物理技能 Physics Skills PX271

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The arbitrary Grassmann algebra element $F \in G_{n}$ can be represented in the forms of the finite series
$$F(y)=F_{0}+F_{i_{1}}^{(1)} y_{i_{1}}+F_{i_{1} i_{2}}^{(2)} y_{i_{1}} y_{i_{2}}+\ldots+F_{i_{1} \ldots i_{n}}^{(n)} y_{i_{1}} \ldots y_{i_{n}},$$
where every summation index takes the values from 1 to $n$. Thanks to the conditions this expansion is broken off. As an example, we consider one-dimensional Grassmann algebra $G_{1}$
$${y, y}=0, \quad y^{2}=0 .$$
For any element $F(y) \in G_{1}$, we have
$$F(y)=F_{0}+y F_{1} .$$

PX271COURSE NOTES ：

In the case of a free field of particles with the spin $1 / 2$, the Lagrangian is given by the expression
$$\mathcal{L}{0}=\bar{\psi}(x)\left[i \gamma^{\mu} \partial{\mu}-m\right] \psi(x),$$
where $S^{c}(x-y)$ is the free Green function of the Dirac field
$$S^{c}(x-y)=\frac{1}{(2 \pi)^{4}} \int d^{4} p \frac{\gamma_{\mu} p^{\mu}+m}{p^{2}-m^{2}+i \epsilon} \exp [-i p(x-y)]$$
and $N$ is an inessential normalization factor.

中微子物理学 Neutrino Physics PX435

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$$M=\left(\begin{array}{cc} 0 & m_{D} \ m_{D}^{T} & m_{M} \end{array}\right)$$
where $m_{D}$ is a general $3 \times k$ complex matrix and $m_{M}$ is a $k \times k$ complex symmetric matrix. $M$ is diagonalized according to
$$M_{\text {diag }}=\left(\begin{array}{cc} m_{\nu} & 0 \ & m_{N} \end{array}\right)=U M U^{T}$$
by the (approximate) matrix
$$U \simeq\left(\begin{array}{cc} 1 & -m_{D} m_{M}^{-1} \ m_{D} m_{M}^{-1} & 1 \end{array}\right)$$
The seesaw formula for the effective light neutrino mass matrix becomes
$$m_{\nu} \simeq-m_{D} m_{M}^{-1} m_{D}^{T}$$
(Minus signs can be removed by field rephasings if necessary.)

PX264 COURSE NOTES ：

$$E_{i}^{\prime}=E_{f}^{\prime}$$
has to hold. Thus,
$$E_{f}=E_{i} \gamma^{2}\left(1-u \cos \theta_{i}\right)\left(1+u \cos \theta_{f}^{\prime}\right),$$
$$\frac{\Delta E}{E}=\gamma^{2}\left[1-\frac{1}{\gamma^{2}}+u\left(\cos \theta_{f}^{\prime}-\cos \theta_{i}\right)-u^{2} \cos \theta_{i} \cos \theta_{f}^{\prime}\right]$$

流体物理学 Physics of Fluids PX264

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$\mathbf{v}=\mathbf{0}$ when $r=a$. Assume the solution takes the form $p=C z, C$ constant, $v_{z}=v_{z}(r)$, and $v_{r}=v_{\theta}=0$. obtain
$$C=\mu \Delta v_{z}=\mu\left(\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial v_{z}}{\partial r}\right)\right)$$
Integration yields
$$v_{z}=-\frac{C}{4 \mu} r^{2}+A \log r+B$$
where $A, B$ are constants. Because we require that the solution be bounded, $A$ must be 0 , because $\log r \rightarrow-\infty$ as $r \rightarrow 0$. Use the no-slip condition to determine $B$ and obtain
$$v_{z}=\frac{C}{4 \mu}\left(a^{2}-r^{2}\right)$$

PX264 COURSE NOTES ：

Let the velocity of $\partial D$ be specified as $\mathbf{V}$, so
$$\mathbf{u} \cdot \mathbf{n}=\mathbf{V} \cdot \mathbf{n}$$
Thus, $\varphi$ solves the Neumann problem:
$$\Delta \varphi=0, \quad \frac{\partial \varphi}{\partial n}=\mathbf{V} \cdot \mathbf{n}$$
If $\varphi$ is a solution, then $\mathbf{u}=\operatorname{grad} \varphi$ is a solution of the stationary homogeneous Euler equations, i.e.,
$$\begin{gathered} \rho(\mathbf{u} \cdot \nabla) \mathbf{u}=-\operatorname{grad} p \ \operatorname{div} \mathbf{u}=0 \ \mathbf{u} \cdot \mathbf{n}=\mathbf{V} \cdot \mathbf{n} \quad \text { on } \partial D \end{gathered}$$

核子物理学 Nuclear Physics PX396

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$$s_{1}=\sqrt{1-s_{3}^{2}} \sin \theta, \quad s_{2}=\sqrt{1-s_{3}^{2}} \cos \theta$$
Then the Lagrangian becomes
$$L=\frac{1}{2 g}\left(\partial_{\mu} \vec{s}\right)^{2}=\frac{1}{2 g}\left(\left(1-s_{3}^{2}\right)\left(\partial_{v} \theta\right)^{2}+\frac{\left(\partial_{\mu} s_{3}\right)^{2}}{1-s_{3}^{2}}\right)$$
If we anticipate that, for small $g, s_{3} \sim \mathrm{O}(\sqrt{g})$, we can approximate $L$ by
$$L=\frac{1}{2 g}\left[\left(\partial_{v} s_{3}\right)^{2}+\left(1-s_{3}^{2}\right)\left(\partial_{v} \theta\right)^{2}+s_{3}^{2}\left(\partial_{\mu} s_{3}\right)^{2}+\cdots\right]$$
Finally, it is convenient to rescale $s_{3}$ to eliminate the overall factor of $g, h(r) \equiv$ $s_{3}(r) / \sqrt{g}$, so
$$L=\frac{1}{2}\left[\left(\partial_{v} h\right)^{2}+\left(\frac{1}{g}-h^{2}\right)\left(\partial_{v} \theta\right)^{2}+g h^{2}\left(\partial_{\mu} h\right)^{2}+\cdots\right]$$

PX396 COURSE NOTES ：

$$m=\frac{1}{a} F(g)$$
However, $m$ is a physical quantity that does not depend on the cutoff scheme, so
$$\frac{\mathrm{d}}{\mathrm{d} a} m=0$$
which implies that $a$ and $g$ must be related. We have already seen that the relation is given by asymptotic freedom in the case of the $\mathrm{O}(3)$ sigma model, and now let’s see that this determines the functional dependence of $F(g)$. Substituting the dimensional-analysis statement for $m$ into the last equation implies that
$$-\frac{1}{a^{2}} F(g)+F^{\prime}(g) \frac{\mathrm{d} g}{\mathrm{~d} a}=0$$
but we have determined how $g$ must depend on the ultraviolet cutoff $a$. Let’s use standard and general notation for that result, the Callan-Symanzik relation,
$$a \frac{\mathrm{d} g}{\mathrm{~d} a}=\beta(g)$$

物理学计划 Physics Project PX319

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$$u=-\ln \left(1-x^{\prime}\right)$$
where $x^{\prime}$ is a random number generated for $x^{\prime} \in[0,1]$. With
$$x=r \cos (\theta)=\sqrt{2 u} \cos (\theta),$$
and
$$y=r \sin (\theta)=\sqrt{2 u} \sin (\theta)$$
We can obtain new random numbers $x, y$ through
$$x=\sqrt{-2 \ln \left(1-x^{\prime}\right)} \cos (\theta),$$
and
$$y=\sqrt{-2 \ln \left(1-x^{\prime}\right)} \sin (\theta),$$
with $x^{\prime} \in[0,1]$ and $\theta \in 2 \pi[0,1]$.

PX264 COURSE NOTES ：

Evaluate thereafter
$$I=\int_{a}^{b} F(x) d x=\int_{a}^{b} p(x) \frac{F(x)}{p(x)} d x$$
by rewriting
$$\int_{a}^{b} p(x) \frac{F(x)}{p(x)} d x=\int_{a}^{b} \frac{F(x(y))}{p(x(y))} d y$$
since
$$\frac{d y}{d x}=p(x) .$$
Perform then a Monte Carlo sampling for
$$\int_{a}^{b} \frac{F(x(y))}{p(x(y))} d y, \approx \frac{1}{N} \sum_{i=1}^{N} \frac{F\left(x\left(y_{i}\right)\right)}{p\left(x\left(y_{i}\right)\right)}$$
with $y_{i} \in[0,1]$,

物理学基础 Physics Foundations PX154

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$\mathbf{d p}=\mathbf{F} d t$
Now let us integrate with respect to time;
$$\mathbf{J}=\int_{t_{1}}^{t_{2}} \mathbf{F} d t$$
however;
$$\mathbf{F}=\frac{d \mathbf{p}}{d t}$$
Substituting for $\mathbf{F}$ from in the integral, we obtain:
$$\mathbf{J}=\int_{t_{1}}^{t_{2}} \frac{d \mathbf{p}}{d t} d t=\int_{p_{1}}^{p_{2}} d \mathbf{p}$$

PX154 COURSE NOTES ：

The distribution functions:
Boltzmann:
$$\bar{n}{i}^{\mathrm{B}}=\frac{N{i}}{A_{i}}=\frac{1}{\alpha \exp \left(+\frac{\varepsilon_{i}}{k_{\mathrm{B}} T}\right)}$$
Bose-Einstein:
$$\bar{n}{i}^{\mathrm{BE}}=\frac{N{i}}{A_{i}}=\frac{1}{\alpha \exp \left(+\frac{\varepsilon_{i}}{k_{\mathrm{B}} T}\right)-1}$$
Fermi-Dirac:
$$\bar{n}{i}^{\mathrm{FD}}=\frac{N{i}}{A_{i}}=\frac{1}{\alpha \exp \left(+\frac{\varepsilon_{i}}{k_{\mathrm{B}} T}\right)+1}$$

统计物理学 Statistical Physics PX366

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When we also define a characteristic temperature, $\Theta_{\mathrm{E}}$, the Einstein-temperature, via
$$\Theta_{\mathrm{E}} \equiv \frac{h v_{\mathrm{E}}}{k_{\mathrm{B}}}$$
We get the following expressions for the molar energy of the crystal and it’s specific heat.
and
$$\begin{gathered} \widetilde{E}^{\mathrm{E}}=\frac{3}{2} R \Theta_{\mathrm{E}}+\frac{3 R \Theta_{\mathrm{E}}}{\exp \left(\frac{\Theta_{\mathrm{E}}}{T}\right)-1} \ \widetilde{C}{V}^{\mathrm{E}}=\left(\frac{\partial \widetilde{E}^{\mathrm{E}}}{\partial T}\right){V}=3 R \frac{\left(\frac{\Theta_{\mathrm{E}}}{T}\right)^{2} \exp \left(\frac{\Theta_{\mathrm{E}}}{T}\right)}{\left[\exp \left(\frac{\Theta_{\mathrm{E}}}{T}\right)-1\right]^{2}} \end{gathered}$$

PX366 COURSE NOTES ：

The density of frequencies, $N(v) \mathrm{d} v$, can be deduced from the density of states
$$N(k) \mathrm{d} k=\frac{k^{2}}{2 \pi^{2}} V \mathrm{~d} k \quad \text { with } k=\frac{2 \pi}{\lambda} \quad \text { and } v \lambda=c$$
Hence, with $k=\frac{2 \pi}{c} v$ and $\mathrm{d} k=\frac{2 \pi}{c} \mathrm{~d} v$
$$N(v) \mathrm{d} v=\frac{4 \pi^{2} v^{2}}{2 \pi^{2} c^{2}} V \frac{2 \pi}{c} \mathrm{~d} v=\frac{4 \pi V v^{2}}{c^{3}} \mathrm{~d} v$$

计算物理学 Computational Physics PX277/PX281

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The corresponding polynomials $P$ are
$$L_{k}(x)=\frac{1}{2^{k} k !} \frac{d^{k}}{d x^{k}}\left(x^{2}-1\right)^{k} \quad k=0,1,2, \ldots,$$
which, up to a factor, are the Legendre polynomials $L_{k}$. The latter fulfil the orthorgonality relation
$$\int_{-1}^{1} L_{i}(x) L_{j}(x) d x=\frac{2}{2 i+1} \delta_{i j},$$
and the recursion relation
$$(j+1) L_{j+1}(x)+j L_{j-1}(x)-(2 j+1) x L_{j}(x)=0 .$$
It is common to choose the normalization condition
$$L_{N}(1)=1 .$$

PX277/PX281 COURSE NOTES ：

$$L_{0}(x)=c_{1}$$
with $c$ a constant. Using the normalization equation $L_{0}(1)=1$ we get that
$$L_{0}(x)=1 .$$
For $L_{1}(x)$ we have the general expression
$$L_{1}(x)=a+b x$$
and using the orthorgonality relation
$$\int_{-1}^{1} L_{0}(x) L_{1}(x) d x=0$$
we obtain $a=0$ and with the condition $L_{1}(1)=1$, we obtain $b=1$, yielding
$$L_{1}(x)=x \text {. }$$

科学传播 Scientific Communication MA262-15

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$$T_{x_{i}, x_{l+1}} \cong 2^{b+1}\left|x_{k}-x_{k+1}\right| .$$
For a sinusoidal input, the input levels are given by:
$$x_{k}=a \operatorname{Sin}\left(\omega k T_{s}+\theta\right) ; x_{k+1}=a \operatorname{Sin}\left(\omega(k+1) T_{s}+\theta\right) .$$
where $a$ is the amplitude of the input sinusoid and $\theta$ is the initial phase, assumed random, since it is not synchronized with the sampler in general. The total number of logic transitions from Eq. (16) is then given by:
$$T_{x_{k}, x_{t+1}} \equiv 2^{b+1} a\left|\operatorname{Sin}\left(\omega T_{s}\right)\right| \operatorname{Sin}\left(\omega(k+1 / 2) T_{s}+\theta\right) \mid .$$

MA260-10 COURSE NOTES ：

negative resistance. Therefore, $R_{E q}$ can be approximated as:
$$R_{E q} \approx \frac{-2}{\operatorname{Re}\left[G_{m, e f f}\right]-\operatorname{Re}\left[g_{\pi}\right]}$$
Equation (2) suggests that $R_{E q}$ degrades rapidly as $\operatorname{Re}\left[\mathrm{g}{\pi}\right]$ approaches $\operatorname{Re}\left[\mathrm{G}{\mathrm{m}, \mathrm{eff}}\right]$, and has a distinct corner frequency which will be revisited later. An expression for the negative to positive transition frequency of $R_{E_{q}}\left(\omega_{t r a n}\right)$ can be calculated by considering the condition $\operatorname{Re}\left[G_{m_{,} e f f}\right]-\operatorname{Re}\left[g_{\pi}\right] \leq 0$, which results in:
$$\omega_{t r a n}=\sqrt{\frac{\left(1+\frac{r_{b}}{r_{\pi}}\right)\left(g_{m}-\frac{1}{r_{\pi}}\right)}{r_{b} C_{\pi}^{2}}}$$
Assuming $r_{b} / r_{\pi} \ll 1, r_{\pi}=\beta / g_{m}$ and $\beta \gg 1$ where $\beta$ is the base to collector current gain, then equation (3) can be simplified to:
$$\omega_{\text {tran }} \approx \sqrt{\frac{\omega_{T}}{r_{b} C_{\pi}}}$$

规范、度量和拓扑结构 Norms, Metrics and Topologies MA260-10

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In a two-dimensional electron gas, due to the relativistic effects, an electric field $\mathbf{E}$ is seen by moving electrons as a magnetic field that couples to the spin in the $x y$ plane. This fact is described by the Bychkov-Rashba Hamiltonian
$$\mathrm{H}{R}=\alpha{R}(|E|) \boldsymbol{\sigma} \cdot\left(\boldsymbol{k} | \times \boldsymbol{e}{z}\right),$$ where $\boldsymbol{k}{|}$is the momentum of an electron, $\boldsymbol{\sigma}$ the vector of Pauli matrices, and $\boldsymbol{e}{z}=(0,0,1)$. Thus, the spin degeneracy is lifted and the energy dispersion has the form $$E{\mathrm{RSO}}=\frac{\hbar^{2}}{2 m^{2}}\left(k_{|} \pm \Delta k\right)-E_{\mathrm{SO}}$$

MA260-10 COURSE NOTES ：

$$\left[\frac{d \sigma(\mathrm{z})}{d z}\right]^{2}-\frac{2 \beta_{0}}{d} \sigma^{2}(\mathrm{z})=\frac{2 \beta_{0}}{d} \sigma^{2}(D)$$
with the boundary conditions
$$\left.\frac{d \sigma(\mathrm{z})}{d z}\right|{z=0}=2 \beta{0} \sigma_{0},\left.\quad \frac{d \sigma(\mathrm{z})}{d z}\right|{z=D}=0$$ Here we have defined $\beta{0}=e^{2} /\left(4 \varepsilon_{0} k_{\perp} A_{0}\right)$. One can rewrite Eq. (6.31) in the form
$$\int_{r_{D}}^{1} \frac{d u}{\left(u^{2}-r_{D}^{2}\right)^{\frac{1}{2}}}=\sqrt{\frac{2 \beta_{0}}{d} D}$$
where $r_{D}=\sigma(D) / \sigma(0)$. Therefore, the potential difference $\Delta V(D)$ between the two ends of a sample of thickness $D$ is
$$\Delta V(D)=2 \mathrm{~A}{0} \sigma{0} \sqrt{2 \beta_{0} d} \frac{1-r_{D}}{\left(1-r_{n}^{2}\right)^{1 / 2}}$$