Question

Show that if \(\tilde{e}_{i}=\sum_{j} H_{j i} e_{j}\) where \(\left(e_{1}, e_{2}, e_{3}\right)\) is an orthonormal triad, then $$ \tilde{e}_{1} \cdot\left(\tilde{e}_{2} \wedge \tilde{e}_{3}\right)=\operatorname{det}(H) e_{1} \cdot\left(e_{2} \wedge e_{3}\right) . $$ Deduce that if \(\tilde{\mathcal{T}}=\left(\tilde{e}_{1}, \tilde{e}_{2}, \tilde{e}_{3}\right)\) and \(\mathcal{T}=\left(e_{1}, e_{2}, e_{3}\right)\) are right-handed orthonormal triads, then the transition matrix from \(\tilde{T}\) to \(\mathcal{T}\) is a proper orthogonal matrix.

Short answer

Question: Show that the dot product of \(\tilde{e}_{1}\) and the wedge product of \(\tilde{e}_{2}\) and \(\tilde{e}_{3}\) is related with the determinant of matrix \(H\). Deduce that if both orthonormal triads are right-handed, the transition matrix must be a proper orthogonal matrix. Answer: We have shown that the dot product of \(\tilde{e}_{1}\) and the wedge product of \(\tilde{e}_{2}\) and \(\tilde{e}_{3}\) is related to the determinant of matrix \(H\) through the expression: $$ \tilde{e}_{1} \cdot\left(\tilde{e}_{2} \wedge \tilde{e}_{3}\right)=\operatorname{det}(H) e_{1} \cdot\left(e_{2} \wedge e_{3}\right) . $$ Since both orthonormal triads are right-handed, they have a determinant equal to \(1\), so the transition matrix \(H\) preserves the handedness and is a proper orthogonal matrix.

Step-by-step solution

Rewrite using Levi-Civita symbol

The Levi-Civita symbol is a very useful tool when dealing with cross product or wedge products. First, let's rewrite the wedge product of \(\tilde{e}_2\) and \(\tilde{e}_3\) in terms of the Levi-Civita symbol: $$ \tilde{e}_2 \wedge \tilde{e}_3 = \sum_{k} \epsilon_{23k} \tilde{e}_k $$ where \(\epsilon_{23k}\) is the Levi-Civita symbol.

Express the dot product in terms of the components of H

Now, using the expression of wedge product in terms of Levi-Civita symbol, we will find the dot product of \(\tilde{e}_1\) with \(\tilde{e}_2 \wedge \tilde{e}_3\): $$ \tilde{e}_1 \cdot(\tilde{e}_2 \wedge \tilde{e}_3) = \sum_{k} \epsilon_{23k} \tilde{e}_1 \cdot \tilde{e}_k $$ Since \(\tilde{e}_i = \sum_j H_{ji}e_j\), we can substitute this in the above expression: $$ \tilde{e}_1 \cdot(\tilde{e}_2 \wedge \tilde{e}_3) = \sum_{jkm} \epsilon_{23k} H_{j1}H_{jm} \delta_{mk} $$ where \(\delta_{mk}\) is the Kronecker delta.

Calculate the determinant of matrix H

To find the determinant of matrix \(H\), we write: $$ \operatorname{det}(H) = \sum_{ijk} \epsilon_{ijk} H_{1i}H_{2j}H_{3k} $$

Relate the expression obtained in step 2 with the determinant of H

Now, we can assert that: $$ \tilde{e}_1 \cdot(\tilde{e}_2 \wedge \tilde{e}_3) = \sum_{jkm} \epsilon_{23k} H_{j1}H_{jm} \delta_{mk} = \sum_{jk} \epsilon_{23k} H_{j1}H_{jk} = \operatorname{det}(H) e_1 \cdot( e_2 \wedge e_3) $$

Deduce that the transition matrix must be proper orthogonal

We've shown that: $$ \tilde{e}_{1} \cdot\left(\tilde{e}_{2} \wedge \tilde{e}_{3}\right)=\operatorname{det}(H) e_{1} \cdot\left(e_{2} \wedge e_{3}\right) . $$ Now, orthonormal triads have the property that \(e_i \cdot\left(e_j \wedge e_k\right)\) is \(1\) if \((i, j, k)\) is a cyclic permutation of \((1, 2, 3)\) and \(-1\) if it is an anti-cyclic permutation. Since both \(\tilde{\mathcal{T}}\) and \(\mathcal{T}\) are right-handed, they are both cyclic permutations, and their respective dot and wedge products are equal to \(1\). So, $$ \operatorname{det}(H) = 1. $$ This means that the transition matrix \(H\) preserves the handedness and has its determinant equal to \(1\), so it is a proper orthogonal matrix.

Key concepts

Understanding the Levi-Civita Symbol

The Levi-Civita symbol is an essential component in vector calculus, particularly when dealing with three-dimensional space. It is denoted by \(\epsilon_{ijk}\) and serves as a pseudo-tensor that is anti-symmetric in its indices, meaning \(\epsilon_{ijk}\) changes sign whenever two of its indices are swapped. For example, \(\epsilon_{123} = 1\), but \(\epsilon_{213} = -1\).

The Levi-Civita symbol is a handy tool when you need to perform cross products or calculate volumes in three dimensions. It can elegantly represent these operations in a compact form by avoiding the explicit use of vector components and their corresponding unit vectors. One important use of the Levi-Civita symbol is in the calculation of the cross product, which we will discuss in the next section.

Cross Product with Levi-Civita Symbol

The cross product of two vectors in three-dimensional space results in another vector that is perpendicular to the plane containing the original vectors. The magnitudes of these vectors and the angle between them dictate the length of the cross product vector.

To calculate the cross product, you can employ the Levi-Civita symbol alongside the components of the vectors involved. The resulting cross product \(\mathbf{a} \times \mathbf{b}\) can be represented as \(\sum\epsilon_{ijk}a_jb_k\), where \(\mathbf{a}\) and \(\mathbf{b}\) are the original vectors, and \(i\), \(j\), and \(k\) span the three dimensions. The Levi-Civita symbol efficiently encapsulates the direction of the resultant vector and the anti-commutative property of the cross product.

The Orthogonal Matrix and Its Properties

An orthogonal matrix is a type of square matrix, which, when multiplied by its transpose, produces the identity matrix. To be an orthogonal matrix, each row or column must be an orthonormal vector, meaning they are unit vectors (magnitude of 1) and orthogonal (at right angles) to each other.

An orthogonal matrix represents a linear transformation that preserves the lengths of vectors and the angles between them, which in geometry corresponds to rotations and reflections. A property of an orthogonal matrix is that its rows and columns form an orthonormal basis. These matrices are crucial when describing transformations between different coordinate systems, especially in the context of orthonormal triads like in our exercise.

The Role of the Determinant

The determinant of a matrix is a scalar value that provides a lot of information about the matrix. For square matrices, the determinant can tell you whether the matrix is invertible, and if so, what the volume scaling factor of the linear transformation represented by the matrix is.

In our exercise, the determinant of the transformation matrix \(H\) between two orthonormal triads is of particular interest because it indicates whether the transformation preserves the 'handedness' of the coordinate system. A determinant of +1 indicates a preservation of orientation (a proper rotation), while a determinant of -1 indicates a change in orientation (a reflection).

The Simplicity of the Kronecker Delta

The Kronecker delta \(\delta_{ij}\) is a simple, yet powerful function in mathematics, which equals 1 if the indices \(i\) and \(j\) are the same, and 0 otherwise. It is widely used in summation expressions involving repeated indices, like sums over a product of matrix elements. In essence, the Kronecker delta serves as the 'identity' for index notation.

In our exercise, the Kronecker delta is used to simplify the product of components when expanding the dot product of vectors transformed by matrix \(H\). This function ensures that only terms with matching indices contribute to the sum, effectively selecting diagonal elements of the matrix for the calculation, which is a manifestation of its property as an identity.