Question

Let \(x=u+v, y=v .\) Find \(d \mathbf{s},\) the a vectors, and \(d s^{2}\) for the \(u, v\) coordinate system and show that it is not an orthogonal system. Hint: Show that the a vectors are not orthogonal, and that \(d s^{2}\) contains \(d u d v\) terms. Write the \(g_{i j}\) matrix and observe that it is symmetric but not diagonal. Sketch the lines \(u=\) const. and \(v=\) const. and observe that they are not perpendicular to each other.

Short answer

The system is not orthogonal because the basis vectors are not orthogonal, the metric tensor contains off-diagonal terms, and the coordinate lines are not perpendicular.

Step-by-step solution

Express coordinates and define basis vectors

Given the transformations, express the coordinates in terms of the new variables:For the coordinate system, define: \( x = u + v \)\( y = v \)Next, determine the basis vectors for each coordinate.

Determine the basis vectors

Compute the partial derivatives to find the basis vectors in the new coordinate system. \( \mathbf{a}_u = \frac{\partial \vec{r}}{\partial u} = (1, 0) \)\( \mathbf{a}_v = \frac{\partial \vec{r}}{\partial v} = (1, 1) \)Here, \( \vec{r} \) is the position vector in the original coordinate system (typically \( (x, y) \)).

Calculate the dot product of basis vectors

Calculate the dot product of the basis vectors to assess orthogonality:\[ \mathbf{a}_u \cdot \mathbf{a}_v = (1, 0) \cdot (1, 1) = 1 \]Since the dot product is non-zero, the vectors are not orthogonal.

Compute the line element \(d s^2\)

Calculate the line element using the basis vectors:\[ d\mathbf{s} = \mathbf{a}_u du + \mathbf{a}_v dv \]\[ d\mathbf{s} \cdot d\mathbf{s} = \mathbf{a}_u \cdot \mathbf{a}_u (du)^2 + 2 \mathbf{a}_u \cdot \mathbf{a}_v du dv + \mathbf{a}_v \cdot \mathbf{a}_v (dv)^2 \]Substitute the dot products:\[ d\mathbf{s} \cdot d\mathbf{s} = (du)^2 + 2 du dv + (dv)^2 \]This contains the \( du dv \) term.

Construct the metric tensor matrix \(g_{ij}\)

The metric tensor (\(g_{ij}\)) can be written as:\[\mathbf{G} = \begin{pmatrix} \mathbf{a}_u \cdot \mathbf{a}_u & \mathbf{a}_u \cdot \mathbf{a}_v \ \mathbf{a}_v \cdot \mathbf{a}_u & \mathbf{a}_v \cdot \mathbf{a}_v \end{pmatrix} \]Substituting the dot products:\[\mathbf{G} = \begin{pmatrix} 1 & 1 \ 1 & 2\end{pmatrix} \]The matrix is symmetric but not diagonal, confirming non-orthogonality.

Sketch the coordinate lines

Sketch the lines for constant \(u\) and \(v\):For constant \(u\): \( x = u + v \rightarrow x - v = u \)For constant \(v\): \( y = v \)Observe that the lines are not perpendicular in the \(xy\) plane.

Key concepts

Coordinate Transformation

In mathematical modeling and geometry, coordinate transformation involves changing from one coordinate system to another. In our given problem, we transform from the original coordinate system (typically denoted as \((x, y)\)) to a new one defined by \(u\) and \(v\). The transformation rules given are \(x = u + v\) and \(y = v\). This redefinition of coordinates helps in simplifying certain problems or providing different perspectives.

To achieve this, we'll express the position vector \( \vec{r} \) in terms of the new variables. Here, the original coordinates \(x\) and \(y\) are written as functions of \(u\) and \(v\). This sets the foundation for finding new basis vectors and understanding the system's geometry.

Basis Vectors

Basis vectors form the foundation of a coordinate system. For a given transformation like \(x = u + v\) and \(y = v\), we define basis vectors in the new coordinate system by computing partial derivatives.

Basis vectors \(\mathbf{a}_u\) and \(\mathbf{a}_v\) are given by:

\[ \mathbf{a}_u = \frac{\partial \vec{r}}{\partial u} = (1, 0) \]
\[ \mathbf{a}_v = \frac{\partial \vec{r}}{\partial v} = (1, 1) \]

Here, \(\vec{r}\) represents the position vector in the original coordinate system.

Each basis vector \(\mathbf{a}_i\) corresponds to the partial derivative of the position vector with respect to the new coordinates. These vectors capture how small changes in \(u\) and \(v\) translate into changes in the original coordinate system.

Dot Product

The dot product of two vectors is a fundamental operation that tells us if the vectors are orthogonal (i.e., perpendicular). For vectors \(\mathbf{a}_u\) and \(\mathbf{a}_v\):

\[ \mathbf{a}_u \cdot \mathbf{a}_v = (1, 0) \cdot (1, 1) = 1 \]

Since the dot product is non-zero, the vectors are not orthogonal.

The non-zero result indicates that \(\mathbf{a}_u\) and \(\mathbf{a}_v\) do not form a right angle. In terms of the coordinate system, this implies that lines of constant \(u\) and \(v\) intersect at non-right angles, confirming that our coordinate system is not orthogonal.

Metric Tensor

The metric tensor \(g_{ij}\) is a key concept in differential geometry, encapsulating the information about the coordinate system's geometry. It's constructed using the dot products of the basis vectors. For our basis vectors \(\mathbf{a}_u\) and \(\mathbf{a}_v\), we have:

\[ \mathbf{G} = \begin{pmatrix} \mathbf{a}_u \cdot \mathbf{a}_u & \mathbf{a}_u \cdot \mathbf{a}_v \ \mathbf{a}_v \cdot \mathbf{a}_u & \mathbf{a}_v \cdot \mathbf{a}_v \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 1 & 2 \end{pmatrix} \]

This matrix is symmetric, which is a general property of metric tensors. However, it's not diagonal, reflecting the non-orthogonality of the coordinate system. Diagonal elements represent the squared lengths of basis vectors, while off-diagonal elements indicate the inner products (or angles). The presence of non-zero off-diagonal elements confirms the coordinate system’s non-orthogonal nature.

Understanding the metric tensor is crucial for analyzing the geometry of any space, helping us to measure distances, angles, and volumes accurately in different coordinate systems.