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The finite orthogonal group obtained by quotienting a crystal space group by its translation subgroup. In three dimensions there are 32 crystallographic point-group types, constrained by preservation of a lattice. It describes macroscopic rotational and reflection symmetry, not the fractional translations of the full space group.

Interstellar shocks arise when relative flow is supersonic: stellar winds, supernova ejecta, colliding clouds and gas moving through strong disturbances can drive them. Neighboring compressive characteristics steepen or a fast flow encounters material it cannot decelerate through upstream sound communication. A thin dissipative front then converts directed kinetic energy into thermal energy and increases entropy.

Integrate mass, normal momentum and total energy conservation through a planar steady shock in its rest frame. With no mass source, negligible external impulse across the thin layer and no radiative energy loss, their normal fluxes are respectivelyFor a perfect gas of constant adiabatic index, $h=\gamma p/[(\gamma -1)\rho ]$. Equality of the fluxes on the two sides yields the three Rankine-Hugoniot conditions in the question. Adiabatic means no energy escapes this control volume; it does not make the shock isentropic.

In the strong-shock limit, let $r={\rho }_2/{\rho }_1=u_1/u_2$ and neglect $p_1$. Momentum conservation gives $p_2={\rho }_1{u}_1^2(1-1/r)$. Substitution into energy conservation givesThe nontrivial compressive root isThe $r=1$ root describes no shock. A monatomic gas has the familiar compression four.

For identical colliding clouds in their center-of-momentum frame, reflection symmetry fixes the contact plane and the uniform compressed material on either side at rest. The unshocked clouds still approach at speeds $\pm V$; their supersonic upstream flow has not received a smooth pressure adjustment ahead of the shock. Let the right shock move to the right at speed $S$. In its own frame the upstream and downstream normal speed magnitudes are $V+S$ and $S$. The same shock compression ratio therefore givesThis is the shock speed in a symmetric cloud collision. For $\gamma =5/3$, the shocks move outward at $V/3$, not at the upstream speed or the relative cloud speed $2V$. As a check, the post-shock pressure is $p_2=(\gamma +1){\rho }_1{V}^2/2$; in the monatomic case $p_2/{\rho }_2=V^2/3$, giving specific internal energy $V^2/2$, exactly the initial bulk kinetic energy per unit cloud mass when brought to rest.

When a shock reaches a cloud's rear boundary, its supply of cold upstream gas ends. If the surrounding pressure is low, the hot compressed gas expands outward and a rarefaction travels back into it, converting thermal energy into motion and cooling the gas. A sustained uniform stationary shocked region is no longer the solution. Interaction with an appreciable exterior medium can instead drive another outward disturbance; the boundary pressure determines that part of the evolution. The stated adiabatic model provides no radiative mechanism that would keep a permanently cold dense sheet at the contact.

Up to a global phase, $x(a-x)$ is real, positive throughout the interior, has no interior nodes of a wavefunction, and is symmetric about $x=a/2$. It is smooth inside the box and has one central maximum. These are the shape and reflection symmetry properties of the lowest particle in a box eigenfunction, $\sin (\pi x/a)$; they make it a plausible variational method trial without making it exact.

There are two carbon-placement isomers, represented by the cyclic base sequences $\mathrm{CCBBB}$ and $\mathrm{CBCBB}$. The carbon vertices are adjacent in the first and separated by one boron vertex in the second. Rotation or reversal of a sequence does not produce another constitutional isomer.

Each arrangement has a reflection symmetry that exchanges a pair of boron sites while leaving the third distinct. Consequently each can give two ${}^{11}\mathrm{B}$ environments, with site populations $2:1$, in a proton-decoupled NMR spectrum. This assumes that the hydrogen and carbonyl arrangements preserve, or rapidly average to, the corresponding skeletal symmetry. Two peaks alone neither choose between the two skeletons nor establish a unique hydrogen arrangement; accidental overlap or fluxionality can also reduce the number of observed signals.

The Sturm oscillation theorem gives three nodes of a wavefunction for the fourth bound state. reflection symmetry and nondegeneracy of one-dimensional bound states make it odd: one node of a wavefunction is at the centre, and the other two lie at opposite positions. In each flat allowed region the wavefunction is a sinusoid, with $k=\sqrt{2m(E-V)}/\hslash$; its wavelength is shortest in the central, deepest region. Outside the well it decays exponentially, with opposite signs on the two sides. Both $\Psi$ and ${\Psi }^{\prime }$ remain continuous at every finite step, as required by wavefunction matching at a finite potential step; only the curvature jumps. In particular, the wavefunction need not vanish at a step.

This original wavefunction matching at a finite potential step illustration uses freely chosen well dimensions and energies. It shows the required parity, node count, wavelength ordering and decaying tails; the source drawing supplies no numerical scale from which to fix exact node positions.

A normal mode has all displacement coordinates proportional to a common time dependence, with fixed relative amplitudes. For a nonzero frequency the time dependence is sinusoidal. A zero-frequency mode is a free displacement or uniform drift rather than an oscillation.

Use the printed mass ordering $M,m,M$. Displacements are measured from the unstretched equilibrium positions, givingEquivalently,This is the generalized eigenvalue problem for normal modes of three masses connected by two equal springs. Substitute $\mathbf{x}=\mathbf{u}{e}^{i\omega t}$ to obtain $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. Reflection symmetry gives an antisymmetric vector $(1,0,-1)$ and symmetric vectors $(1,b,1)$. The former has ${\omega }^2=k/M$. For a symmetric vector, the end and centre equations give $M{\omega }^2=k(1-b)$ and $m{\omega }^{2}b=2k(b-1)$. Their solutions are $b=1$, $\omega =0$, or $b=-2M/m$, ${\omega }^2=k/M+2k/m$. HenceThese are all three eigenmodes. The last two keep the centre of mass fixed and are orthogonal in the mass-weighted inner product. The general zero-mode contribution is $(A+Bt)(1,1,1)$, accounting for both a centre-of-mass shift and a conserved total momentum. There are no fixed end springs in this system.

The original amplitude profiles illustrate $M=2m$; the symmetric internal shape is rescaled for display. The formula above gives every positive mass ratio.

Two further physical examples are standing-wave modes of a string fixed at both ends, with spatial shapes $\sin (n\pi x/L)$ and frequencies $n\pi c/L$, and electrical normal modes of two coupled LC circuits, in which the capacitor charges oscillate at fixed relative amplitudes. Independent normal mode coordinates simplify both systems in the same way as the spring-chain eigenvectors.

Put the positive plate at $z=d/2$ and the negative plate at $z=-d/2$. Reflection through $z=0$, combined with reversal of the two plate charges, makes the tangential fields cancel at every point of the mid-plane. Thus $E_x=E_y=0$ at the corner as well as inside the capacitor.

For the uniform-charge estimate at a capacitor corner, assemble four identical square pairs around their common corner. Their normal fields there are equal by rotation symmetry and add to the central field $\sigma /{ϵ}_0$ of the larger plate pair. With the bulk parallel-plate relation $\sigma \simeq {ϵ}_{0}V/d$, the conventional corner estimate isThis quarter-field result uses prescribed uniform surface charge, not symmetry alone for finite equipotential conductors. Real conducting plates redistribute charge near their edges; $d\ll a$ justifies the bulk relation away from the edges but does not remove the local corner fringe. For that exact conductor problem only the zero tangential components follow from the stated reflection symmetry; its normal coefficient requires a fringing-field boundary-value solution. The standard parallel-plate formula explicitly neglects edge effects.^[ref]

A normal mode is a small-oscillation motion with every coordinate proportional to one common sinusoid, at a fixed amplitude ratio and relative phase. Expanding an arbitrary small motion in normal mode coordinates decouples its linear equations into independent oscillators. With mechanical mass matrix $\mathsf{M}$ and mechanical stiffness matrix $\mathsf{K}$, the generalized eigenvalue equation is $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$.

Horizontal gravitational restoring force on each bob is $-m_i(g/l)x_i$. The two springs add $k(x_2-x_1)$ on the first bob, $k(x_1-2x_2+x_3)$ on the middle bob, and $k(x_2-x_3)$ on the last. Dividing by the bob masses gives the stated equations withReflection symmetry separates an antisymmetric end motion from two symmetric motions. The three coupled pendulums with equal end masses have independent modesCall their positive frequencies ${\omega }_0,{\omega }_a,{\omega }_s$ respectively. The middle amplitude in the last mode ensures $m{x}_1+M{x}_2+m{x}_3=0$. The modes are orthogonal in the mass-weighted inner product, rather than generally in the unweighted coordinate dot product.

The initially stationary displacement decomposes asThere are no sine terms because every initial velocity is zero. ThereforeThese give $(0,0,d)$ at $t=0$ and solve all three coupled equations. The spring constant is assumed positive; for $k=0$ the frequencies become degenerate and the same superposition reduces to uncoupled pendulum motion.

The original example uses $M=2m$ and $k/m=g/l$ to display the mode vectors and resulting superposition. Those illustrative parameter ratios are not additional supplied data.

Two identical cold clouds approach at speeds $\pm V$ in their center-of-momentum frame. Reflection symmetry makes the uniformly compressed gas stationary. A shock moving outward at $S$ sees upstream speed $V+S$ and downstream speed $S$; their ratio equals the strong adiabatic compression $(\gamma +1)/(\gamma -1)$. Solving gives the displayed speed. For a monatomic gas $S=V/3$. Finite-cloud breakout ends the steady upstream supply and requires a subsequent boundary-release calculation.

For equal lengths, end masses $m$, middle mass $M$, and equal springs between adjacent bobs, reflection symmetry gives mode vectors $(1,1,1)$, $(1,0,-1)$ and $(1,-2m/M,1)$. Their squared frequencies are displayed above. These vectors are orthogonal in the inner product weighted by $\mathrm{diag}(m,M,m)$. Equal end masses, rather than equality of all three masses, suffice for this symmetry reduction.