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In vacuum, a photon of wave vector $\mathbf{k}$ carries momentum $\mathbf{p}=\hslash \mathbf{k}$ and energy $E=\hslash c|\mathbf{k}|$. In a direct optical interband transition this momentum is generally small compared with relevant crystal momenta, motivating the nearly vertical-transition approximation in wave vector space. Electromagnetic momentum in matter requires additional conventions.

The photon number flux density is the number of photons crossing unit normal area per unit time. A monochromatic beam of light intensity $I$ and frequency $f$ has ${\Phi }_N=I/(hf)$ because each photon carries energy $hf$. Its units are ${\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. For a mixed spectrum, integrate the spectral light intensity divided by the corresponding photon energy.

A photon rocket converts stored mass-energy into a collimated beam of photons. In its instantaneous rest frame, energy conservation gives $c^{2}dM=-d{E}_{\gamma }$, and momentum conservation gives $Mcd\eta =d{E}_{\gamma }/c$. Therefore a monotone change in rapidity costs $M_f/M_i=e^{-|\Delta \eta |}$. The rocket's rest mass includes its fuel and stored internal energy; inefficiency or an angular spread of the exhaust increases the fuel requirement.

Absorption of momentum $\mathbf{p}$ followed by emission of ${\mathbf{p}}^{\prime }$ changes a free atom's momentum by $\mathbf{p}-{\mathbf{p}}^{\prime }$. For isotropic emission of fixed magnitude, its mean vector is zero. The energy change then separates into a directional work term and two positive recoil terms. If absorption directions are also isotropic and uncorrelated with the initial momentum, the work term averages away. Equal incoming and outgoing photon magnitudes give two recoil energies per cycle.

The peak of a measured radiation spectrum associated with full deposition of the incident photon energy. Its width reflects detector resolution; other interactions form partial-energy continua.

For an ultrarelativistic proton of energy $E$ colliding head-on with a photon of energy $ϵ$, the invariant squared collision energy is approximately $(m_p{c}^2{)}^2+4Eϵ$. Producing a final proton and neutral pion requires at least $(m_p{c}^2+m_{\pi }{c}^2{)}^2$. Thus $E_{\mathrm{th}}=m_{\pi }{c}^2(2m_p{c}^2+m_{\pi }{c}^2)/(4ϵ)$. A broad isotropic photon background turns this threshold into an energy-loss feature depending on the photon distribution and source distance, not an absolute ceiling for all arriving protons.

Processes limiting light-induced damage when absorbed excitation exceeds safe photochemical use. Energy dissipation, regulated electron flow and repair have complementary roles.

Photorespiration is the metabolic route processing products of RuBisCO oxygenation and recovering part of their carbon. It consumes energy and releases some carbon dioxide; under particular conditions it also provides a sink for energy and reducing power.

Photosynthesis captures light energy and transforms it into chemical forms that a biological cell can use. In plants, algae and cyanobacteria, this supports carbon fixation: incorporation of carbon from carbon dioxide into chemical compounds used to build living material. Capturing light and incorporating carbon are distinct processes coupled through metabolism. See OpenStax Biology 2e, 8.1: overview of photosynthesis.

For oxygenic photosynthesis, a simplified net equation writes a six-carbon sugar, glucose, as the representative product:This is net stoichiometry, not a one-step mechanism. A worked bookkeeping check counts six carbon chemical atoms, twelve hydrogen chemical atoms, and eighteen oxygen chemical atoms on each side: the product side has six oxygen chemical atoms in the sugar and twelve in molecular oxygen. The light label supplies energy, not additional chemical atoms.

Crucially, the released molecular oxygen comes from water, not carbon dioxide. Cancelling water made and consumed in different steps hides this distinction in the net equation. The Calvin cycle supplies a net three-carbon triose phosphate, such as glyceraldehyde 3-phosphate, for subsequent sugar synthesis; it does not directly export glucose. See OpenStax Biology 2e, 8.2: light-dependent reactions and OpenStax Biology 2e, 8.3: using light energy to make organic molecules.

The useful achievement is connecting incoming light to both usable energy and new biological material. ATP is an immediate energy currency; NADPH supplies electrons for synthesis; carbon dioxide supplies the carbon incorporated into carbohydrates. These roles are complementary. Conservation of energy still holds: photosynthesis transforms energy, never creates it, and some absorbed energy is dissipated as heat. A plant also carries out cellular respiration, including in biological cells that perform photosynthesis. Gas exchange reflects both processes. Oxygenic photosynthesis remains central to interpreting terrestrial molecular oxygen as an atmospheric biosignature gas. See OpenStax Biology 2e, 8.3: photosynthesis and respiration.

Production of useful molecules using photosynthesis to supply fixed carbon, energy or reducing power. Engineering a Cytochrome P450 at a photosynthetic membrane illustrates how an introduced biosynthetic reaction can draw on light-driven electron transfer.^[ref]

An organized light-harvesting and reaction-center assembly supporting photosynthetic energy conversion.

An antenna transfers excitation to a reaction centre, where charge separation initiates electron transfer. This distinguishes capture of a photon from conservation of its energy in a redox reaction. In oxygenic photosynthesis, two photosystems act in a connected electron-transfer chain. ^[ref]

Many ionic crystals have high melting temperatures compared with small molecular solids, because reorganizing their strongly interacting ions costs energy. But melting leaves a dense liquid with interactions; it does not perform the complete separation defining a lattice energy. At fixed pressure, the transition occurs when the Gibbs free energies of the solid and liquid match. The relevant enthalpy and entropy differences both matter, so lattice energy alone is not a universal ordering of melting temperatures. Strong covalent bonds in extended networks can also give high melting temperatures.^[ref]

A polymer exhibiting piezoelectricity. A compliant, low-density film can accept larger bending or strain than a brittle ceramic, while low permittivity can yield a large voltage coefficient despite a modest charge coefficient. Extractable energy still depends on mechanical input, capacitance and electrical loading.

Adding positive quadratic confinement $t\int h^2/2$ to the membrane energy changes its height spectrum to $k_{B}T/(\sigma q^2+t)$. The continuous height-shift symmetry is explicitly broken. Connected correlation function decay on the finite length $\xi$ and the height-difference variance saturates for every internal dimension, although its saturation value can depend on a microscopic cutoff. A negative confinement coefficient gives an unstable quadratic model rather than pinning.

Under the hypotheses of steady planar ideal-MHD field-line invariants, the specific enthalpy relation and the velocity projection of momentum give $\mathbf{B}\cdot \nabla Q=0$. The horizontal total energy flux, including the Poynting vector, is $k{\mathbf{B}}_{h}Q$. The magnetic correction accounts for work exchanged between the fluid and the field.

For a planar nematic director $\mathbf{n}=(\cos \chi ,\sin \chi )$ and constant amplitude ${\lambda }_0$, the nematic order parameter isDifferentiation gives $\nabla \cdot Q={\lambda }_0(-\sin 2\chi {\chi }_x+\cos 2\chi {\chi }_y,\cos 2\chi {\chi }_x+\sin 2\chi {\chi }_y)$. The coefficient matrix is orthogonal, proving the identity. Hence $K|\nabla \cdot Q{|}^2/2$ is the angle-gradient energy with effective modulus $K{\lambda }_0^2$. The full tensor-gradient norm instead obeys ${\sum }_{ijk}({\partial }_k{Q}_{ij}{)}^2=2{\lambda }_0^2|\nabla \chi {|}^2$ in this normalization; the two local densities must not be confused.

A rigid rotor restricted to one angular coordinate $\varphi$. Single-valued spatial states $e^{im\varphi }$ have $m\in \mathbb{Z}$ and energy $E_m={\hslash }^2{m}^2/(2I)$, with moment of inertia $I$. The zero-angular-momentum level has state degeneracy one; every nonzero $|m|$ has two directions. For labelled molecular ends its high-temperature rotational partition function is $\sqrt{\pi T/{\theta }_{\mathrm{rot}}}$, where ${\theta }_{\mathrm{rot}}={\hslash }^2/(2I{k}_{\mathrm{B}})$. Equivalent orientations and nuclear spin statistics require a further molecular counting convention. One quadratic angular momentum gives rotational heat capacity $k_{\mathrm{B}}/2$ per molecule, rather than $k_{\mathrm{B}}$ for an unrestricted linear rigid rotor.

The Planck constant $h$ connects the frequency $\nu$ of an electromagnetic field mode with a single photon's energy, $E=h\nu$. Its physical dimension is action, or energy multiplied by time. It sets the scale of quantum mechanics: the de Broglie wavelength of a particle is $\lambda =h/p$, and discrete energy levels often have spacings involving $h$.

In the International System of Units, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ exactly. Fixing this value defines the kilogram through already defined units of length and time; it does not remove experimental uncertainty from a practical mass measurement.^[ref] The reduced Planck constant $\hslash =h/(2\pi )$ is convenient when angular frequency replaces ordinary frequency.

Planck's law gives the spectral radiance of an ideal blackbody in thermal equilibrium at absolute temperature $T>0$. It predicts how much power is carried in each small frequency or wavelength interval, per projected area and per solid angle. The whole curve is determined by $T$: no material-specific parameter enters this ideal Planck function.

Writing $\nu$ for frequency in hertz and $\lambda$ for wavelength in metres, the two forms of the Planck function areHere $h$ is the Planck constant, $k_B$ the Boltzmann constant, and $c$ the speed of light in vacuum. Their exact values in the International System of Units are $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. ^[ref]

The units matter: $B_{\nu }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, whereas $B_{\lambda }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$. Both are spectral radiance, rather than radiant exitance or energy density. To report $B_{\lambda }$ per micrometre, multiply its per-metre value by $10^{-6}$; a micrometre-wide interval is $10^{-6}$ metres wide.

For a computed example, at $T=6000\mathrm{K}$ and $\lambda =0.500\mu \mathrm{m}$, the exponent is $hc/(\lambda k_{B}T)\approx 4.79592$. Substitution into Planck's law gives $B_{\lambda }\approx 3.17569\times 10^{13}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$, or $3.17569\times 10^7\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ per micrometre. At the same wavelength, the illustrative temperatures $3000\mathrm{K}$ and $4500\mathrm{K}$ give $2.60268\times 10^5$ and $6.37793\times 10^6$ in the latter units. These are ideal predictions for stipulated temperatures.

The key change of variables is $\nu =c/\lambda$. Corresponding positive intervals carry the same power, so $B_{\lambda }d\lambda =B_{\nu }|d\nu |$ andEqual-width wavelength intervals have unequal widths in frequency; the factor $c/{\lambda }^2$ compensates for this. Merely replacing the horizontal-axis labels of $B_{\nu }$ by $c/\nu$ does not produce $B_{\lambda }$. This explains the frequency and wavelength peaks of Planck's law.

A short derivation uses the Boltzmann distribution for a single electromagnetic mode with thermal energies $nh\nu$, $n=0,1,\dots$. Put $q=e^{-h\nu /(k_{B}T)}$. The geometric series gives $Z={\sum }_{n\ge 0}{q}^n=(1-q{)}^{-1}$ and mean thermal energy $h\nu q/(1-q)$. The electromagnetic mode density in a box, including both independent polarization of an electromagnetic wave states of a photon, is $8\pi {\nu }^2/c^3$ per volume per unit frequency. Multiplication gives spectral energy density $u_{\nu }=8\pi h{\nu }^3/[c^3(e^{h\nu /(k_{B}T)}-1)]$; isotropy gives $B_{\nu }=c{u}_{\nu }/(4\pi )$. This counts thermal excitation energy, excluding zero-point energy. ^[ref]

Planck's law connects quantum mechanics to a measurable radiation curve. Its small-$h\nu /(k_{B}T)$ limit is the Rayleigh-Jeans law, $B_{\nu }\approx 2k_{B}T{\nu }^2/c^2$; its large-$h\nu /(k_{B}T)$ limit is the Wien approximation, $B_{\nu }\approx (2h{\nu }^3/c^2)e^{-h\nu /(k_{B}T)}$. The suppression at high frequency makes the total energy density finite. Integrating the curve gives the Stefan–Boltzmann law, while locating its peak gives Wien's displacement law. Thus one formula predicts both the spectral shape and the total emission.

In a homogeneous lossless dielectric, let $\mathbf{D}=\mathbf{\epsilon }\mathbf{E}$ with a real symmetric permittivity tensor, and $\mathbf{B}=\mu \mathbf{H}$ with scalar positive magnetic permeability. A source-free plane electromagnetic wave with angular frequency $\omega \ne 0$, real wavevector $\mathbf{k}$ and electric field amplitude $\mathbf{e}$ satisfiesThe second equation follows from Gauss's law. It makes the electric displacement field transverse; the electric field itself need not be transverse in an anisotropic medium.

If $\mathbf{e}$ is an eigenvector of the permittivity tensor with nonzero eigenvalue ${\epsilon }_e$, then ${\epsilon }_e\mathbf{k}\cdot \mathbf{e}=0$, so $\mathbf{e}\perp \mathbf{k}$. The first equation reduces to the dispersion relation $|\mathbf{k}{|}^2={\omega }^2\mu {\epsilon }_e$. For a propagating solution the time-averaged Poynting vector is $⟨\mathbf{S}⟩=|\mathbf{e}{|}^2\mathbf{k}/(2\omega \mu )$ and is parallel to the wavevector.

A general anisotropic eigenmode need not have its electric field along a principal dielectric direction. Its energy flux can then depart from the wavevector, producing spatial walk-off. The nonzero-${\epsilon }_e$ assumption matters: at a zero permittivity eigenvalue, Gauss's law alone no longer forces $\mathbf{k}\cdot \mathbf{e}=0$.^[ref]