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1 物理代写| Contravariant and Covariant Components of a Vector 相对论代考

物理代写| Contravariant and Covariant Components of a Vector 相对论代考

物理代写

3.3 Contravariant and Covariant Components of a Vector
About the notation, we have used the same case for both kinds of components as that of the vector field. The $V^{i}$ are called “contravariant” components of a vector. $^{1}$ Similarly, $V_{i}$ are called “covariant” components of a vector. In fact, we follow a
${ }^{1}$ We generally omit the phrase “components of”, as the meaning is clear from the context even without that phrase.
42
3 Tensor Algebra
similar convention even for the basis vectors but as observed it is opposite to the components of a vector.

Let us now consider coordinate transformations. Consider another coordinate system $x^{\prime i}$. In the common region where $x^{i}$ and $x^{i i}$ are both defined and $x^{\prime i}$ are functions of $x^{j}$, that is $x^{\prime i}\left(x^{j}\right)$ and vice-versa. The only condition is that the Jacobian matrix of the transformation $\partial x^{\prime i} / \partial x^{j}$ should be non-singular so that the transformations are locally invertible – they form a valid coordinate system. Contravariant and covariant components of vectors and corresponding basis vectors transform differently under coordinate transformations.

Derivation 2 Find how the basis vectors change under the coordinate transformation $x^{i} \rightarrow x^{i i}=x^{i i}\left(x^{j}\right)$

Coordinate transformations change the components of a vector, but not the vector itself. Hence in the primed coordinates
$$
\mathbf{e}{i}^{\prime}:=\frac{\partial \mathbf{x}}{\partial x^{\prime i}}=\sum{j=1}^{n} \frac{\partial \mathbf{x}}{\partial x^{j}} \frac{\partial x^{j}}{\partial x^{\prime i}}=\sum_{j=1}^{n} \frac{\partial x^{j}}{\partial x^{\prime i}} \mathbf{e}{j} $$ To find how the other set of basis vectors, $\mathrm{e}^{i}:=\nabla x^{i}$, transform we notice that $$ \mathbf{e}^{i i} \cdot \mathrm{d} \mathbf{x}=\mathrm{d} x^{i i}=\sum{j=1}^{n} \frac{\partial x^{i i}}{\partial x^{j}} \mathrm{~d} x^{j}=\sum_{j=1}^{n} \frac{\partial x^{i i}}{\partial x^{j}} \mathbf{e}^{j} \cdot \mathrm{d} \mathbf{x}
$$
If this identity has to hold for any arbitrary $\mathrm{d} \mathbf{x}$, one should have
$$
\mathbf{e}^{i i}=\sum_{j=1}^{n} \frac{\partial x^{i i}}{\partial x^{j}} \mathbf{e}^{j} ; \quad \mathbf{e}^{i i} \cdot \mathbf{e}_{j}=\frac{\partial x^{\prime i}}{\partial x^{j}}
$$
Derivation 3 Find how the components of a vector $\mathbf{V}$ transform for the change of coordinate $x^{i} \rightarrow x^{i i}=x^{\prime i}\left(x^{j}\right)$.

Since a vector does not change in the new coordinates, only its components do,
$$
\mathbf{V}=\sum_{i=1}^{n} V^{i} \mathbf{e}{i}=\sum{i=1}^{n} V_{i} \mathbf{e}^{i}=\sum_{i=1}^{n} V^{i} \mathbf{e}{i}^{\prime}=\sum{i=1}^{n} V_{i}^{\prime} \mathbf{e}^{i i}
$$
Then using eqn. $(3.2 .7,3.2 .8,3.2 .5,3.3 .1 \& 3.3 .3)$ one can write
$$
\begin{aligned}
V^{i i} &=\mathbf{V} \cdot \mathbf{e}^{i i}=\sum_{k=1}^{n} V^{k} \mathbf{e}{k} \cdot \sum{j=1}^{n} \frac{\partial x^{i i}}{\partial x^{j}} \mathbf{e}^{j}=\sum_{j=1}^{n} \frac{\partial x^{i i}}{\partial x^{j}} V^{j} \
V_{i}^{\prime} &=\mathbf{V} \cdot \mathbf{e}{i}^{\prime}=\sum{k=1}^{n} V_{k} \mathbf{e}^{k} \cdot \sum_{j=1}^{n} \frac{\partial x^{j}}{\partial x^{\prime i}} \mathbf{e}{j}=\sum{j=1}^{n} \frac{\partial x^{j}}{\partial x^{\prime i}} V_{j}
\end{aligned}
$$
If we take $\mathbf{e}_{i}$ as the basis, then we see that the covariant components transform in the same way as this basis while the contravariant components transform with the inverse matrix. This is the historical reason for this nomenclature.

Physical quantities for one, do not change under coordinate transformations, but components of vectors and tensors do. That is why physical quantities appear as pairs of covariant and contravariant components summed over the same index, like in Eq. (3.3.4), because this procedure constructs coordinate independent quantities. This motivates us to consider the summation signs, in, e.g., Eq. (3.3.4), redundant. Therefore, whenever we see the same index twice in an expression, one in the subscript and one in the superscript, a summation over all allowed values of the index (normally 1 to $n$ or 0 to 3 for space-time coordinates) is implied, unless explicitly specified otherwise. This is called Einstein’s summation convention, a very useful shorthand, which some people jokingly say is Einstein’s only contribution to mathematics!

For example, using the summation convention one can succintly write the transformation properties of the covariant and contravariant components as,
$$
V^{\prime i}=\frac{\partial x^{i i}}{\partial x^{j}} V^{j} ; \quad V_{i}^{\prime}=\frac{\partial x^{j}}{\partial x^{\prime i}} V_{j}
$$
The summation convention is used in the rest of this book, unless, otherwise specified.

物理代写| Contravariant and Covariant Components of a Vector 狭义相对论代考

物理代考

3.3 向量的逆变和协变分量
关于符号，我们对两种分量都使用了与向量场相同的情况。 $V^{i}$ 被称为向量的“逆变”分量。 $^{1}$ 类似地，$V_{i}$ 被称为向量的“协变”分量。事实上，我们遵循一个
${ }^{1}$ 我们通常会省略短语“components of”，因为即使没有该短语，其含义也可以从上下文中清楚地看出。
42
3 张量代数
即使对于基向量也有类似的约定，但正如观察到的，它与向量的分量相反。

现在让我们考虑坐标变换。考虑另一个坐标系 $x^{\prime i}$。在 $x^{i}$ 和 $x^{ii}$ 都被定义并且 $x^{\prime i}$ 是 $x^{j}$ 的函数的公共区域中，即 $x^{ \prime i}\left(x^{j}\right)$ 反之亦然。唯一的条件是变换 $\partial x^{\prime i} / \partial x^{j}$ 的雅可比矩阵应该是非奇异的，以便变换是局部可逆的——它们形成一个有效的坐标系。向量的逆变分量和协变分量以及对应的基向量在坐标变换下变换不同。

推导2 求坐标变换下基向量如何变化 $x^{i} \rightarrow x^{i i}=x^{i i}\left(x^{j}\right)$

坐标变换会改变向量的分量，但不会改变向量本身。因此在素坐标
$$
\mathbf{e}{i}^{\prime}:=\frac{\partial \mathbf{x}}{\partial x^{\prime i}}=\sum{j=1}^{n} \frac{\partial \mathbf{x}}{\partial x^{j}} \frac{\partial x^{j}}{\partial x^{\prime i}}=\sum_{j=1} ^{n} \frac{\partial x^{j}}{\partial x^{\prime i}} \mathbf{e}{j} $$ 为了找出另一组基向量 $\mathrm{e}^{i}:=\nabla x^{i}$ 是如何变换的，我们注意到 $$ \mathbf{e}^{ii} \cdot \mathrm{d} \mathbf{x}=\mathrm{d} x^{ii}=\sum{j=1}^{n} \frac{\partial x ^{ii}}{\partial x^{j}} \mathrm{~d} x^{j}=\sum_{j=1}^{n} \frac{\partial x^{ii}}{\部分 x^{j}} \mathbf{e}^{j} \cdot \mathrm{d} \mathbf{x}
$$
如果这个恒等式必须为任意 $\mathrm{d} \mathbf{x}$ 持有，则应该有
$$
\mathbf{e}^{ii}=\sum_{j=1}^{n} \frac{\partial x^{ii}}{\partial x^{j}} \mathbf{e}^{j} ; \quad \mathbf{e}^{i i} \cdot \mathbf{e}_{j}=\frac{\partial x^{\prime i}}{\partial x^{j}}
$$
推导 3 找出向量 $\mathbf{V}$ 的分量如何随着坐标的变化 $x^{i} \rightarrow x^{ii}=x^{\prime i}\left(x^{j }\右）$。

由于向量在新坐标中没有改变，只有它的分量会改变，
$$
\mathbf{V}=\sum_{i=1}^{n} V^{i} \mathbf{e}{i}=\sum{i=1}^{n} V_{i} \mathbf{ e}^{i}=\sum_{i=1}^{n} V^{i} \mathbf{e}{i}^{\prime}=\sum{i=1}^{n} V_ {i}^{\prime} \mathbf{e}^{ii}
$$
然后使用eqn。 $(3.2 .7,3.2 .8,3.2 .5,3.3 .1 \& 3.3 .3)$ 一个可以写
$$
\开始{对齐}
V^{ii} &=\mathbf{V} \cdot \mathbf{e}^{ii}=\sum_{k=1}^{n} V^{k} \mathbf{e}{k} \ cdot \sum{j=1}^{n} \frac{\partial x^{ii}}{\partial x^{j}} \mathbf{e}^{j}=\sum_{j=1}^ {n} \frac{\partial x^{ii}}{\partial x^{j}} V^{j} \
V_{i}^{\prime} &=\mathbf{V} \cdot \mathbf{e}{i}^{\prime}=\sum{k=1}^{n} V_{k} \mathbf {e}^{k} \cdot \sum_{j=1}^{n} \frac{\partial x^{j}}{\partial x^{\prime i}} \mathbf{e}{j }=\sum{j=1}^{n} \frac{\partial x^{j}}{\partial x^{\prime i}} V_{j}
\end{对齐}
$$
如果我们以$\mathbf{e}_{i}$ 为基，那么我们看到协变分量的变换方式与此基相同，而逆变分量随逆矩阵变换。这就是这种命名法的历史原因。

一个物理量在坐标变换下不会改变，但向量和张量的分量会改变。这就是为什么物理量显示为在同一索引上求和的成对的协变和逆变分量，就像在等式中一样。 (3.3.4)，因为这个程序构造了坐标独立量。这促使我们考虑求和符号，例如，等式。 (3.3.4)，多余的。因此，每当我们在一个表达式中两次看到相同的索引时，一个在下标中，一个在上标中，索引的所有允许值（通常为 1 到 $n$ 或 0 到 3 对于时空坐标）的总和是暗示，除非另有明确说明。这就是所谓的爱因斯坦求和约定，一种非常有用的速记，有人戏称这是爱因斯坦对数学的唯一贡献！

例如，使用求和约定可以简洁地将协变和逆变分量的变换属性写为：
$$
V^{\prime i}=\frac{\partial x^{i i}}{\partial x^{j}} V^{j} ; \quad V_{i}^{\prime}=\frac{\partial x^{j}}{\partial x^{\prime i}} V_{j}
$$
除非另有说明，否则本书其余部分均使用求和约定。