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Chemical equilibrium of $n+{\nu }_e\leftrightarrow p+e^{-}$, with negligible electron chemical potential, gives the stated neutron-to-proton ratio. Positive electron-neutrino degeneracy reduces the neutron fraction, whereas negative degeneracy increases it. The primordial helium mass fraction then responds through neutron-limited helium synthesis while protons outnumber neutrons; a sufficiently neutron-rich mixture instead becomes proton limited, so that trend must not be extrapolated indefinitely. Extra neutrino energy density also accelerates cosmic expansion, changing weak freeze-out and the time available for neutron decay. These effects can compete; a quantitative Big Bang nucleosynthesis calculation must account for the altered reaction rates as well as the altered expansion rate.

The spectral radiation energy density $u(\omega )$ is defined per unit angular frequency and physical volume. Thermal Planck law convention excludes the vacuum zero-point term. A density per ordinary frequency instead obeys $u_{\omega }=u_{\nu }/(2\pi )$, evaluated at $\nu =\omega /(2\pi )$; Einstein coefficients multiplying these densities must be converted correspondingly.

Radiation pressure is the normal momentum flux transferred by electromagnetic radiation to a surface. A photon travelling in vacuum carries momentum $p_{\gamma }=E_{\gamma }/c=h\nu /c$, where $E_{\gamma }$ is its energy, $\nu$ its frequency, $h$ the Planck constant and $c$ the speed of light. Photon absorption transfers that momentum; specular reflection reverses its normal component. This lets light exert a force even when almost no energy becomes heat in the receiver.

Two useful formulas describe different preparations. A directed beam normally incident on a stationary perfect absorber gives pressure $I/c$, and on a stationary perfect mirror gives $2I/c$. Here $I$ is light intensity, the incident power per beam cross-sectional area perpendicular to propagation, in $\mathrm{W}{\mathrm{m}}^{-2}$. An isotropic photon gas has pressure $u/3$, where $u=U/V$ is its radiation internal energy per volume, in $\mathrm{J}{\mathrm{m}}^{-3}$. Stored energy density and travelling energy flux must be distinguished: an isotropic field can have zero net energy flux and positive radiation pressure.

The physical appeal is that the same momentum conservation connects a single photon collision, a mirror's measurable force, and a radiation-filled cavity's equation of state. The directional calculation below complements radiation pressure on a perfectly absorbing surface; the equilibrium spectrum and its temperature dependence belong to blackbody radiation and the photon gas.

For radiation in vacuum with isotropy, energy is distributed equally per solid angle: write $u_{\Omega }=u/(4\pi )$ for energy density per unit solid angle. Let $\theta$ be the angle between a ray's path and the wall normal. Its normal velocity $c\cos \theta$ determines the arrival rate, while normal momentum per energy is $\cos \theta /c$. Their product is ${\cos }^2\theta$. For a perfectly reflecting cavity wall,The leading two is the normal momentum reversal; the incoming hemisphere contains half of the field's directional energy density. This is also the normal component of the bulk pressure tensor, $P_{nn}={\int }_{4\pi }{u}_{\Omega }{\cos }^2\theta d\Omega =u/3$. Isotropy makes all three diagonal components equal, while their sum is $u$ because the squared direction components sum to one. The factor one third therefore depends on three-dimensional angular geometry, not on the shape of the blackbody radiation spectrum.

A perfectly absorbing wall exposed to this incoming hemisphere receives only $u/6$ from absorption. If it is in thermal equilibrium with the isotropic radiation, its outgoing emission supplies another $u/6$ of recoil, recovering $u/3$. A cold absorber before equilibrium is a different boundary condition. Similarly, the one-way incident energy flux is $cu/4$, whereas the net energy flux across a plane in the undisturbed isotropic field is zero. Dividing $cu/4$ by $c$ misses the extra normal-momentum projection and does not give the cavity pressure.

For a stipulated isotropic radiation energy density $u=6\mu \mathrm{J}{\mathrm{m}}^{-3}$, the bulk pressure is $2\mu \mathrm{Pa}$ regardless of its spectral shape. In thermal equilibrium, blackbody radiation supplies $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$, so $p_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$; Stefan–Boltzmann law defines $\sigma$. The equilibrium photon gas derives this same equation of state from its free energy.^[ref]

Reduced Planck units choose $\hslash =c=8\pi G=1$. With the reduced Planck mass defined by $M_{\mathrm{Pl}}=(8\pi G{)}^{-1/2}$ after setting $\hslash =c=1$, this convention gives $M_{\mathrm{Pl}}=1$ and the Einstein-Hilbert action coefficient $M_{\mathrm{Pl}}^2/2=1/2$.

For a spatially flat Friedmann-Lemaître-Robertson-Walker metric, the Friedmann equation is then $3H^2=\rho$, where $\rho$ is the total energy density, including any vacuum contribution. With spatial-curvature parameter $k$, it is instead $3(H^2+k/a^2)=\rho$. Setting the unreduced Planck mass to one would set $G=1$, leaving the factor $8\pi$ in the gravitational equation. Keeping these conventions distinct prevents spurious factors in gravitational and inflationary calculations.

At fixed temperature, a photon gas has constant energy density $u$ and radiation pressure $p=u/3$. An increase $\Delta V$ in volume requires both $\Delta U=u\Delta V$ and expansion work $p\Delta V$, giving absorbed heat $4u\Delta V/3$.

A homogeneous canonical scalar field contributes energy density $\rho ={\dot{\phi }}^2/2+V$ and pressure $P={\dot{\phi }}^2/2-V$. When it is the only source in a spatially flat FLRW metric, the Friedmann equation is $3M_{\mathrm{Pl}}^2{H}^2=\rho$, and its evolution follows the scalar field equation of motion. Positive energy density implies $w\ge -1$ because $\rho +P={\dot{\phi }}^2\ge 0$, even if the scalar potential is negative.

For a canonical real scalar field with metric signature $(-,+,+,+)$ and scalar potential $V$, variation of its minimally coupled action gives the displayed equation. A homogeneous field in an FLRW metric therefore obeys $\ddot{\phi }+3H\dot{\phi }+V^{\prime }(\phi )=0$ in natural units. Its energy density is ${\dot{\phi }}^2/2+V$ and its pressure is ${\dot{\phi }}^2/2-V$, which satisfy the cosmological continuity equation.

The local Skyrme baryon density is $\mathcal{B}(x)=-{ϵ}_{ijk}\mathrm{tr}(L_i{L}_j{L}_k)/(24{\pi }^2)$, with $L_i=U^{†}{\partial }_{i}U$ and $B=\int \mathcal{B}{d}^{3}x$. It is invariant under global isorotations and transforms as a scalar under proper spatial rotations, so its contours reveal the geometric shape of a Skyrmion. For the rational map approximation for Skyrmions, $\mathcal{B}=-f^{\prime }{\sin }^{2}f{J}_R/(2{\pi }^2{r}^2)$, where $J_R$ is the angular Jacobian of a rational map. Integrating $J_R$ over the sphere gives $4\pi \mathrm{deg}R$, recovering the integer total charge. A decreasing profile and holomorphic angular map give a nonnegative density, but a general field may have regions of negative density; positivity is not a general topological theorem. The density is different from the energy density.

The slow-roll approximation describes a canonical inflaton whose potential energy dominates its kinetic energy and whose field acceleration is small compared with Hubble friction:If the inflaton also dominates the energy density and cosmological spatial curvature is negligible, the Friedmann equation and inflaton equation of motion reduce towhere $M_{\mathrm{Pl}}=(8\pi G{)}^{-1/2}$ is the reduced Planck mass and $c=\hslash =1$. The second equation describes a damping-controlled drift down the inflaton potential.^[ref]

The potential slow-roll parameters give useful consistency checks, ${ϵ}_V\ll 1$ and $|{\eta }_V|\ll 1$, along the traversed field interval. Small initial kinetic energy alone does not guarantee that these inequalities persist. For $V=m^2{\varphi }^2/2$ and $\varphi >0$, the two approximate equations give $\dot{\varphi }\simeq -\sqrt{2/3}m{M}_{\mathrm{Pl}}$, so the field decreases nearly linearly while the Hubble parameter slowly changes. Cosmic inflation ends at ${ϵ}_H=1$, where the first Hubble slow-roll parameter equals unity; estimating that endpoint by ${ϵ}_V\simeq 1$ extrapolates the approximation to its boundary. Cosmic inflation can also occur outside ordinary slow roll, so the two terms are not interchangeable.

For number density $n$, rest-frame energy density $\rho$ and pressure $p$, the enthalpy per particle is $ϵ=(\rho +p)/n$. In an isentropic barotropic fluid, $d\rho =ϵdn$ and $dp=ndϵ$. It differs from enthalpy per unit mass by the particle mass when that mass is fixed. Using $ϵ$ as thermodynamic variable gives $n=p^{\prime }(ϵ)$ and $\rho =ϵp^{\prime }(ϵ)-p(ϵ)$.

Energy density per unit spectral coordinate, such as $U_{\nu }$ per frequency or $U_{\lambda }$ per wavelength. Conservation of energy in corresponding bins requires $U_{\lambda }=U_{\nu }|d\nu /d\lambda |$; therefore their peak positions need not map by $\lambda =c/\nu$.

A stationary mode shape does not imply absence of energy. Write $A=2a$ and $y=A\sin (kx)\cos (\omega t)$. For the ideal wave equation on a string, the kinetic energy and elastic potential energy per unit reference length add toHere $P$ is signed power crossing a position on the string, positive to the right; $e$ is an energy density per length, not per volume. Differentiation and $\mu y_{tt}=\tau y_{xx}$ give the local conservation of energy identity ${\partial }_{t}e+{\partial }_{x}P=0$. Substitution into $P$ givesIt can be nonzero away from a node of a standing wave or an antinode, but its time average over an oscillation period is zero. Energy flows locally back and forth even though this pure standing wave transports no net energy across a section over a complete cycle. At a node of a standing wave, $y_t=0$, so kinetic energy density is zero, but $y_x$ need not vanish: elastic potential energy density can be positive there.

For a fixed-end normal mode, integration over $0\le x\le L$ givesThe last equality uses $\tau k^2=\mu {\omega }^2$. At a straight-string snapshot, $U=0$ while $K=E$; at an extreme snapshot, $K=0$ while $U=E$. This is normal-mode energy exchange. A driven real string with damping requires continuing input power to replace lost energy, so a visually stationary pattern does not establish zero average supplied power.

Time-independent crossed electric and magnetic fields can carry a nonzero energy flux $\mathbf{S}=\mathbf{E}\times \mathbf{H}$. A steady coaxial circuit is an example: radial electric and azimuthal magnetic fields carry energy axially from source to load. Static energy density does not imply zero flux.

The Stefan-Boltzmann constant converts the fourth power of absolute temperature into blackbody radiant exitance:Here $k_B$ is the Boltzmann constant, $h$ the Planck constant and $c$ the speed of light. In the current International System of Units, these three constants have exact defining values, so $\sigma$ is exact as an expression in them and $\pi$. The displayed finite decimal is an approximation, not an exact decimal or a measured uncertainty. The 2022 recommended-constants table lists an ellipsis and an exact uncertainty for $\sigma$.^[ref]

A dimensional check uses $[k_B]=\mathrm{J}{\mathrm{K}}^{-1}$, $[h]=\mathrm{J}\mathrm{s}$ and $[c]=\mathrm{m}{\mathrm{s}}^{-1}$. Consequently $[k_B^4/(h^3{c}^2)]=\mathrm{J}{\mathrm{s}}^{-1}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}=\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, exactly the units needed in the Stefan–Boltzmann law.

For an equilibrium photon gas, the energy density is $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$. The radiation pressure is $p_{\mathrm{rad}}=u/3=4\sigma T^4/(3c)$. These describe the radiation filling an equilibrium cavity; $u$ has units $\mathrm{J}{\mathrm{m}}^{-3}$, whereas the outward radiant exitance through a small opening is $cu/4$. The factor $1/4$ comes from averaging propagation directions and their projected area over a radiation field with isotropy, not from reducing the opening's emissivity.

A perfect fluid with equation of state $P=\rho$ is called a stiff fluid. For positive energy density, its constant equation-of-state parameter is $w=1$. The cosmological continuity equation gives $\rho \propto a^{-6}$. In a flat universe dominated by this fluid, the Friedmann equation then gives $a(t)\propto (t-t_B{)}^{1/3}$. This statement assumes a nonzero fluid density and negligible other components.

The magnetic energy stored in a linear solenoid of constant self-inductance $L$ isin the long-solenoid approximation. The first formula is an inductor relation; the second identifies its storage with magnetic energy in the interior magnetic field, under the vacuum-core approximation. This agreement joins the electrical circuit description to the spatial magnetic field description. The expression $B^2/(2{\mu }_0)$ is the vacuum magnetic energy density. ^[ref].

A short derivation starts from the terminal electric potential difference $v_L=LdI/dt$, with positive electric current entering the terminal at higher potential. The absorbed power is $v_{L}I$, and integrating from zero electric current givesThe result describes recoverable storage in the inductor; Joule heating in the wire is a separate expenditure of energy. ^[ref].

For the long-coil example, $L\approx 4.9348\mathrm{mH}$ and $I=0.50\mathrm{A}$ give $U\approx 6.1685\times 10^{-4}\mathrm{J}=0.61685\mathrm{mJ}$. Substituting the ideal $B\approx 1.256637\mathrm{mT}$ and volume $A\ell \approx 9.8175\times 10^{-4}{\mathrm{m}}^3$ into $B^{2}A\ell /(2{\mu }_0)$ gives the same value. Doubling electric current quadruples stored magnetic energy. In a real coil, reducing electric current can return this energy to an electrical circuit, or release it as Joule heating in a closed RL circuit.

For a static rest-frame electromagnetic stress-energy tensor with no Poynting flux, a Lorentz transformation to a laboratory where it moves along $+x$ gives $u={\gamma }^2(u_0+{\beta }^2{T}_{\ast }^{xx})$ and $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})$. Thus energy flux includes stress as well as rest energy density. In an ideal parallel-plate capacitor, $T_{\ast }^{xx}=+u_0$ for motion perpendicular to the electric field, but $T_{\ast }^{xx}=-u_0$ for motion parallel to it. The former has $S_x=2{\gamma }^{2}v{u}_0$; the latter has zero interior flux, with work at the moving plates transferring energy at the boundaries. The field subsystem need not have an integrated four-momentum independent of simultaneity surface; include material support stresses for a closed system.

In a spatially flat universe dominated by a positive constant energy density $ϵ$, the Friedmann equations give $H^2=8\pi Gϵ/(3c^2)$. On the expanding branch $a(t)=a_i{e}^{H(t-t_i)}$, so the scale factor doubles after $(\log 2)/H$. This concerns a linear length scale; the enclosed physical volume grows by a factor of eight over the same interval.

A homogeneous canonical scalar field with positive scalar potential has energy density ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and pressure $P_{\varphi }={\dot{\varphi }}^2/2-V$. When its potential energy dominates its kinetic energy, the equation-of-state parameter approaches $-1$. If other components and spatial curvature are negligible, the Friedmann acceleration equation gives accelerating expansion. Small initial kinetic energy is not enough to ensure long cosmic inflation; the field must also remain on a sufficiently flat part of the scalar potential, as measured by the potential slow-roll parameters.