### 物理代写|理论力学代写theoretical mechanics代考|PHYSICS3544

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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|理论力学代写theoretical mechanics代考|Vector Fields

If a vector is associated with each position, we speak of a vector field. With scalar fields, a scalar is associated with each position. The vector field $\mathbf{a}(\mathbf{r})$ is only continuous at $\mathbf{r}{0}$ if all paths approaching $\mathbf{r}{0}$ have the same limit. For scalar fields, this is already an essentially stronger requirement than in one dimension.

Instead of drawing a vector field with arrows at many positions, it is often visualized by a set of field lines: at every point of a field line the tangent points in the direction of the vector field. Thus $\mathbf{a} | \mathrm{d} \mathbf{r}$ and $\mathbf{a} \times \mathrm{d} \mathbf{r}=\mathbf{0}$.

For a given vector field many integrals can be formed. In particular, we often have to evaluate integrals over surfaces or volumes. In order to avoid double or triple integral symbols, the corresponding differential is often written immediately after the integral symbol: $\mathrm{d} V$ for the volume, df for the surface integral, e.g., $\int \mathrm{d} \mathbf{f} \times \mathbf{a}$ instead of $-\int \mathbf{a} \times$ df (in this way the unnecessary minus sign is avoided for the introduction of the curl density or rotation on p. 13). Here df is perpendicular to the related surface element. However, the sign of df still has to be fixed. In general, we consider the surface of a volume $V$, which will be denoted here by $(V)$. Then df points outwards. Corresponding to $(V)$, the edge of an area $A$ is denoted by $(A)$.
An important example of a scalar integral is the line integral $\int \mathrm{d} \mathbf{r} \cdot \mathbf{a}(\mathbf{r})$ along a given curve $\mathbf{r}(t)$. If the parameter $t$ determines the points on the curve uniquely, then the line integral $$\int \mathrm{d} \mathbf{r} \cdot \mathbf{a}(\mathbf{r})=\int \mathrm{d} t \frac{\mathrm{d} \mathbf{r}}{\mathrm{d} t} \cdot \mathbf{a}(\mathbf{r}(t))$$
is an ordinary integral over the scalar product $\mathbf{a} \cdot \mathrm{d} \mathbf{r} / \mathrm{d} t$. Another example of a scalar integral is the surface integral $\int \mathrm{df} \cdot \mathbf{a}(\mathbf{r})$ taken over a given area $A$ or over the surface $(V)$ of the volume $V$.

Besides the scalar integrals, vectorial integrals like $\int \mathrm{d} V \mathbf{a}, \int \mathrm{d} \mathbf{f} \times \mathbf{a}$, and $\int \mathrm{d} \mathbf{r} \times \mathbf{a}$ can arise, e.g., the $x$-component of $\int \mathrm{d} V$ a is the simple integral $\int \mathrm{d} V a_{x}$.

Different forms are also reasonable through differentiation: vector fields can be deduced from scalar fields, and scalar fields (but also vector fields and tensor fields) from vector fields. These will now be considered one by one. Then the operator $\nabla$ will always turn up. The symbol $\nabla$, an upside-down $\Delta$, resembles an Ancient Greek harp and hence is called nabla, after W. R. Hamilton (see 122).

The gradient of a scalar function $\psi(\mathbf{r})$ is the vector field
$$\operatorname{grad} \psi \equiv \nabla \psi, \quad \text { with } \nabla \psi \cdot \mathrm{d} \mathbf{r} \equiv \mathrm{d} \psi \equiv \psi(\mathbf{r}+\mathrm{d} \mathbf{r})-\psi(\mathbf{r})$$
This is clearly perpendicular to the area $\psi=$ const. at every point and points in the direction of $\mathrm{d} \psi>0$ (see Fig. 1.4). The value of the vector $\nabla \psi$ is equal to the derivative of the scalar function $\psi(\mathbf{r})$ with respect to the line element in this direction. In Cartesian coordinates, we thus have
$$\nabla \psi=\mathbf{e}{x} \frac{\partial \psi}{\partial x}+\mathbf{e}{y} \frac{\partial \psi}{\partial y}+\mathbf{e}{z} \frac{\partial \psi}{\partial z}=\left(\mathbf{e}{x} \frac{\partial}{\partial x}+\mathbf{e}{y} \frac{\partial}{\partial y}+\mathbf{e}{z} \frac{\partial}{\partial z}\right) \psi$$

Here $\partial \psi / \partial x$ is the partial derivative of $\psi(x, y, z)$ with respect to $x$ for constant $y$ and $z$. (If other quantities are kept fixed instead, then special rules have to be considered, something we shall deal with in Sect. 1.2.7.)
The gradient is also obtained as a limit of a vectorial integral:
$$\nabla \psi=\lim {V \rightarrow 0} \frac{1}{V} \int{(V)} \text { df } \psi(\mathbf{r})$$
If we take a cube with infinitesimal edges $\mathrm{d} x, \mathrm{~d} y$, and $\mathrm{d} z$, we have on the right-hand side as $x$-component $(\mathrm{d} x \mathrm{~d} y \mathrm{~d} z)^{-1}{\mathrm{~d} y \mathrm{~d} z \psi(x+\mathrm{d} x, y, z)-\mathrm{d} y \mathrm{~d} z \psi(x, y, z)}=$ $\partial \psi / \partial x$, and similarly for the remaining components. Hence, also
$$\int_{V} \mathrm{~d} V \nabla \psi-\int_{(V)} \mathrm{d} \mathbf{f} \psi$$
because a finite volume can be divided into infinitesimal volume elements, and for continuous $\psi$, contributions from adjacent planes cancel in pairs. With this surface integral the gradient can be determined even if $\psi$ is not differentiable (singular) at individual points-the surface integral depends only upon points in the neighbourhood of the singular point, where everything is continuous. (In Sect. 1.1.12, we shall consider the example $\psi=1 / r$.)

## 物理代写|理论力学代写theoretical mechanics代考|Divergence

While a vector field has been derived from a scalar field with the help of the gradient, the divergence associates a scalar field with a vector field:
$$\operatorname{div} \mathbf{a} \equiv \nabla \cdot \mathbf{a} \equiv \lim {V \rightarrow 0} \frac{1}{V} \int{(V)} \mathbf{d f} \cdot \mathbf{a}$$

For the same cube as in the last section, the right-hand expression yields
\begin{aligned} \frac{1}{\mathrm{~d} x \mathrm{~d} y \mathrm{~d} z} & {\left[\mathrm{~d} y \mathrm{~d} z\left{a_{x}(x+\mathrm{d} x, y, z)-a_{x}(x, y, z)\right}\right.} \ &+\mathrm{d} z \mathrm{~d} x\left{a_{y}(x, y+\mathrm{d} y, z)-a_{y}(x, y, z)\right} \ &\left.+\mathrm{d} x \mathrm{~d} y\left{a_{z}(x, y, z+\mathrm{d} z)-a_{z}(x, y, z)\right}\right]=\frac{\partial a_{x}}{\partial x}+\frac{\partial a_{y}}{\partial y}+\frac{\partial a_{z}}{\partial z} \end{aligned}
as suggested by the notation $\nabla$. a, i.e., a scalar product between the vector operator $\nabla$ and the vector $\mathbf{a}$. With this we have also proven Gauss’s theorem
$$\int_{V} \mathrm{~d} V \nabla \cdot \mathbf{a}=\int_{(V)} \mathrm{d} \mathbf{f} \cdot \mathbf{a}$$
since for any partition of the finite volume $V$ into infinitesimal ones and for a continuous vector field $\mathbf{a}$, the contributions of adjacent planes cancel in pairs. The integrals here may even enclose points at which a (r) is singular (see Fig. 1.5 left). We shall discuss this in more detail in Sect. 1.1.12.

The integral $\int$ df $\cdot \mathbf{a}$ over an area is called the $\int u x$ of the vector field $\mathbf{a}(\mathbf{r})$ through this area (even if $\mathbf{a}$ is not a current density). In this picture, the integral over the closed area $(V)$ describes the source strength of the vector field, i.e., how much more flows into $V$ than out. The divergence is therefore to be understood as a source density. A vector field is said to be source-free if its divergence vanishes everywhere. (If the source density is negative, then “drains” predominate.)

The concept of a field-line tube is also useful (we discussed field lines in Sect. 1.1.4). Its walls are everywhere parallel to a (r). Therefore, there is no flux through the walls, and the flux through the end faces is equal to the volume integral of $\nabla \cdot \mathbf{a}$. For a source-free vector field $(\nabla \cdot \mathbf{a}=0)$, the flux flowing into the field-line tube through one end face emerges again from the other.

## 物理代写|理论力学代写theoretical mechanics代考|Vector Fields

$$\int \mathrm{d} \mathbf{r} \cdot \mathbf{a}(\mathbf{r})=\int \mathrm{d} t \frac{\mathrm{d} \mathbf{r}}{\mathrm{d} t} \cdot \mathbf{a}(\mathbf{r}(t))$$

$$\operatorname{grad} \psi \equiv \nabla \psi, \quad \text { with } \nabla \psi \cdot \mathrm{d} \mathbf{r} \equiv \mathrm{d} \psi \equiv \psi(\mathbf{r}+\mathrm{d} \mathbf{r})-\psi(\mathbf{r})$$

$$\nabla \psi=\mathbf{e} x \frac{\partial \psi}{\partial x}+\mathbf{e} y \frac{\partial \psi}{\partial y}+\mathbf{e} z \frac{\partial \psi}{\partial z}=\left(\mathbf{e} x \frac{\partial}{\partial x}+\mathbf{e} y \frac{\partial}{\partial y}+\mathbf{e} z \frac{\partial}{\partial z}\right) \psi$$

$$\nabla \psi=\lim V \rightarrow 0 \frac{1}{V} \int(V) \mathrm{df} \psi(\mathbf{r})$$

$(\mathrm{d} x \mathrm{~d} y \mathrm{~d} z)^{-1} \mathrm{~d} y \mathrm{~d} z \psi(x+\mathrm{d} x, y, z)-\mathrm{d} y \mathrm{~d} z \psi(x, y, z)=\partial \psi / \partial x$ ，对于其余的组件也是如此。因此，也
$$\int_{V} \mathrm{~d} V \nabla \psi-\int_{(V)} \mathrm{d} \mathbf{f} \psi$$

## 物理代写|理论力学代写theoretical mechanics代考|Divergence

$\$ \$$loperatorname{div} \mathbf{a} \equiv \backslash nabla \backslash cdot \backslash mathbf { a } lequiv \backslash \lim {V \backslash Irightarrow 0} \backslash frac {1} V} \backslash Iint {(\mathrm{V})} \mathrm{~ I m a t h b f { d f } ~ \ c d o t ~ \ m a t h b f { a }} \ \$$

Ibegin ${a l i g n} \backslash$ frac ${1} \backslash \backslash m a t h r m{\sim d} x \backslash m a t h r m{\sim d}$ 和 $\backslash$ mathrm ${\sim d}$ Z \& ${\backslash \backslash$ eft $\backslash \backslash m a t h r m{\sim d}$ 和 $\backslash$ mathrm ${\sim d}}$ Z $\backslash$ eft ${a$

$$\int_{V} \mathrm{~d} V \nabla \cdot \mathbf{a}=\int_{(V)} \mathrm{d} \mathbf{f} \cdot \mathbf{a}$$

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