Use axes adapted to the two orthogonal rotations in the diagram:For a linearly oscillating Hertzian dipole with real unit axis $\mathbf{d}$ and total time-averaged radiated power $P$, the angular distribution of electric dipole radiation isThe normalization follows from $\int {\sin }^2\vartheta d\Omega =8\pi /3$. Equal readings therefore implyHere $d_z\ne 0$, since otherwise all components would vanish. Reversing the axis is immaterial, so choose $d_z>0$. The complete family isDefine $a=\sqrt{2}-1$, $b=-\sqrt{2}-1$, and $D=1+(s_B\sqrt{2}-1{)}^2+(s_C\sqrt{2}-1{)}^2$. Then $d_z^2=1/D$, and the absolute intensity fixesFor the measured distance and intensity, the allowed axes and powers are:

The intensity data do not select a unique orientation or total power. This is the dipole-axis ambiguity from three intensity measurements: the readings constrain squared projections and lose their signs. For example, the first branch makes an angle $30.36^{\circ }$ with $A$, but the other branches are equally consistent with the three measurements. A further suitable observation direction or polarization measurement is needed to remove the ambiguity. The calculation uses radiation-zone intensities, as appropriate to the Hertzian dipole model.

Let $I_0$ be the intensity from either fibre alone at the observation point, assuming equal output intensities, matching polarizations and negligible drift. The screen samples the relative phase $\Phi (X)$ between the two outputs. For an ideal monochromatic laser, first-order optical coherence is complete andThe interference fringes are approximately equally spaced in the small-angle far field, with a slowly varying diffraction envelope. If the output separation is $a$ and the screen distance is $D$, the interference fringe spacing is approximately ${\lambda }_{0}D/a$. Equal phases at the two input ends do not ensure a central bright fringe: the extra fibre contributes carrier phase $2\pi n_{\mathrm{eff}}(l_2-l_1)/{\lambda }_0$ in addition to the free-space optical path difference.

For a finite linewidth, the time-averaged interference signal becomesHere $g^{(1)}$ may be defined for the slowly varying field envelope, with the carrier phase already included in $\Phi$. A realistic laser therefore gives reduced interferometric visibility; for $V=1/2$, its extrema are $I_0$ and $3I_0$.

For a Lorentzian line shape of frequency full width at half maximum $\Delta \nu$, temporal optical coherence givesThe relevant index is the group index. Using the intended nondispersive approximation $n_g\simeq n_{\mathrm{eff}}=2$,This delayed-fibre linewidth measurement assumes the loss of contrast comes from linewidth. Unequal output intensities, polarization mismatch or mechanical phase fluctuations also lower interferometric visibility. If dispersion matters, the stated phase index alone does not specify the necessary group index.

At an air–glass interface, resolve the incident unpolarized light into components perpendicular ($s$) and parallel ($p$) to the plane of incidence. The Fresnel equations give the parallel reflection coefficientAt Brewster's angle, $r_p=0$. Together with Snell's law, this gives ${\theta }_i+{\theta }_t=\pi /2$ and $\tan {\theta }_i=n_2/n_1$. Thus, taking the incident medium to be air,The remaining reflected component has linear polarization perpendicular to the plane of incidence. This assumes a transparent, nonmagnetic glass with negligible absorption.

The symmetric group $S_3$ is generated by transpositions, each of order two. A group homomorphism to ${\mathbb{Z}}_3$ must map every transposition to zero, because the only target element annihilated by two is zero. Hence the map to ${\mathbb{Z}}_3$ is necessarily trivial.

For $S_3\to {\mathbb{Z}}_2$, the 3-cycles map to zero because their order divides three. All transpositions are conjugate group elements, and a map to an abelian group takes conjugate elements to the same value. That common value is either zero or one. Both choices work: they give the trivial map and the sign homomorphism, with an odd permutation sent to $1$ and an even permutation sent to $0$.

A group homomorphism from the additive cyclic group ${\mathbb{Z}}_6$ is determined by the image $t$ of its generator $1$, with the consistency condition $6t=0$ in the target. Every element of either target satisfies this condition. ThereforeandThese formulas are well-defined because changing $n$ by six changes $tn$ by a target multiple of its modulus. They list respectively two and three maps.

Choose $g\ne e$. By Lagrange's theorem, its cyclic subgroup has order dividing the prime 101. It is not of order one, so $⟨g⟩=G$. Thus the group is cyclic, and hence abelian. In particular, every pair of elements is $g^i,g^j$, and $g^i{g}^j=g^{i+j}=g^{j+i}=g^j{g}^i$.

A group homomorphism $f:G\to H$ preserves the operation: $f(ab)=f(a)f(b)$ for all $a,b\in G$. A group isomorphism is a bijective group homomorphism; its inverse also preserves the operation. The kernel of a group homomorphism is $\ker f=\{g\in G:f(g)=e_H\}$, a normal subgroup of $G$.

Write the dihedral group as $D_6=⟨r,s\mid r^6=s^2=1,srs=r^{-1}⟩$, where $r$ rotates through $\pi /3$. There are six rotations and six reflections, so the order is 12. Conjugation by $s$ reverses a rotation, and conjugation by $r$ sends $r^{j}s$ to $r^{j+2}s$. Consequently the conjugacy classes and their sizes areThe last two classes distinguish the reflection axes through opposite vertices and through opposite edge midpoints. Their sizes sum to 12.

Use group axioms from two-sided local inverse identities. Choose any $a$ in the nonempty semigroup and put $e_R=a_L^{-1}a$, $e_L=a{a}_R^{-1}$. The given identities say $b{e}_R=b$ and $e_{L}b=b$ for every $b$. Therefore $e_L{e}_R=e_L$ by the right-identity property, and $e_L{e}_R=e_R$ by the left-identity property. Thus $e_L=e_R=e$ is a two-sided identity element.

For an arbitrary $a$, set $b=a_L^{-1}$ and $c=a_R^{-1}$. Inserting $e$ into the original identities gives $ba=e$ and $ac=e$. By associativity,Thus every element has a two-sided inverse element. Closure and associativity were assumed, so all the group axioms hold. No cancellation or identity was used before it had been established.

For the first Cayley table, associativity fails: $(a\ast a)\ast b=1\ast b=b$, whereas $a\ast (a\ast b)=a\ast c=d$. It therefore does not define a group.

For the second table, assign the residues $0,1,2,3,4$ to $1,a,c,b,d$, respectively. Every table entry agrees with addition modulo an integer modulo five. Thus this table defines the cyclic group of order five, with identity $1$ and inverses $a^{-1}=d$, $c^{-1}=b$. The identification with a known group verifies associativity as well as closure and the inverse axioms; inspecting only the identity row and inverses would not suffice.

The disk still has only three degrees of freedom and hence three normal modes: two centre coordinates and one in-plane rotation. Five springs alter the restoring quadratic form but add no disk coordinates. The fivefold rotational symmetry makes the translational stiffness proportional to the two-dimensional identity matrix, and makes translation-rotation coupling vanish. Therefore the two translation modes are always degenerate; the rotational mode is separate, with possible accidental equality of its frequency for particular parameters. The general formulas in small oscillations of a disk attached to prestretched radial springs apply with $N=5$.

The quadratic Lagrangian is diagonal in $x,y,\theta$, so the Euler-Lagrange equations give two pure translations and one pure rotation. Their angular frequencies areA horizontal translation has $(x,y,\theta )=(A\cos {\omega }_{x}t,0,0)$; a vertical translation has $(0,A\cos {\omega }_{y}t,0)$; and the rotational mode has $(0,0,A\cos {\omega }_{\theta }t)$. Any fixed-direction translation is another choice in the same two-dimensional degenerate normal mode space. The rotation keeps the centre fixed.

Let $\mathbf{q}=(x,y)$ be the centre displacement and $\theta$ the small disk rotation. Put $d=l-l_0>0$. The moment of inertia of a uniform disc is $I=m{a}^2/2$, obtained from $I={\int }_0^a\rho r^2(2\pi rdr)$ and $m=\rho \pi a^2$. Thus the kinetic energy is $T=m|\dot{\mathbf{q}}{|}^2/2+m{a}^2{\dot{\theta }}^2/4$.

For one spring, let ${\mathbf{n}}_j$ point radially towards its fixed anchor and let ${\mathbf{t}}_j$ be the perpendicular unit vector. The moving attachment is $\mathbf{q}+aR(\theta ){\mathbf{n}}_j$, while the anchor is $(a+l){\mathbf{n}}_j$. Expanding their separation to second order gives the spring extension beyond natural length:Insert this in the elastic potential energy $V=(k/2){\sum }_j(s_j-l_0{)}^2$. For the three symmetric directions, ${\sum }_j{\mathbf{n}}_j={\sum }_j{\mathbf{t}}_j=0$ and ${\sum }_j({\mathbf{n}}_j\cdot \mathbf{q}{)}^2={\sum }_j({\mathbf{t}}_j\cdot \mathbf{q}{)}^2=3|\mathbf{q}{|}^2/2$. Dropping the constant equilibrium energy givesTherefore the quadratic Lagrangian isThe prestretch terms cannot be omitted: they supply transverse spring stiffness and the entire rotational restoring torque, as derived generally in small oscillations of a disk attached to prestretched radial springs.

Let $J_{ij}={\int }_B{x}_i{x}_j{d}^{3}x$. Rotational symmetry makes $J$ an isotropic second-rank tensor, so $J_{ij}=C{\delta }_{ij}$. Contracting the indices and integrating in spherical coordinates givesThe isotropic second moment of a ball therefore givesOnly the scalar part of $M$ contributes; its antisymmetric tensor and symmetric-traceless parts give zero.

Take the trace first: the antisymmetric tensor has zero diagonal and $X$ is traceless, so $T_{ii}=3t$. Taking symmetric and antisymmetric parts then gives the scalar, symmetric-traceless and axial tensor decomposition:The Kronecker delta satisfies ${\delta }_{ii}=3$, which verifies $X_{ii}=0$. Substitution verifies the original decomposition and also proves its uniqueness.

For a rotation matrix $R\in SO(3)$, the Cartesian tensor component transformation law isAn isotropic tensor is unchanged under every such rotation, so its components satisfy $T_{ijkl}=R_{ia}{R}_{jb}{R}_{kc}{R}_{ld}{T}_{abcd}$ for every $R\in SO(3)$. The repeated indices are summed by the Einstein summation convention.

Write $q=i\omega$. A constant-coefficient linear ordinary differential equation has transfer function $1/P(q)$, where $P(q)={\sum }_{n=1}^c{a}_n{q}^n$. The specified simple poles correspond to $q=0,-2,1$. A minimal-order choice is thereforeThus one possible system isup to a common nonzero multiplicative constant. The root $q=1$ permits an exponentially growing homogeneous solution; the pole data therefore describe an unstable system. This does not prevent the requested algebraic reconstruction of its transfer function.