# 电动力学和光学|PHYS3035/PHYS3935 Electrodynamics and Optics代写 sydney代写

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The students are cultivated as high-quality innovative professionals. They are well developed in the
aspects of morality, intelligence, physique, and aesthetic. They possess the basic theoretical knowledge in
the fields of Materials Science and Physics，and are well trained in the applied research, technological
development, and engineering.

$$G(q, \omega)=\frac{1}{q^{2}-q_{0}^{2}}\left(U-\frac{q q}{q_{0}^{2}}\right)$$
which sometimes is useful. Like the Huygens propagator, the dyadic Green function is also singular for $q=q_{0}$.

The plane-wave representation of the dyadic Green function for the magnetic field is obtained by inserting the Fourier integral transformation
$$\boldsymbol{G}{\mathrm{M}}(\boldsymbol{R} ; \omega)=(2 \pi)^{-3} \int{-\infty}^{\infty} \boldsymbol{G}_{\mathrm{M}}(q, \omega) \mathrm{e}^{i \boldsymbol{q} \cdot \boldsymbol{R}} \mathrm{d}^{3} q$$

Calculations analogous to those carried out to determine $\boldsymbol{G}(\boldsymbol{R} ; \omega)$ result in
$$\boldsymbol{G}{\mathrm{M}}(\boldsymbol{q}, \omega)=\frac{q}{q{0}} \frac{1}{q^{2}-q_{0}^{2}} \boldsymbol{U} \times \hat{\boldsymbol{q}}$$
an expression which also is singular for $q=q_{0}$. The folding theorem in $\boldsymbol{r}$-space gives when applied to
$$\boldsymbol{B}(q, \omega)=\frac{i \mu_{0} \omega}{c_{0}} \boldsymbol{G}{\mathrm{M}}(q, \omega) \cdot J(q, \omega)$$ A current density parallel to the $q$-direction does not give rise to a magnetic field, and the magnetic field generated by a current density perpendicular to $\hat{q}$ always lies in a plane perpendicular to $\hat{q}$. To prove the above-mentioned claims we expand the unit dyad after a triple set of orthogonal unit vectors $\hat{q}{\perp}^{(1)}, \hat{q}{\perp}^{(2)}$, and $\hat{q}$, with $\hat{\boldsymbol{q}}{1}^{(1)} \times \hat{\boldsymbol{q}}{2}^{(2)}=\hat{\boldsymbol{q}}$, i.e., $$\boldsymbol{U}=\hat{\boldsymbol{q}}{\perp}^{(1)} \hat{q}{\perp}^{(1)}+\hat{\boldsymbol{q}}{\perp}^{(2)} \hat{\boldsymbol{q}}_{\perp}^{(2)}+\hat{q} \hat{q}$$

## PHYS3035/PHYS3935 COURSE NOTES ：

In the mixed representation, the electric field, $\boldsymbol{E}(z ; q |, \omega)$, from a current density distribution, $\boldsymbol{J}(z: q |, \omega)$, is outside the distribution given by
$$E\left(z ; q_{|}, \omega\right)=i \mu_{0} \omega \int_{-\infty}^{\infty} G\left(z-z^{\prime} ; q_{1}, \omega\right) \cdot J\left(z^{\prime} ; q_{|}, \omega\right) \mathrm{d} z^{\prime}$$
where
$$G\left(Z ; q_{|}, \omega\right)=\Gamma\left(s g n Z ; q_{1}, \omega\right) \mathrm{e}^{i x_{\perp}|Z|}$$
with
$$\Gamma\left(\operatorname{sgn} Z: q_{|}, \omega\right)=\frac{i}{2 q_{0}^{2} \kappa \perp}\left[q_{0}^{2} U-q_{|} q_{|}-\kappa_{\perp}^{2} \hat{z} \bar{z}-\left(q_{|}^{\hat{z}}+\hat{z} q_{|}\right) \kappa_{\perp} \operatorname{sgn} Z\right]$$
Let us now assume that the source current density is nonvanishing only on a plane sheet located at $z^{\prime}=z_{0}$. Thus,
$$J\left(z^{\prime} ; q_{|}, \omega\right)=J_{0}\left(q_{|}, \omega\right) \delta\left(z^{\prime}-z_{0}\right)$$

# 物理学|PHYS1002 Physics 1 (Fundamentals)代写 monash代写

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This course is a real-time, online course where the instructor and students meet via web conferencing tools, at scheduled days and times. Instructors and students share information, ideas and learning experiences in a virtual course environment. Participation in synchronous courses requires students to have reliable, high-speed internet access, a computer (ideally with a webcam), and a headset with a microphone.

Let $Q_{B} Q_{R}, T_{R}$ and $W$, respectively, denote, the heat drawn by the refrigerator from the body at temperature $T$, out of which $Q_{R}$ is rejected in the room at temperature $T_{R}$ and a work $W$ is performed on the refrigerator. Now,
$$Q_{R}=Q_{B}+W \text { and } Q_{B}=Q_{R}-W$$
And the Coefficient of performance of the refrigerator, $(\mathrm{COF})=\frac{W}{Q_{R}}=\frac{\left(T_{R}-T\right)}{T_{R}}$ or
$$Q_{R}=\frac{W T_{R}}{\left(T_{R}-T\right)}$$

Substituting this value of $Q_{R}$ in Eq. S-7.10.1 gives:
\begin{aligned} &Q_{B}=Q_{R}-W=\frac{W T_{R}}{\left(T_{R}-T\right)}-W=W\left(\frac{T}{\left(T_{R}-T\right)}\right) \ &W=\left(\frac{\left(T_{R}-T\right)}{T}\right) Q_{B} \end{aligned}
shows that in order to draw a quantity of heat $Q_{B}$ from the body at temperature $T$ the amount of work required to be done is $W$ whichis equal to $\left(\frac{\left(T_{R}-T\right)}{T}\right) Q_{B}$.

## PHS2061 COURSE NOTES ：

The thermal capacity (at constant volume) is defined as $C_{V} \equiv\left(\frac{\partial U}{\partial T}\right){V}$, which in the present case is $K T^{3}$. Using the relation $\left(\frac{\partial S}{\partial T}\right){V}=\frac{1}{T}\left(\frac{\partial U}{\partial T}\right){V}=\frac{K T^{3}}{T}=K T^{2}$, and hence, $$\Delta S=\int{T_{R}}^{0} K T^{2} d T=-\frac{K}{3} T_{R}^{3}$$
Negative sign indicates that this is the decrease in the entropy.
Change in internal energy of the body may be calculated using the relation
$$d U=C d T \quad \text { or } \quad \Delta U=\int_{T_{R}}^{0} K T^{3} d T=-\frac{K}{4} T_{R}^{4}$$

# 量子和热物理学|PHS2061 Quantum and thermal physics代写 monash代写

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The equation of state for Van der Waals gas is,
$$P=\frac{R T}{(v-b)}-\frac{a}{v^{2}}$$
Therefore, $\left(\frac{\partial P}{\partial T}\right){v}=\frac{R}{(v-b)}$ and $\left(\frac{\partial P}{\partial v}\right){T}=\frac{2 a(v-b)^{2}-R T v^{3}}{(v-b)^{2} v^{3}}$

$c_{P}-c_{v}=-T \frac{\left[\left(\frac{\partial P}{\partial T}\right){v}\right]^{2}}{\left(\frac{\partial P}{\partial v}\right){T}}=\frac{R}{1-2 a(v-b)^{2} / R T v^{3}}$
Also,
$$\kappa_{T}=-\frac{1}{v}\left(\frac{\partial v}{\partial P}\right){T}=-\frac{1}{v}\left(\frac{\partial P}{\partial v}\right){T}^{-1}=\frac{(v-b)^{2} v^{2}}{R T v^{3}-2 a(v-b)^{2}}$$
But $\kappa_{s}=\frac{\kappa_{T}}{\gamma}$ and $\gamma=1.66$ for a monatomic gas
Hence,
Similarly,
\begin{aligned} &\kappa_{s}=0.6 \kappa_{T}=0.6 \frac{(v-b)^{2} v^{2}}{R T v^{3}-2 a(v-b)^{2}} \ &\beta=\frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_{v}=\frac{R v^{2}(v-b)}{R T v^{3}-2 a(v-b)^{2}} \end{aligned}

## PHS2061 COURSE NOTES ：

Since
$$d F=d U-T d s-S d T$$
The difference in the value of $F$ of two nearby equilibrium states of an open system may be given by
$$d F=-P d V-S d T+\mu d \mathbb{N}$$
$7.61$
Also
$$d F=\left(\frac{\partial F}{\partial V}\right){T, \mathbb{N}} d V+\left(\frac{\partial F}{\partial T}\right){V, \mathbb{N}} d T+\left(\frac{\partial F}{\partial \mathbb{N}}\right){V, T} d \mathbb{N}$$ Comparison of $$\mu=\left(\frac{\partial F}{\partial \mathbb{N}}\right){V, T}$$

# 工程物理学|PHS1002 Physics for engineering代写 monash代写

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An exact first-degree first-order ODE is one of the form
$$A(x, y) d x+B(x, y) d y=0 \quad \text { and for which } \quad \frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}$$
In this case $A(x, y) d x+B(x, y) d y$ is an exact differential, $d U(x, y)$ . In other words
$$A d x+B d y=d U=\frac{\partial U}{\partial x} d x+\frac{\partial U}{\partial y} d y$$

from which we obtain
\begin{aligned} &A(x, y)=\frac{\partial U}{\partial x} \ &B(x, y)=\frac{\partial U}{\partial y} \end{aligned}
Since $\partial^{2} U / \partial x \partial y=\partial^{2} U / \partial y \partial x$ we therefore require
$$\frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}$$

## PHS1002 COURSE NOTES ：

Substituting $y=v x$ we obtain
$$v+x \frac{d v}{d x}=v+\tan v .$$
Cancelling $v$ on both sides, rearranging and integrating gives
$$\int \cot v d v=\int \frac{d x}{x}=\ln x+c_{1}$$
But
$$\int \cot v d v=\int \frac{\cos v}{\sin v} d v=\ln (\sin v)+c_{2}$$
so the solution to the $\mathrm{ODE}$ is $y=x \sin ^{-1} A x$, where $A$ is a constant.

# 基础物理学|PHS1001 Foundation physics代写 monash代写

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$$\int_{a V} v_{i} n_{j} \mathrm{~d} A=\int_{V} \frac{\partial v_{i}}{\partial x_{j}} \mathrm{~d} V .$$
GrEEN’s identities also follow from adequate application of the Divergence Theorem. The first GREEN’s identity is obtained by choosing in $f=\phi v$. This yields
\begin{aligned} \int_{\partial V} \phi v \cdot n \mathrm{~d} A &=\int_{V} \operatorname{div}(\phi v) \mathrm{d} V \ &=\int_{V}{\phi \operatorname{div} v+v \operatorname{grad} \phi} \mathrm{d} V . \end{aligned}

If we substitute here $v=\operatorname{grad} \psi$ and observe that div $v=\operatorname{div} \operatorname{grad} \psi=\Delta \psi$, where $\Delta$ is the LAPLACE operator, then
$$\int_{V}{\operatorname{grad} \phi \cdot \operatorname{grad} \psi} \mathrm{d} V=-\int_{V} \phi \Delta \psi \mathrm{d} V+\int_{\partial V} \phi \frac{\partial \psi}{\partial n} \mathrm{~d} A$$
which is GREEN’s first identity.
If in the roles of $\phi$ and $\psi$ are interchanged, we obtain
$$\int_{V}{g r a d \psi \cdot \operatorname{grad} \phi} \mathrm{d} V=-\int_{V} \psi \Delta \phi \mathrm{d} V+\int_{\partial V} \psi \frac{\partial \phi}{\partial n} \mathrm{~d} A$$

## PHS1001 COURSE NOTES ：

which, when combined, yields
$$\int_{\mathcal{C}} \phi \mathrm{d} x=-\int_{A_{C}}(\operatorname{grad} \phi) \times n \mathrm{~d} A$$
If $\phi$ in is the ith component of a vector field, then it also implies
$$\int_{\mathcal{C}} v \otimes \mathrm{d} x=-\int_{A_{\mathcal{C}}}(\operatorname{grad} v) \times \boldsymbol{n} \mathrm{d} A$$
A further interesting formula is obtained by selecting
$$e_{i} \times \int_{\mathcal{C}} v_{i} \mathrm{~d} x=\oint_{\mathcal{C}} v \times \mathrm{d} x=-\int_{A_{C}} \underbrace{e_{i} \times \operatorname{grad} v_{i}}{-\text {curl } v} \times n \mathrm{~d} A$$ so that $$\oint{\mathcal{C}} v \times \mathrm{d} x=\int_{A_{\mathcal{C}}}(\operatorname{curl} v) \times \boldsymbol{d} A$$

# 数学建模|MTH2040 Mathematical modelling代写 monash代写

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If $|\alpha|=\nu$, then $\lambda_{1}=\lambda_{2}=-\alpha$ and the corresponding generalized eigenspace is spanned by
$$z^{1}=\left[\begin{array}{c} 1 \ -\alpha \end{array}\right], \quad z^{2}=\left[\begin{array}{l} 0 \ 1 \end{array}\right]$$
where $z^{1}$ is a real eigenvector and $z^{2}$ is a generalized eigenvector of second order. By (34) the corresponding (generalized) real eigenmotions are
$$x^{1}(t)=e^{-\alpha t}\left[\begin{array}{c} 1 \ -\alpha \end{array}\right], \quad x^{2}(t)=e^{-\alpha t}\left[\begin{array}{l} 0 \ 1 \end{array}\right]+t e^{-\alpha t}\left[\begin{array}{c} 1 \ -\alpha \end{array}\right]$$

If $|\alpha|<v$ (hence $z^{2}=\overline{z^{1}}$ ) then $\lambda_{1,2}=-\alpha \pm x \sqrt{\nu^{2}-\alpha^{2}}$ and the corresponding real modes are by $(44),(45)$ and $(50)$
$$x^{1}(t)=e^{-\alpha t}\left(\cos \left(\sqrt{u^{2}-\alpha^{2}} t\right)\left[\begin{array}{c} 1 \ -\alpha \end{array}\right]-\sin \left(\sqrt{u^{2}-\alpha^{2}} t\right)[\sqrt{0}]\right)$$

## MTH2040 COURSE NOTES ：

is called the Fibonacci sequence and plays a role in various fields of mathematics. The eigenvalues of the matrix in (59) are $\lambda_{1,2}=(1 \pm \sqrt{5}) / 2$ and the corresponding real modes are
$$z^{1}(t)=\left(\frac{1+\sqrt{5}}{2}\right)^{t}\left[\begin{array}{c} (1+\sqrt{5}) / 2 \ 1 \end{array}\right], \quad z^{2}(t)=\left(\frac{1-\sqrt{5}}{2}\right)^{t}\left[\begin{array}{c} (1-\sqrt{5}) / 2 \ 1 \end{array}\right]$$
The solution of the initial value problem is of the form $x(t)=\alpha_{1} z^{1}(t)+\alpha_{2} z^{2}(t)$ for some $\left(\alpha_{1}, \alpha_{2}\right) \in \mathbb{R}^{2}$. Since $x(0)=[1,0]^{\top}$ we obtain $\alpha_{1}=1 / \sqrt{5}, \alpha_{2}=-1 / \sqrt{5}$ and hence the following analytic expression for the Fibonacci sequence
$$\xi(t)=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{t+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{t+1}\right]$$

# 高级工程数学|ENG2005 Advanced engineering mathematics代写 monash代写

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$$\lambda^{2}+4 \lambda+3=0,$$
with the roots $\lambda_{1}=-1$ and $\lambda_{2}=-3$, so the complementary function is
$$y_{c}(x)=C_{1} e^{-x}+C_{2} e^{-3 x} \text {. }$$
The nonhomogeneous term $e^{-x}$ is contained in the complementary function, so by Step 3 (b) in Table $6.2$ we must seek a particular integral of the form
$$y_{p}(x)=A x e^{-x} .$$

Substituting the expression for $y_{p}(x)$ into the differential equation gives $\left(-2 A e^{-x}+A x e^{-x}\right)+4\left(A e^{-x}-A x e^{-x}\right)+3 A x e^{-x}=e^{-x}, \quad$ or $2 A e^{-x}=e^{-x}$, showing that $A=1 / 2$. So, in this case, the particular integral is $y_{P}(x)=(1 / 2) x e^{-x}$ and the general solution is
$$y(x)=C_{1} e^{-x}+C_{2} e^{-3 x}+(1 / 2) x e^{-x} .$$
The initial condition $y(0)=2$ will be satisfied if
$$2=C_{1}+C_{2},$$
and the initial condition $y^{\prime}(0)=1$ will be satisfied if
$$1 / 2=-C_{1}-3 C_{2}$$

## ENG2005 COURSE NOTES ：

$$\lambda^{2}+2 \lambda+1=0$$
with the repeated root $\lambda=-1$. Thus, the complementary function is
$$y_{c}(x)=C_{1} e^{-x}+C_{2} x e^{-x} .$$
Two linearly independent solutions are thus
$$y_{1}(x)=e^{-x} \text { and } y_{2}(x)=x e^{-x} \text {, }$$
while the nonhomogeneous term is $f(x)=x e^{-x}$. The Wronskian
$$W(x)=\left|\begin{array}{ll} y_{1} & y_{2} \ y_{1}^{\prime} & y_{2}^{\prime} \end{array}\right|=e^{-x}\left(e^{-x}-x e^{-x}\right)+e^{-x} x e^{-x}=e^{-2 x}$$
so substituting in shows that the particular integral is
$$y_{p}(x)=-e^{-x} \int x^{2} \mathrm{~d} x+x e^{-x} \int x \mathrm{~d} x=\frac{1}{6} x^{3} e^{-x} .$$
The general solution is
$$y(x)=C_{1} e^{-x}+C_{2} x e^{-x}+\frac{1}{6} x^{3} e^{-x} .$$
This result could, of course, have been found by the method of undetermined coefficients.

# 工程数学|ENG1005 Engineering mathematics代写 monash代写

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For $f(x)=3 x^{3}-x$ we have
\begin{aligned} f(-x) &=3(-x)^{3}-(-x)=-3 x^{3}+x \ &=-\left(3 x^{3}-x\right)=-f(x) \end{aligned}
So $3 x^{3}-x$ is an odd function.
For $f(x)=\frac{x^{2}}{1+x^{2}}$
$$f(-x)=\frac{(-x)^{2}}{1+(-x)^{2}}=\frac{x^{2}}{1+x^{2}}=f(x)$$
so this is even.

If $f(x)=\frac{2 x}{x^{2}-1}$ then
$$f(-x)=\frac{2(-x)}{(-x)^{2}-1}=-\frac{2 x}{x^{2}-1}=-f(x)$$
so $f(x)$ is odd.
If $f(x)=\frac{x^{2}}{x+1}$ we have
$$f(-x)=\frac{(-x)^{2}}{-x+1}=\frac{x^{2}}{1-x}$$
This is not equal to $f(x)$ or $-f(x)$ and so this function is neither odd nor even.

## ENG1005 COURSE NOTES ：

$$y=f(x)=\frac{1}{x-2} \quad(x \neq 2)$$
If $y=4$, find $x$.
We have
$$4=\frac{1}{x-2}$$
so
$$4(x-2)=1=4 x-8$$
and therefore
$$4 x=9$$
and
$$x=\frac{9}{4}$$

# 微分方程与建模|MTH2032 Differential equations with modelling代写 monash代写

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Let us denote the steady-state number of infectives by $\hat{I}$ and the steadystate number of susceptibles by $\hat{S}$, where $\hat{I}$ and $\hat{S}$ are constants. Then,
$$I_{k+1}=I_{k}=\hat{I} \text { and } S_{k+1}=S_{k}=\hat{S}$$
are the steady-state solutions. Substituting (4) into (3) we obtain
$$\hat{I}=f \hat{S} \hat{I}$$

$$\hat{S}=\hat{S}-f \hat{S} \hat{I}+B,$$
or
\begin{aligned} \hat{I}(1-f \hat{S}) &=0 \ f \hat{S} \hat{I}-B &=0 . \end{aligned}
Our aim is to solve for $\hat{I}$ and $\hat{S}$. From equation (5a) there are two cases to be considered: $\hat{I}=0$ or $\hat{S}=1 / f$.

## MTH2032 COURSE NOTES ：

$$Y_{k+1}=(1-r) Y_{k}$$
The solutions of the linear differenceare only approximate solutions of but they have the advantage that they can be found in closed form.
The closed-form solution of is
$$Y_{k}=(1-r)^{k} Y_{0}$$
where $Y_{0}$ can be calculated from the initial population $N_{0}$ using. Hence an approximate solution to the original is given by
$$N_{k} \simeq K+K(1-r)^{k} Y_{0}$$

# 多变量微积分|MTH2010 – Multivariable calculus代写 monash代写

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As the picture illustrates, a point $X=(x, y)$ lies on $\ell$ if and only if the vector $\mathbf{X}$ is of the form $\mathbf{P}{\mathbf{0}}+t \mathbf{v}$, for some (positive or negative) scalar $t$. (For the point $X$ shown, $t=2$.) Therefore, points on the line are those that satisfy the vector equation $$(x, y)=\left(x{0}, y_{0}\right)+t(a, b)$$

for some real number $t$. This means that we can think of the line as the image of the vectorvalued function $\mathbf{X}: \mathbb{R} \rightarrow \mathbb{R}^{2}$ defined for all real numbers $t$ by $\mathbf{X}(t)=\left(x_{0}, y_{0}\right)+t(a, b)$. In the language of parametric equations, $\ell$ can be parametrized by
$$x(t)=x_{0}+a t ; \quad y(t)=y_{0}+b t ; \quad-\infty<t<\infty .$$
As $t$ ranges through all real numbers, $\mathbf{X}(t)$ traces out the entire line, which is infinite in both directions.

## MTH2010 COURSE NOTES ：

The line is determined either by the vector-valued function
$$\mathbf{X}(t)=(0.28,-0.96,1.67)+t(.96,0.28,0.33)$$
or, equivalently, by the parametric equations
$$x=0.28+0.96 t ; \quad y=-0.96+0.28 t ; \quad z=1.67+0.33 t,$$
whichever we please. For good measure, here is yet another equivalent description:
$$\mathbf{X}(t)=(0.28+0.96 t) \mathbf{i}+(-0.96+0.28 t) \mathbf{j}+(1.67+0.33 t) \mathbf{k}$$
Notice that time $t=0$ corresponds to the point $(0.28,-0.96,1.67)$-no matter the form.