# 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}

# 物理数学|PH40073 Mathematical physics代写

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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 $\nabla f=\frac{y}{z} \hat{\mathbf{e}}{x}+\frac{x}{z} \hat{\mathbf{e}}{y}-\frac{x y}{z^{2}} \hat{\mathbf{e}}_{z}$

## 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
$$\nabla \cdot \mathbf{E}=\frac{r^{2}-3 x^{2}}{r^{5}}+\frac{r^{2}-3 y^{2}}{r^{5}}+\frac{r^{2}-3 z^{2}}{r^{5}}=0 .$$

# 数学物理 Mathematical Physics MATHS5073_1 /MATHS4107_1

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Now approaching this problem using vector notation, we note that the torque about the fulcrum due to $F_{1}$ is $\tau_{1}=r_{1} \times F_{1}$. Here $r_{1}$ is a vector from the fulcrum to the point of application of $\mathbf{F}{1}$, and $$\tau{1}=\mathbf{r}{1} \times \mathbf{F}{1}=r_{1} F_{1} \sin \theta_{1} \hat{\mathbf{n}}=d_{1} F_{1} \hat{\mathbf{n}}$$
where $\hat{\mathbf{n}}$ is a unit vector normal to, and out of the plane of the figure. A similar analysis of the torque $\tau_{2}$ reveals that it is
$$\tau_{2}=\mathbf{r}{2} \times \mathbf{F}{2}=-r_{2} F_{2} \sin \theta_{2} \hat{\mathbf{n}}=-d_{2} F_{2} \hat{\mathbf{n}}$$
the minus sign occurring because $\hat{\mathbf{n}}$ has the same meaning as before but this torque is directed into the plane of the figure. We now observe that $\theta_{2}=\pi-\theta_{1}$, and therefore $\sin \theta_{2}=\sin \theta_{1}$. At equilibrium the sum of these torques is zero.

## MATHS5073_1 /MATHS4107_1 COURSE NOTES ：

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 $\nabla f=\frac{y}{z} \hat{\mathbf{e}}{x}+\frac{x}{z} \hat{\mathbf{e}}{y}-\frac{x y}{z^{2}} \hat{\mathbf{e}}_{z}$