ON and OFF receptive-field subregions distinguish locations by response polarity. In the classic mapping experiment, a bright spot in an ON region increased firing during illumination; illumination in an OFF region suppressed maintained firing and/or evoked a discharge when the light disappeared. If maintained firing was very low, suppression could be hard to see. The recorded decrease alone did not distinguish direct inhibition at a synapse from reduced excitation. ^[ref] These are functional regions in visual space, not compartments of a cell body.

Phi-fourth theory has a scalar interaction $-\lambda {\varphi }^4/4!$. In four dimensions its coupling is classically marginal, and its one-loop four-point function receives bubble corrections in the three Mandelstam channels.

A free real scalar field with zero mass has the displayed wave equation in Minkowski spacetime. In curved spacetime its dynamics also depend on curvature coupling: a minimally coupled massless scalar field and a conformally coupled scalar field need not have the same cosmological modes.

Cover the circle by two arcs whose intersection has two components and take the product with the projective plane. The Mayer–Vietoris sequence has maps $(a,b)\mapsto (a+b,-a-b)$ on the two copies of the homology of the plane. Its kernels and cokernels, together with $H_1({\mathbb{RP}}^2;\mathbb{Z})=\mathbb{Z}/2$, give the displayed groups; all higher groups vanish. The product is a closed nonorientable three-manifold, but is homotopy equivalent to a noncompact oriented four-manifold.

Choose a complex structure $J$ on ${\mathbb{R}}^{2m}$, with $J^2=-I$. Its induced map on $\mathbb{R}{P}^{2m-1}$ has no fixed point: $Jx=\lambda x$ for a real nonzero vector would give ${\lambda }^2=-1$. The maps $\cos tI+\sin tJ$, for $0\le t\le \pi /2$, give a homotopy from the identity to $J$, because they are invertible throughout. Thus this fixed-point-free map is homotopic to the identity.

For $n\ge 2$, a nonsurjective self-map of Real projective space ${\mathbb{RP}}^n$ has zero mod-two degree of a map between closed manifolds. In the mod-two cohomology ring of real projective space, the nonzero degree-one class $a$ has $a^n\ne 0$. If pullback fixed $a$, it would fix $a^n$, contradicting zero degree. Thus the map is trivial on the fundamental group and lifts to $S^n$. A point missed by the original map has both its inverse images missed by the lift. The lift therefore takes values in a punctured sphere, which is contractible, and is null-homotopic. Composing its contraction with the covering projection proves the conclusion.

The quotient of the three-sphere by its antipodal map is a closed oriented three-manifold. Equivalently it is the lens space obtained by integer surgery of coefficient two on the unknot. Its fundamental group is cyclic of order two, from the sphere double covering. This makes it a basic example for Jones polynomial surgery invariants.

// Target: knot-theory.bigb

The outward-normal-first volume form on the unit sphere pulls back under the antipodal map by $(-1{)}^{n+1}$. For $n\ge 1$, the sphere is connected, and its orientation descends to the twofold quotient exactly when this deck transformation preserves it. Thus Real projective space is orientable precisely in odd positive dimensions; the zero-dimensional case is one point and is also orientable.

Map each real into $(0,1)$, choose its canonical binary expansion, and flatten the resulting countable array of bits using a bijection $j:{\mathbb{N}}^2\to \mathbb{N}$. Interpret these bits as ternary digits zero and one. At a first differing position, the leading difference exceeds the sum of all possible tail differences, so this is an injection from real sequences to a bounded real interval. The bijection $j(n,k)=(n+k-2)(n+k-1)/2+n$ lists successive finite diagonals.

A real interval contains every real number lying between any two of its elements.

An energy profile plots an explicitly specified energy against a reaction coordinate. Identify whether the vertical axis is potential energy, enthalpy, or Gibbs free energy; these descriptions are related but not interchangeable. A potential-energy profile describes a path through configurations, while a free-energy description also accounts for the relevant thermodynamic conditions.

For the two chosen profiles, the reaction energy barriers from the reactants are $80$ and $50\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. From the reaction products, the same maxima are $100$ and $70\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ higher. Both paths have endpoint difference $-20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The barrier is measured from the relevant starting endpoint, not from an arbitrary zero of potential energy.

The appeal of the diagram is that it separates two useful questions: which endpoint is lower, and which path is easier to traverse? Catalysis changes the route through transition states while retaining the endpoint thermodynamics of the same overall chemical reaction. A real catalytic route may have several steps and reaction intermediates; the two simple curves do not specify such a reaction mechanism. The plotted potential-energy difference alone is not an equilibrium constant or a calculated rate constant. See Chemistry 2e, 12.7: alternative reaction routes and catalysis.

A potential energy surface assigns potential energy to the configurations of the reacting chemical atoms, for a specified electronic description. A reaction coordinate selects a path through this many-dimensional configuration space; a potential-energy profile then records the potential energy along that path. Different paths can share endpoints but cross different barriers. See the IUPAC Gold Book: potential-energy reaction surface.

In a clear terrestrial atmosphere, Rayleigh scattering by air molecules redirects some sunlight toward an observer looking away from the Sun. Shorter wavelengths are preferentially scattered, enriching the scattered light in blue compared with the incoming sunlight. Along the longer atmospheric path to the low Sun, more short-wavelength light is removed from the direct beam, helping produce red sunsets. ^[ref].

Violet light has an even shorter wavelength than blue light, so the Rayleigh scattering wavelength dependence alone cannot predict the perceived sky color. The incoming solar optical spectrum and the eye's differing response to wavelengths both matter; human vision is less sensitive to violet than to blue. ^[ref]. Multiple scattering and other atmospheric particles also affect the observed sky. Cloud droplets are generally too large for this simple Rayleigh scattering model: Mie scattering describes spherical droplets, with much weaker visible wavelength selectivity, while repeated scattering helps make clouds appear white. ^[ref].

For incident unpolarized light and the isotropic electric dipole response of Rayleigh scattering by a small sphere, let $\theta$ be the angle between the incident beam's propagation direction and the outgoing direction. The differential scattering cross-section and its relative angular pattern arewhere $d\Omega$ is an element of solid angle. Averaging over the two transverse directions of incident polarization of an electromagnetic wave gives this pattern. Forward scattering at $0^{\circ }$ and backward scattering at $180^{\circ }$ are equally strong: $F=1$. Sideways scattering at $90^{\circ }$ is weakest, but nonzero: $F=1/2$. ^[ref].

The normalization is consistent becauseThus integrating $d\sigma /d\Omega$ gives the total Rayleigh scattering cross-section $\sigma$. The plotted $F$ has maximum one; it is not itself normalized to unit solid angle integral.

For a single electric dipole driven by incident linear polarization, the angular distribution of electric dipole radiation instead has a ${\sin }^2\alpha$ factor, where $\alpha$ is measured from the electric dipole moment direction, parallel to the incident electric field. That direction is perpendicular to the incident beam. Confusing $\alpha$ with the beam angle $\theta$ would incorrectly predict zero forward scattering. Averaging the dipole patterns over incident unpolarized light yields the displayed $F(\theta )$.

For a fixed sphere and fixed refractive index, both the Rayleigh scattering cross-section and scattering efficiency are proportional to ${\lambda }^{-4}$. Halving wavelength multiplies either by $16$, as long as the shorter wavelength still satisfies $ka\ll 1$. If optical dispersion changes $m$, the material factor changes too; the pure fourth-power comparison then needs correction.

For equal incident light intensities at blue wavelength $450\mathrm{nm}$ and red wavelength $650\mathrm{nm}$, the same toy sphere givesBlue light therefore produces about $4.35$ times the scattered power. This compares the particle's response at two wavelengths; it does not assert that an actual sky has this ratio of observed blue and red light intensities.

Consider a fixed, homogeneous, nonabsorbing, nonmagnetic sphere of radius $a$ in vacuum. Let $m$ be its relative refractive index, a fixed real number; in vacuum this equals the sphere's refractive index. Treat $m$ as independent of wavelength over the range being compared, neglecting optical dispersion. Air can be approximated by vacuum here because its refractive index is close to one. These assumptions describe a toy particle, rather than a literal model of an individual air molecule.

The Rayleigh scattering cross-section $\sigma$ is an effective area: the total scattered power is $P_{\mathrm{sca}}=I_{\mathrm{inc}}\sigma$, where $I_{\mathrm{inc}}$ is the incident light intensity. For $ka\ll 1$ its leading approximation isThe units are area, since $k^4{a}^6$ has units ${\mathrm{m}}^{-4}{\mathrm{m}}^6={\mathrm{m}}^2$. This Rayleigh scattering cross-section measures scattering, not conversion of the incident electromagnetic radiation into heat. The definition follows ^[ref].

To see the mechanism, the nearly uniform applied electric field produces an electric dipole moment with peak magnitudewhere $E_0$ is the incident electric field peak wave amplitude and ${ϵ}_0$ is vacuum permittivity. This is the sphere's response in electrostatics, with its permittivity divided by vacuum permittivity equal to $m^2$ under the nonmagnetic assumption. ^[ref]. The averaged electric dipole radiation power is $P_{\mathrm{sca}}={\omega }^4{p}_0^2/(12\pi {ϵ}_0{c}^3)$, with angular frequency $\omega =ck$ and speed of light $c$. Dividing by $I_{\mathrm{inc}}={ϵ}_{0}c{E}_0^2/2$ gives the displayed Rayleigh scattering cross-section. Thus doubling radius multiplies the electric dipole moment by eight and the scattered power by $64$, provided the larger sphere still satisfies $ka\ll 1$. ^[ref].

A thin liquid lining inside a rigid cylindrical tube has pressure perturbation $p_1=-\sigma (a^{-2}{h}_1+h_{1,xx})$ at leading order in $h_0/a$. Lubrication theory gives flux $q=-h_0^3{p}_x/(3\mu )$, so mass conservation gives the displayed growth rate. Long waves with $0<k^2{a}^2<1$ grow, while short waves are stabilized by curvature. The maximum occurs at $k^2{a}^2=1/2$.

For a smooth inviscid parallel shear profile with an exponentially growing mode, let $U_s$ be its velocity at an inflection point. The real and imaginary parts of the integrated Rayleigh equation for inviscid shear flow imply $\int U^{\prime \prime }(U-U_s)|\psi {|}^2/|U-c{|}^2=-\int (|{\psi }^{\prime }{|}^2+k^2|\psi {|}^2)<0$. Consequently $U^{\prime \prime }(U-U_s)<0$ somewhere in the channel. This strengthens Rayleigh's inflection-point theorem but is still only a necessary condition, not a sufficient instability criterion.

For a normal mode of an inviscid parallel flow, the perturbation pressure is $p=-(U-c){\varphi }^{\prime }+U^{\prime }\varphi$. At an interface where $U$ is continuous but $U^{\prime }$ jumps, continuity of normal velocity and pressure gives continuity of $\varphi$ and the displayed flux. Consequently $(U-c)[{\varphi }^{\prime }]=[U^{\prime }]\varphi$. A sinuous mode of a planar jet may therefore have a cusp at a symmetric velocity-profile corner: even parity alone does not force its one-sided derivative to vanish.

Decaying normal modes of this continuous piecewise-linear profile satisfy ${\varphi }^{\prime \prime }-k^2\varphi =0$ between the corners. Pressure matching at a piecewise-linear shear interface yields an odd varicose mode of a planar jet with real speed $c=(1-e^{-2k})/(2k)$ and an even sinuous mode of a planar jet whose speeds satisfyIts reduced discriminant factors as $(2k-3-e^{-2k}-4e^{-k})(2k-3-e^{-2k}+4e^{-k})$. The second factor is positive for $k>0$, while the first is strictly increasing and crosses zero once. Thus exactly the long-wave even branch is exponentially unstable, for $0<k<k_c$, where $2k_c-3-e^{-2k_c}-4e^{-k_c}=0$.