Tag Archives: 巴斯大学代写

高能天体物理学|PH40113 High energy astrophysics代写

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高能天体物理学|PH40113 High energy astrophysics代写


The rigorous theory of Cherenkov radiation (more correctly called Cherenkov-Vavilov radiation after its co-discoverers) was developed by Frank and Tamm from which the basic formulae are derived [6].
From this theory the emission formula is derived:
$$
\begin{array}{cc}
\mathrm{d} E / \mathrm{d} t=\left(e^{2} / c^{2}\right) & \int \sin ^{2} \theta \omega \mathrm{d} \omega . \
\text { I } & \text { II } \quad \text { III }
\end{array}
$$
The basic components of this equation were expounded by Jelley [11]. The mechanism is quite different from other, more familiar, radiation mechanisms such as synchrotron radiation or bremsstrahlung. The three factors in the Cherenkov formula for radiation yield can be understood by analogy with the single elementary classical dipole.



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PH40113 COURSE NOTES ：

$$
\delta u=s(\varphi) \int_{0}^{\varphi} c(\psi) q(\psi) \mathrm{d} \psi-c(\varphi) \int_{0}^{\varphi} s(\psi) q(\psi) \mathrm{d} \psi
$$
where $s(\varphi)$ and $c(\varphi)$ are two independent solutions of the homogenous equation $f_{, \rho \varphi}+(1-3 \zeta \cos \varphi) f=0$. These are the so-called Mathieu functions:
$$
s(\varphi)=\sin \varphi+O(\zeta) ; \quad c(\varphi)=\cos \varphi+O(\zeta)
$$
The O-terms are actually series in $\sin n \varphi$ or $\cos n \varphi$ of order $\zeta$ or smaller, whose explicit expression we fortunately don’t need. The whole solution is now



相对论的宇宙学|PH40112 Relativistic cosmology代写

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相对论的宇宙学|PH40112 Relativistic cosmology代写


It is actually easier to solve $G_{\alpha \beta}=0$ than $R_{\alpha \beta}=0$. From (4.20) we find:
$$
r\left(1-\mathrm{e}^{-2 \lambda}\right)=b \quad \rightarrow \quad \mathrm{e}^{2 \lambda}=\frac{1}{1-b / r}
$$
with $b$ constant. Substitute that in $G_{11}=0$ :
$$
2 \nu^{\prime}=\frac{b / r^{2}}{1-b / r} \quad \rightarrow \quad \mathrm{e}^{2 \nu}=A(1-b / r)
$$
since the expression on the left can be integrated to $2 \nu=\log (1-b / r)+$ const. The constant $A$ must be 1 because must be the Lorentz metric for $r \rightarrow \infty$. We insert these results :
$$
\mathrm{d} s^{2}=(1-b / r) c^{2} \mathrm{~d} t^{2}-\frac{\mathrm{d} r^{2}}{1-b / r}-r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{d} \varphi^{2}\right)
$$



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PH40112 COURSE NOTES ：

$$
\frac{e^{2}}{1-r_{s} / r}-\frac{\dot{r}^{2}}{1-r_{s} / r}-\frac{h^{2}}{r^{2}}=\kappa
$$
which we may write as
$$
\dot{r}^{2}=e^{2}-V(r) ; \quad \text { with } \quad V(r)=\left(1-\frac{r_{s}}{r}\right)\left(\frac{h^{2}}{r^{2}}+\kappa\right) .
$$
For $\kappa=0$ (massless particles), $V$ has a maximum at $1.5 r_{\mathrm{s}}$. For massive particles it is necessary to distinguish between a high and a low angular momentum $h$, measured by the parameter $a \equiv r_{\mathrm{s}}^{2} / 2 h^{2}$. For $a>\frac{1}{6}$ (low angular momentum) $V(r)$ increases monotonously, and for $a<\frac{1}{6}$ (high angular momentum) $V(r)$ has two extrema at



光电子学|PH40086 Photonics代写

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光电子学|PH40086 Photonics代写


$$
\left(\overleftrightarrow{P}{\perp s} \mathcal{E}{s}(\omega)\right)=\left(\mathcal{N}{\mathrm{s}}^{\text {tord }}(\omega)\right)^{2} \overleftrightarrow{I}{\mathrm{s}}\left(\overleftrightarrow{P}{\perp \mathrm{s}} \mathcal{E}{\mathrm{s}}(\omega)\right)
$$
for
$$
\overleftrightarrow{I}{\mathrm{s}} \stackrel{\text { def }}{=} \overleftrightarrow{P}{\perp s}\left(\overleftrightarrow{\epsilon}{c}(\omega)\right)^{-1} \overleftrightarrow{P}{\perp s}
$$
Therefore,
$$
\overleftrightarrow{I}{\mathrm{s}}=\frac{1}{\left(\operatorname{Nord}{\mathrm{s}}(\omega)\right)^{2}} \stackrel{\leftrightarrow}{P}{\perp \mathrm{s}} $$ Since and $$ \mathbf{a} \cdot\left(\overleftrightarrow{P}{\perp \mathbf{s}} \mathbf{b}\right)=\left(\overleftrightarrow{P}_{\perp \mathbf{s}} \mathbf{a}\right) \cdot \mathbf{b} \quad \forall \mathbf{a}, \mathbf{b} \in \mathbb{C}^{3}
$$



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PH40086 COURSE NOTES ：

fulfills the conditions
$$
\begin{aligned}
&\jmath(\mathbf{x}, t)=0 \text { for }(|\mathbf{x}|, t) \notin[0,+R] \times[0,+\infty) \
&\Longrightarrow \quad \mathcal{R}(\mathbf{x}, t ; \jmath)=0 \quad \text { for } t \notin\left[0, \frac{|\mathbf{x}|-R}{c}\right]
\end{aligned}
$$
and
$$
\hat{H} \tilde{\mathcal{R}}(\mathbf{x}, \omega ; \widetilde{\jmath})=-i \omega \mu_{0} \tilde{\mathcal{J}}(\mathbf{x}, \omega)
$$
for all sufficiently well-behaved $\boldsymbol{\jmath}$, where
$$
\begin{aligned}
\widetilde{\mathcal{R}}(\mathbf{x}, \omega ; \tilde{\jmath}) & \stackrel{\text { def }}{=} \frac{1}{\sqrt{2 \pi}} \int \mathcal{R}(\mathbf{x}, t ; \jmath) e^{+i \omega t} \mathrm{~d} t \
&=\int \overleftrightarrow{r}\left(\mathbf{x}-\mathbf{x}^{\prime}, \omega\right) \tilde{\mathcal{J}}\left(\mathbf{x}^{\prime}, \omega\right) \mathrm{d} V_{\mathbf{x}^{\prime}}
\end{aligned}
$$



纳米科技|PH40085 Nanoscience代写

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这是一份bath巴斯大学PH40085作业代写的成功案

纳米科技|PH40085 Nanoscience代写


$$
\lambda=\frac{2 \pi r}{n}=\frac{h}{p}=\frac{h}{m v}
$$
Solve for $v$ to get (express $v$ in terms of $r$ )
$$
v=\frac{n h}{2 \pi m r}=\frac{n \hbar}{m r}
$$
Replace this into the main equation $(3.3)$
$$
\frac{n^{2} \hbar^{2}}{m r}=\frac{q^{2}}{4 \pi \varepsilon_{0}}
$$
Rearrange this to get
$$
r=\frac{4 \pi \epsilon_{0} n^{2} \hbar^{2}}{m q^{2}}
$$
If $n=1$ (the lowest orbit) this gives us the Bohr radius
$$
a_{0}=\frac{4 \pi \epsilon_{0} \hbar^{2}}{m q^{2}}
$$



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PH40085 COURSE NOTES ：


From the previous chapter we know that the Bohr radius of the electron-hole pair is
$$
a B=\frac{4 \pi \epsilon \epsilon_{0} \hbar^{2}}{\mu q^{2}}
$$
where $\mu$ is the exciton reduced mass (i used $m$ last time, sorry for the notation change).

Usually folks like to write the exciton Bohr radius in terms of the standard textbook Bohr radius for the electron, $a_{o}$. Recall that
$$
a_{o}=\frac{4 \pi \epsilon_{o} \hbar^{2}}{m_{o} q^{2}}
$$
Therefore the bulk exciton Bohr radius can be re-written as
$$
a_{B}=\frac{4 \pi \epsilon_{o} \hbar^{2}}{m_{o} q^{2}} \frac{\epsilon m_{o}}{\mu}
$$



高级量子理论|PH40084 Advanced quantum theory代写

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物理数学|PH40073 Mathematical physics代写


Derivation. Starting with a three-component wave function $\phi_{i}$ describing a massive spin-1 free particle in its rest frame, two possible rest-frame covariant forms exist: a covariant four-vector $\phi_{\mu}=\left(0, \phi_{i}\right)$ and a rank-two antisymmetric (field) tensor $\phi_{\mu \nu}$ given by (recall the angular-momentum tensor operators $L_{\mu v}$ and $J_{\mu v}$ in Chapter 3) $\phi_{0 i}=-\phi_{i 0}=\partial_{t} \phi_{i}$ and $\phi_{00}=\phi_{i j}=0$. In a general frame, the boosted form of $\phi_{\mu v}$ can be obtained from $\phi_{m}$ as
$$
\phi_{\mu v}=\partial_{\mu} \phi_{v}-\partial_{v} \phi_{\mu^{}} $$ The free-particle dynamical relation between $\phi_{\mu}$ and $\phi_{\mu v}$ is called the Proca equation: $$ \partial^{v} \phi_{\mu v}=m^{2} \phi_{\mu^{}}
$$
Owing to the antisymmetric structure of $\phi_{\mu \nu}$, the derivative of (4.38) implies the subsidiary condition
$$
\partial^{\mu} \phi_{\mu}=0
$$



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PH40084 COURSE NOTES ：


Furthermore, since $H_{0}$ must be an observable hermitian operator, so must $\alpha_{i}$ and $\beta$ be hermitian:
$$
\alpha_{i}^{\dagger}=\alpha_{i}, \quad \beta^{\dagger}=\beta .
$$
The adjoint row bispinor $\psi^{\dagger}$ can be combined with the column bispinor $\psi$ to form a positive definite probability density
$$
\rho=\psi^{\dagger} \psi=\sum_{\sigma=1}^{4} \psi_{\sigma}^{*} \psi_{\sigma},
$$
now naturally linked with the hermitian probability current density
$$
\mathbf{j}=\psi^{\dagger} \boldsymbol{\alpha} \psi .
$$



物理数学|PH40073 Mathematical physics代写

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这是一份bath巴斯大学PH40073作业代写的成功案

物理数学|PH40073 Mathematical physics代写


Sometimes the determination of the gradient is straightforward. Given
$$
f(x, y, z)=\frac{x y}{z},
$$
we easily find
$$
\frac{\partial f}{\partial x}=\frac{y}{z}, \quad \frac{\partial f}{\partial y}=\frac{x}{z}, \quad \frac{\partial f}{\partial z}=-\frac{x y}{z^{2}}
$$
leading to $\nabla f=\frac{y}{z} \hat{\mathbf{e}}{x}+\frac{x}{z} \hat{\mathbf{e}}{y}-\frac{x y}{z^{2}} \hat{\mathbf{e}}_{z}$



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PH40073 COURSE NOTES ：

Dropping the factor $q / 4 \pi \varepsilon_{0}$, and taking
$$
E_{x}=\frac{x}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}},
$$
we get
$$
\frac{\partial E_{x}}{\partial x}=\frac{1}{r^{3}}-\frac{3}{2} \frac{2 x^{2}}{r^{5}}=\frac{r^{2}-3 x^{2}}{r^{5}}
$$
Adding to this the analogous expressions for $\partial E_{y} / \partial y$ and $\partial E_{z} / \partial z$, we get
$$
\nabla \cdot \mathbf{E}=\frac{r^{2}-3 x^{2}}{r^{5}}+\frac{r^{2}-3 y^{2}}{r^{5}}+\frac{r^{2}-3 z^{2}}{r^{5}}=0 .
$$



银河系和宇宙学入门|PH30111 Galaxies and introduction to cosmology代写

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这是一份bath巴斯大学PH30111作业代写的成功案

银河系和宇宙学入门|PH30111 Galaxies and introduction to cosmology代写


$$
\begin{aligned}
&r=R+x \
&\theta=\Omega t+y
\end{aligned}
$$
where $\Omega$ is the angular velocity for a circular path:
$$
\Omega^{2}=\frac{1}{R} \frac{\partial U(R, 0)}{\partial r}
$$
In polar coordinates the equations of motion are written as$$
\begin{aligned}
\ddot{r}-\dot{\theta}^{2} r &=-\frac{\partial U}{\partial r}, \
r \ddot{\theta}+2 \dot{r} \dot{\theta} &=0 \
\ddot{z} &=-\frac{\partial U}{\partial z} .
\end{aligned}
$$



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PH30111 COURSE NOTES ：

and inserting this into the first equation, we have
$$
\ddot{x}-2 \Omega(a-2 x \Omega)-\Omega^{2} x=-x \frac{\partial^{2} U(R, 0)}{\partial r^{2}} .
$$
This equation can be written in the form
$$
\ddot{x}+\kappa^{2}\left(x-x_{0}\right)=0
$$
and
$$
\dot{y}=-\frac{2 \Omega x}{R},
$$
with
$$
\kappa^{2}=\frac{\partial^{2} U(R, 0)}{\partial r^{2}}+3 \Omega^{2}=R \frac{\mathrm{d} \Omega^{2}}{\mathrm{~d} R}+4 \Omega^{2} .
$$



广义相对论|PH30101 General relativity代写

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广义相对论|PH30101 General relativity代写


the corresponding null geodesics are
$$
\begin{aligned}
&\gamma_{X}: x^{A A^{\prime}} \equiv x_{0}^{A A^{\prime}}+\lambda \bar{X}^{A} X^{A^{\prime}} \
&\gamma_{Y}: x^{A A^{\prime}} \equiv x_{1}^{A A^{\prime}}+\mu \bar{Y}^{A} Y^{A^{\prime}}
\end{aligned}
$$
If these intersect at some point, say $x_{2}$, one finds
$$
x_{2}^{A A^{\prime}}=x_{0}^{A A^{\prime}}+\lambda \bar{X}^{\mathcal{A}} X^{A^{\prime}}=x_{1}^{A A^{\prime}}+\mu \bar{Y}^{A} Y^{A^{\prime}}
$$
where $\lambda \mu \in R$. Hence
$$
x_{2}^{A A^{\prime}} \bar{Y}{A} X{A^{\prime}}=x_{0}^{A A^{\prime}} \bar{Y}{A} X{A^{\prime}}=x_{1}^{A A^{\prime}} \bar{Y}{A} X{A^{\prime}}
$$



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PH30101 COURSE NOTES ：

This enables one to define the spinor field
$$
p^{A} \equiv p^{A A^{\prime} B^{\prime}} \pi_{A^{\prime}} \pi_{B^{\prime}}
$$
and the patching function
$$
f^{A} \equiv p^{A} g\left(p_{B} \omega^{B}, \pi_{B^{\prime}}\right)
$$
and the function
$$
F\left(x^{a}, \pi_{A^{\prime}}\right) \equiv g\left(i p_{A} x^{A C^{\prime}} \pi_{C^{\prime}}, \pi_{A^{\prime}}\right)
$$



网络|PH30098 Networks代写

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超导技术|PH30079 Superconductivity代写


$$
E=-\sum_{i j} p_{i j} \ln p_{i j}
$$
where
$$
p_{i j}=t_{i j} / O_{i}
$$
In $p_{i j}, t_{i j}$ is the number of commuters between districts $i$ and $j$, while $O_{i}$ is the outflows of district $i$.



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PH30098 COURSE NOTES ：

$y_{i u}+y_{i j} \geq 2 y_{i j}$ for all $i, j$
$\sum_{i j} \sum_{j \dot{i}} y_{i j}=k(k-1)$
$y_{i j} \in{0,1}$ for all $i, j$
The first constraint implies that if $y_{i j}=1$, then $y_{u}=y_{j}=1$. The second constraint imposes the correct number of connections. Consequently, if $y_{i i}=y_{i j}=1$, then $y_{i j}=1$.




超导技术|PH30079 Superconductivity代写

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超导技术|PH30079 Superconductivity代写


$$
\mu_{\mathrm{dc}}=-\left(V_{\mathrm{s}}+V_{\mathrm{h}}+V_{\mathrm{c}}\right) \frac{B_{\mathrm{app}}}{\mu_{0}} .
$$
For field cooling, the magnetic field is trapped in the open hole, while surface currents shield the superconductor itself and the enclosed cavity from this field, which gives for the magnetic moment
$$
\mu_{\mathrm{fc}}=-\left(V_{\mathrm{s}}+V_{\mathrm{c}}\right) \frac{B_{\mathrm{app}}}{\mu_{0}} .
$$
Associated with the effective magnetic moment (5.16) there is an effective magnetization $M_{\text {eff }}$ defined by Eq. (5.5) in terms of the total volume $(5.15)$
$$
M_{\mathrm{eff}}=\frac{\mu_{\mathrm{eff}}}{V_{\mathrm{T}}}=\chi_{\mathrm{eff}} \frac{B_{\mathrm{app}}}{\mu_{0}} .
$$



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PH30079 COURSE NOTES ：

A quantitative measure of the degree of granularity of a sample is its porosity $P$, which is defined by
$$
P=\left(1-\rho / \rho_{x^{-r a y}}\right),
$$
where the density $\rho$ of the sample is
$$
\rho=\frac{m}{V_{\mathrm{T}}}
$$
and the $x=r a y$ density is calculated from the expression
$$
\rho_{\mathrm{x}-\mathrm{ray}}=\frac{[\mathrm{MW}]}{V_{0} N_{\mathrm{s}}}
$$