Question

Show that in a general coordinate system with variables x₁, x₂, x₃, the contravariant basis vectors are given by

${\mathit{a}}^{\mathbf{i}}\mathbf{=}\mathbf{\nabla }{\mathit{x}}_{\mathbf{i}}\mathbf{=}\mathit{i}\frac{\mathbf{\partial }{\mathbf{x}}_{\mathbf{i}}}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\mathit{j}\frac{\mathbf{\partial }{\mathbf{x}}_{\mathbf{i}}}{\mathbf{\partial }\mathbf{y}}\mathbf{+}\mathit{k}\frac{\mathbf{\partial }{\mathbf{x}}_{\mathbf{i}}}{\mathbf{\partial }\mathbf{z}}$

Hint:Write the gradient in terms of its covariant components and the basis

vectors to get$\mathbf{\nabla }\mathit{u}\mathbf{=}{\mathit{a}}^{\mathbf{j}}\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }{\mathbf{x}}_{\mathbf{j}}}$and let$\mathit{u}\mathbf{=}{\mathit{x}}_{\mathbf{i}}$ .

Short answer

The equation is proved.

${\mathrm{a}}^{\mathrm{i}}=\frac{\partial {\mathrm{x}}_{\mathrm{i}}}{\partial \mathrm{x}}{\hat{\mathrm{e}}}_{\mathrm{x}}+\frac{\partial {\mathrm{x}}_{\mathrm{i}}}{\partial \mathrm{y}}{\hat{\mathrm{e}}}_{\mathrm{y}}+\frac{\partial {\mathrm{x}}_{\mathrm{i}}}{\partial \mathrm{z}}{\hat{\mathrm{e}}}_{\mathrm{z}}$

Step-by-step solution

Given information.

A general coordinate system is defined with contravariant basis vectors.

Definition of covariance and contravariance.

The components of a vector relative to a tangent bundle basis are covariant in differential geometry if they change with the same linear transformation as the basis. If they change as a result of the inverse transformation, they are contravariant.

Utilize the given formula to solve.

Utilize the formula $\frac{\partial {\mathrm{x}}_{\mathrm{i}}}{\partial {\mathrm{x}}_{\mathrm{k}}}={\mathrm{\delta }}_{\mathrm{k}}^{\mathrm{i}}$to solve the question.

Make appropriate substitution in the gradient equation.

Substitute $\mathrm{u}={\mathrm{x}}_1$in the equation of gradient in the contravariant basis.

$$\begin{gathered} \nabla \mathrm{u}={\mathrm{a}}^{\mathrm{k}}\frac{\partial \mathrm{u}}{\partial {\mathrm{x}}_{\mathrm{k}}} \\ \nabla {\mathrm{x}}_{\mathrm{i}}={\mathrm{a}}^{\mathrm{k}}\frac{\partial {\mathrm{x}}_{\mathrm{i}}}{\partial {\mathrm{x}}_{\mathrm{k}}} \\ \nabla {\mathrm{x}}_{\mathrm{i}}={\mathrm{a}}^{\mathrm{i}} \end{gathered}$$

Evaluate the previous equation.

From the previous equation, make an inference.

${\mathrm{a}}^{\mathrm{i}}=\frac{\partial {\mathrm{x}}_{\mathrm{i}}}{\partial \mathrm{x}}{\hat{\mathrm{e}}}_{\mathrm{x}}+\frac{\partial {\mathrm{x}}_{\mathrm{i}}}{\partial \mathrm{y}}{\hat{\mathrm{e}}}_{\mathrm{y}}+\frac{\partial {\mathrm{x}}_{\mathrm{i}}}{\partial \mathrm{z}}{\hat{\mathrm{e}}}_{\mathrm{z}}$