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Take the dot product of the orthogonal basis expansion with ${\mathbf{u}}_j$. All terms except the $j$th vanish by orthogonality, leavingThereforeThe basis is not assumed normalized, so its squared lengths must remain in the denominators. Applying $S$ and using $S{\mathbf{u}}_i={\lambda }_i{\mathbf{u}}_i$ gives

The defining condition for an antisymmetric matrix forcesIts dot product with $(x,y{)}^T$ is zero, and its Euclidean norm is $|a|\sqrt{x^2+y^2}$. Thus it combines a right-angle rotation with a uniform dilation by $|a|$. More precisely, the rotation is clockwise for $a>0$ and anticlockwise for $a<0$. A direction-independent description is a clockwise right-angle rotation followed by multiplication by the signed scalar $a$.

A second application givesGeometrically this is a half-turn combined with a dilation by $a^2$. Because $A$ is nonzero, $a\ne 0$.

For a constant vector field, the flux integral is its dot product with the vector area. Thus

The Kronecker delta restricts the Einstein summation convention to equal indices, giving the dot product

For the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$, the product rule givesThe mixed dot products cancel. The divergence theorem therefore yields Green's second identity,The fields and boundary are assumed sufficiently regular. A singular Green function can be treated by excising a small ball around its source and taking the radius to zero, or using distributions.

The coordinate tangent vectors are the partial derivatives of the position vector:Their pairwise dot products vanish. Their lengths, the scale factors of orthogonal coordinates, are $1,r,1+r\sin \alpha$. Since $R<1$, the last is positive throughout the body. Thus these circular torus coordinates are orthogonal wherever $r>0$; at $r=0$ the cross-section angle degenerates, as in ordinary polar coordinates.

The ordered Jacobian determinant isIts sign is positive for $r>0$, so the volume element is $dV=r(1+r\sin \alpha )drd\alpha d\beta$. The coordinate degeneracy on the central circle has zero volume and does not obstruct volume integration.

Use the product rule for the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The mixed dot products cancel. Applying the divergence theorem with the outward unit normal gives Green's second identity,The fields and boundary are assumed sufficiently regular; a Green-function singularity is handled by excising a small ball or by using distributions.

Apply the product rule to the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The two mixed dot products cancel. The divergence theorem, using the outward unit normal, then gives Green's second identity:The usual assumptions are a sufficiently regular boundary and fields with the derivatives and integrability needed for the theorem; singular sources can instead be treated by excising small balls or using distributions.

In orthonormal Cartesian coordinates, a rank-$n$ tensor has components transforming under a change of axes $R$ asRepeated indices are summed. This is the tensor component transformation law; for rank one it is precisely the transformation law of a vector.

Use the vector triple product to write the double cross product asBoth dot products are invariant under an orthogonal change of axes, while $u^{\prime }=Ru$ and ${\Omega }^{\prime }=R\Omega$. ThereforeHence $W$ is a vector. This argument also avoids the sign issue for an individual cross product under an orientation-reversing orthogonal transformation: the double product has two such factors, whose signs cancel.

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.

The ellipse bounding the surface is $u=1$. With the upward normal, its positively oriented boundary is traversed by increasing $v$ from zero to $2\pi$:The given components are Cartesian components, so their ordinary dot product with this tangent isTherefore Stokes theorem givesA direct check uses $F_x=2y/a$ and $F_y=-2x/a$, hence $(\nabla \times \mathbf{F}{)}_z=-4/a$, and multiplies by the ellipse area $\pi a^2\sinh 1\cosh 1$. The third component cannot affect either calculation because the surface and its boundary lie at fixed $z$. Equivalently one may apply Stokes to the smooth planar field $(2y/a,-2x/a,0)$, avoiding any need to differentiate the elliptic-coordinate expression for the third component at its foci.

Write the position vector as $\mathbf{r}(\rho ,\varphi ,z)=(\rho \cos \varphi ,\rho \sin \varphi ,z)$. Its coordinate tangent vectors areTheir pairwise dot products vanish. Their lengths, the scale factors of orthogonal coordinates, areThus the cylindrical coordinate system is orthogonal wherever it is a nonsingular coordinate chart. At $\rho =0$, the angular tangent vector vanishes and the azimuth is undefined; this is the usual coordinate singularity on the axis, not a failure of the Euclidean geometry.

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 columns of $R_z$ are an orthonormal basis, since ${\cos }^2\varphi +{\sin }^2\varphi =1$ and the first two columns have zero dot product. Thus $R_z^T{R}_z=I$: it is an orthogonal matrix. Consequentlyand hence $\boxed{|{\mathbf{v}}^{\prime }|=|\mathbf{v}|}$. This is the preservation of the Euclidean norm by a rotation matrix.

Differentiate the position vector with respect to the three coordinates to obtain the coordinate basis vectors:The chain rule gives $d\mathbf{x}={\mathbf{h}}_{u}du+{\mathbf{h}}_{v}dv+{\mathbf{h}}_{\varphi }d\varphi$. The scale factors of orthogonal coordinates are the corresponding norms:Direct dot products give ${\mathbf{h}}_u\cdot {\mathbf{h}}_v=uv-uv=0$, ${\mathbf{h}}_u\cdot {\mathbf{h}}_{\varphi }=0$, and ${\mathbf{h}}_v\cdot {\mathbf{h}}_{\varphi }=0$. Thus the parabolic coordinates are orthogonal wherever the coordinate vectors are nondegenerate. Their degeneracy on $uv=0$ is the usual axis degeneracy of the azimuthal coordinate.

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 $(0,0,\pm 1)$.

Let $a=\sinh 1$ and $c=\cosh 1$. The region is the first-octant portion of the ellipsoid $(x^2+y^2)/a^2+z^2/c^2\le 1$. In cylindrical coordinates, write $\rho =\sqrt{x^2+y^2}$ and $0\le \varphi \le \pi /2$. Away from the axis, the divergence isThe field is singular on the axis, so applying the classical divergence theorem there without qualification would be invalid. Instead remove $\rho <\epsilon$ and apply it to the remaining region. On the added cylindrical boundary the unit normal is $-{\mathbf{e}}_{\rho }$, andIts area element is $\epsilon d\varphi dz$, so its flux is $O(\epsilon |\log \epsilon |)\to 0$. The axial component of the field has zero dot product with this normal. The volume integral of the divergence is locally integrable, because its $1/\rho$ factor cancels the cylindrical measure. Thus removing a singular axis in a flux integral gives the desired improper surface integral with no extra axial term.

The divergence theorem now reduces the calculation toThe integral over $s$ is the area of a quarter unit disk. All boundary fluxes are interpreted as convergent improper integrals at the axis.

Choose an orthonormal basis with $\mathbf{n}={\mathbf{e}}_3$. A half-turn about $\mathbf{n}$ kills all mixed axial-transverse components. Commutation with a quarter-turn makes the transverse block a scalar identity matrix plus an antisymmetric matrix. The resulting oriented-axis invariant Cartesian tensor, in coordinate-free form, iswith real constants $a,d,h$. The Kronecker delta, the outer product $n_i{n}_j$, and the displayed Levi-Civita symbol contraction are invariant under every proper rotation about $\mathbf{n}$, so this form is also sufficient. In the chosen basis,The antisymmetric part must be retained unless further symmetry is imposed.

Decompose the electric field as $\mathbf{E}={\mathbf{E}}_{\perp }+{\mathbf{E}}_{\parallel }$, where ${\mathbf{E}}_{\parallel }=(\mathbf{n}\cdot \mathbf{E})\mathbf{n}$. The electric current density isFor a nonzero real electric field, the alignment of current and field in an axisymmetric conductor gives the complete classification of collinearity:

For completeness, if ${\mathbf{E}}_{\perp }\ne 0$ and $\mathbf{J}=\lambda \mathbf{E}$, taking the dot product with $\mathbf{n}\times {\mathbf{E}}_{\perp }$ forces $h=0$. The transverse equation then gives $\lambda =a$, while a nonzero axial component gives $\lambda =d$. This proves there are no missing directions. Zero current is included as the zero scalar multiple; if “parallel” requires a direction for $\mathbf{J}$, exclude cases where that scalar is zero. For a strictly anisotropic crystal the last possibility is absent.

For a holomorphic function, the derivative supplied by complex differentiability must have the same value along real and imaginary increments. Those two computations giveEquating real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if the first partial derivatives are continuous near a point and obey these equations there, real differentiability giveswhich proves complex differentiability under this regularity assumption.

The gradients of the real and imaginary parts have dot productand $|\nabla u{|}^2=|\nabla v{|}^2=|f^{\prime }(z){|}^2$. At every point with $f^{\prime }(z)\ne 0$, the two level sets are regular curves, their gradients are nonzero normal vectors, and their tangent directions are therefore perpendicular.

The printed phrase "at every point" needs this regular-point qualification. At a critical point with $f^{\prime }(z)=0$, the level sets may cross or fail to have a unique tangent. For example, $f(z)=z^2$ gives $u=x^2-y^2$ and $v=2xy$: at the origin the zero level sets are respectively the diagonals and the coordinate axes. Thus the valid conclusion is $\boxed{\text{orthogonality at every regular intersection, where }{f}^{\prime }(z)\ne 0}$.

Let $\mathbf{u}$ be the boost velocity, and decompose the fields into components parallel and perpendicular to it. The parallel components are unchanged. The perpendicular components obeyAll vectors inside these expressions are perpendicular to $\mathbf{u}$. Thus $|\mathbf{u}\times {\mathbf{B}}_{\perp }{|}^2=u^2{B}_{\perp }^2$, and similarly for ${\mathbf{E}}_{\perp }$. In the difference of squares, the mixed terms cancel becauseIt follows thatAdding the unchanged parallel components provesFor the mixed dot product, expand directly. The terms ${\mathbf{E}}_{\perp }\cdot (\mathbf{u}\times {\mathbf{E}}_{\perp })$ and $(\mathbf{u}\times {\mathbf{B}}_{\perp })\cdot {\mathbf{B}}_{\perp }$ vanish. The remaining cross-product identity isHenceParallel-perpendicular mixed dot products are zero in both frames, soRotations also preserve these two expressions, completing the proof for proper Lorentz transformations. The second expression is a pseudoscalar under spatial parity, but is invariant under the boosts at issue here.

These results also follow from tensor contractions. In the convention used below, $F_{\mu \nu }{F}^{\mu \nu }=2(B^2-E^2/c^2)$ and $F_{\mu \nu }{\tilde{F}}^{\mu \nu }=-4\mathbf{E}\cdot \mathbf{B}/c$. The numerical sign of the latter depends on the chosen orientation of the dual electromagnetic field tensor, while its invariance under proper Lorentz transformations does not.