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For a right-handed regular system of orthogonal coordinates, the gradient in orthogonal coordinates gives $\nabla u_i={\mathbf{e}}_i/h_i$. Thus $\varphi {\mathbf{e}}_1=(\varphi h_1)\nabla u_1$, and the curl product rule givesAlso $\nabla u_2\times \nabla u_3={\mathbf{e}}_1/(h_2{h}_3)$ and the divergence of this cross product is zero, by the divergence-of-a-cross-product identity and the vanishing curl of each gradient. Applying the divergence product rule to $\varphi {\mathbf{e}}_1=(\varphi h_2{h}_3)(\nabla u_2\times \nabla u_3)$ givesCyclic permutation and addition give the full component formulas for any vector field. This derivation accounts for the varying coordinate basis without differentiating its unit vectors explicitly. The factors $h_i>0$ are scale factors of orthogonal coordinates; the chart must avoid coordinate singularities.

For $a>0$, $u\ge 0$, $0\le v<2\pi$ and real $z$, these orthogonal coordinates have coordinate tangents $r_u=(a\sinh u\cos v,a\cosh u\sin v,0)$, $r_v=(-a\cosh u\sin v,a\sinh u\cos v,0)$ and $r_z=(0,0,1)$. Their scale factors of orthogonal coordinates are $h_u=h_v=a\sqrt{{\sinh }^{2}u+{\sin }^{2}v}$ and $h_z=1$; the ordered Jacobian determinant is positive away from the two focal lines. Positive constant $u$ gives confocal elliptical cylinders, while generic constant $v$ gives a hyperbolic-cylinder half-branch selected by the signs of $x$ and $y$. The limiting $u=0$ surface is the strip $y=0$, $|x|\le a$, with arbitrary $z$. Constant $z$ gives horizontal planes.

For a regular chart of orthogonal coordinates $\mathbf{r}(q)$, define $h_i=|{\partial }_{q_i}\mathbf{r}|>0$ and ${\mathbf{e}}_i=h_i^{-1}{\partial }_{q_i}\mathbf{r}$. The tangent unit vectors form an orthonormal basis. Matching $d\Phi =\nabla \Phi \cdot d\mathbf{r}$ with $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{q}_i$ proves the formula. Dividing by a scale factor converts a derivative per coordinate increment into a derivative per physical distance in that direction.

For focal distance $a>0$, $\xi \ge 0$ and $0\le \theta \le \pi$, the displayed prolate spheroidal coordinates are orthogonal coordinates. The three coordinate tangent vectors have lengths $h_{\xi }=h_{\theta }=a\sqrt{{\sinh }^2\xi +{\sin }^2\theta }$ and $h_{\varphi }=a\sinh \xi \sin \theta$. Their product gives the Jacobian determinant $a^3({\sinh }^2\xi +{\sin }^2\theta )\sinh \xi \sin \theta$. A positive constant $\xi$ gives a prolate spheroid with equatorial semiaxis $a\sinh \xi$, polar semiaxis $a\cosh \xi$, and foci $(0,0,\pm a)$. The surface degenerates to the focal segment at $\xi =0$; this coordinate degeneracy is distinct from a singularity of the surrounding Euclidean space.

Orthogonal coordinates have pairwise orthogonal coordinate tangent directions. Their metric is diagonal, with scale factors of orthogonal coordinates $h_i=|\partial \mathbf{x}/\partial q_i|$. Spherical coordinates and cylindrical coordinates are examples.

These orthogonal coordinates extend two confocal families of parabolas parallel to an axis. In the displayed scaling their scale factors of orthogonal coordinates are $h_u=h_v=2\sqrt{u^2+v^2}$ and $h_w=1$, so the Laplacian is $\Delta \varphi =({\varphi }_{uu}+{\varphi }_{vv})/[4(u^2+v^2)]+{\varphi }_{ww}$ away from $u=v=0$. With $u\ge 0$, signed $v$ distinguishes the upper and lower half-planes; the two signs at $u=0$ describe the same negative $x$ half-axis. Fixed positive $u^2$ or $v^2$ describes a parabolic cylinder. This differs from the rotational three-dimensional parabolic coordinates, whose coordinate surfaces are elliptic paraboloids.

These plane orthogonal coordinates use two families of confocal parabolas. On the region $y\ge 0$, take $u,v\ge 0$ and defineThen $x=(u^2-v^2)/2$ and $y=uv$. The coordinate tangent vectors $(u,v)$ and $(-v,u)$ have zero dot product, and both have length $\sqrt{u^2+v^2}$. Thus the Jacobian determinant is $u^2+v^2$, the area element is $(u^2+v^2)dudv$, and these orthogonal coordinates degenerate only at the origin. Fixed $u=a>0$ gives $x=(a^2-y^2/a^2)/2$; fixed $v=b>0$ gives $x=(y^2/b^2-b^2)/2$. This normalization is a rescaling of the plane section of parabolic cylindrical coordinates.

Without a restricted parameter domain, simultaneous sign reversal $(u,v)\mapsto (-u,-v)$ preserves the image, so these planar parabolic coordinates are globally two-to-one. The inverse function theorem supplies local inverse branches away from the origin. The nonnegative quadrant is one useful branch chart for the upper half-plane.

Vector identity. Work in Cartesian coordinates and use the Einstein summation convention. Contraction of the two Levi-Civita symbols givesThus the convective acceleration identity isEach term is a geometrically defined vector field: the directional derivative of a vector along itself, the gradient of a scalar field, and a cross product with the curl. Equality in a Cartesian frame is equality of these vectors, so it holds in every coordinate system. When using a varying coordinate basis, derivatives of the varying basis vectors must be included; differentiating only the component functions would describe a different operation.

Spherical metric. LetDifferentiating gives the coordinate tangent vectorsTheir pairwise dot products vanish, proving that spherical coordinates are orthogonal coordinates. Their lengths are the scale factors of orthogonal coordinates:These formulas use the usual chart $r>0,0<\theta <\pi$; the origin and polar axis are coordinate singularities.

Azimuthal field. Write the speed as $U(r)$ to distinguish it from the velocity field $\mathbf{u}=U(r)e_{\varphi }$. Using the curl in spherical coordinates,Also $\frac{1}{2}\nabla |\mathbf{u}{|}^2=U{U}^{\prime }{e}_r$. Since $e_{\varphi }\times e_r=e_{\theta }$ and $e_{\varphi }\times e_{\theta }=-e_r$,The $U{U}^{\prime }$ terms cancel in the convective acceleration identity, givingThis centripetal acceleration of an azimuthal flow points toward the polar axis. A streamline is a circle of cylindrical radius $r\sin \theta$, so its acceleration is $-U^2{e}_{\rho }/(r\sin \theta )$, with $e_{\rho }=\sin \theta e_r+\cos \theta e_{\theta }$. This independent geometric argument recovers both nonzero components and explains why $U^{\prime }$ cannot appear: the particle travels at fixed $r$.

For $r>0$, the chain rule gives $\partial r/\partial x_i=x_i/r$. Differentiating the radial factor gives $d\varphi /dr=e^{ikr}(ikr-1)/(4\pi r^2)$, so in Cartesian coordinatesThe origin is excluded because the scalar field is singular there.

The Stokes theorem states ${\int }_S(\nabla \times \mathbf{F})\cdot \mathbf{n}dS={\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{\ell }$ for a sufficiently smooth vector field and oriented surface with piecewise smooth boundary. The boundary orientation follows the right-hand rule relative to $\mathbf{n}$.

For the convective acceleration identity, use the Levi-Civita symbol contraction ${ϵ}_{ijk}{ϵ}_{k\ell m}={\delta }_{i\ell }{\delta }_{jm}-{\delta }_{im}{\delta }_{j\ell }$. It givesRearranging proves the required vector identity.

For orthogonal coordinates, define the scale factors of orthogonal coordinates $h_i=|\partial \mathbf{r}/\partial q_i|$ and tangent unit vectors ${\mathbf{e}}_i=h_i^{-1}\partial \mathbf{r}/\partial q_i$. They form an orthonormal basis, and $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{q}_i$. Equating $d\psi ={\sum }_i{\partial }_{q_i}\psi d{q}_i$ with $\nabla \psi \cdot d\mathbf{r}$ identifies each component. Thus the gradient in orthogonal coordinates isThis holds on a regular coordinate chart where all scale factors are positive.

In Cartesian coordinates, $\nabla \Phi =({\Phi }_x,{\Phi }_y,{\Phi }_z)$. For orthogonal coordinates, let $\mathbf{r}(q_1,q_2,q_3)$ be the position, $h_i=|\partial \mathbf{r}/\partial q_i|$ the scale factors of orthogonal coordinates, and ${\mathbf{e}}_i=h_i^{-1}\partial \mathbf{r}/\partial q_i$ the corresponding unit vectors. Since $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{q}_i$ and $d\Phi =\nabla \Phi \cdot d\mathbf{r}={\sum }_i{\Phi }_{q_i}d{q}_i$, comparison of independent increments gives the gradient in orthogonal coordinates:For the hyperbolic parametrization of oblate spheroidal coordinates, the three coordinate tangent vectors areTheir pairwise dot products are zero. Their squared Euclidean norms are ${\sinh }^{2}R+{\sin }^2\theta$, ${\sinh }^{2}R+{\sin }^2\theta$, and ${\cosh }^{2}R{\cos }^2\theta$, respectively. Thus the coordinate surfaces meet orthogonally wherever this is a regular coordinate chart, andThe last expression uses the usual latitude range $|\theta |<\pi /2$; without that range a scale factor is its absolute value. Axial or ring degeneracies are excluded from a regular chart.

A complex $3\times 3$ matrix is diagonalizable exactly when it has a basis of three linearly independent eigenvectors. If all its eigenvalues are one, a diagonalization would have $D=I$ and hence $A=PI{P}^{-1}=I$. Therefore the specified matrix, which is not a scalar identity matrix, is not diagonalizable.

For the displayed real symmetric matrix, choose the orthogonal matrixIts columns form an orthonormal eigenbasis, andIn the orthogonal coordinates $u=(x+y)/\sqrt{2}$, $v=(x-y)/\sqrt{2}$, the quadric surface has equation $u^2+3v^2+3z^2=1$. It is a prolate spheroid, with long semiaxis one in direction $(1,1,0)$ and two equal short semiaxes $1/\sqrt{3}$. The figure gives its principal-axis sketch.

A diagonalizable matrix can certainly have exactly two distinct eigenvalues: $\mathrm{diag}(1,1,2)$ is already diagonal and has an eigenbasis despite its repeated eigenvalue.

For a twice continuously differentiable vector field in Cartesian coordinates, the curl and divergence arePut $\mathbf{\omega }=\nabla \times \mathbf{v}$. Using the Levi-Civita symbol, the Kronecker delta contraction and the product rule,The last term vanishes because the divergence of a curl is zero: the symmetric mixed derivatives contract against an antisymmetric symbol. Thus the curl of a cross product identity becomes

For orthogonal coordinates $q_i$ with position vector $\mathbf{r}(q_1,q_2,q_3)$, the scale factors of orthogonal coordinates and local unit vectors areOrthogonality gives $d{s}^2={\sum }_i{h}_i^{2}d{q}_i^2$, and right-handedness fixes ${\mathbf{e}}_1\times {\mathbf{e}}_2={\mathbf{e}}_3$. In cylindrical coordinates, $\mathbf{r}=(\rho \cos \varphi ,\rho \sin \varphi ,z)$, soThe radial vector points away from the axis; the azimuthal vector is tangent in the direction of increasing $\varphi$; ${\mathbf{e}}_z$ is the Cartesian positive $z$ direction. The azimuthal basis is undefined on the coordinate axis itself.

For the axial velocity profile, the cylindrical coordinates formula givesThe associated cross product isThis field has only a radial component depending on $\rho$, so its curl is zero. Also $\nabla \cdot \mathbf{u}={\partial }_z{u}_z=0$. The identity above therefore yieldsIn fact both directional derivatives vanish: the first differentiates $u_z{\mathbf{e}}_z$ only in the azimuthal direction, and the second differentiates $\mathbf{\omega }$ only in the $z$ direction. All physical vector fields extend smoothly through $\rho =0$, despite the coordinate-basis singularity.

Differentiate the position vector, keeping the other coordinates fixed in each derivative:The first two have zero dot product, since their two planar contributions cancel, and both are orthogonal to ${\mathbf{h}}_z$. Hence these are orthogonal coordinates away from coordinate degeneracies. Their scale factors, which are the lengths of these coordinate basis vectors, areFor example $|{\mathbf{h}}_u{|}^2=a^2[{\sinh }^{2}u{\cos }^{2}v+{\cosh }^{2}u{\sin }^{2}v]=a^2({\sinh }^{2}u+{\sin }^{2}v)$. The unit basis vectors are ${\mathbf{h}}_i/h_i$ when $h_i\ne 0$; the given coordinate basis vectors themselves are not normalized. The Jacobian determinant is $a^2({\sinh }^{2}u+{\sin }^{2}v)>0$ except on the focal lines. These are elliptic cylindrical coordinates.

Differentiate the position vector with respect to the three coordinates:Their pairwise dot products vanish. For example, ${\mathbf{r}}_{\rho }\cdot {\mathbf{r}}_u=-\sinh \rho \cosh \rho \sin u\cos u+\sinh \rho \cosh \rho \sin u\cos u=0$. Thus these are orthogonal coordinates wherever the coordinate tangents are nonzero. Their scale factors of orthogonal coordinates areThe absolute Jacobian determinant is their product, givingThe ordered coordinate system is left handed: ${\mathbf{r}}_{\rho }\cdot ({\mathbf{r}}_u\times {\mathbf{r}}_v)=-h_{\rho }{h}_u{h}_v$. The volume measure uses the absolute value. The axis and the focal circle contain coordinate degeneracies, as expected for oblate spheroidal coordinates. This is the hyperbolic parametrization of oblate spheroidal coordinates. The local TeX has an OCR substitution ${\sinh }^{2}u$ in the volume formula; the original PDF correctly prints ${\sin }^{2}u$.

The Jacobian determinant is the oriented scalar triple product of the three coordinate basis vectors:Here ${\mathbf{h}}_u\times {\mathbf{h}}_v=(u^2+v^2)(-\sin \varphi ,\cos \varphi ,0)$, soThe sign is nonnegative on the stated range, and the absolute Jacobian determinant also equals $h_u{h}_v{h}_{\varphi }$ as expected for orthogonal coordinates.

The azimuthal vector ${\mathbf{h}}_{\varphi }$ is automatically orthogonal to the other two. Their remaining dot product isThus orthogonal coordinates require $g{g}^{\prime }=\sinh \xi \cosh \xi$, which integrates to $g^2={\sinh }^2\xi +C$. The initial value gives $C=1$, and the smooth branch with $g(0)=1$ isThe corresponding coordinate-vector norms are $|{\mathbf{h}}_{\xi }|=|{\mathbf{h}}_{\theta }|=\sqrt{{\sinh }^2\xi +{\sin }^2\theta }$ and $|{\mathbf{h}}_{\varphi }|=\sinh \xi \sin \theta$, as in the hyperbolic parametrization of prolate spheroidal coordinates. Coordinate degeneracies on the axis and focal segment are understood.

Eliminating the angular coordinates on the requested surface givesIt is a prolate spheroid with long axis $z$, equatorial radius $\sinh 1$, polar semiaxis $\cosh 1$, and foci $(0,0,\pm 1)$.

If orthogonal coordinates $q_i$ satisfythen $h_i=|\partial \mathbf{x}/\partial q_i|$ are their scale factors.

These factors determine the diagonal metric and volume element of orthogonal coordinates.