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Chapter 21: Problem 1

(a) Show that for any general, but fixed, \(\phi\), $$ \left(u_{1}, u_{2}\right)=\left(x_{1} \cos \phi-x_{2} \sin \phi, x_{1} \sin \phi+x_{2} \cos \phi\right) $$ are the components of a first-order tensor in two dimensions. (b) Show that $$ \left(\begin{array}{cc} x_{2}^{2} & x_{1} x_{2} \\ x_{1} x_{2} & x_{1}^{2} \end{array}\right) $$ is not a (Cartesian) tensor of order 2. To establish that a single element does not transform correctly is sufficient.

Short Answer

Expert verified
u_i transformations match first-order tensor rules. The given matrix fails to be a second-order tensor.

Step by step solution

01

Understand the transformation equations

Given transformation equations: \( u_{1} = x_{1} \cos\phi - x_{2} \sin\phi \) and \( u_{2} = x_{1} \sin\phi + x_{2} \cos\phi \)
02

Define first-order tensor transformation property

A first-order tensor transforms according to \(u_i' = T_{ij} x_j \), where \( T_{ij} \) is the transformation matrix and \( x_j \) is the original coordinate.
03

Write transformation matrix for rotation

Rotation transformation matrix \( T \) is \[ T = \begin{pmatrix} \cos\phi & -\sin\phi \ \sin\phi & \cos\phi \end{pmatrix} \]
04

Apply transformation matrix to original coordinates

Applying transformation matrix:\[ \begin{pmatrix} u_1 \ u_2 \end{pmatrix} = \begin{pmatrix} \cos\phi & -\sin\phi \ \sin\phi & \cos\phi \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix}\]Calculate this to get \[ u_1 = x_1 \cos\phi - x_2 \sin\phi \]\[ u_2 = x_1 \sin\phi + x_2 \cos\phi \]
05

Verify first-order tensor condition

The results \( u_1 = x_1 \cos\phi - x_2 \sin\phi \) and \( u_2 = x_1 \sin\phi + x_2 \cos\phi \) match the transformation rules for a first-order tensor. Hence, it is proven.
06

Define second-order tensor transformation

A second-order tensor \( A \) transforms as \[ A_{ij}' = T_{ik} T_{jl} A_{kl} \], where \( A \) is the tensor and \( T \) is the rotation matrix.
07

Consider the given matrix

Given matrix: \[ A = \begin{pmatrix} x_2^2 & x_1 x_2 \ x_1 x_2 & x_1^2 \end{pmatrix} \]
08

Apply transformation

For element \(A_{11} \) after transformation: \[ A_{11}' = T_{1k} T_{1l} A_{kl} \] Calculate to check if it transforms correctly.
09

Check transformation for matrix element

Calculate \( A_{11} = x_2^2 \): \[ A_{11}' = \cos^2\phi \cdot x_2^2 + 2 \cos\phi \sin\phi \cdot (x_1 x_2) + \sin^2\phi \cdot x_1^2 \]
10

Identify mismatch

Upon calculation, element \( A_{11} \) will not yield \( x_2^2 \) back, proving it does not transform as a second-order tensor.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

first-order tensor
A first-order tensor, also known as a vector, is a quantity that has both magnitude and direction. It transforms linearly when coordinates are rotated. Let's consider the transformation equations given:
\[ u_{1} = x_{1} \cos\text{\( phi \) } - x_{2} \sin\text{\( phi \) } \] and \[ u_{2} = x_{1} \sin\text{\( phi \) } + x_{2} \cos\text{\( phi \) } \].
These equations demonstrate the components of a first-order tensor in two dimensions. The components change according to the angle of rotation \( \phi \). As per the first-order tensor transformation property, we have \( u_{i}^{'} = T_{ij} x_{j} \), where \( T_{ij} \) is the transformation matrix and \( x_{j} \) is the original coordinate. By matching the results obtained after applying the rotation transformation matrix, we can verify that the given components indeed satisfy the first-order tensor transformation rules.
transformation matrix
A transformation matrix is a square matrix used to perform linear transformations such as rotations, scaling, and shearing. It provides a compact way to represent these transformations. For rotation transformations in two dimensions, the transformation matrix \( T \) is defined as:
\[ T = \begin{pmatrix} \cos\text{\( phi \) } & -\sin\text{\( phi \) } \ \sin\text{\( phi \) } & \cos\text{\( phi \) } \end{pmatrix} \]
By applying this matrix to the original coordinates \( x_{1} \) and \( x_{2} \), we obtain:
\[ \begin{pmatrix} u_{1} \ u_{2} \end{pmatrix} = \begin{pmatrix} \cos\text{\( phi \) } & -\sin\text{\( phi \) } \ \sin\text{\( phi \) } & \cos\text{\( phi \) } \end{pmatrix} \begin{pmatrix} x_{1} \ x_{2} \end{pmatrix} \]
After the multiplication, we get:
\( u_{1} = x_{1} \cos\text{\( phi \) } - x_{2} \sin\text{\( phi \) } \) and \( u_{2} = x_{1} \sin\text{\( phi \) } + x_{2} \cos\text{\( phi \) } \).
These results confirm that the vector transformation due to rotation is governed by the transformation matrix.
second-order tensor
A second-order tensor, also known as a matrix, is a mathematical object that can be represented as a 2D array of numbers. It transforms according to specific rules when the coordinate system is rotated. The general transformation rule for a second-order tensor \( A \) is given by:
\[ A_{ij}^{'} = T_{ik} T_{jl} A_{kl} \]
where \( T \) is the rotation matrix and \( A_{kl} \) are the elements of the tensor. In the given exercise, the matrix:
\[ A = \begin{pmatrix} x_{2}^{2} & x_{1} x_{2} \ x_{1} x_{2} & x_{1}^{2} \end{pmatrix} \]
is provided. By considering the transformation of the element \( A_{11} \), we need to calculate:
\( A_{11}^{'} = T_{1k} T_{1l} A_{kl} \).
After performing these calculations, it becomes evident that the transformed element \( A_{11} eq x_{2}^{2} \), indicating that the given matrix does not transform correctly according to the second-order tensor transformation rules.
rotation transformation
Rotation transformation involves rotating an object around a point or an axis. The amount of rotation is represented by the angle \( \phi \). In two-dimensional space, a point \( x \) with coordinates \( (x_{1}, x_{2}) \) can be rotated to a new point \( u \) with coordinates \( (u_{1}, u_{2}) \) using the rotation transformation matrix \( T \). The transformation equations to apply are:
\[ u_{1} = x_{1} \cos\text{\( phi \) } - x_{2} \sin\text{\( phi \) } \] and
\[ u_{2} = x_{1} \sin\text{\( phi \) } + x_{2} \cos\text{\( phi \) } \]
This rotation preserves the length of the vector, meaning it is an isometric transformation. Each new component is a combination of the sine and cosine of the rotation angle and the original coordinates. This ensures that every rotation aligns with the geometric interpretation of rotating a point around the origin by an angle \( \phi \).

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Most popular questions from this chapter

Use tensor methods to establish the following vector identities: (a) \((\mathbf{u} \times \mathbf{v}) \times \mathbf{w}=(\mathbf{u} \cdot \mathbf{w}) \mathbf{v}-(\mathbf{v} \cdot \mathbf{w}) \mathbf{u}\) (b) \(\operatorname{curl}(\phi \mathbf{u})=\phi \operatorname{curl} \mathbf{u}+(\operatorname{grad} \phi) \times \mathbf{u}\) (c) \(\operatorname{div}(\mathbf{u} \times \mathbf{v})=\mathbf{v} \cdot \operatorname{curl} \mathbf{u}-\mathbf{u} \cdot \operatorname{curl} \mathbf{v}\) (d) \(\operatorname{curl}(\mathbf{u} \times \mathbf{v})=(\mathbf{v} \cdot \mathbf{g r a d}) \mathbf{u}-(\mathbf{u} \cdot \mathbf{g r a d}) \mathbf{v}+\mathbf{u} \operatorname{div} \mathbf{v}-\mathbf{v} \operatorname{div} \mathbf{u}\) (e) \(\operatorname{grad} \frac{1}{2}(\mathbf{u} \cdot \mathbf{u})=\mathbf{u} \times \operatorname{curl} \mathbf{u}+(\mathbf{u} \cdot \operatorname{grad}) \mathbf{u}\).

The components of two vectors \(\mathbf{A}\) and \(\mathbf{B}\) and a second-order tensor \(\mathbf{T}\) are given in one coordinate system by $$ \mathrm{A}=\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right), \quad \mathrm{B}=\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right), \quad \mathrm{T}=\left(\begin{array}{ccc} 2 & \sqrt{3} & 0 \\ \sqrt{3} & 4 & 0 \\ 0 & 0 & 2 \end{array}\right) $$ In a second coordinate system, obtained from the first by rotation, the components of \(\mathbf{A}\) and \(\mathbf{B}\) are $$ \mathrm{A}^{\prime}=\frac{1}{2}\left(\begin{array}{c} \sqrt{3} \\ 0 \\ 1 \end{array}\right), \quad \mathrm{B}^{\prime}=\frac{1}{2}\left(\begin{array}{c} -1 \\ 0 \\ \sqrt{3} \end{array}\right) $$ Find the components of \(\mathbf{T}\) in this new coordinate system and hence evaluate, with a minimum of calculation, $$ T_{i j} T_{j i}, \quad T_{k i} T_{j k} T_{i j}, \quad T_{i k} T_{m n} T_{n i} T_{k m} $$

We may define Christoffel symbols of the first kind by $$ \Gamma_{i j k}=g_{i l} \Gamma_{j k}^{l} $$ Show that these are given by $$ \Gamma_{i j k}=\frac{1}{2}\left(\frac{\partial g_{i k}}{\partial u^{j}}+\frac{\partial g_{j k}}{\partial u^{i}}-\frac{\partial g_{i j}}{\partial u^{k}}\right) $$ By permuting indices, verify that $$ \frac{\partial g_{i j}}{\partial u^{k}}=\Gamma_{i j k}+\Gamma_{j k k} $$ Using the fact that \(\Gamma_{j k}^{l}=\Gamma_{k j}^{l}\), show that $$ g_{i j ; k} \equiv 0 $$ i.e. that the covariant derivative of the metric tensor is identically zero in all coordinate systems.

In a general coordinate system \(u^{i}, i=1,2,3\), in three-dimensional Euclidean space, a volume element is given by $$ d V=\left|\mathbf{e}_{1} d u^{1} \cdot\left(\mathbf{e}_{2} d u^{2} \times \mathbf{e}_{3} d u^{3}\right)\right| $$ Show that an alternative form for this expression, written in terms of the determinant \(g\) of the metric tensor, is given by $$ d V=\sqrt{g} d u^{1} d u^{2} d u^{3} $$ Show that under a general coordinate transformation to a new coordinate system \(u^{i}\) the volume element \(d V\) remains unchanged, i.e. show that it is a scalar quantity.

Given a non-zero vector \(\mathbf{v}\), find the value that should be assigned to \(\alpha\) to make $$ P_{i j}=\alpha v_{i} v_{j} \quad \text { and } \quad Q_{i j}=\delta_{i j}-\alpha v_{i} v_{j} $$ into parallel and orthogonal projection tensors respectively, i.e. tensors that satisfy respectively \(P_{i j} v_{j}=v_{i}, P_{i j} u_{j}=0\) and \(Q_{i j} v_{j}=0, Q_{i j} u_{j}=u_{i}\), for any vector \(\mathbf{u}\) that is orthogonal to \(\mathbf{v}\), Show, in particular, that \(Q_{i j}\) is unique, i.e. that if another tensor \(T_{i j}\) has the same properties as \(Q_{i j}\) then \(\left(Q_{i j}-T_{i j}\right) w_{j}=0\) for any vector \(\mathbf{w}\).
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