# 电动力学、狭义和广义相对论|PHYS3100 Electrodynamics, Special and General Relativity UWA代写

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\begin{aligned} M_{o} c^{2} &=E^{\prime}+Q \ 0 &=p^{\prime}-Q / c, \end{aligned}
where $E^{\prime}$ and $p^{\prime}$ label the relativistic energy and momentum of the final (unexcited) atom of mass $M_{a}^{\prime}$. Let us say that we detect the photon and measure its energy $Q$. So, we need to eliminate $E^{\prime}$ and $p^{\prime}$ . A slick way to do this is to use the energy-momentum relation $E^{\prime 2}=p^{2} c^{2}+M_{o}^{2} c^{4}$. Rearranging gives
$$\begin{gathered} E^{\prime}=M_{o} c^{2}-Q \ p^{\prime}=Q / c \end{gathered}$$
$\mathrm{SO}$,
$$E^{\prime 2}-p^{\prime 2} c^{2}=\left(M_{o} c^{2}-Q\right)^{2}-Q^{2}=M_{o}^{\prime 2} c^{4}$$
Finally,
$$M_{o}^{2} c^{4}-2 M_{o} c^{2} Q=M_{0}^{22} c^{4}$$

## PHYS3100 COURSE NOTES ：

Let us write this in terms of the energy difference between the initial and final atoms, taken at rest,
$$\Delta E \equiv M_{0} c^{2}-M_{0}^{\prime} c^{2} .$$
We focus on $\Delta E$ because this would be the energy difference between discrete quantum energy levels of the atom. One could calculate such a difference from first principles in quantum mechanics. Solving for $M_{a}^{\prime} c^{2}$,
$$M_{0} c^{2}-\Delta E=M_{0}^{\prime} c^{2}$$
Squaring gives
$$M_{0}^{2} c^{4}-2 M_{0} c^{2} \Delta E+(\Delta E)^{2}=M_{0}^{\prime 2} c^{4} .$$
Combining this with Eq. (6.41) gives
$$Q=\Delta E\left(1-\frac{\Delta E}{2 M_{0} c^{2}}\right)$$

# 现代和前沿物理学|PHYS1200 Modern and Frontier Physics UWA代写

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A particle is moving with a speed of modulus $v$ and components $\left(v_{x}, v_{y}, v_{z}\right)$. What is the modulus $v^{\prime}$ of the velocity for an observer moving with a speed $w$ along the $x$ axis? Comment the result as $v$ and/or $w$ approach $c$.
Answer: Applying the relativistic laws for the addition of velocities we find
$$v^{\prime 2}=v_{x}^{\prime 2}+v_{y}^{\prime 2}+v_{z}^{\prime 2}=\frac{\left(v_{x}-w\right)^{2}+\left(1-w^{2} / c^{2}\right)\left(v_{x}^{2}+v_{y}^{2}\right)}{\left(1-v_{x} w / c^{2}\right)^{2}}$$
and after some simple algebra
$$v^{\prime 2}=c^{2}\left(1-\frac{\left(1-v^{2} / c^{2}\right)\left(1-w^{2} / c^{2}\right)}{\left(1-v_{x} w / c^{2}\right)^{2}}\right)$$
It is interesting to notice that as $v$ and/or $w$ approach $c$, also $v^{\prime}$ approaches $c$ (from below): that verifies that a body moving with $v=e$ moves with the same velocity in every reference frame (invariance of the speed of light).

## PHYS1200 COURSE NOTES ：

To derive the equation of motion, let us notice that, in the frame instantaneously at rest with the particle, the velocity goes from 0 to adt in the interval $d \tau$, so that $v$ changes into $(v+a d \tau) /\left(1+\operatorname{vad} \tau / c^{2}\right)$ in the same interval, meaning that $d v / d \tau=a\left(1-v^{2} / c^{2}\right)$. Using previous equations, it is easy to derive
$$\frac{d u}{d t}=\frac{d u}{d v} \frac{d v}{d \tau} \frac{d \tau}{d t}=a$$
which can immediately be integrated, with the initial condition $u(0)=v(0)=0$, as $u(t)=a t$. Using the relation between $u$ and $v$ we have
$$v(t)=\frac{a t}{\sqrt{1+a^{2} t^{2} / c^{2}}} .$$
That gives the variation of velocity, as observed in the laboratory, for a uniformly accelerated motion: for $t \ll c / a$ one recovers the non-relativistic result, while for $t \gg c / a$ the velocity reaches asymptotically that of light. The dependence of $v$ on $t$ can be finally integrated, using the initial condition $x(0)=0$, giving
$$x(t)=\frac{c^{2}}{a}\left(\sqrt{1+\frac{a^{2} t^{2}}{c^{2}}}-1\right) .$$

# 古典和前沿物理学|PHYS1100 Classical and Frontier Physics UWA代写

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$$y_{j}(t)=\sum_{i=1}^{n} \int_{-\infty}^{+\infty} h_{i j}(\tau) x_{i}(t-\tau) d \tau$$
The cross correlation between outputs $y_{j}(t)$ and $y_{k}(t)$ is defined as follows:
$$C_{j k}(\tau)=\lim {T \rightarrow \infty} \frac{1}{2 T} \int{-T}^{+T} y_{j}(t) y_{k}(t+\tau) d t$$
From the definition of $C_{j k}(\tau)$ it is seen that:
$$C_{k j}(\tau)=\left\langle y_{k}(t) y_{j}(t+\tau)\right\rangle$$
where:
$$\langle()\rangle=\lim {T \rightarrow \infty} \frac{1}{2 T} \int{-T}^{+T}() d t$$

## PHYS1100 COURSE NOTES ：

The next step is true only under the condition that the process is ergodic. In this case, the last equation can be written as:
\begin{aligned} G_{j k}(\omega)=& \lim {T \rightarrow \infty} \frac{1}{2 T}\left(\int{-T}^{+T} y_{j}(u) e^{i \omega u} d u\right) \ & \times\left(\int_{-T}^{+T} y_{k}(v) e^{-i \omega v} d v\right) \end{aligned}
This last relation can then be written:
$$G_{j k}(\omega)=\lim {T \rightarrow \infty} \frac{\bar{y}{j}^{}(T, \omega) \bar{y}{k}(T, \omega)}{2 T}$$ where: $$\bar{y}{j}(T, \omega)=\int_{-T}^{+T} y_{j}(t) e^{i \omega t} d t$$
and $y_{j}^{}$ is the complex conjugate of $y_{j}$.
$$\bar{y}{k}(T, \omega)=\int{-T}^{+T} y_{k}(t) e^{-i \omega t} d t$$

# 当代物理学专题|PHYS3012 Topics in Contemporary Physics UWA代写

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which is equivalent to
$$\frac{d^{3} \hat{\gamma}}{d t^{3}} \mp \sqrt{-1} \kappa_{2} \frac{d^{2} \hat{\gamma}}{d t^{2}}+\left(\kappa_{1}^{2}+1\right) \frac{d \hat{\gamma}}{d t} \mp \sqrt{-1} \kappa_{2} \hat{\gamma}=0$$
where the double signs take the same signature. Since its characteristic equation
$$\lambda^{3} \mp \sqrt{-1} \kappa_{2} \lambda^{2}+\left(\kappa_{1}^{2}+1\right) \lambda \mp \sqrt{-1} \kappa_{2}=0$$
should have pure imaginary solutions, by setting $\lambda=\sqrt{-1} \Lambda$ we have
$$\Lambda^{3} \mp \kappa_{2} \Lambda^{2}-\left(\kappa_{1}^{2}+1\right) \Lambda \pm \kappa_{2}=0,$$
which turns to
$$\left(\Lambda \mp \frac{\kappa_{2}}{3}\right)^{3}-\frac{1}{3}\left(3 \kappa_{1}^{2}+\kappa_{2}^{2}+3\right)\left(\Lambda \mp \frac{\kappa_{2}}{3}\right) \mp \frac{\kappa_{2}}{27}\left(9 \kappa_{1}^{2}+2 \kappa_{2}^{2}-18\right)=0 .$$

## PHYS3012 COURSE NOTES ：

Finally, in (c) is supposed that non-autonomus vector field $\xi_{h}^{t}$ on $M$ is integrable up to time one, i.e. there exists a family of curves $h \rightarrow f_{h}^{t}$ in the diffeomorphism group of $M$ with $f_{0}^{t}=i d$ and
$$\frac{d}{d h} f_{h}^{t}=\xi_{h}^{t} \circ f_{h^{}}^{t}$$ Then since Lie derivative $\mathcal{L}{\xi{h}^{t}}$ satisfies $\mathcal{L}{\xi{h}^{t}}=d i_{\xi_{h}^{t}}+i_{\xi_{h}^{t}} d$ and the forms $\omega_{t}^{h}$ are closed, one concludes that
$$\frac{d}{d h}\left(\left(f_{h}^{t}\right)^{} \omega_{t}^{h}\right)=\left(f_{h}^{t}\right)^{}\left(\mathcal{L}{\xi{h}^{t}} \omega_{t}^{h}+\frac{d}{d h} \omega_{t}^{h}\right)=\left(f_{h}^{t}\right)^{}\left(d i_{\xi_{h}} \omega_{t}^{h}+\sigma_{t}\right)=0$$
for all $h \in[0,1]$. This shows that $f^{t}:=f_{h}^{t}$ fulfills the requirement $\left(f^{t}\right)^{} \tilde{\omega}{t}=$ $\tilde{\omega}{t}^{a}$ since
$$\left(f^{t}\right)^{} \tilde{\omega}{t}-\tilde{\omega}{t}^{a}=\left(f_{1}^{t}\right)^{} \tilde{\omega}{t}^{1}-\left(f{0}^{t}\right)^{} \tilde{\omega}{t}^{0}=\int{0}^{1} \frac{d}{d h}\left(\left(f_{h}^{t}\right)^{*} \omega_{t}^{h}\right) d h=0 .$$
Putting the parameter $t=1$ and $\varepsilon \rightarrow 0$ we obtain the existing part of the theorem.

# 物理数学|PHYS3011 Mathematical Physics UWA代写

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From the last equation, we get, by differentiation with respect to $\beta_{s}^{i}(t), i \geq 2$, and restricting the result to $\gamma$ (i.e., putting $t=t_{0}$ ), the equation
$$\sum_{k=2}^{\operatorname{dim}{R} M} \Gamma{(j k)}^{i}(\gamma(s))\left[x^{k}(\gamma(s))-f^{k}(g(\xi))\right]=0$$
for $i \geq 1, j \geq 2$ and every $\gamma(s)$. Hence
$$\Gamma_{(j k)}^{i}(\gamma(s))=0 \quad \text { for } i \geq 1, j, k \geq 2 \text {, and } s \in J_{U^{*}}^{\prime}$$
Moreover, by construction
$$0=\left.\nabla_{\dot{\gamma}} E_{i}^{\prime}\right|{\gamma(s)}=\dot{\gamma}^{j}(s) \nabla{E_{j}^{\prime}} E_{i}^{\prime}=\left.\dot{\gamma}^{j}(s) \Gamma^{k}{ }{i j}(\gamma(s)) E{k}^{\prime}\right|_{\gamma(s)}$$

## PHYS3011 COURSE NOTES ：

\begin{aligned} &x^{1}(q)=x^{1}(p)+\ln \left(1+y^{1}(q)\right) \ &x^{i}(q)=x^{i}(p)+y^{i}(q) \quad \text { for } i \geq 2 \end{aligned}
from which the Riemannian normal coordinates can be expressed as
\begin{aligned} &y^{1}(q)=\exp \left(x^{1}(q)-x^{1}(p)\right)-1 \ &y^{i}(q)=x^{i}(q)-x^{i}(p) \quad \text { for } i \geq 2 . \end{aligned}

# 量子计算|PHYS3005 Quantum Computation UWA代写

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The sudden removal of the shutter marks the beginning of a “quantum race” where the particles run along the positive $x$-axis. In order to elucidate the spreading of the signal all one has to do is to calculate $\psi(x, t)$ starting with
$$\psi_{0}(x)=\psi(x, t=0)=\Theta(-x) \mathrm{e}^{\mathrm{i} / x}$$
Using one finds
$$\langle x \mid \psi(t \geq 0)\rangle=M(x ; k ; h t / m)$$
where the Moshinsky function $\mathrm{M}$ is defined in terms of the complementary error function ,
$$M(x ; k ; \tau)=\frac{1}{2} \exp \left(\mathrm{i} k x-\mathrm{i} k^{2} \tau / 2\right) \operatorname{erfc}\left[\frac{x-k \tau}{(2 \mathrm{i} \tau)^{1 / 2}}\right]$$
with $\mathrm{i}^{1 / 2}=\exp (\mathrm{i} \pi / 4)$.
An interesting property of is revealed when we evaluate the particle number probability. Introducing $u=(h k t / m-x) /(\pi h t / m)^{1 / 2}$, we obtain

## PHYS3005 COURSE NOTES ：

$$\nabla \cdot \mathbf{j}(\mathbf{r})=-\frac{2}{h} \mathfrak{S}\left[\boldsymbol{\sigma}(\mathbf{r})^{} \psi_{\mathrm{sc}}(\mathbf{r})\right]$$ where $\mathfrak{S}[x]$ stands for the imaginary part of $x$. Thus, the inhomogeneity $\sigma(\mathbf{r})$ acts as a source for the particle current $j(r)$. By integration over the source volume, and inserting we obtain a bilinear expression for the total particle current $J(E)$, i. e., the total scattering rate: $$J(E)=-\frac{2}{h} \mathfrak{J}\left[\int \mathrm{d}^{3} r \int \mathrm{d}^{3} r^{\prime} \sigma(\mathbf{r})^{} G\left(\mathbf{r}, \mathbf{r}^{\prime} ; E\right) \sigma\left(\mathbf{r}^{\prime}\right)\right]$$
Some important identities concerning the total current $J(E)$ are most easily recognized in a formal Dirac bra-ket representation. In view we may express $J(E)$ by
$$J(E)=-\frac{2}{h} S[\langle\sigma|G| \sigma\rangle]=\frac{2 \pi}{\hbar}\langle\sigma|\delta(E-H)| \sigma\rangle,$$

# 天体物理学和空间科学|PHYS3003 Astrophysics and Space Science UWA代写

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with $\sigma(M)$ and $\rho(M)$ given. Thus, subclumps of mass $f M$ having mean density $f^{1-3 \alpha} \rho(M)$ typically collide at velocity $\sigma(M)$ (i.e. the internal velocity dispersion of their parent clump) and the resulting shock-compressed layer typically fragments into subsubclunps of mass $f^{2} M$ having internal velocity dispersion $f^{2 \beta} \sigma(M)$. We can therefore substitute $m_{\text {FRAG }} \rightarrow f^{2} M$, $\sigma_{\mathrm{s}} \rightarrow f^{2 \beta} \sigma(M), \rho_{\mathrm{O}} \rightarrow f^{1-3 \alpha} \rho(M)$ and $v_{\mathrm{O}} \rightarrow \sigma(M)$ in Eqn. 1, to obtain
$$f^{14 \beta+3 \alpha-5}\left(\frac{M}{M_{\bullet}}\right)^{6 \beta+3 \alpha-3}=\frac{G^{3} M_{*}^{2} \rho_{\bullet}}{\sigma_{\bullet}}$$
Since this must be true for any $f$ and any $M$, both the exponents on the lefthand side of Eqn. 5 must be zero, which in combination with gives
FOR TURBULENT HIGH-MASS CLUMPS:
$\alpha \simeq \frac{1}{2}$
$$\beta \simeq \frac{1}{4}, \quad \gamma \simeq \frac{7}{4}$$

## PHYS3003 COURSE NOTES ：

If we now require that $t_{\text {COUPLE }} \lesssim t_{\mathrm{FF}}=\left(3 \pi / 32 G \rho_{\mathrm{s}}\right)^{1 / 2}$, we obtain
$$P_{\mathrm{S}} \equiv \rho_{\mathrm{s}} a_{\mathrm{s}}^{2} \gtrsim P_{\mathrm{COUPLE}} \simeq \frac{64 G}{3}\left[\frac{r_{\mathrm{D}} \rho_{\mathrm{D}}}{Z_{\mathrm{D}}}\right]^{2} \simeq 10^{5} \mathrm{~cm}^{-3} \mathrm{~K} k_{\mathrm{B}}$$
where $k_{\mathrm{B}}$ is the Boltzmann constant. We emphasize two important aspects of this result. (i) Since in our model of dynamically triggered star formation $P_{\mathrm{S}}=\rho_{\mathrm{s}} a_{\mathrm{s}}^{2}=P_{\text {RMM }}=\rho_{\mathrm{o}} v_{\mathrm{O}}^{2}$, this is in effect a threshold for the ram pressure which will trigger rapid star formation in a turbulent cloud. (ii) $10^{5} \mathrm{~cm}^{-3} \mathrm{~K} k_{\mathrm{B}}$ is typical of the turbulent pressures found in star-forming molecular clouds; for a self-gravitating cloud, it corresponds to a critical surface-density
$$\Sigma_{\text {COUPLE }} \simeq\left(\frac{P_{\text {COUPLE }}}{G}\right)^{1 / 2} \simeq 1.4 \times 10^{-2} \mathrm{~g} \mathrm{~cm}^{-2},$$
or equivalently to a critical column-density $N_{\mathrm{H}{2}} \simeq 4 \times 10^{21} \mathrm{H}{2} \mathrm{~cm}^{-2}$.

# 电动力学和相对论|PHYS3002 Electrodynamics and Relativity UWA代写

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Show that the 4-velocity
$$u^{\mu}=\frac{d x^{\mu}}{d \tau}=\left(\frac{c}{\sqrt{1-v^{2} / c^{2}}}, \frac{\mathbf{v}}{\sqrt{1-v^{2} / c^{2}}}\right)=(c \gamma, \mathbf{v} \gamma)$$
Show that the condition for such a motion is
$$w^{\mu} w_{\mu}=\text { constant }=-w_{0}^{2}$$
where $w_{0}$ is the usual three dimensional acceleration.
Show that in a fixed frame (b) reduces to
$$\frac{d}{d t} \frac{v}{\sqrt{1-v^{2} / c^{2}}}=w_{0}$$
Show that
$$\begin{gathered} x=\frac{c^{2}}{w_{0}}\left(\sqrt{1+\frac{w_{0}^{2} t^{2}}{c^{2}}}-1\right) \ v=\frac{w_{0} t}{\sqrt{1+w_{0}^{2} t^{2} / c^{2}}} \end{gathered}$$

## PHYS3002 COURSE NOTES ：

a) Write the Lagrangian and the Hamiltonian, and rewrite them in terms of the variables
$$\xi=(\varphi+\theta) / \sqrt{2} \quad \eta=(\varphi-\theta) / \sqrt{2}$$
b) Find an integral of motion other than the energy, and show that its Poisson bracket with $\mathcal{H}$ is zero.
c) Integrate the equations of motion with these initial conditions at $t=$ 0 :
$$\theta=-\frac{\pi}{4}, \quad \varphi=+\frac{\pi}{4}, \quad \dot{\theta}=\dot{\varphi}=0$$

# 量子力学和原子物理学|PHYS3001 Quantum Mechanics and Atomic Physics UWA代写

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The direct comparison of angle resolved Auger spectra yields further results. Hemmers determined the energy resolved Auger spectrum for two different polar angles: $\theta=0^{\circ}$ and $\theta=54.7^{\circ}$. It follows from (3.116) for the Auger angular distribution
$$\begin{gathered} I\left(0^{\circ}\right)=\frac{I_{0}}{4 \pi}(1+\beta) \ I\left(54.7^{\circ}\right)=\frac{I_{0}}{4 \pi} \end{gathered}$$

## PHYS3001 COURSE NOTES ：

The first term in corresponds to the electric dipole $E 1$ approximation with the transition operator (in the length form of the long-wave length approximation)
$$T_{\lambda}^{E 1}=D_{\lambda}=\sum_{n}\left(r_{n}\right){\lambda}=\sqrt{\frac{4 \pi}{3}} \sum{n} r_{n} Y_{1 \lambda}\left(\theta_{n}, \phi_{n}\right) .$$
The ‘first-order’ corrections to the dipole approximation are given by the second, magnetic dipole $M 1$ term with the transition operator
$$T_{\lambda}^{M 1}=-\mathrm{i} M_{\lambda}=-\mathrm{i} \frac{\alpha}{2} \sum_{n}\left[\left(\ell_{n}\right){\lambda}+2\left(s{n}\right){\lambda}\right],$$ and by the third, electric quadrupole $E 2$ term with the transition operator $$T{\lambda}^{E 2}=\frac{\mathrm{i} \alpha \omega}{2 \sqrt{3}} Q_{\lambda}=\frac{\mathrm{i} \alpha \omega}{2 \sqrt{3}} \sqrt{\frac{4 \pi}{5}} \sum_{n} r_{n}^{2} Y_{2 \lambda}\left(\theta_{n}, \phi_{n}\right) .$$

# 许多粒子体系|PHYS2002 Many Particle Systems代写 UWA代写

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The behavior of this equation is regular in the limit $B \rightarrow 0:$ the flow of $g_{k}$ is effectively negligible for $B \leqslant D^{-2}$. Hence we can replace by the following simpler equation:
$$\frac{\mathrm{d} g_{k}}{\mathrm{~d} B}=-\frac{\rho N g_{k}^{2}}{2 B} \mathrm{e}^{-2 B e_{k}^{2}}$$
with the initial condition now posed at $B=D^{-2}$ :
$$g_{k}\left(B=D^{-2}\right)=\frac{g}{N}$$
This equation can be integrated easily
$$\frac{1}{N}\left(\frac{1}{g_{k}(B)}-\frac{1}{g_{k}\left(D^{-2}\right)}\right)=\frac{\rho}{2} \int_{D^{-2}}^{B} \mathrm{~d} B^{\prime} \frac{\exp \left(-2 B^{\prime} \epsilon_{k}^{2}\right)}{B^{\prime}}$$
yielding
$$g_{k}(B=\infty)=\frac{1}{N} \frac{g}{1+\rho g \operatorname{Ei}\left(1,2 \epsilon_{k}^{2} / D^{2}\right) / 2}$$

$$g_{k^{\prime} k}(B)=\frac{g_{\mathrm{IR}}(B)}{N} \mathrm{e}^{-B\left(e_{k^{\prime}}-e_{k}\right)^{2}},$$
and derive the scaling equation for $g_{\mathrm{IR}}(B)$ by putting it into the flow equation (2.103) for $k^{\prime}=k=0$. This gives
$$\frac{\mathrm{d} g_{\mathrm{IR}}}{\mathrm{d} B}=-2 g_{\mathrm{IR}}^{2} \rho \int_{0}^{D} \mathrm{~d} \epsilon \epsilon \mathrm{e}^{-2 B \epsilon^{2}}$$
$$\frac{\mathrm{d} g_{\mathrm{IR}}}{\mathrm{d} B}=-\frac{\rho g_{\mathrm{IR}}^{2}}{2 B}$$
with the initial condition posed at $B=D^{-2}$ :
$$g_{\mathrm{IR}}\left(B=D^{-2}\right)=\frac{g}{N}$$
This is exactly the scaling equation derived from a conventional scaling with the identification $B=D_{\text {eff }}^{-2}$. It therefore agrees with the exact solution.