This problem beautifully illustrates how a non-orthogonal coordinate system impacts fundamental geometric measurements. The most important result is the non-zero off-diagonal term in the metric tensor, $g_{12}=1$, which is the defining characteristic of a non-orthogonal system, confirming that the new basis vectors are not perpendicular. Furthermore, the diagonal element $g_{22}=2$ shows the basis vector $E_2$ is not normalized (it has a length of $\sqrt{2}$ ). This non-trivial metric structure means that the formula for the length of a curve must include a cross-term $\left(2 \frac{d y^1}{d t} \frac{d y^2}{d t}\right)$, which accounts for the angle between the axes. If the system were Cartesian, this term would vanish, simplifying the line element back to the standard Pythagorean formula.

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✍️Mathematical Proof

$\complement\cdots$Counselor

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• Non-Orthogonality is Quantified by Off-Diagonal Terms: The most crucial result is the non-zero off-diagonal term of the metric tensor ( $g_{12}=g_{21}=1$ ). A Cartesian (orthogonal) system would only have non-zero terms on the main diagonal. The presence of $g_{12}=1$ directly confirms that the basis vectors, $E_1=e_1$ and $E_2=e_1+e_2$, are not perpendicular to each other ( $E_1 \cdot E_2 \neq 0$ ).
• Non-Unit Basis Vectors: The diagonal component $g_{22}=2$ indicates that the basis vector $E_2$ is not a unit vector. Its length (scale factor) is $\sqrt{g_{22}}=\sqrt{2}$. In contrast, $g_{11}=1$ means $E_1$ is a unit vector.
• The Inverse Metric is Essential for Dual Bases: The inverse metric tensor, $g ^{a b}= \left(\begin{array}{cc}2 & -1 \\ -1 & 1\end{array}\right)$, is non-trivial. In general relativity and advanced tensor analysis, this matrix is used to raise indices and, critically, to find the dual basis vectors $E^a$. The dual basis vectors, $E^1$ and $E^2$, are orthogonal to the coordinate lines, but they are not the same as the tangent basis vectors $E_1$ and $E_2$.
• Line Length Requires Cross-Terms: The final expression for the length of a curve, $L= \int \sqrt{\ldots} d t$, includes the cross-term $2 \frac{d y^1}{d t} \frac{d y^2}{d t}$. This term is necessary because the coordinate lines are non-orthogonal. If the coordinate system were orthogonal ( $g_{12}=0$ ), this cross-term would vanish, and the line element would simplify to the familiar Pythagorean form (just the squared derivatives).

✍️Mathematical Proof

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1. Derivation of Tensor Transformation Properties for Mixed Tensors
2. The Polar Tensor Basis in Cartesian Form
3. Verifying the Rank Two Zero Tensor
4. Tensor Analysis of Electric Susceptibility in Anisotropic Media
5. Analysis of Ohm's Law in an Anisotropic Medium
6. Verifying Tensor Transformations
7. Proof of Coordinate Independence of Tensor Contraction
8. Proof of a Tensor's Invariance Property
9. Proving Symmetry of a Rank-2 Tensor
10. Tensor Symmetrization and Anti-Symmetrization Properties
11. Symmetric and Antisymmetric Tensor Contractions
12. The Uniqueness of the Zero Tensor under Specific Symmetry Constraints
13. Counting Independent Tensor Components Based on Symmetry
14. Transformation of the Inverse Metric Tensor
15. Finding the Covariant Components of a Magnetic Field
16. Covariant Nature of the Gradient
17. Christoffel Symbol Transformation Rule Derivation
18. Contraction of the Christoffel Symbols and the Metric Determinant
19. Divergence of an Antisymmetric Tensor in Terms of the Metric Determinant
20. Calculation of the Metric Tensor and Christoffel Symbols in Spherical Coordinates
21. Christoffel Symbols for Cylindrical Coordinates
22. Finding Arc Length and Curve Length in Spherical Coordinates
23. Solving for Metric Tensors and Christoffel Symbols
24. Metric Tensor and Line Element in Non-Orthogonal Coordinates

🧄Proof and Derivation-1

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