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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.

Apply the product rule in Einstein notation, using the Levi-Civita symbol to express the cross product:The third line uses a cyclic permutation of indices in the first term and an interchange of two indices in the second. Rearranging proves the divergence of a cross product identity:

The divergence of a cross product identity and the vacuum Maxwell equations implyThe divergence theorem therefore gives the required balance withAn additive time-independent constant in $\lambda$ is immaterial. This is the vacuum Poynting theorem in the normalization of the given equations: equivalently ${\partial }_t\lambda +\nabla \cdot (\mathbf{E}\times \mathbf{B})=0$. The order of the cross product accounts for the plus sign in the stated integral identity.

Dividing the coordinate tangents by their scale factors of orthogonal coordinates gives an orthonormal basisEvaluate the Cartesian cross products componentwise:ConsequentlyThe ordered cylindrical basis is right-handed: its scalar triple product ${\mathbf{e}}_{\rho }\cdot ({\mathbf{e}}_{\varphi }\times {\mathbf{e}}_z)$ equals $+1$.

Let an upward force $P$ act along $x=0$ and a downward parallel force $Q$ along $x=d>0$. An equivalent resultant force must preserve both total force and torque. Its signed vertical component is $R=P-Q$, and taking torques about the first line gives $R{x}_R=-Qd$. ThusIts magnitude is $|P-Q|$ and its direction agrees with the larger force. Its line lies outside the two original lines on the side of that larger force; for example, $P>Q$ gives $x_R<0$. The products of the original forces and their perpendicular distances from this line are equal.

If $P=Q\ne 0$ and $d\ne 0$, the total force vanishes but the torque is $-Pd$. No single force is equivalent to this system: it is a force couple of magnitude $Pd$. Its torque is independent of the reference point because translating that point changes the torque by the cross product of the shift and the zero resultant force. Equal opposite forces on the same line instead cancel completely.

The eye is rotated by six extraocular muscles. The rectus muscles approach it from behind; the oblique muscles have oblique lines of pull, with the superior oblique muscle tendon redirected by an orbital pulley. Attachments on the sclera give lever arms about the rotation centre. In the straight-ahead position their actions are as follows; vertical and torsional contributions vary with gaze direction.^[ref]

Coordinated contraction selects the desired rotation and balances unwanted components. This illustrates torque control; the full ocular geometry is not simply the parallel-force pair of the first calculation.

For a field point in the plane of the loop, both the directed line element $d\mathbf{s}$ and the displacement $\mathbf{r}-\mathbf{s}$ lie in that plane. Their cross product is normal to it, so every term of the Biot-Savart law is normal, and their integral is normal as well. This argument also allows the field to vanish at particular points.

Choose the loop in the $xy$-plane with centre at the origin and let positive current be counterclockwise as seen from $+z$. On its lower side, write $\mathbf{s}=(t,-a,0)$, $-a\le t\le a$, soThat side contributesAll four contributions have the same direction and size by rotation. Thuswhere $\mathbf{n}$ follows the right-hand rule for the current.

The elementary elimination step replaces row $i$ by row $i$ minus a multiple $m_{ij}$ of an earlier pivot row $j$. With coordinate vectors $e_i$, its premultiplication matrix isand its inverse is $I+m_{ij}{e}_i{e}_j^T$, since $(e_i{e}_j^T{)}^2=0$. Choose $m_{ij}$ as the current entry below the pivot divided by the current pivot. Required pivots are assumed nonzero, consistently with elimination without row interchanges.

For dimension 4, first eliminate positions $(2,1),(3,1),(4,1)$, then $(3,2),(4,2)$, then $(4,3)$. ThusThe rightmost operation acts first, and every multiplier is computed from the matrix at its own stage. For general $n$, the numbers of eliminations in successive columns are $n-1,n-2,\dots ,1$, totaling $n(n-1)/2$; a zero multiplier simply gives an identity operation.

The product $E$ is unit lower triangular and $U=EA$ is upper triangular. Its inverse $L=E^{-1}$ is unit lower triangular, soThe stored multiplier in position $(i,j)$ is the corresponding subdiagonal entry of $L$, as follows by reversing the pivot-row subtractions column by column.

If $A$ is upper Hessenberg, only the first subdiagonal can be nonzero below the diagonal. At stage $j$ only position $(j+1,j)$ needs elimination. Subtracting row $j$ cannot create lower entries further down because its entries before its pivot are already zero. Hence only $n-1$ operations are needed. The inverse factors occur in increasing column order, and their cross products vanish, givingwhich is lower bidiagonal; $U$ is upper triangular, possibly dense above its diagonal.

If the Hessenberg matrix is also symmetric, it is tridiagonal. At each elimination step the pivot row has no entries past its immediate superdiagonal, so the updated matrix retains that upper bandwidth. Thus $L$ is lower bidiagonal and $U$ is upper bidiagonal, with at most two nonzero entries in each row, as required.

Use the source orientation, anticlockwise viewed from $+z$. The position and oriented line element areThe Lorentz force on the electric current element is $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$. Evaluating the cross product givesEach component integrates to zero around $0\le \theta <2\pi$, so ${\mathbf{F}}_{\mathrm{net}}=0$. Equivalently, uniform magnetic flux density permits $I\oint d\mathbf{l}\times \mathbf{B}=I(\oint d\mathbf{l})\times \mathbf{B}=0$. Uniformity is essential; a nonuniform magnetic field can exert a net force on a magnetic dipole.

For a steady thin-wire electric current, the Biot-Savart law iswhere the source element follows positive current. For a circular loop, source points are $(R\cos \varphi ,R\sin \varphi ,0)$ and the observer is $(0,0,z)$. Their cross product with $d{\mathbf{\ell }}^{\prime }$ has axial component $R^{2}d\varphi$; transverse components integrate to zero. Thus the on-axis magnetic field of a circular current loop, multiplied by the number of coincident turns, isTake positive current counterclockwise as seen from $+z$. magnetic field lines pass upward through the loop and return around its exterior, closing rather than terminating. They fan outward above its centre and converge below it. The original numerical Biot-Savart law integration below illustrates these directions; the wire singularities are masked.

The axisymmetric magnetic field near a regular axis follows from Gauss's law for magnetism. In cylindrical coordinates, regularity gives $B_{\rho }(0,z)=0$, while symmetry gives $B_z(\rho ,z)=B_0(z)+O({\rho }^2)$. Integrate ${\partial }_{\rho }(\rho B_{\rho })=-\rho B_0^{\prime }(z)$ from the axis:A formal $C(z)/\rho$ term is excluded by regularity. For $z>0$ the axial field decreases upward, so the radial component is outward, consistent with the sketch.

On the small receiving coil, $d\mathbf{\ell }=ad\varphi \hat{\mathbf{\varphi }}$. The Lorentz force $I_{2}d\mathbf{\ell }\times \mathbf{B}$ has axial component $-I_{2}a{B}_{\rho }(a,d)d\varphi$. Contributions from the axial field cancel radially around the loop. ThereforeWith both currents giving magnetic dipole moments along $+z$ and $d>0$, this is attraction. Reversing one current reverses the force. The small coil has $m_2=N_2{I}_2\pi a^2$, so the result is exactly the leading magnetic dipole force $F_z=m_2{B}_0^{\prime }(d)$. Equivalently $U=-{\mathbf{m}}_2\cdot {\mathbf{B}}_0$. The approximation requires $a\ll R$ and negligible variation over the receiving coil; corrections begin at relative order $a^2/(R^2+d^2)$.

For a linear nondispersive medium with real absolute permittivity ${ϵ}_a=ϵ{ϵ}_0$ and permeability ${\mu }_0$, instantaneous field energy density and energy flux arewith local balance ${\partial }_{t}u+\nabla \cdot \mathbf{N}=-\mathbf{J}\cdot \mathbf{E}$. energy stored locally and energy passing through a surface are distinct: integrating the balance relates their rates and the power given to charges.

Between the coaxial conductors, cylindrical Gauss and Ampère laws giveThe inner line charge is $\lambda =2\pi {ϵ}_{a}V/\ln (b/a)$, so the capacitance per length is $C=2\pi {ϵ}_a/\ln (b/a)$. Integrating magnetic energy over the annulus gives $\int B^2(2\pi rdr)/(2{\mu }_0)={\mu }_0{I}^2\ln (b/a)/(4\pi )=L{I}^2/2$, hence $L={\mu }_0\ln (b/a)/(2\pi )$. ThusThese are per-length parameters for ideal conductors, excluding internal magnetic penetration.

Now let the voltage and current amplitudes be $V_0,I_0$ and multiply both transverse fields by $e^{i(kz-\omega t)}$. Their divergences vanish in the annular dielectric. The two nontrivial curl equations reduce toConsequently both $k^2={\omega }^2{\mu }_0{ϵ}_a$ and $V_0=Z_0{I}_0$ are required for a forward wave. Arbitrary independent static $V$ and $I$ cannot acquire the same forward-wave factor and still satisfy Maxwell's equations. This TEM travelling-wave voltage-current relation and the dispersion constraint identify the transverse electromagnetic mode and its speed $v_p=1/\sqrt{LC}$.

For the travelling wave, $⟨N_z⟩=\mathrm{Re}(V_0{I}_0^{\ast })/[4\pi r^2\ln (b/a)]$. Integrating over the dielectric annulus yieldsThis is the familiar circuit power, carried here through the dielectric rather than through the ideal metal. In root mean square notation it is $V_{\mathrm{rms}}{I}_{\mathrm{rms}}$ for the forward lossless wave; the corresponding static fields give $VI$ without a time-average factor. The average line energy is $(C|V_0{|}^2+L|I_0{|}^2)/4=C|V_0{|}^2/2$, whose product with wave speed agrees with the power.

Finite conductor resistance requires a small longitudinal electric field to drive current. Its cross product with the azimuthal magnetic field produces a radial Poynting component into each conductor: inward radially at the inner conductor and outward radially into the outer conductor. This Poynting flux into lossy coaxial conductors explains why forward power decreases along the cable as energy becomes heat in the metal. The mode is then only approximately TEM; strictly zero longitudinal electric field would be inconsistent with resistive current.

The field law. In the Biot-Savart law, $d\mathbf{B}$ is the contribution to the magnetic flux density at the observation point from a directed wire element $d\mathbf{\ell }$ carrying steady conventional electric current $I$. The vector $\mathbf{R}$ points from that element to the observation point, $R=|\mathbf{R}|$, and $\hat{\mathbf{R}}=\mathbf{R}/R$. The constant ${\mu }_0$ is the vacuum permeability; the cross product sets the direction by the right-hand rule. Integrating the elements gives the total field, in teslas.

The moment of the shown loop. The arrows in the original PDF run counterclockwise viewed from $+z$. Thus the magnetic dipole moment isIts units are $\mathrm{A}{\mathrm{m}}^2$. Reversing the current reverses both this moment and the field.

Magnetic field and force. A steady electric current produces a magnetic field. In vacuum the Biot-Savart law givesThe Lorentz force on the charge carriers transfers force to the material carrying them. The resulting force on a current element of another circuit is $d{\mathbf{F}}_2=I_{2}d{\mathbf{\ell }}_2\times {\mathbf{B}}_1$, or, for a continuous current distribution, the magnetic force per unit volume is $\mathbf{f}=\mathbf{J}\times \mathbf{B}$.

For two long parallel straight wires a distance $r$ apart, Ampère's law gives $B_1={\mu }_0{I}_1/(2\pi r)$. ThereforeThe force is attractive for parallel currents and repulsive for opposite currents, by the cross product rule. Its units are $\mathrm{N}{\mathrm{m}}^{-1}$. This approximation neglects wire thickness and end effects; a complete circuit includes its return current. Element-by-element forces need not form an isolated Newton's-third-law pair, but complete stationary circuits have balanced mutual forces when the entire system is included.

Energy description. For two linear coupled circuits the magnetic energy isHere $q$ is a mechanical coordinate and $M$ the signed mutual inductance. At prescribed currents, the generalized force is $I_1{I}_2\partial M/\partial q$ if the self-inductances do not vary. This positive derivative is obtained from magnetic coenergy, or by including the electrical sources' work; differentiating $W_B$ with a minus sign at fixed currents would omit that source work. At prescribed flux linkages the force instead equals the negative derivative of stored magnetic energy with those linkages fixed. The geometry thus determines both inductance and mechanical force.

For a steady filamentary electric current $I$ in vacuum, the Biot-Savart law isThe line element points with conventional electric current, and the cross product fixes the magnetic flux density direction by the right-hand rule.

As an application, an infinitely long straight wire has circular symmetry about its axis. At perpendicular distance $s$,Its direction follows the right-hand rule. The following part applies the same law to a circular current loop.

One application is the slow steady precession of a rapidly spinning gyroscope supported at one point. Let its centre of mass be a distance $\ell$ from the pivot along its spin axis, at angle $\alpha$ to the upward vertical. Write $J\simeq I_1\omega$ for its dominant spin angular momentum. In a reference frame precessing steadily about the vertical with angular velocity ${\mathbf{\Omega }}_p$, the spin angular momentum is approximately stationary, soThe gravitational torque has magnitude $mg\ell \sin \alpha$. The cross product on the right has magnitude ${\Omega }_{p}J\sin \alpha$, giving, for a nonvertical axis,Its direction follows from the torque equation. This is the slow precession approximation: the additional angular momentum associated with precession is small compared with the spin contribution.

Use the divergence identity for a cross product, followed by Faraday's law of induction and the Ampère-Maxwell equation:With the electromagnetic energy density of part (b), these giveIntegrate over a fixed volume $V$ and apply the divergence theorem, choosing $d\mathbf{S}$ outward:This is Poynting's theorem. The outward electromagnetic energy flux equals minus the power transferred to matter and minus the rate at which electromagnetic energy accumulates inside. For the dissipative electrical conductor in this question, the matter term is Ohmic heating.

Choose the electric current counterclockwise as viewed from $+z$; reversing it reverses the magnetic field. Parameterize the circular wire byThe cross product in the Biot-Savart law isConsequently the transverse terms integrate to zero over $0\le \varphi \le 2\pi$, while the axial term is constant:ThusThe direction is $+z$ on both halves of the axis for this electric current orientation, including $z<0$; it is not “away from the plane” on both sides. The right-hand rule gives the same direction.

In magnetostatics, Ampère's circuital law isIt relates magnetic circulation around a closed curve to the steady current through a spanning surface. It is especially useful when symmetry makes the tangential field constant on a chosen path: an infinite straight wire gives $B={\mu }_{0}I/(2\pi r)$, while an ideal long solenoid gives approximately $B={\mu }_{0}nI$ inside.

The Biot-Savart law gives the field contribution from the actual current geometry:The cross product determines the direction by the right-hand rule. Integrating around a circular loop of radius $a$ gives $B={\mu }_{0}I/(2a)$ at its centre. A finite wire segment can also be integrated directly, even though it lacks enough symmetry for an Ampère-loop shortcut. The full steady-current circuit includes a return path; a segment calculation specifies that segment's contribution.

Parametrize the surface of revolution byThe magnitude of the cross product ${\mathbf{r}}_x\times {\mathbf{r}}_{\theta }$ is $y\sqrt{1+y^{\prime 2}}$. Integrating the surface area element over $\theta$ givesPositivity of $y$ removes an otherwise necessary absolute value in the area element.

In vacuum the source-free equations areAdding the divergence equations with weights $1,ic$ gives the complex divergence equation. Adding the curl equations in the same way gives $\nabla \times \mathbf{W}=(i/c){\mathbf{W}}_t$. Conversely real and imaginary parts of these two complex equations recover all four real equations.

For the cosine ansatz, differentiation yields divergence $(\mathbf{\Omega }\cdot \mathbf{K})\sin (\omega t-\mathbf{K}\cdot \mathbf{x})$ and curl $-(\mathbf{\Omega }\times \mathbf{K})\sin (\omega t-\mathbf{K}\cdot \mathbf{x})$. The time derivative is $-\omega \mathbf{\Omega }\sin (\omega t-\mathbf{K}\cdot \mathbf{x})$. Since that sine is not identically zero, the required conditions are preciselyDotting the second equation with $\mathbf{\Omega }$ gives $\mathbf{\Omega }\cdot \mathbf{\Omega }=0$ because $\omega \ne 0$. This dot product is complex bilinear, not Hermitian. Taking the cross product again with $\mathbf{K}$ and using transversality gives $-|\mathbf{K}{|}^2\mathbf{\Omega }=-{\omega }^2\mathbf{\Omega }/c^2$. Nonzero amplitude therefore requires $\boxed{|\mathbf{K}{|}^2={\omega }^2/c^2}$.

Taking real and imaginary parts yields the full conditionsThey imply $\mathbf{e}\cdot \mathbf{b}=0$ and $|\mathbf{e}|=c|\mathbf{b}|$, together with the dispersion relation. Conversely a nonzero transverse $\mathbf{e}$, this dispersion relation and $\mathbf{b}=\mathbf{K}\times \mathbf{e}/\omega$ satisfy all the conditions.

The physical fields are $\mathbf{E}=\mathbf{e}\cos (\omega t-\mathbf{K}\cdot \mathbf{x})$ and $\mathbf{B}=\mathbf{b}\cos (\omega t-\mathbf{K}\cdot \mathbf{x})$. Each oscillates along one fixed real direction, so both fields are linearly polarized. Complex encoding of the two fields here does not mean either real field is circularly polarized.

For a normal constrained extremal, introduce a constant Lagrange multiplier and use $L=F-\lambda G$. The Euler-Lagrange equation isWhen both integrands have no explicit $x$ dependence, the Beltrami identity gives $L-y^{\prime }{L}_{y^{\prime }}=\text{constant}$, as follows by differentiating this expression and using the Euler-Lagrange equation.

Parametrize the surface by $(f(z)\cos \theta ,f(z)\sin \theta ,z)$. The cross product of its two coordinate tangents has magnitude $f\sqrt{1+(f^{\prime }{)}^2}$, so integration in $\theta$ gives $A=2\pi {\int }_a^{b}f\sqrt{1+(f^{\prime }{)}^2}dz$. Integrating the cross-sectional disk areas gives $V=\pi {\int }_a^b{f}^{2}dz$.

Dividing the augmented functional $A-\lambda V$ by $\pi$, useThe Euler-Lagrange equation isThe Beltrami identity becomes $2f/\sqrt{1+(f^{\prime }{)}^2}-\lambda f^2=C$. At an endpoint $f\to 0$, the absolute value of the first term is at most $2f$, so $C=0$, even if the profile has an infinite endpoint slope. For a nondegenerate regular closed surface take $f>0$ on $(a,b)$. Then $\lambda >0$, and, with $R=2/\lambda$,Combining this identity with the Euler-Lagrange equation yields $f^{\prime \prime }=-R^2/f^3$. Consequently $(f^2{)}^{\prime \prime }=2((f^{\prime }{)}^2+f{f}^{\prime \prime })=-2$. The two zero endpoint values uniquely determineThe positive radius profile is therefore a semicircle and generates a sphere, centred at height $(a+b)/2$, with radius $R=(b-a)/2$. In particularThe graph is smooth in the interior; its vertical slopes at the two poles are compatible with a smooth spherical surface. Its volume is $4\pi R^3/3$, the volume associated with this stationary candidate.

For minimality, the three-dimensional isoperimetric inequality says that every closed surface enclosing volume $V$ has $A^3\ge 36\pi V^2$, with equality for a sphere. The displayed extremal attains equality, proving that it minimizes area at its fixed enclosed volume. The uniqueness above concerns the regular connected positive profile with the two stated endpoint heights.