# 广义相对论 General Relativity PHYS5390M01

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$$g\left(\lambda^{3} \gamma, \lambda \pi_{A^{\prime}}\right)=\lambda^{-1} g\left(\gamma, \pi_{A^{\prime}}\right) \forall \lambda \in C-{0} .$$
This enables one to define the spinor field
$$p^{A} \equiv p^{A A^{\prime} B^{\prime}} \pi_{A^{\prime}} \pi_{B^{\prime}} \quad,$$
and the patching function
$$f^{A} \equiv p^{A} g\left(p_{B} \omega^{B}, \pi_{B^{\prime}}\right),$$
and the function
$$F\left(x^{a}, \pi_{A^{\prime}}\right) \equiv g\left(i p_{A} x^{A C^{\prime}} \pi_{C^{\prime}}, \pi_{A^{\prime}}\right) .$$

## PHYS5390M01COURSE NOTES ：

and such that
$$\psi_{A^{\prime} B^{\prime} C^{\prime}}=\mathcal{D}{A A^{\prime}} \gamma{B^{\prime} C^{\prime}}^{A}$$
The second potential is a spinor field symmetric in its unprimed indices
$$\rho_{C^{\prime}}^{A B}=\rho_{C^{\prime}}^{(A B)}$$
subject to the equation
$$\mathcal{D}^{C C^{\prime}} \rho_{C^{\prime}}^{A B}=0$$
and it yields the $\gamma_{B^{\prime} C^{\prime}}^{A}$ ‘ potential by means of
$$\gamma_{B^{\prime} C^{\prime}}^{A}=\mathcal{D}{B B^{\prime}} \rho{C^{\prime}}^{A B}$$
If we introduce the spinor fields $v_{C^{\prime}}$ and $x^{B}$ obeying the equations
$$\begin{gathered} \mathcal{D}^{A C^{\prime}} \nu_{C^{\prime}}=0 \ \mathcal{D}{A C^{\prime}} \chi^{A}=2 i \nu{C^{\prime}} \end{gathered}$$

# 量子现象 Physics 4- Quantum Phenomena PHYS231101

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To calculate $R$ and $T$ we need to evaluate the current $J$ for $z<0$ and for $z>0$. Using the wavefunction from we have
$$J(z<0)=-\frac{|C|^{2} g \hbar k_{1}}{m^{}}\left(1-|r|^{2}\right)=J_{i}-J_{\mathrm{r}}$$ Similarly, using the wavefunction from eq. $(1.39)$, $$J(z>0)=-\frac{|C|^{2} q \Phi k_{2}}{m^{}}|t|^{2}=J_{1}$$
Hence from
\begin{aligned} &R=J_{1} / J_{1}=|r|^{2} \ &T=J_{3} / J_{1}=|r|^{2} k_{2} / k_{1} \end{aligned}

## PHYS231101COURSE NOTES ：

Continuity Equation: Consider a single electron whose probability density is given
$$n(\mathbf{r}, t)=\Psi *(\mathbf{r}, t) \Psi(\mathbf{r}, t)$$
and whose probability current density is given
$$\mathbf{J}(\mathbf{r}, t)=-\frac{i q h}{2 m *}[(\nabla \Psi r) * \Psi-\Psi *(\nabla \Psi)]$$

# 高能量天体物理学 High Energy Astrophysics PHYS201501

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$$H=\frac{p^{2}}{2 m_{e}}-V(r)$$
Making the usual substitution: $\boldsymbol{p}=-i \hbar \nabla$ we get the relevant form of (114):
$$\left(-\frac{\hbar^{2}}{2 m_{e}} \nabla^{2}-V(r)\right) \psi(\boldsymbol{r})=E \psi(\boldsymbol{r})$$
It is convenient to use atomic units where the natural unit of length is the Bohr radius: $a_{0} \equiv \hbar^{2} / m e^{2}=0.52910^{-8} \mathrm{~cm}$, and the natural unit of energy is twice the Rydberg constant: $e^{2} / a_{0} \equiv 2 R y=27.2 \mathrm{eV}=4.3610^{-11}$ erg. In these units, $e=\hbar=m=1$.
Equation (119) then takes the form:
$$\left(\frac{1}{2} \nabla^{2}+E+V(r)\right) \psi(\boldsymbol{r})=0$$

## PHYS201501COURSE NOTES ：

$$\frac{1}{2} m v^{2}=\frac{Z e^{2}}{2 r}$$
For the ground-state:
$$r \simeq \frac{a_{0}}{Z},$$
and thus:
$$v \simeq\left(\frac{Z^{2} e^{2}}{m a_{0}}\right)^{1 / 2}=(Z \alpha) c$$

# 纳米技术入门 Introduction to Nanotechnology PHYS131101

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This is often written in the abbreviated form,
$$i \hbar \frac{\partial}{\partial t} \psi(\mathbf{r}, t)=\left[-\frac{\hbar^{2}}{2 m} \nabla^{2}+V(\mathbf{r})\right] \psi(\mathbf{r}, t),$$
and the momentum operator $\hat{p}$ itself as
$$\hat{\mathbf{p}}=\frac{\hbar}{i}\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)=\frac{\hbar}{i} \boldsymbol{\nabla} .$$

## PHYS131101COURSE NOTES ：

$$n(\mathbf{r}, t) \propto I(\mathbf{r}, t) \propto|\mathbf{E}(\mathbf{r}, t)|^{2} .$$
Naturally, the total number of photons $N(t)$ at time $t$ is found by integrating over the entire volume $\mathcal{V}$,
$$N(t)=\int_{\mathcal{V}} d \mathbf{r} n(\mathbf{r}, t) .$$
Returning to the quantum wavefunction $\psi(\mathbf{r}, t)$ of a single particle we postulate in analogy withthat the intensity $|\psi(\mathbf{r}, t)|^{2}$ of the wavefunction is related to the particle density $n$ and write
$$n(\mathbf{r}, t) \propto I(\mathbf{r}, t) \propto|\psi(\mathbf{r}, t)|^{2} .$$
But since we have only one particle the analogy reduces to
$$1=\int_{\mathcal{V}} d \mathbf{r} n(\mathbf{r}, t)$$

# 数学 Maths 2 PHYS130001/PHYS238001

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Let $b_{\alpha \beta \mid \sigma}:=\partial_{\sigma} b_{\alpha \beta}-\Gamma_{\alpha \sigma}^{\tau} b_{\tau \beta}-\Gamma_{\beta \sigma}^{\tau} b_{\alpha \tau}$ denote the first-order covariant derivatives of the curvature tensor, defined here by means of its covariant components. Show that these covariant derivatives satisfy the Codazzi-Mainardi identities
$$b_{\alpha \beta \mid \sigma}=b_{\alpha \sigma \mid \beta}$$
which are themselves equivalent to the relations (Thm. 2.8-1)
$$\partial_{\sigma} b_{\alpha \beta}-\partial_{\beta} b_{\alpha \sigma}+\Gamma_{\alpha \beta}^{\tau} b_{\tau \sigma}-\Gamma_{\alpha \sigma}^{\tau} b_{\tau \beta}=0$$
Hint: The proof is analogous to that given in for establishing the relations $\left.b_{\beta}^{\tau}\right|{\alpha}=\left.b{\alpha}^{\tau}\right|_{\beta}$.

## PPHYS130001/PHYS238001COURSE NOTES ：

$u_{i}^{\varepsilon}\left(x^{\varepsilon}\right)=u_{i}(\varepsilon)(x)$ for all $x^{\varepsilon}=\pi^{\varepsilon} x \in \bar{\Omega}^{\varepsilon}$,
where $\pi^{c}\left(x_{1}, x_{2}, x_{3}\right)=\left(x_{1}, x_{2}, \varepsilon x_{3}\right)$. We then assume that there exist constants $\lambda>0, \mu>0$ and functions $f^{i}$ independent of $\varepsilon$ such that
$$\begin{gathered} \lambda^{\varepsilon}=\lambda \text { and } \mu^{\varepsilon}=\mu, \ f^{i, \varepsilon}\left(x^{\varepsilon}\right)=\varepsilon^{p} f^{i}(x) \text { for all } x^{\varepsilon}=\pi^{e} x \in \Omega^{\varepsilon}, \end{gathered}$$

# 震动与热物理 Vibrations & Thermal Phys (JH) PHYS128001

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It can also be written
$$\frac{D \rho}{D t}+\rho \nabla \cdot \mathbf{u}=0$$
where a widely-used abbreviation for the differential operator is
$$\frac{D}{D t} \equiv \frac{\partial}{\partial t}+\mathbf{u} . \nabla$$
An important special case is that of an incompressible fluid in which there is no change of density following the motion of the fluid. If a small volume of fluid is labeled, then the fluid bearing that label always has the same density wherever it is in the future, although its neighbors may have different densities. The idea can be expressed in mathematical form by considering a short interval of time $\delta t$; the idea of incompressibility means that
$$\rho(\mathbf{r}+\mathbf{u} \delta t, t+\delta t)=\rho(\mathbf{r}, t)+O\left(\delta t^{2}\right)$$

## PHYS128001COURSE NOTES ：

$$\frac{\partial S}{\partial t}+\mathbf{u} \cdot \nabla S=\nabla . \kappa \nabla S+q$$
The special case when $\rho, c_{p}$, and $k$ are all constants reduces the equation to
$$\frac{\partial T}{\partial t}+\mathbf{u} . \nabla T=\kappa \nabla^{2} T+\frac{q}{\rho c_{p}}$$
If, in addition, the entire medium is moving with a constant velocity $U$ in the direction of the positive $x$ axis,
$$\frac{\partial T}{\partial t}+U \frac{\partial T}{\partial x}=\kappa \nabla^{2} T+\frac{q}{\rho c_{p}}$$

# 震动与波浪 Vibrations & Waves PHYS125001

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$$A=\frac{F}{2 \pi \sqrt{4 \pi^{2} m^{2}\left(f^{2}-f_{0}^{2}\right)^{2}+b^{2} f^{2}}}$$
Show that the maximum occurs not at $f_{\mathrm{a}}$ but rather at the frequency
$$f_{\text {res }}=\sqrt{f_{o}^{2}-\frac{b^{2}}{8 \pi^{2} m^{2}}}=\sqrt{f_{o}^{2}-\frac{1}{2} \mathrm{FWHM}^{2}}$$

## PHYS125001COURSE NOTES ：

\begin{aligned} a &=F / m \ &=4 T h / \mu w^{2} . \end{aligned}
The time required to move a distance $b$ under constant acceleration $a$ is found by solving $h=\frac{1}{2} a t^{2}$ to yield
\begin{aligned} t &=\sqrt{2 b / a} \ &=w \sqrt{\frac{\mu}{2 T}} . \end{aligned}
Our final result for the velocity of the pulses is
\begin{aligned} |v| &=w / t \ &=\sqrt{\frac{2 T}{\mu}} . \end{aligned}

# 计算 Computing 1 PHYS122001

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A relation $R$ over a set $A$ is said to be complete over $A$ iff for all $a, b \in A$, either $(a, b)$ $\in R$ or $(b, a) \in R$. A poset that is also complete is often called a linear (or total) ordering.

Clearly the relation $\leq$ over $\mathbf{N}$ is complete and so a linear ordering, since for all $m, n \in \mathrm{N}$ either $m \leq n$ or $n \leq m$. So is the usual lexicographic ordering of words in a dictionary.

On the other hand, whenever a set $A$ has more than one element, then the relation $\subseteq$ over $\mathscr{P}(A)$ is not complete. Reason: take any two distinct $a, b \in A$, and consider the singletons ${a},{b}$; they are both elements of $\mathscr{P}(A)$, but neither ${a} \subseteq{b}$ nor ${b} \subseteq{a}$ because $a \neq b$.

## PHYS122001COURSE NOTES ：

Yes: for every $a \in \mathbf{Z}$ there is a unique $b \in \mathbf{Z}$ with $b=|a|$.
No: for two reasons. First, $\operatorname{dom}(R)=\mathbf{N} \subset \mathbf{Z}$. Second, even within this domain, the ‘at most one’ condition is not satisfied, since $a$ can be positive or negative.
Yes: for every $a \in \mathbf{Z}$ there is a unique $b \in \mathbf{Z}$ with $b=a^{2}$.
No: same reasons as for (ii).
Yes: for every $a \in \mathbf{Z}$ there is a unique $b \in \mathbf{Z}$ with $b=a+1$.
Yes: every $x \in \mathbf{Z}$ is of the form $a+1$ for a unique $a \in \mathbf{Z}$, namely for $a=x-1$.

# 物理 Physics 2 PHYS121001/PHYS231009

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\begin{aligned} &F_{x} / m=-\frac{v^{2}}{r} \cos \theta \text {, so } \ &F_{x}=-m \frac{v^{2}}{r} \cos \theta . \end{aligned}
Since our goal is an equation involving the period, it is natural to eliminate the variable $v=$ circumference $/ T=2 \pi r / T$, giving
$$F_{x}=-\frac{4 \pi^{2} m r}{T^{2}} \cos \theta .$$
The quantity $r \cos \theta$ is the same as $x$, so we have
$$F_{x}=-\frac{4 \pi^{2} m}{T^{2}} x$$

## PHYS121001COURSE NOTES ：

$$\frac{F_{r}}{F_{t}}=\frac{2 \pi m}{b f}\left(f^{2}-f_{r e s}^{2}\right)$$
This is the ratio of the wasted force to the useful force, and we see that it becomes zero when the system is driven at resonance.
The amplitude of the vibrations can be found by attacking the equation $|F|=b v=2 \pi b A f$, which gives
$$A=\frac{\left|F_{\mathrm{t}}\right|}{2 \pi b f} .$$
However, we wish to know the amplitude in terms of $|\boldsymbol{F}|$, not $\left|F_{\mathrm{t}}\right|$. From now on, let’s drop the cumbersome magnitude symbols. With the Pythagorean theorem, it is easily proven that
$$F_{t}=\frac{F}{\sqrt{1+\left(\frac{F_{t}}{F_{t}}\right)^{2}}},$$

# 多变量方法 Multivariate Methods MATH5745M01

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$$0=g_{x}^{0}\left(x-x_{0}\right)+g_{y}^{0}\left(\phi(x)-\phi\left(x_{0}\right)\right)+E_{g}\left(x, \phi(x), x_{0}, \phi\left(x_{0}\right)\right)$$
Now divide through by $x-x_{0}$ to get
$$0=g_{x}^{0}+g_{y}^{0}\left(\frac{\phi(x)-\phi\left(x_{0}\right)}{x-x_{0}}\right)+\frac{E_{g}\left(x, \phi(x), x_{0}, \phi\left(x_{0}\right)\right)}{x-x_{0}}$$
Thus,
$$g_{y}^{0}\left(\frac{\phi(x)-\phi\left(x_{0}\right)}{x-x_{0}}\right)=-g_{x}^{0}-\frac{E_{g}\left(x, \phi(x), x_{0}, \phi\left(x_{0}\right)\right)}{x-x_{0}}$$
and assuming $g_{x}^{0} \neq 0$ and $g_{y}^{0} \neq 0$, we can solve to find
$$\frac{\phi(x)-\phi\left(x_{0}\right)}{x-x_{0}}=-\frac{g_{x}^{0}}{g_{y}^{0}}-\frac{1}{g_{y}^{0}} \frac{E_{g}\left(x, \phi(x), x_{0}, \phi\left(x_{0}\right)\right)}{x-x_{0}}$$

Define $h(t)=u\left(x_{0}, t\right)$. The second order Taylor expansion of $h$ is then
$$h(t)=h\left(t_{0}\right)+h^{\prime}\left(t_{0}\right)\left(t-t_{0}\right)+\frac{1}{2} h^{\prime \prime}\left(c_{t}\right)\left(t-t_{0}\right)^{2}$$
where $c_{t}$ is some point between $t_{0}$ and $t$. Using the chain rule for functions of two variables, it is easy to see $h\left(x_{0}\right)=u\left(x_{0}, y_{0}\right), h^{\prime}(t)=u_{t}\left(x_{0}, t\right)$ and $h^{\prime \prime}(t)=u_{t t}\left(x_{0}, t\right)$. Hence, we can rewrite the expansion as
$$u(x, t)=u\left(x_{0}, t_{0}\right)+u_{t}\left(x_{0}, t_{0}\right)\left(t-t_{0}\right)+\frac{1}{2} u_{t t}\left(x_{0}, c_{t}\right)\left(t-t_{0}\right)^{2} .$$
We can then write these another way by letting $\Delta t=t-t_{0}$. This gives
$$u\left(x_{0}, t_{0}+\Delta t\right)=u\left(x_{0}, t_{0}\right)+u_{t}\left(x_{0}, t_{0}\right) \Delta t+\frac{1}{2} u_{t t}\left(x_{0}, c_{t}\right)(\Delta t)^{2} .$$