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# 物理代写| Example Oblique Coordinate System on a Plane 相对论代考

## 物理代写

We will work out the formal algebra presented above for a two-dimensional toy case. We construct a coordinate system, where the constant ” $X$ ” and ” $Y$ ” lines are not orthogonal, but they are at an angle $\omega$, as shown in Fig. 3.2. This example has also been described in Narlikar $(2010)$, however here we give more details of the calculation.

Utilising the fact that vectors do not change under coordinate transformations, only their components do, we will compute all the quantities in an orthogonal Cartesian coordinate system. We use the lower case letters, $x$ and $y$, to represent this coordinate system. The $x$-axis is aligned with the $X=0$ line and the $y$-axis is orthogonal to the $x$-axis maintaining the right-hand rule at the common origin $\mathrm{O}$. In Fig. $3.2$, the $x$-axis (and the unit vector $\hat{\mathbf{i}}$ ) is coincident with the $\mathbf{e}_{X}$ direction and the $y$-axis (and the unit vector $\hat{\mathbf{j}}$ ) is coincident with the $\mathrm{e}^{Y}$ direction (reason for this overlap of axes will become clear in a moment).

An arbitrary point $\mathrm{P}$, which has the Cartesian coordinates $(x, y)$, will have coordinates $(X, Y)$ in the oblique coordinates following the relation
One can write the above coordinate transformation as
\begin{aligned} y &=Y \sin \omega \ x &=X+Y \cos \omega \ \Rightarrow \quad \mathbf{x} &:=\hat{\mathbf{i}} x+\hat{\mathbf{j}} y=\hat{\mathbf{i}}(X+Y \cos \omega)+\hat{\mathbf{j}} Y \sin \omega \end{aligned}
Thus we get the (covariant) basis for the contravariant components as
\begin{aligned} &\mathbf{e}{X}=\frac{\partial \mathbf{x}}{\partial X}=\hat{\mathbf{i}} \ &\mathbf{e}{Y}=\frac{\partial \mathbf{x}}{\partial Y}=\hat{\mathbf{i}} \cos \omega+\hat{\mathbf{j}} \sin \omega \end{aligned}
3.7 Example: Oblique Coordinate System on a Plane
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The (contravariant) basis for the covariant components can be expressed as
\begin{aligned} \mathrm{e}^{X} &=\nabla X=\hat{\mathbf{i}}-\hat{\mathbf{j}} \cot \omega \ \mathrm{e}^{Y} &=\nabla Y=\hat{\mathbf{j}} / \sin \omega . \end{aligned}
Following the above expressions, the direction of the basis vectors have been marked in Fig. 3.2. Although the directions have been marked, the units along different axes are different, as reflected in the amplitude of the basis vectors. For example, unit of length along $\mathrm{e}^{Y}$ is $\left|\mathrm{e}^{Y}\right|=1 / \sin \omega$, which is reflecting the fact that unit change in $Y$ corresponds to $\left|\mathrm{e}^{Y}\right|$ change in the Cartesian coordinates. Similarly, unit change along the covariant $X$ axis (which we will call $x_{X}$ ), is equivalent to $\left|\mathbf{e}{X}\right|=1 / \sin \omega$ change in the Cartesian coordinates. It is easy to see that $\mathbf{e}^{X} \cdot \mathbf{e}{X}=\mathbf{e}^{Y} \cdot \mathbf{e}{Y}=1$ and $\mathbf{e}^{X} \cdot \mathbf{e}{Y}=\mathbf{e}^{Y} \cdot \mathbf{e}_{X}=0$.

For an arbitrary point P, with Cartesian position vector $\mathbf{x}:=\hat{\mathbf{i}} x+\hat{\mathbf{j}} y$, the contravariant and covariant components with respect to the tilted coordinate system are given by
$$x^{X}=\mathbf{x} \cdot \mathbf{e}^{X}=x-y \cot \omega=X,$$
These are in the units of the corresponding basis vectors, $\mathrm{e}^{X}, \mathrm{e}^{Y}, \mathrm{e}{X}$ and $\mathrm{e}{Y}$ respec tively (not the ones which were used for scalar product). The norm of the vector,
$\qquad x_{Y}=\mathbf{x} \cdot \mathbf{e}{Y}=x \cos \omega+y \sin \omega=X \cos \omega+$ orm of the $$x^{X} x{X}+x^{Y} x_{Y}=x^{2}+y^{2}=|\mathbf{x}|^{2}$$
is indeed consistent with the corresponding Cartesian value.
Now let us represent the contravariant and covariant components in Fig. 3.2. One can do this is two ways. First, the easiest, express $\mathbf{e}{X} x^{X}, \mathbf{e}{Y} x^{Y}, \mathbf{e}^{X} x_{X}$ and $\mathbf{e}^{Y} x_{Y}$ in terms of $x, y, \hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ and plot on the figure, which will show the respective vectors can be represented by $\mathrm{OA}, \mathrm{OB}, \mathrm{OC}$ and $\mathrm{OD}$. Note that, the vectors appear to have different lengths from the actual coordinates values due to the non-unit length of the basis vectors.

## 物理代考

$$\开始{对齐} y &=Y \sin \omega \ x &=X+Y \cos \omega \ \Rightarrow \quad \mathbf{x} &:=\hat{\mathbf{i}} x+\hat{\mathbf{j}} y=\hat{\mathbf{i}}(X+Y \cos \omega )+\hat{\mathbf{j}} Y \sin \omega \end{对齐}$$

$$\开始{对齐} &\mathbf{e}{X}=\frac{\partial \mathbf{x}}{\partial X}=\hat{\mathbf{i}} \ &\mathbf{e}{Y}=\frac{\partial \mathbf{x}}{\partial Y}=\hat{\mathbf{i}} \cos \omega+\hat{\mathbf{j}} \罪\欧米茄 \end{对齐}$$
3.7 示例：平面上的斜坐标系
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$$\开始{对齐} \mathrm{e}^{X} &=\nabla X=\hat{\mathbf{i}}-\hat{\mathbf{j}} \cot \omega \ \mathrm{e}^{Y} &=\nabla Y=\hat{\mathbf{j}} / \sin \omega 。 \end{对齐}$$

$$x^{X}=\mathbf{x} \cdot \mathbf{e}^{X}=x-y \cot \omega=X,$$

$\qquad x_{Y}=\mathbf{x} \cdot \mathbf{e}{Y}=x \cos \omega+y \sin \omega=X \cos \omega+$ 的形式 $$x^{X} x{X}+x^{Y} x_{Y}=x^{2}+y^{2}=|\mathbf{x}|^{2}$$

Matlab代写