Question

Do Problems for the ( \(u, v\) ) coordinate system if \(x=u(1-v), y=u \sqrt{2 v-v^{2}}\). Find \(d s^{2},\) the scale factors, the area element, the vector \(d \mathbf{s},\) the unit basis vectors, and the covariant and contravariant basis vectors.

Short answer

Transform equations: x = u(1-v), y = u√(2v - v²). Metric tensor components: g₁₁, g₁₂, g₂₂. ds²: g₁₁du² + 2g₁₂dudv + g₂₂dv². Scale factors: h_u, h_v. Area element: h_u h_v du dv.

Step-by-step solution

Express the transformation equations

The coordinates transformation is given by: \( x=u(1-v) \) and \( y=u \sqrt{2v-v^{2}} \).

Calculate partial derivatives of the coordinates

Find the partial derivatives of \( x \) and \( y \) with respect to \( u \) and \( v \): \( \frac{\partial x}{\partial u} = 1-v \)\( \frac{\partial x}{\partial v} = -u \)\( \frac{\partial y}{\partial u} = \sqrt{2v-v^{2}} \)\( \frac{\partial y}{\partial v} = \frac{u(2-v)}{2\sqrt{2v-v^{2}}} \)

Calculate the metric tensor components

Calculate the metric tensor components using the partial derivatives:\[ g_{11} = \left(\frac{\partial x}{\partial u}\right)^2 + \left(\frac{\partial y}{\partial u}\right)^2 \]\[ g_{12} = g_{21} = \frac{\partial x}{\partial u}\frac{\partial x}{\partial v} + \frac{\partial y}{\partial u}\frac{\partial y}{\partial v} \]\[ g_{22} = \left(\frac{\partial x}{\partial v}\right)^2 + \left(\frac{\partial y}{\partial v}\right)^2 \]

Evaluate metric tensor components

Using the partial derivatives calculated:\( g_{11} = (1-v)^2 + (\sqrt{2v-v^2})^2 = 1 - 2v + v^2 + 2v - v^2 = 1 \)\( g_{12} = g_{21} = (1-v)(-u) + (\sqrt{2v-v^2})(\frac{u(2-v)}{2\sqrt{2v-v^2}}) = -u(1-v) + u(2-v)/2 = -u + uv + u - v\)\( g_{22} = (-u)^2 + (\frac{u(2-v)}{\sqrt{2v-v^2}})^2/4 = u^2 + u^2(2-v)^2/4 = u^2(1+(2-v)^2/4) \)

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

The line element is given by:\[ ds^2 = g_{11} du^2 + 2g_{12} dudv + g_{22} dv^2\]Substitute the metric tensor components to find \(ds^2\).

Determine the scale factors

The scale factors \( h_u \) and \( h_v \) are related to the metric tensor by:\[ h_u = \sqrt{g_{11}} \]\[ h_v = \sqrt{g_{22}} \]

Find the area element

The area element \( dA \) is given by:\[ dA = h_u h_v du dv \]

Calculate the vector ds

The vector \(d \mathbf{s}\) is given by:\[ d \mathbf{s} = \left(h_u\frac{du}{ds}, h_v\frac{dv}{ds}\right) \]

Determine the unit basis vectors

The unit basis vectors are:\( \mathbf{e}_u = \frac{1}{h_u}\left(\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}\right) \)\( \mathbf{e}_v = \frac{1}{h_v}\left(\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}\right) \)

Find the covariant and contravariant basis vectors

The covariant basis vectors are found from the partial derivatives:\( \mathbf{e}_u \) and \( \mathbf{e}_v \).The contravariant basis vectors are obtained using the inverse of the metric tensor.

Key concepts

Metric Tensor

In physics, the metric tensor is a crucial concept when dealing with coordinate transformations. It helps us understand the geometry of different coordinate systems.
When transforming coordinates, like from Cartesian \((x, y)\) to another set \((u, v)\), the metric tensor provides the coefficients that describe how distances and angles are calculated in the new system.
The metric tensor's components are derived from the partial derivatives of the transformation equations. For the given transformation \((x = u(1-v))\) and \((y = u \sqrt{2v-v^2}))\), the components are:
\[\begin{aligned} g_{11} &= (1-v)^2 + (\frac{\text{d}y}{\text{d}u})^2,\ g_{12} &= g_{21} = \frac{\text{d}x}{\text{d}u}\frac{\text{d}x}{\text{d}v} + \frac{\text{d}y}{\text{d}u}\frac{\text{d}y}{\text{d}v},\ g_{22} &= \frac{\text{d}x}{\text{d}v}^2 + \frac{\text{d}y}{\text{d}v}^2 \end{aligned}\]
This tensor tells us how lengths transform under the new coordinates.

Scale Factors

Scale factors simplify the understanding of how measurements change between coordinate systems.
In our transformation, the scale factors \(h_u \) and \(h_v\) are directly related to the metric tensor's diagonal components \((g_{11} \text{and} \g_{22}))\).
These are calculated as:
\[\begin{aligned} h_u &= \sqrt{g_{11}}, \ h_v &= \sqrt{g_{22}} \end{aligned} \]
Scale factors essentially 'stretch' or 'compress' measurements along the coordinate axes.
For instance, \((h_u) \) and \((h_v)\) help us determine how differentials such as \(ds\) and \(dA\) relate between the coordinate systems.

Basis Vectors

Basis vectors are the building blocks for vector spaces in different coordinate systems.
When coordinates transform, so do the basis vectors. Covariant basis vectors are obtained directly from partial derivatives of the coordinate transformation equations.
For our coordinates \((u, v)\), the covariant basis vectors \((\mathbf{e}_u \text{and} \mathbf{e}_v))\) are:
\[\begin{aligned} \mathbf{e}_u &= \frac{1}{h_u}\begin{pmatrix} \frac{\text{d}x}{\text{d}u} & \frac{\text{d}x}{\text{d}u} \end{pmatrix},\ \mathbf{e}_v &= \frac{1}{h_v}\begin{pmatrix} \frac{\text{d}x}{\text{d}v} & \frac{\text{d}x}{\text{d}v} \end{pmatrix} \end{aligned} \]
Contravariant basis vectors are found using the inverse of the metric tensor.
Understanding these vectors is crucial for describing any vector or change in vector in the transformed coordinate system.

Area Element

The area element \((dA))\) represents the infinitesimal area in the transformed coordinate system.
It is found by multiplying the scale factors of each coordinate:
\[\begin{aligned} dA = h_u \cdot h_v \cdot du \cdot dv \end{aligned} \]
For our transformation, the area element is:
\[\begin{aligned} dA = \sqrt{g_{11}} \cdot \sqrt{g_{22}} \cdot du \cdot dv = du dv \end{aligned} \]
This represents how a tiny area element in the \((x, y)\) coordinates is transformed in the \((u, v)\) system.
By understanding the area element, we can perform integrations over areas in the transformed coordinate system, ensuring accurate calculations.