Jacobian

Relevant parts to questions...

• Using $\frac{\partial x_i^{\prime }}{\partial x_j}=L_{ij}$ and $\frac{\partial x_i}{\partial x_j^{\prime }}=L_{ji}$ to convert index-derivatives into entries of the Coordinate Transformation Matrix.
• Using these inside the chain rule: $\frac{\partial f}{\partial x_i^{\prime }}=\frac{\partial f}{\partial x_j}\frac{\partial x_j}{\partial x_i^{\prime }}=L_{ij}\frac{\partial f}{\partial x_j}$.
• Using them to prove that $\nabla f$ is a vector and $\frac{\partial u_i}{\partial x_j}$ is a rank-2 tensor.

The Jacobian entries are the partial derivatives of one coordinate system with respect to another. They tie the Coordinate Transformation Matrix $L_{ij}$ directly to calculus-style index manipulation.

Starting from $x_i^{\prime }=L_{ik}{x}_k$ and differentiating:

$$\frac{\partial x_i^{\prime }}{\partial x_j}=\sum _k\left(\frac{\partial L_{ik}}{\partial x_j}{x}_k+L_{ik}\frac{\partial x_k}{\partial x_j}\right)=\sum _k{L}_{ik}{\delta }_{kj}=L_{ij}$$

Because $L$ is constant in Cartesian coordinates, $\frac{\partial L_{ik}}{\partial x_j}=0$, and the only surviving term comes from $\frac{\partial x_k}{\partial x_j}={\delta }_{kj}$. The same argument, applied to the inverse transformation $x_i=L_{ji}{x}_j^{\prime }$, gives the second identity:

$$\boxed{\frac{\partial x_i^{\prime }}{\partial x_j}=L_{ij},\frac{\partial x_i}{\partial x_j^{\prime }}=L_{ji}}$$

Properties

• Constant in Cartesian coordinates::$L_{ij}$ does not depend on position, so its derivatives vanish. This is exactly the property that fails in curvilinear coordinates, and the reason Christoffel Symbols exist.
• Forward and inverse swap the indices::the two Jacobians are each other’s transposes, matching $L^T=L^{-1}$.
• Orthogonality-inherited identity::$\frac{\partial x_i^{\prime }}{\partial x_j}\frac{\partial x_k^{\prime }}{\partial x_j}=L_{ij}{L}_{kj}={\delta }_{ik}$.

Applications

1. Chain-rule conversion of derivatives, using the Jacobian::$\frac{\partial f}{\partial x_i^{\prime }}=\frac{\partial f}{\partial x_j}\frac{\partial x_j}{\partial x_i^{\prime }}=L_{ij}\frac{\partial f}{\partial x_j}$.
2. Proving the Gradient is a vector, via the chain rule::$[\nabla f{]}_i^{\prime }=\frac{\partial f}{\partial x_i^{\prime }}=L_{ij}\frac{\partial f}{\partial x_j}=L_{ij}[\nabla f{]}_j$.
3. Proving $\frac{\partial u_i}{\partial x_j}$ is a rank-2 tensor::combining the product rule with $\frac{\partial L_{ik}}{\partial x_j^{\prime }}=0$ and the chain rule gives $\frac{\partial u_i^{\prime }}{\partial x_j^{\prime }}=L_{ik}{L}_{j\ell }\frac{\partial u_k}{\partial x_{\ell }}$.

Cartesian-only argument

The step $\frac{\partial L_{ik}}{\partial x_j^{\prime }}=0$ relies on $L$ being constant. In curvilinear coordinates (spherical, cylindrical, etc.) the transformation varies with position, and this derivation breaks. That is exactly the motivation for generalised coordinates and covariant differentiation.

Convert a derivative; pick up an $L$.