物理代写| Tensors of Higher Rank 相对论代考代写 - 代考代写：100%准时可靠您的作业代写专家

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1 物理代写| Tensors of Higher Rank 相对论代考

物理代写| Tensors of Higher Rank 相对论代考

物理代写

3.5.1 Transformation Laws
From this point onwards, we will not write the basis vectors $\mathbf{e}{i}$ or $\mathbf{e}^{i}$ explicitly. Just the contravariant and covariant components of vectors/tensors and the metric would take care of the algebra. That is why we often drop or abuse the phrase 46 3 Tensor Algebra “components of” and use contravariant or convariant vector/tensor to mean that they are components of those vectors/tensors. There are also mixed tensors with both covariant and contravariant indices, e.g., the Kronecker Delta symbol $\delta{j}^{i} .$

Components of tensors transform following similar rules as those of vectors. We will first consider second rank tensors and indicate how the transformation law generalises to tensors of higher rank. Under a coordinate transformation $x^{i} \rightarrow x^{\prime i}=$ $x^{i}\left(x^{j}\right)$, components of a contravariant tensor $T^{i j}$, covariant tensor $T_{i j}$ and a mixed tensor $T_{j}^{i}$ transform respectively as:
$$
\begin{aligned}
T^{\prime k l} &=T^{i j} \frac{\partial x^{\prime k}}{\partial x^{i}} \frac{\partial x^{\prime l}}{\partial x^{j}} \
T_{k l}^{\prime} &=T_{i j} \frac{\partial x^{i}}{\partial x^{\prime k}} \frac{\partial x^{j}}{\partial x^{\prime l}} \
T_{l}^{\prime k} &=T_{j}^{i} \frac{\partial x^{\prime k}}{\partial x^{i}} \frac{\partial x^{j}}{\partial x^{\prime l}}
\end{aligned}
$$
The pattern is clear. If a tensor has more contravariant indices then more matrices of the form Eq. (3.3.8) enter into the transformation law. In case of covariant indices the inverse of the matrix given in Eq. (3.3.9) enters. Thus one may write down the tensor transformation law for a tensor with $m$ contravariant indices and $n$ covariant indices. Such a tensor is of rank $m+n$.

We give below two examples of tensors of rank 2, contravariant, covariant and mixed which we have encountered before but perhaps not recognised them as such.
Derivation 4 Show that $g_{i j}$ and $g^{i j}$ transform respectively as covariant and con-
travariant tensors under coordinate transformation $x^{i} \rightarrow x^{i i}=x^{i j}\left(x^{i}\right)$
Using the results obtained in Derivation 2, it is straightforward to show that
$$
g_{i j}^{\prime}=\mathrm{e}{i}^{\prime} \cdot \mathrm{e}{j}^{\prime}=\left(\frac{\partial x^{i}}{\partial x^{\prime k}} \mathrm{e}{k}\right) \cdot\left(\frac{\partial x^{j}}{\partial x^{\prime \prime}} \mathrm{e}^{l}\right)=\frac{\partial x^{i}}{\partial x^{\prime k}} \frac{\partial x^{j}}{\partial x^{\prime \prime}} g{k l},
$$
$g^{\prime i j}=\mathrm{e}^{\prime i} \cdot \mathrm{e}^{\prime j}=\left(\frac{\partial x^{i}}{\partial x^{k}} \mathrm{e}{k}\right) \cdot\left(\frac{\partial x^{\prime j}}{\partial x^{l}} \mathrm{e}^{l}\right)=\frac{\partial x^{i}}{\partial x^{k}} \frac{\partial x^{\prime j}}{\partial x^{l}} g^{k l} .$ Another way is to use the quotient law. But then the argument is somewhat involved. We may start from Eq. $(3.4 .3)$ that since ds $^{2}$ is a scalar and $\mathrm{d} x^{i}$ is an arbi- trary contravariant vector then $g{i j}$ must be a covariant tensor of rank $2 .$ Not true!
Although $\mathrm{d} x^{i}$ is an arbitrary contravariant vector $\mathrm{d} x^{i} \mathrm{~d} x^{j}$ is not arbitrary because it
is symmetric. This argument will only show that the symmetric part of $g_{i j}$ is a tensor
See Exercise $3.5 b$. However, if in addition, we impose the symmetry on $g_{i j}$ then the
proof is complete. A similar reasoning shows that $g^{i j}$ must be a contravariant tensor
of rank 2 .
3.5 Tensors of Higher Rank
47
Derivation 5 Show that Kronecker Delta $\delta_{j}^{i}$ transform as a mixed tensor.
Since the transformed $\delta_{j}^{i}$ symbols match $\delta_{j}^{\prime i}$ in the new coordinates,
$$
\delta_{j}^{i} \frac{\partial x^{\prime k}}{\partial x^{i}} \frac{\partial x^{j}}{\partial x^{\prime \prime}}=\frac{\partial x^{\prime k}}{\partial x^{\prime}}=\delta_{l}^{\prime k}
$$
A theorem, which is often useful is the Quotient Rule. There are two of them: one for the outer product and one for the inner product. First consider the outer product of tensors $A_{i j}$ and $B^{k}$ say, $C_{i j}^{k}=A_{i j} B^{k}$. Then if $C_{i j}^{k}$ and $B^{k}$ are tensors, then $A_{i j}$ is a tensor. We have taken here a simple case but this result is true for tensors of arbitrary rank.

Now consider the inner product $C_{i}=A_{i j} B^{j}$, where we are given that $C_{i}$ and $B^{j}$ are tensors. Then $A_{i j}$ is a tensor if this equation holds for arbitrary $B^{j}$ (Exercise 3.5). Again this statement is true for tensors of arbitrary rank.
3.5.2 Symmetric and Anti-symmetric Tensors
A tensor $T_{\ldots i \ldots j} \ldots$ is symmetric in the indices $i$ and $j$, if
ed to several indices for te ensor is totally symmetric dices. It is totally antisym of the indices and remain a symmetric part, $T_{(i j)}$, hake this point clear.

物理代考

3.5.1 转化规律
从现在开始，我们不会明确地写出基向量 $\mathbf{e}{i}$ 或 $\mathbf{e}^{i}$。只有向量/张量和度量的逆变和协变分量才能处理代数。这就是为什么我们经常放弃或滥用这句话 46 3 张量代数 “组成部分”，并使用逆变或协变向量/张量来表示它们是这些向量/张量的组成部分。还有具有协变和逆变索引的混合张量，例如，克罗内克三角洲符号 $\delta{j}^{i} .$

张量的分量变换遵循与向量相似的规则。我们将首先考虑二阶张量，并指出变换定律如何推广到更高阶的张量。在坐标变换 $x^{i} \rightarrow x^{\prime i}=$ $x^{i}\left(x^{j}\right)$ 下，逆变张量 $T^{ij }$，协变张量 $T_{ij}$ 和混合张量 $T_{j}^{i}$ 分别变换为：
$$
\开始{对齐}
T^{\prime kl} &=T^{ij} \frac{\partial x^{\prime k}}{\partial x^{i}} \frac{\partial x^{\prime l}}{ \部分 x^{j}} \
T_{kl}^{\prime} &=T_{ij} \frac{\partial x^{i}}{\partial x^{\prime k}} \frac{\partial x^{j}}{\部分 x^{\prime l}} \
T_{l}^{\prime k} &=T_{j}^{i} \frac{\partial x^{\prime k}}{\partial x^{i}} \frac{\partial x^{ j}}{\partial x^{\prime l}}
\end{对齐}
$$
图案很清晰。如果张量具有更多的逆变指数，则更多的矩阵形式为等式。 (3.3.8) 进入转化规律。在协变指数的情况下，方程式中给出的矩阵的逆矩阵。 (3.3.9) 进入。因此，可以写下具有 $m$ 逆变索引和 $n$ 协变索引的张量的张量变换定律。这样的张量的秩为 $m+n$。

我们在下面给出了 2 阶张量的两个例子，逆变的、协变的和混合的，我们之前遇到过但可能没有认识到它们。
推导 4 证明 $g_{i j}$ 和 $g^{i j}$ 分别变换为协变和 con-
坐标变换下的易变张量 $x^{i} \rightarrow x^{i i}=x^{i j}\left(x^{i}\right)$
使用推导 2 中获得的结果，很容易证明
$$
g_{ij}^{\prime}=\mathrm{e}{i}^{\prime} \cdot \mathrm{e}{j}^{\prime}=\left(\frac{\partial x ^{i}}{\partial x^{\prime k}} \mathrm{e}{k}\right) \cdot\left(\frac{\partial x^{j}}{\partial x^{ \prime \prime}} \mathrm{e}^{l}\right)=\frac{\partial x^{i}}{\partial x^{\prime k}} \frac{\partial x^{j }}{\partial x^{\prime \prime}} g{kl},
$$
$g^{\prime ij}=\mathrm{e}^{\prime i} \cdot \mathrm{e}^{\prime j}=\left(\frac{\partial x^{i}}{\部分 x^{k}} \mathrm{e}{k}\right) \cdot\left(\frac{\partial x^{\prime j}}{\partial x^{l}} \mathrm{e }^{l}\right)=\frac{\partial x^{i}}{\partial x^{k}} \frac{\partial x^{\prime j}}{\partial x^{l} } g^{kl} .$ 另一种方法是使用商法。但随后的论点有些涉及。我们可以从方程式开始。 $(3.4 .3)$ ，因为 ds $^{2}$ 是一个标量并且 $\mathrm{d} x^{i}$ 是一个 arbi- trary 逆变向量，则 $g{i j}$ 必须是秩为 $2 的协变张量。$ 不正确！
虽然 $\mathrm{d} x^{i}$ 是一个任意的逆变向量 $\mathrm{d} x^{i} \mathrm{~d} x^{j}$ 不是任意的，因为它
是对称的。这个论点只会表明 $g_{i j}$ 的对称部分是张量
请参阅练习 $3.5 b$。但是，如果另外，我们对 $g_{i j}$ 施加对称性，那么
证明完成。类似的推理表明 $g^{i j}$ 必须是逆变张量
等级 2 。
3.5 高阶张量
47
推导 5 证明 Kronecker Delta $\delta_{j}^{i}$ 变换为混合张量。
由于转换后的 $\delta_{j}^{i}$ 符号在新坐标中与 $\delta_{j}^{\prime i}$ 匹配，
$$
\delta_{j}^{i} \frac{\partial x^{\prime k}}{\partial x^{i}} \frac{\partial x^{j}}{\partial x^{\prime \prime}}=\frac{\partial x^{\prime k}}{\partial x^{\prime}}=\delta_{l}^{\prime k}
$$
一个经常有用的定理是商数法则。其中有两种：一种用于外积，一种用于内积。首先考虑张量 $A_{i j}$ 和 $B^{k}$ 的外积，比如 $C_{i j}^{k}=A_{i j} B^{k}$。那么如果 $C_{i j}^{k}$ 和 $B^{k}$ 是张量，那么 $A_{i j}$ 是张量。我们在这里举了一个简单的例子，但是这个结果对于任意等级的张量是正确的。

现在考虑内积 $C_{i}=A_{i j} B^{j}$，其中 $C_{i}$ 和 $B^{j}$ 是张量。如果这个方程对任意 $B^{j}$ 成立，那么 $A_{i j}$ 就是一个张量（练习 3.5）。同样，这个陈述对于任意等级的张量也是正确的。
3.5.2 对称和反对称张量
张量 $T_{\ldots i \ldots j} \ldots$ 在索引 $i$ 和 $j$ 中是对称的，如果
ed to te ensor 的几个指标是完全对称的骰子。它与索引完全相反，并且保持对称部分，$T_{(i j)}$，清楚这一点。