}}{M^{2}} t$$
where $R$ is the radius of the initial orbit of the $\mu$ mesion, being
$$R \approx \frac{\hbar^{2}}{m e^{2}}$$
At the $\mu$ ground state the radius of its orbit is the Bohr radius of the mesic atom
$$r_{0}=\frac{\hbar^{2}}{M e^{2}},$$

# 电磁学 Electromagnetism PHYS201

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The special case of a point charge at the origin, for which $\rho=q \delta(\mathbf{r})$ and $\mathbf{E}=q\left(\mathbf{r} / r^{3}\right)$, shows that $\nabla \cdot\left(\mathbf{r} / r^{3}\right)$ acts as if
$$\nabla \cdot \frac{\mathbf{r}}{r^{3}}=4 \pi \delta(\mathbf{r})$$
yields an equation for the electrostatic potential $\phi:$
$$\nabla \cdot \mathbf{E}=-\nabla \cdot \Gamma \phi=4 \pi \rho \quad \text { or } \quad \nabla^{2} \phi=-4 \pi \rho .$$
This is known as Poisson’s equation. In a portion of space where $\rho=0$, becomes
$$\nabla^{2} \phi=0$$

## PHYS201COURSE NOTES ：

$$\psi_{0}(\mathbf{x}, 0)=e^{i \mathbf{k}{0}-\left(\mathbf{x}-\mathbf{x}{0}\right)} h\left(\mathbf{x}-\mathbf{x}{0}\right)+\text { c.c. }$$ where $$h\left(\mathbf{x}-\mathbf{x}{0}\right)=\int d \mathbf{q} a(\mathbf{q}) e^{i \mathbf{q} \cdot\left(\mathbf{x}-\mathbf{x}_{0}\right)}$$

# 凝聚态物质物理学 Condensed Matter Physics PHYS202

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Therefore we have demonstrated how we can always construct a unitary representation by the transformation:
$$\hat{A}{x}=d^{-1 / 2} U^{-1} A{x} U d^{1 / 2}$$
where
\begin{aligned} &H=\sum_{x=1}^{h} A_{x} A_{x}^{\dagger} \ &d=\sum_{x=1}^{h} \hat{A}{x} \hat{A}{x}^{\dagger} \end{aligned}
and where $U$ is the unitary matrix that diagonalizes the Hermitian matrix $H$ and $\hat{A}{x}=U^{-1} A{x} U$.

## PHYS202COURSE NOTES ：

However, if $c=0$ then we cannot write but instead we have to consider $M M^{\dagger}=0$
$$\sum_{k} M_{i k} M_{k j}^{\dagger}=0=\sum_{k} M_{i k} M_{j k}^{}$$ for all $i j$ elements. In particular, for $i=j$ we can write $$\sum_{k} M_{i k} M_{i k}^{}=\sum_{k}\left|M_{i k}\right|^{2}=0$$
Therefore each element $M_{i k}=0$ so that $M$ is a null matrix. This completes proof of the case $\ell_{1}=\ell_{2}$ and $M=\mathcal{O}$.

Finally we prove that for $\ell_{1} \neq \ell_{2}$, then $M=\mathcal{O}$. Suppose that $\ell_{1} \neq \ell_{2}$, then we can arbitrarily take $\ell_{1}<\ell_{2}$. Then $M$ has $\ell_{1}$ columns and $\ell_{2}$ rows. We can make a square $\left(\ell_{2} \times \ell_{2}\right)$ matrix out of $M$ by adding $\left(\ell_{2}-\ell_{1}\right)$ columns of zeros

# 复杂函数 Complex Functions MATH243

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$$\frac{d w}{d z}=\frac{\delta w}{8 x}=\frac{1}{i} \frac{\delta v}{\delta y}$$
follow
und
$$\frac{\delta}{\delta c}\left(\frac{d w}{d z}\right)=\frac{1}{i} \frac{\delta^{2} w}{\delta x \delta y}$$
$$\frac{\delta}{\delta y}\left(\frac{d v}{d z}\right)=\frac{\partial^{2} v}{8 x \delta y}$$
sonsequently
$$\frac{\delta}{\delta y}\left(\frac{d u t}{d z}\right)=i \frac{\delta}{\delta x}\left(\frac{d u t}{d z}\right) \text {, }$$

## MATH243COURSE NOTES ：

$$w^{3}-w+z=0$$
If, for brevity, we put
$$p=\sqrt[3]{\frac{1}{2}\left(-z-\sqrt{z^{2}-\frac{4}{27}}\right)}, q=\sqrt[3]{\frac{1}{2}\left(-z+\sqrt{\left.z^{2}-\frac{4}{27}\right)}\right.},$$
and the two imaginary cube roots of unity
$$\frac{-1+i \sqrt{3}}{2}=\alpha, \frac{-1-i \sqrt{3}}{2}=\alpha^{2}$$
Cardan’s formula gives for the three roots of the above

# 经典力学 Classical Mechanics MATH228

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$$v=c \widehat{\boldsymbol{r}}+(c t) \Omega \widehat{\boldsymbol{\theta}}=c(\widehat{\boldsymbol{r}}+\Omega t \hat{\boldsymbol{\theta}})$$
and
$$\boldsymbol{a}=\left(0-(c t) \Omega^{2}\right) \widehat{\boldsymbol{r}}+(0+2 c \Omega) \widehat{\boldsymbol{\theta}}=c \Omega(-\Omega t \widehat{\boldsymbol{r}}+2 \widehat{\boldsymbol{\theta}}) .$$
The speed of the particle at time $t$ is thus given by $|v|=c\left(1+\Omega^{2} t^{2}\right)^{1 / 2}$. To find the angle between $v$ and $a$, consider
\begin{aligned} v \cdot a &=c^{2} \Omega(-\Omega t+2 \Omega t)=c^{2} \Omega^{2} t \ &>0 \end{aligned}
for $t>0$. Hence, for $t>0$, the angle between $v$ and $a$ is acute.

## MATH228COURSE NOTES ：

Suppose a particle $P$ moves in any manner around the circle $r=b$, where $r, \theta$ are plane polar coordinates. Then the velocity and acceleration vectors of $P$ are given by
\begin{aligned} &v=v \widehat{\theta} \ &a=-\left(\frac{v^{2}}{b}\right) \widehat{\boldsymbol{r}}+\dot{v} \widehat{\boldsymbol{\theta}} \end{aligned}
where $v(=b \dot{\theta})$ is the circumferential velocity of $P$.

# 波浪现象 Wave Phenomena PHYS103

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$$\left(\frac{\partial}{\partial t}+\boldsymbol{u} \cdot \nabla\right) \epsilon-\frac{p}{\rho^{2}}\left(\frac{\partial}{\partial t}+\boldsymbol{u} \cdot \nabla\right) \rho=0$$
In the presence of energy sources or heat transport, the right-hand side of this equation would not be zero. For an ideal gas, with $\epsilon=p /[\rho(\gamma-1)]$, this equation reduces to a particularly useful form:
$$\left(\frac{\partial}{\partial t}+\boldsymbol{u} \cdot \nabla\right) p-c_{s}^{2}\left(\frac{\partial}{\partial t}+\boldsymbol{u} \cdot \nabla\right) \rho=0 .$$

## PHYS103COURSE NOTES ：

$$\nabla \cdot \boldsymbol{E}=4 \pi k_{1} \rho_{c}$$
the absence of magnetic monopoles,
$$\nabla \cdot B=0$$
$$\nabla \times \boldsymbol{E}=-k_{3} \frac{\partial \boldsymbol{B}}{\partial t},$$
and Maxwell’s generalization of Ampere’s Law
$$\nabla \times \boldsymbol{B}=\frac{k_{2}}{k_{1} k_{3}} \frac{\partial \boldsymbol{E}}{\partial t}+4 \pi \frac{k_{2}}{k_{3}} \boldsymbol{J} .$$

# 热物理学和物质属性 Thermal Physics and Properties of Matter  PHYS102

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$$\Delta_{\text {vit }} H=\Delta_{\text {fus }} H\left(T_{\mathrm{g}}\right)=\Delta_{\text {fus }} H\left(T_{\mathrm{m}}\right)+\frac{3}{2} \frac{\Delta_{\text {fus }} H\left(T_{\mathrm{m}}\right)}{T_{\mathrm{m}}}\left(\frac{2}{3} T_{\mathrm{m} \mathrm{}}-T_{\mathrm{m}}\right)$$
or
$$\Delta_{\text {vit }} H=\frac{1}{2} \Delta_{\text {fus }} H\left(T_{\mathrm{m}}\right)$$

## PHYS102COURSE NOTES ：

$$R_{i} \leq \frac{x+y+2(1-x-y)}{1-x-y}=\frac{2-x-y}{1-x-y} \leq R_{i+1}$$
Then the mole fractions, $x\left(\mathrm{Q}^{i}\right)$, of particular $\mathrm{Q}$-units are calculated from the mass balance equations:
$$(1-x-y)\left[x\left(\mathrm{Q}^{i}\right)(i / 2+4-i)+x\left(\mathrm{Q}^{i+1}\right)(i / 2+7 / 2-i)\right]=2-x-y$$
$$x\left(Q^{i}\right)+x\left(Q^{i+1}\right)=1$$
Unfortunately, the above approach can be used for the rough orientation purposes only. In fact, Q-units disproportionate and the Q-distribution is given by the equilibrium of the disproportionation reactions of the type:
$$2 \mathrm{Q}^{n} \leftrightarrow \mathrm{Q}^{n+1}+\mathrm{Q}^{n-1}, \quad n=1,2,3$$

# 热物理学和物质属性 Thermal Physics and Properties of Matter  PHYS102

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$$\Delta_{\text {vit }} H=\Delta_{\text {fus }} H\left(T_{\mathrm{g}}\right)=\Delta_{\text {fus }} H\left(T_{\mathrm{m}}\right)+\frac{3}{2} \frac{\Delta_{\text {fus }} H\left(T_{\mathrm{m}}\right)}{T_{\mathrm{m}}}\left(\frac{2}{3} T_{\mathrm{m} \mathrm{}}-T_{\mathrm{m}}\right)$$
or
$$\Delta_{\text {vit }} H=\frac{1}{2} \Delta_{\text {fus }} H\left(T_{\mathrm{m}}\right)$$

## PHYS102COURSE NOTES ：

$$R_{i} \leq \frac{x+y+2(1-x-y)}{1-x-y}=\frac{2-x-y}{1-x-y} \leq R_{i+1}$$
Then the mole fractions, $x\left(\mathrm{Q}^{i}\right)$, of particular $\mathrm{Q}$-units are calculated from the mass balance equations:
$$(1-x-y)\left[x\left(\mathrm{Q}^{i}\right)(i / 2+4-i)+x\left(\mathrm{Q}^{i+1}\right)(i / 2+7 / 2-i)\right]=2-x-y$$
$$x\left(Q^{i}\right)+x\left(Q^{i+1}\right)=1$$
Unfortunately, the above approach can be used for the rough orientation purposes only. In fact, Q-units disproportionate and the Q-distribution is given by the equilibrium of the disproportionation reactions of the type:
$$2 \mathrm{Q}^{n} \leftrightarrow \mathrm{Q}^{n+1}+\mathrm{Q}^{n-1}, \quad n=1,2,3$$