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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 $r$.

Compare the total differential $d\varphi ={\sum }_i{\partial }_{u_i}\varphi d{u}_i$ with $d\varphi =\nabla \varphi \cdot d\mathbf{r}$ and $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{u}_i$. The orthonormality of the coordinate basis gives the gradient in orthogonal coordinates:To obtain the curl without neglecting the variation of the basis vectors, write $\varphi {\mathbf{e}}_1=(\varphi h_1)\nabla u_1$. The curl of a gradient vanishes, and the product rule givesLikewise $\nabla u_2\times \nabla u_3={\mathbf{e}}_1/(h_2{h}_3)$ because the system is right-handed. This cross product has zero divergence: $\nabla \cdot (\nabla u_2\times \nabla u_3)=\nabla u_3\cdot (\nabla \times \nabla u_2)-\nabla u_2\cdot (\nabla \times \nabla u_3)=0$. The divergence product rule applied to $(\varphi h_2{h}_3)(\nabla u_2\times \nabla u_3)$ now givesCyclic permutation and linearity give the general component formulas described by divergence and curl from orthogonal coordinate gradients:andIn cylindrical coordinates, $(h_r,h_{\theta },h_z)=(1,r,1)$. The specified field has no $z$ dependence or $z$ component, soIt is an irrotational vector field for every $r>0$ exactly when $\boxed{p=3,\alpha =4}$. This statement is local to an angular coordinate chart: the component proportional to $\theta$ is not single-valued under $\theta \mapsto \theta +2\pi$.

For proper rotations the Levi-Civita symbol is invariant, so contracting the two rank-one tensors ${\partial }_j$ and $v_k$ gives $c_i={ϵ}_{ijk}{\partial }_j{v}_k$ with transformation $c_i^{\prime }=R_{i\ell }{c}_{\ell }$. Thus the curl is a vector under proper rotations. Under an orientation-reversing orthogonal change it acquires an additional determinant sign: it is an axial vector when $v$ is a polar vector.

Commuting smooth partial derivatives and using epsilon contraction gives the curl of the curl identity:Now take the curl of Faraday's law and use the vacuum Ampère-Maxwell equation: $\nabla \times \nabla \times E=-{\partial }_t(\nabla \times B)=-c^{-2}{E}_{tt}$. Since $\nabla \cdot E=0$, the identity gives $\Delta E-c^{-2}{E}_{tt}=0$. Taking the curl of the Ampère-Maxwell equation gives the same result for $B$, using $\nabla \cdot B=0$. Hence both satisfy the electromagnetic wave equation

The radial component of the curl in spherical coordinates isIt contains no $f_r$, so $R$ cannot contribute. Substituting the tangential components givesThe mixed partial derivatives of smooth $G$ agree, so its terms cancel. Multiplication by $r$, to take the dot product with the position vector $\mathbf{r}=r{\mathbf{e}}_r$, does not change this independence from $R,G$.

The outward area element is $d\mathbf{S}=a^2\sin \theta d\theta d\varphi {\mathbf{e}}_r$. ThereforeSplit the sphere into the northern cap $0\le \theta \le \alpha$ and the southern cap $\alpha \le \theta \le \pi$. Their induced boundary orientations are $+C$ and $-C$, respectively. The Stokes theorem applies to ${\mathbf{A}}^{(+)}$ on the north and ${\mathbf{A}}^{(-)}$ on the south. Adding their fluxes gives $2\pi (1-\cos \alpha )+2\pi (1+\cos \alpha )=4\pi$, agreeing with direct integration.

There is no contradiction with the flux of a globally smooth curl through a closed surface: neither potential is regular on the entire sphere. On their overlap they differ by $2{\mathbf{e}}_{\varphi }/(r\sin \theta )=2\nabla \varphi$, a local gauge transformation. The longitude is not a globally single-valued scalar, so that difference can have nonzero circulation around $C$.

For a differentiable vector field in Cartesian coordinates, the divergence and curl arewith summation over repeated indices. In components the curl is $({\partial }_y{F}_z-{\partial }_z{F}_y,{\partial }_z{F}_x-{\partial }_x{F}_z,{\partial }_x{F}_y-{\partial }_y{F}_x)$. The contraction of two Levi-Civita symbols and the product rule giveRecognizing the directional derivatives gives the curl of a cross product identity

In Cartesian coordinates, use the Levi-Civita symbol and the Einstein summation convention to expandThe contraction of the two symbols uses the Kronecker delta, and the final step is the product rule. Thus the convective acceleration identity isPut $\mathbf{\omega }=\nabla \times \mathbf{u}$. The curl of a gradient vanishes, and the curl of a cross product identity givesSince the divergence of a curl is zero, taking the curl of the first identity yieldsTwice continuous differentiability suffices for the mixed-derivative identities used here.

Away from the axis, cylindrical coordinates put the field in the line vortex formThe only potentially nonzero component of its curl isThe other components vanish because there is no axial dependence and no radial or axial velocity. Thus the curl is zero for $\rho >0$; the field is singular on the axis, so a classical curl there is not defined.

For a spanning surface avoiding that axis, the smoothness hypotheses of Stokes theorem hold and give ${\oint }_C\mathbf{v}\cdot d\mathbf{r}=0$. If the enclosed planar region contains the axis, the singularity prevents applying that theorem directly. Excise a small disk about the origin. The resulting annulus has zero curl, so its outer circulation equals the counterclockwise circulation on the small circle:This sign assumes positive, counterclockwise orientation. Clockwise orientation gives $-\kappa$; more generally the value is $\kappa$ times the winding number about the origin. A curve through the origin is excluded because its integral is singular. In the distribution interpretation the missing vorticity is concentrated on the axis as $\kappa \delta (x)\delta (y){\mathbf{e}}_z$, explaining the nonzero circulation despite zero classical curl away from it.

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.

Using the Levi-Civita symbol and the Einstein summation convention in Cartesian coordinates, the product rule givesThe cyclic permutation in the first term preserves the sign, whereas exchanging the first two indices in the second reverses it. Recognizing the two curls proves

Write the curl and cross product with the Levi-Civita symbol, then use its contraction identity:Expanding with the product rule givesRecognize the divergence in the first and third terms, and the directional derivatives in the other two. The curl of a cross product identity is thereforeThe derivatives act on both vector fields. No assumption of vanishing divergence or spatially constant fields has been made.

An axial vector transforms under an orthogonal matrix $Q$ asFor a proper rotation, $\det Q=1$, so this is the usual polar vector rule. Under an orientation-reversing change, the extra sign distinguishes the two. In particular, spatial inversion leaves an axial vector unchanged while reversing a polar vector.

Let $C_i=(\nabla \times u{)}_i={ϵ}_{ijk}{\partial }_j{u}_k$. The determinant identity for the Levi-Civita symbol isIt follows by multiplying the three-index determinant identity by $Q_{ir}$ and using orthogonality. The derivative transformation from part (ii) then givesHence the curl of a polar vector field is an axial vector field. The Levi-Civita determinant factor is exactly the extra sign required under reflections.

Direct differentiation givesThe curl is $(0,0,b-a)$, so the preceding convention givesPut $h=(a+b)/2$. The normalized principal axes and eigenvalues of the symmetric tensor $s$ areThe planar axes lie along the two angle bisectors of the coordinate axes. If $h\ne 0$, one planar direction extends and the perpendicular direction contracts, with no change along the third axis. The labels need not be ordered by size when $h<0$. If $a+b=0$, then $s=0$, all principal values vanish and any orthonormal frame is a principal frame; the flow is pure local rotation in the plane.

Take the closed surface with its outward orientation. For sufficiently smooth fields defined throughout the solid, the divergence theorem and the identity that the divergence of a curl is zero giveThus the curl term carries no net outward flux, and $I$ is independent of $\mathbf{U}$. The hole in the solid torus does not invalidate the divergence theorem; singularities of $\mathbf{U}$ inside the integration region would invalidate the smoothness argument and must be excluded.

Take any constant vector $\mathbf{a}$. Its curl is zero, so part (a) and the divergence theorem givewhere $\mathbf{n}$ is the outward unit normal vector. The scalar triple product identity rewrites the last integrand as $\mathbf{a}\cdot (\mathbf{F}\times \mathbf{n})$. Since $\mathbf{a}$ is arbitrary, equality of these scalar products implies the vector identity

First compute the curl directly:Reflection symmetry in $x$ makes the last component integrate to zero. The middle component gives the hemisphere's volume, soFor an independent surface integral, the closed boundary consists of the curved hemisphere and its flat base. The right-hand side in part (b) is ${\int }_S\mathbf{n}\times \mathbf{F}dS$.

On the curved part use spherical coordinates with $0\le \theta \le \pi /2$ and $0\le \varphi <2\pi$. There,The last component has zero angular integral. The first two components giveOn the base, $z=0$ and the outward normal vector is $-{\mathbf{e}}_z$. Thus $\mathbf{n}\times \mathbf{F}=(x^2+y^2,0,0)$. In polar coordinates its integral isThe first components cancel, and therefore both required sides equal

For an oriented piecewise smooth surface $S$ with boundary curve $C=\partial S$, and a continuously differentiable vector field $\mathbf{F}$ near $S$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal, $dS$ is positive surface area, and $d\mathbf{r}$ is the directed tangent line element. The positive traversal of $C$ is fixed by the right-hand rule relative to $\mathbf{n}$. Thus the flux of the curl through the surface equals the circulation along its oriented boundary.

The divergence of a cross product and $\nabla \times \mathbf{a}=0$ give $\nabla \cdot (\mathbf{a}\times \mathbf{A})=-\mathbf{a}\cdot (\nabla \times \mathbf{A})$. On the boundary the scalar triple product gives $(\mathbf{a}\times \mathbf{A})\cdot \hat{\mathbf{n}}=-\mathbf{a}\cdot (\hat{\mathbf{n}}\times \mathbf{A})$. Thus the divergence theorem givesBecause $\mathbf{a}$ is arbitrary, the curl identity follows:

Gauge freedom. The magnetic vector potential satisfies $\mathbf{B}=\nabla \times \mathbf{A}$. The transformationleaves $\mathbf{B}$ unchanged because the curl of a gradient is zero. For time-dependent fields, transform the electric scalar potential as $\varphi \mapsto \varphi -{\partial }_t\chi$ as well, so that $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ is unchanged. The scalar $\chi$ must have the regularity and single-valuedness appropriate to the physical region and its boundary conditions. Local values of $\mathbf{A}$ therefore depend on the gauge transformation.

Gauge and nonzero component. In the standard axisymmetric Coulomb gauge representative for the long-solenoid approximation,The divergence vanishes since $A_{\varphi }$ is independent of the azimuth. Inside, this agrees with the usual symmetric gauge representative $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. The Coulomb gauge condition alone does not force the other two components to vanish; the symmetry, regularity and customary boundary choice select this representative from the remaining gauge freedom.

Radial dependence. Write the interior magnetic field as $B_0\hat{\mathbf{z}}$, where $B_0={\mu }_{0}nI$ for a vacuum solenoid of solenoid turn density $n$. Its exterior magnetic field vanishes in the infinitely long idealization. Applying Stokes theorem to a coaxial circle gives $2\pi r{A}_{\varphi }(r)=\Phi (r)$. Therefore the Coulomb-gauge vector potential of a long solenoid isIt grows linearly inside, is continuous at $R$, and decays as $r^{-1}$ outside. Although its exterior curl is zero, its circulation is the enclosed magnetic flux $\Phi =B_0\pi R^2$. No regular single-valued gauge function removes it throughout the exterior while preserving this circulation.

Electric field of the rotating charged surface. For clarity retain absolute permittivity $ϵ$ and permeability $\mu$ in the cylinder formulae; setting them to unity gives the rationalized-unit convention in the question. If the stated unity means relative material constants in SI, use ${ϵ}_0,{\mu }_0$ instead. Assume prescribed charge densities in the laboratory frame and the nonrelativistic, steady-rotation model.

The axisymmetric surface charge is stationary as a distribution even while the material rotates. A Gaussian cylinder of length $L$ has enclosed charge zero for $r<R$ and $2\pi RL\sigma$ for $r>R$. Gauss's law therefore givesThe radial electric field jumps from zero to $\sigma /ϵ$ at the surface and then decays as $r^{-1}$. Its direction reverses for negative $\sigma$. The angular velocity does not enter this electric-field result for a specified surface charge.

Magnetic field of the volume-charged cylinder. A shell of radius $a$ and thickness $da$ has surface charge $\rho da$, azimuthal surface current $d{K}_{\varphi }=\rho \omega ada$, and the field of an infinite solenoid: $d{B}_z=\mu \rho \omega ada$ inside that shell and zero outside it. At radius $r<R$, only shells with $a>r$ contribute. Thus the magnetic field of a rotating charged solid cylinder isWith no applied background magnetic field, $\mathbf{B}=0$ outside. The profile is parabolic, largest on the axis and zero at the boundary. The sign follows $\rho \omega$. As a check, the current density is $J_{\varphi }=\rho \omega r$, and Ampère's circuital law gives $-d{B}_z/dr=\mu J_{\varphi }$, as required.

Electron impulse when the current is switched off. Choose positive $I$ so the initial magnetic flux is ${\Phi }_i={\mu }_{0}nI\pi R^2$ along $+z$. Faraday's law of induction gives, outside the solenoid,For an electron of electric charge $q=-e$, treat its position as fixed while the short pulse acts. The inductive impulse outside a switched solenoid isConsequently the nonrelativistic final velocity isThe induced electric field circulates in the positive azimuthal direction while a positive flux decreases, so the electron accelerates in the negative direction. Once the pulse has passed, there is no exterior field and it travels along the tangent at the impulse point, rather than continuing in a circular orbit. If the resulting momentum is relativistic, use ${\mathbf{v}}_f=c{\mathbf{p}}_f/\sqrt{m_e^2{c}^2+|{\mathbf{p}}_f{|}^2}$ instead. The sudden-switch result assumes negligible displacement during the pulse and the usual quasistatic electromagnetic induction approximation; “instantaneous” switching does not mean instantaneous propagation of fields to arbitrary distance.

One can check the sign using conserved canonical angular momentum: axial symmetry conserves $p_{\varphi }^{\mathrm{can}}=L_z+qr{A}_{\varphi }$. Initially $L_z=0$, and finally $A_{\varphi }=0$, so $L_z=q{\Phi }_i/(2\pi )=-e{\Phi }_i/(2\pi )$.

Interferometric measurement at constant current. Use a coherent electron beam split into two paths passing on opposite sides of a shielded long solenoid, then recombine them to form interference fringes. Keep the current constant during the passage and suppress leakage fields on the paths. In the Aharonov-Bohm effect, the phase difference due to the magnetic vector potential is the closed-loop integralThe overall sign reverses if the path orientation is reversed. The displacement in fringe periods is $\Delta \phi /(2\pi )$, so the measured fringe shift is linear in $I$ and repeats whenA gauge transformation adds a gradient whose closed-loop integral vanishes for a single-valued $\chi$. Endpoint changes in electron wavefunction phase cancel in the interference measurement. The measured Aharonov-Bohm phase is therefore gauge invariant: it determines the flux, or equivalently the circulation of $\mathbf{A}$. In the chosen Coulomb gauge this implies $A_{\varphi }(d)={\mu }_{0}nI{R}^2/(2d)$, but the experiment does not select or measure a unique local vector potential independently of gauge.

Far field of the rotating sphere. Interpret uniform charge as uniform volume charge of total $Q$, and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. The magnetic dipole moment of a rigidly rotating charge distribution isIts axial component is $m_z=(\rho \omega /2)\int r_{\perp }^2{d}^{3}r$. For a solid sphere, $\int r_{\perp }^2{d}^{3}r=8\pi R^5/15$ and $\rho =3Q/(4\pi R^3)$, soThe magnetic field of a rotating charged spherical body is a static magnetic dipole field at large distances:In particular, $|\mathbf{B}|={\mu }_0|\mathbf{m}|\sqrt{1+3{\cos }^2\vartheta }/(4\pi r^3)$. Both the angular mean of the magnitude and its root mean square therefore scale as $\boxed{r^{-3}}$; the latter is $B_{\mathrm{rms}}=\sqrt{2}{\mu }_0|\mathbf{m}|/(4\pi r^3)$. The angular average of the vector is zero, which is different from average strength. Steady rotation leaves the charge and current distributions stationary, so this is magnetostatics, not a radiative $r^{-1}$ field. If uniform charge instead means a thin spherical surface, replace the dipole moment by $Q{R}^2\mathbf{\omega }/3$; the angular pattern and $r^{-3}$ scaling are unchanged.

Empirical equivalence must be specified relative to a domain and an account of what counts as observable. Two theories may fit all evidence collected so far while predicting different future results; that is ordinary underdetermination by a finite record. A stronger equivalence requires the same observable consequences for every permitted experiment in the stated domain. For stochastic predictions, matching a set of possible outcomes is not enough: the theories must assign the same outcome probability distributions under corresponding preparations and testing assumptions.

A clear physical example at the level of descriptions occurs in electromagnetism. Write the electric field and magnetic field using a scalar potential $\varphi$ and magnetic vector potential $\mathbf{A}$:For a smooth, single-valued function $\chi$, the gauge transformationleaves both fields unchanged: the curl of a gradient vanishes, and the mixed derivatives cancel in the electric-field expression. With the same charges, boundary conditions and force law, these descriptions give the same classical measurements. Agreement is structural, not an accidental fit to a few readings. However, many physicists regard such descriptions as representing one physical situation, so this alone is not an example of two genuinely incompatible physical theories. The smooth global transformation assumption also matters; equality of local fields is not an unrestricted license to identify every globally different potential configuration.

A second example exhibits possible surplus structure. For particles interacting by Newtonian forces depending only on relative positions, a common Galilean transformationleaves relative positions and accelerations unchanged. Accordingly, adding a privileged but completely uncoupled state of uniform rest does not alter any internal mechanical prediction. Rival choices of that unobservable rest state cannot be distinguished by these experiments. This is a deliberately specified completion of the mechanical model: it does not show that rotation or acceleration is undetectable, or that Newtonian mechanics and special relativity are empirically equivalent. The latter give different high-speed predictions. Whether undetectable absolute rest makes a genuinely different theory or a redundant representation is precisely one philosophical issue raised by the example.

Thus there are concrete cases of empirical redundancy, while the existence of equally well-motivated, fully distinct rivals with identical consequences across all physics is a much stronger claim. Historical theories often agree only in limited regimes or after additional assumptions. One must not convert that restricted agreement into an equivalence theorem for every possible test.

The philosophical relevance does not depend on finding a complete historical pair. A constructed rival can expose the logical limits of evidence, although an artificial extension need not deserve the same credibility as an economical account. If two specified models have equal likelihoods for every admissible data set $D$, thenfor finite positive prior odds. The Bayes factor is one; the data do not discriminate between them. Choices about simplicity or interpretation would be additional considerations. Constructive empiricism treats empirical adequacy as the relevant belief commitment in acceptance, while a scientific realist must explain what further warrants belief in the unobservable structure. The equivalence argument identifies a limit of evidential discrimination; it does not establish that every theoretical claim is equally good or that every unobservable claim is false.^[ref]