电磁学|PH30077/PH30078 Electromagnetism 2/Magnetism代写

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这是一份bath巴斯大学PH30077/PH30078作业代写的成功案

电磁学|PH30077/PH30078 Electromagnetism 2代写


$$
\frac{\epsilon-1}{\epsilon+2}=\frac{4 \pi}{3} n \alpha
$$
where $\alpha$ is the atomic polarizability and $n$ the number of atoms per unit volume. This formula predicts that the measurable quantity
$$
\frac{(\epsilon+2) n}{\epsilon-1}
$$
for a given substance should be approximately independent of external parameters, such as pressure and temperature. Note that weak coupling between the atoms corresponds to small $n \alpha$, so that
$$
\epsilon-1 \cong 4 \pi n \alpha .
$$



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PH30077/PH30078 COURSE NOTES :

$$
\mathbf{F}{\mathrm{Con} p}=p \mathbf{B}=-\frac{I}{c} \oint d \mathbf{I}^{\prime} \times p \frac{\left(\mathbf{r}^{\prime}-\mathbf{r}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}} $$ We recognize $$ p \frac{\left(\mathbf{r}^{\prime}-\mathbf{r}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}}=\mathbf{B}{p}\left(\mathbf{r}^{\prime}\right)
$$
where $\mathbf{B}{p}\left(\mathbf{r}^{\prime}\right)$ is the magnetic field that would be produced at $\mathbf{r}^{\prime}$ by the hypothetical pole at $\mathbf{r}$. Thus, $$ \mathbf{F}{C \text { on } p}=-\frac{I}{c} \oint d \mathbf{l}^{\prime} \times \mathbf{B}_{p}\left(\mathbf{r}^{\prime}\right),
$$




激光物理学|PH30032 Laser physics代写

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这是一份bath巴斯大学PH30032作业代写的成功案

激光物理学|PH30032 Laser physics代写

Now
$$
\frac{T(0,0, t)}{T(0,0, \infty)}=\frac{4(k t)^{1 / 2}}{w \pi} .
$$
Let us assume that the approximation $w \gg 2(k T)^{1 / 2}$ holds if
$$
w / 2(k t)^{1 / 2}=5 .
$$
For times such that this inequality holds we have
$$
\frac{T(0,0, t)}{T(0,0, \infty)} \sim 0.3 \text {. }
$$
In other words, we reached one-third of the final temperature in the following times: $\mathrm{Al}, 5.4 \times 10^{-6} \mathrm{sec} ; 304 \mathrm{~S} . \mathrm{S}$., $7.4 \times 10^{-5} \mathrm{sec}$; and carbon phenolic, $1 \times 10^{-3} \mathrm{sec}$.



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PH30032 COURSE NOTES :

$$
t_{c}=\left(\alpha^{-1}\right)^{2} / \kappa
$$
Thus the cooling rate at the surface is
$$
\Delta T(z) / t_{\mathrm{c}}=(1-R) \alpha^{3} I_{0} K t_{\mathrm{p}} / \rho C .
$$
In many applications it is necessary to obtain $T=T_{\mathrm{m}}$. By comparison it is clear that the cooling rate for case 1 will always be very high. In fact,
$$
d T / d t=T_{\mathrm{m}} / t_{\mathrm{p}},
$$
and if one has sufficient energy to obtain $T_{\mathrm{m}}$ in, say, a $10^{-9} \sec$ pulse, one will have
$$
d T / d t \approx 10^{12 \circ} \mathrm{C} / \text { sec. }
$$




数学方法|PH30025 Mathematical methods代写

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这是一份bath巴斯大学PH30025作业代写的成功案

数学方法|PH30025 Mathematical methods代写

Suppose the complex numbers $z_{1}$ and $z_{2}$ are represented inby the points $P_{1}$ and $P_{2}$ respectively. Also, let $A$ be the point $z=1$. Then, with $O P_{2}$ as base corresponding to $\mathrm{OA}$, the triangle $\mathrm{OP}{2} \mathrm{P}$ is constructed to be similar to triangle $\mathrm{OAP}{1}$. Now,
$$
\begin{gathered}
\mathrm{OP} / r_{2}=\mathrm{OP} / \mathrm{OP}{2}=\mathrm{OP}{1} / \mathrm{OA}=r_{1} / 1 \
\mathrm{OP}=r_{1} r_{2} .
\end{gathered}
$$
Also
$$
\angle \mathrm{POP}{2}=\angle \mathrm{P}{1} \mathrm{OA}=\theta_{1}
$$
and so
$$
\angle \mathrm{POA}=\theta_{1}+\theta_{2} \text {. }
$$


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PH30025 COURSE NOTES :

$\operatorname{Cosh}^{-1} x$ may be expressed in terms of logarithms as follows:
If
$$
y=\cosh ^{-1} x
$$
then
$x=\cosh y=1 / 2\left(\mathrm{e}^{y}+\mathrm{e}^{-y}\right)$
and so
$$
e^{2 y}-2 x e^{y}+1=0 \text {. }
$$
Regarding this as a quadratic equation in $\mathrm{e}^{y}$,
$$
e^{y}=x \pm\left(x^{2}-1\right)^{1 / 2}
$$



凝聚态物理学|PH20017/PH20063 Condensed matter physics 1代写

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这是一份bath巴斯大学PH20017/PH20063作业代写的成功案

凝聚态物理学|PH20017/PH20063 Condensed matter physics 1代写

We consider the case when the transport relaxation time defined as
$$
\frac{1}{\tau(E)} \equiv 2 \pi n_{\mathrm{im}} \sum_{\boldsymbol{k}^{\prime}}(1-\cos \Theta) V_{\mathrm{sc}}^{2}\left(\boldsymbol{k}^{\prime}-k\right) \delta\left(E_{\boldsymbol{k}^{\prime}}-E_{k}\right)
$$
depends only on energy $E$. Here $\Theta$ is the angle between $v^{\prime}=\partial E_{\boldsymbol{h}^{\prime}} / \partial \boldsymbol{k}^{\prime}$ and $v$. Then keeping the terms linear in the electric field $E$, the Boltzmann equation for the function $F$ becomes
$$
F \cdot v=-e \tau(E)\left[E \cdot v-F \cdot\left(v \times B \cdot \nabla_{k}\right) v\right]
$$
If the magnetic field is sufficiently weak ( $\omega \tau \ll 1 ; \omega \propto B$ is the Larmour frequency), one can keep only the terms linear in $B$ with the following result for the non-equilibrium part of the distribution function:
$$
F \cdot v=-e \tau(E)\left[E \cdot v+e \tau(E) E \cdot\left(v \times B \cdot \nabla_{k}\right) v\right]
$$
Using the current density
$$
\boldsymbol{j}=-2 e \sum_{k} v f(\boldsymbol{r}, \boldsymbol{k}, t)=2 e \sum_{k} v \frac{\partial n_{k}}{\partial E} v \cdot \boldsymbol{F}
$$
one obtains the longitudinal conductivity $\sigma_{x x}$,
$$
\sigma_{x x}=-2 e^{2} \sum_{k} \frac{\partial n_{k}}{\partial E} \tau(E) v_{x}^{2}
$$


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PPH20017/PH20063 COURSE NOTES :

$$
A(\boldsymbol{r})=-\frac{\nabla \Phi}{e^{}} $$ Then the magnetic flux becomes $$ \Phi_{\mathrm{B}}=\oint_{C} \mathrm{~d} \boldsymbol{l} \cdot \boldsymbol{A}(\boldsymbol{r})=\frac{\delta \Phi}{e^{}}
$$
Here $\delta \Phi$ is a change of the phase in the round trip along the contour. The wavefunction is single-valued if $\delta \Phi=2 \pi p$ where $p=0,1,2, \ldots .$ Hence, the flux is quantized ( $\Phi_{\mathrm{B}}=p \Phi_{0}$ ) and the flux quantum (in ordinary units)
$$
\Phi_{0}=\frac{\pi \hbar c}{e}=2.07 \times 10^{-7} \mathrm{G} \mathrm{cm}^{2}
$$



电磁学|PH20014/PH20061 Electromagnetism 1代写

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这是一份bath巴斯大学PH20014/PH20061作业代写的成功案

电磁学|PH20014/PH20061 Electromagnetism 1代写

which is simply the Klein-Gordon equation
$$
\left[\eta^{a b} p_{a} p_{b}+m_{e f f}^{2}\right] \psi=0
$$
with a mass term
$$
m_{e f f}^{2}=\frac{2 \mathcal{E}}{D+1} m^{2}
$$
Non-stationary states are superpositions of Klein-Gordon states with different values of $m_{e f f}^{2}$. It can be shown that as long as we only superimpose states with $\varepsilon>0$, i.e. $m_{e f f}^{2}>0$ non-tachyonic, the wavepacket follows a timelike trajectory.
Next, consider minisuperspace models of the form


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PPH20014/PH20061 COURSE NOTES :

$$
\phi=\frac{q}{|\mathbf{r}-\mathbf{b}|}+\delta \phi
$$
wherc
$$
\delta \phi=-q \frac{a}{b} \frac{(\epsilon-1)}{\epsilon+1} I
$$
with
$$
I=\frac{1}{\left(1+y^{2}-2 y \cos \theta\right)^{1 / 2}}-\frac{1}{\gamma y^{1 / \gamma}} \int_{0}^{\gamma} d y^{\prime} \frac{\left(y^{\prime}\right)^{1 / \gamma-1}}{\left(1+y^{\prime 2}-2 y^{\prime} \cos \theta\right)^{1 / 2}}
$$



量子物理学入门|PH10048 Introduction to quantum physics代写

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这是一份bath巴斯大学PH10048作业代写的成功案

量子物理学入门|PH10048 Introduction to quantum physics代写

where $c_{\in}$ is a complex constant to be chosen such that $U_{E}$ is unitary. Begin with the relativistic particle action, where
$$
\psi\left(x^{\prime}, \tau+\epsilon\right)=\int d^{D} x \mu_{\epsilon} \exp \left[-c_{\epsilon} m \sqrt{-\eta_{\mu \nu} \Delta x^{\mu} \Delta x^{\nu}}\right] \psi(x, \tau)
$$
with measure
$$
\mu_{\epsilon}^{-1}=\int d^{D} x \exp \left[-c_{\epsilon} m \sqrt{-\eta_{\mu \nu} \Delta x^{\mu} \Delta x^{\nu}}\right]
$$
so that $U_{E} \rightarrow 1$ as $\in \rightarrow 0$ Comparing to the corresponding expression for a free nonrelativistic particle
$$
\psi\left(x^{\prime}, \tau+\epsilon\right)=\int d^{D} x \mu_{\epsilon} \exp \left[m \frac{\delta_{i j} \Delta x^{i} \Delta x^{j}}{(-i \epsilon \hbar)}\right] \psi(x, t)
$$


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PH10048 COURSE NOTES :

which is simply the Klein-Gordon equation
$$
\left[\eta^{a b} p_{a} p_{b}+m_{e f f}^{2}\right] \psi=0
$$
with a mass term
$$
m_{e f f}^{2}=\frac{2 \mathcal{E}}{D+1} m^{2}
$$
Non-stationary states are superpositions of Klein-Gordon states with different values of $m_{e f f}^{2}$. It can be shown that as long as we only superimpose states with $\varepsilon>0$, i.e. $m_{e f f}^{2}>0$ non-tachyonic, the wavepacket follows a timelike trajectory.
Next, consider minisuperspace models of the form



物理学的数学方法|PH10007/PH20107 Mathematical methods for physics 1代写

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这是一份bath巴斯大学PH10007/PH20107作业代写的成功案

物理学的数学方法|PH10007/PH20107 Mathematical methods for physics 1代写

We can see immediately that
$$
\frac{\partial f}{\partial x}=2 x+3 y, \quad \frac{\partial f}{\partial y}=3 x, \quad \frac{d y}{d x}=\frac{1}{\left(1-x^{2}\right)^{1 / 2}}
$$
and so, using (5.8) with $x_{1}=x$ and $x_{2}=y$,
$$
\begin{aligned}
\frac{d f}{d x} &=2 x+3 y+3 x \frac{1}{\left(1-x^{2}\right)^{1 / 2}} \
&=2 x+3 \sin ^{-1} x+\frac{3 x}{\left(1-x^{2}\right)^{1 / 2}}
\end{aligned}
$$
Obwiously the same expression would have resulted if we had substituted for $y$ from the start, but the above method often produces results with reduced calculation, particularly in more complicated examples.


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PH20107 COURSE NOTES :

In the above notation, $A(x, y)=3 y$ and $B(x, y)=x$ and so
$$
\frac{\partial A}{\partial y}=3, \quad \frac{\partial B}{\partial x}=1 .
$$
As these are not equal it follows that the differential is inexact.
Determining whether a differential containing many variable $x_{1}, x_{2}, \ldots, x_{n}$ is exact is a simple extension of the above. A differential containing many variables can be written in general as
$$
d f=\sum_{i=1}^{n} g_{i}\left(x_{1}, x_{2}, \ldots, x_{n}\right) d x_{i}
$$
and will be exact if
$$
\frac{\partial g_{i}}{\partial x_{j}}=\frac{\partial g_{j}}{\partial x_{i}} \quad \text { for all pairs } i, j
$$
There will be $\frac{1}{2} n(n-1)$ such relationships to be satisfied.



电与磁|PH10006/PH10051 Electricity & magnetism代写

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这是一份bath巴斯大学PH10006/PH10051作业代写的成功案

电与磁|PH10006 Electricity & magnetism代写

Question: My old camcorder runs off of a big lead-acid battery that is labeled 12 volts, $4 \mathrm{AH}$. The “AH” stands for ampere-hours. What is the maximum amount of energy the battery can store?
Solution: An ampere-hour is a unit of current multiplied by a unit of time. Current is charge per unit time, so an ampere-hour is in fact a funny unit of charge.
$$
(1 \mathrm{~A})(1 \mathrm{hour})=(1 \mathrm{C} / \mathrm{s})(3600 \mathrm{~s})
$$
Naw $3600 \mathrm{C}$ is a huge number of
Now $3600 \mathrm{C}$ is a huge number of charged particles, but the total loss of potential energy will just be their total charge multiplied by the voltage difference across which they move:
$$
\begin{aligned}
\Delta P E_{\text {etec }} &=q \Delta V \
&=(3600 \mathrm{C})(12 \mathrm{~V}) \
&=43 \mathrm{~kJ}
\end{aligned}
$$


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PH10006/PH10051 COURSE NOTES :

Question: A charged balloon falls to the ground, and its charge begins leaking off to the Earth. Suppose that the charge on the balloon is given by $q=a e^{-\Delta t}$. Find the current as a function of time, and interpret the answer.
Solution: Taking the derivative, we have
$$
\begin{aligned}
I &=\frac{\mathrm{d}}{\mathrm{d} t}\left(a e^{-b t}\right) \
&=-a b e^{-b t}
\end{aligned}
$$
The exponential function approaches zero as the exponent gets more and more negative. This means that both the charge and the current are decreasing in magnitude with time. It makes sense that the charge approaches zero, since the balloon is losing its charge. It also makes sense that the current is decreasing in magnitude, since charge cannot flow at the same rate forever without overshooting zero.



量子物理学入门|PH10001 Introduction to quantum physics代写

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这是一份bath巴斯大学PH10001作业代写的成功案

量子物理学入门|PH10001 Introduction to quantum physics代写

$$
E_{k}^{1} \equiv \bar{V}{k k}, $$ and $$ \psi{k}^{1}=-\sum_{j \neq{k}} \frac{\psi_{j}^{0}}{E_{k}^{0}-E_{j}^{0}} V_{j k}
$$
Here, ${k}$ denotes the modified degenerate states.
The second order correction is obtained by solving Eq. (59),
$$
\psi_{k}^{2}=-\frac{1}{E_{k}^{0}-H_{0}}\left(E_{k}^{0} \psi_{k}^{0}-V \psi_{k}^{1}\right)
$$


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PH10001 COURSE NOTES :

$$
\Psi=\sum_{l} C_{l} \Phi_{l}
$$
The individual “configuration” Slater determinants, $\Phi_{l}$, are constructed from spin-orbitals, $\phi_{j}$, given by
$$
\phi_{j}=|\alpha\rangle \sum_{\lambda} C_{\lambda j} \chi_{\lambda} \text { or }|\beta\rangle \sum_{\lambda} C_{\lambda j} \chi_{\lambda} \text {, }
$$
where $\chi_{\lambda}$ is a spatial basis function. The $\chi_{\lambda}$ generally fall into two classes:

  • Slater-type orbitals (STOs)
    $$
    \chi_{\lambda}=N_{n l m \zeta} Y_{l m}(\theta, \phi) r^{n-1} \exp (-\zeta r),
    $$



数学的工业应用|MA50284 Industrial applications of mathematics代写

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The unit is the second part of a bespoke package that covers the applied mathematics aspect of the training, focussing on concrete contexts of industrial application of mathematics. The first unit (Mathematical Modelling for Industry), delivered in Semester 1, provides the relevant theoretical foundations.

这是一份Bath巴斯大学MA50284作业代写的成功案

数学的工业应用|MA50284 Industrial applications of mathematics代写

$\quad E=\frac{1}{2} \rho v^{2}+\frac{1}{\gamma-1} P$,
where $\gamma$ is the ratio of the specific heats. For helium, $\gamma=c_{\mathrm{p}} / c_{\mathrm{v}}=5 / 3$.

When the pulse of gas is initiated, we assume that the shock tube is in thermal equilibrium with a room temperature of $T_{0}=300 \mathrm{~K}$ and that the gas is at rest, $v_{0}=0$. Across the valve is a jump of pressure:
$\quad P(x, 0)= \begin{cases}P_{\mathrm{L}}=4 \mathrm{~atm}, & xx_{0},\end{cases}$


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MA50284 COURSE NOTES :

$\frac{\mathrm{d} \rho}{\rho}+\frac{\mathrm{d} v}{v}+\frac{\mathrm{d} A}{A}=0$.
In the same steady one-dimensional limit, using , the conservation of momentum reduces to
$v \mathrm{~d} v+\frac{1}{\rho} \mathrm{d} P=v \mathrm{~d} v+\frac{c^{2}}{\rho} \mathrm{d} \rho=0$.
Eliminating $\mathrm{d} \rho / \rho$ from and rearranging give the expression
$$
\frac{\mathrm{d} A}{\mathrm{~d} v}=\frac{A}{v}\left(M^{2}-1\right),
$$
where the Mach Number $M=v / c$.