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In electrostatics, the electric potential $\Phi$ tells us the interaction potential energy per unit test electric charge in the electric field of prescribed sources. Its most directly useful quantity is an electric potential difference: how much work per unit electric charge an external agent must supply to move that test electric charge slowly between two points. This replaces the bookkeeping of a vector force along a path with the difference of two scalar values. The test electric charge must be small enough not to rearrange the sources appreciably.

For motion from $A$ to $B$, write $\Delta \Phi =\Phi (B)-\Phi (A)$. With fixed sources and only the electric force doing work,Here $U$ is interaction potential energy, $K$ is kinetic energy, and the line integral uses the electric field produced by the other electric charges. For slow transport with unchanged kinetic energy, the external work is $W_{\mathrm{ext}}=\Delta U$, so $\Delta \Phi =W_{\mathrm{ext}}/q$ for nonzero $q$. This external work has the opposite sign to the work done by the electric field. The SI unit is the volt,A volt measures energy per unit electric charge, not energy itself. For example, crossing an electric potential difference of $+5\mathrm{V}$ changes the potential energy of $q=+2\mathrm{nC}$ by $+10\mathrm{nJ}$, but changes that of $q=-2\mathrm{nC}$ by $-10\mathrm{nJ}$. Both test electric charges encounter the same electric potential difference; their opposite signs reverse the energy change.^[ref]

The electrostatic potential is a scalar function of position. The electric field is a conservative vector field, and all three components follow from the negative gradient of the electric potential:The electric field points toward decreasing electric potential. A positive test electric charge feels electric force in that direction; a negative test electric charge feels electric force in the opposite direction, since $\mathbf{F}=q\mathbf{E}$. A freely accelerated negative test electric charge therefore moves toward higher electric potential while lowering its potential energy $U=q\Phi$.^[ref]

Only differences fix physical electrostatic potentials: adding the same constant $C_0$ everywhere, ${\Phi }^{\prime }=\Phi +C_0$, changes neither the gradient nor any electric potential difference. Choosing an electrical reference potential is this elementary gauge transformation. Setting zero at infinity is convenient for finitely many localized point charges; it is not available for an ideal uniform electric field extending to infinity. There we can choose a finite reference point. Assigning zero at ground is a convention, whereas physically grounding an electrical conductor also allows exchange of electric charge to impose a boundary value. Zero electric potential at a point never by itself implies zero electric field.^[ref]

For a fixed point charge $Q$ at the origin in vacuum, choose zero at infinity. Coulomb's law giveswhere ${\epsilon }_0$ is the vacuum permittivity. Use the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $Q=+2\mathrm{nC}$. At $r=0.20\mathrm{m}$,A test electric charge $q=-3\mathrm{nC}$ has $U=-2.697\times 10^{-7}\mathrm{J}$ and radial electric force $F_r=-1.3485\times 10^{-6}\mathrm{N}$, directed inward. Moving it inward from $r_A=0.20\mathrm{m}$ to $r_B=0.10\mathrm{m}$ raises the electric potential from $89.9$ to $179.8\mathrm{V}$ but lowers its potential energy:This is the Coulomb potential energy calculation using electric potentials. The source's singular self-field at $r=0$ is excluded; when finding the force on a source point charge, use only the electric field of the other sources.^[ref]

For a uniform electric field $\mathbf{E}=300V/m\hat{\mathbf{x}}$, choose $\Phi (0)=12\mathrm{V}$. ThenMoving from $x_A=0$ to $x_B=0.04\mathrm{m}$ gives $\Delta \Phi =-12\mathrm{V}$. For $q=+2\mathrm{nC}$ the change in potential energy is $-24\mathrm{nJ}$ and the electric field does $+24\mathrm{nJ}$ of work. For $q=-2\mathrm{nC}$, the same prescribed displacement gives $\Delta U=+24\mathrm{nJ}$: the external agent must supply that work for slow transport. This uniform electric field approximates the interior of a parallel-plate capacitor sufficiently far from the edges; the fringing field of a capacitor limits the approximation.^[ref]

An equipotential is a level set $\Phi =\text{constant}$. On a smooth equipotential with nonzero gradient, every tangent displacement has $d\Phi =\nabla \Phi \cdot d\mathbf{l}=0$, so the electric field is normal to the equipotential. Thus moving a test electric charge along an equipotential requires no net electrical work. Around a point charge, the equipotentials are spheres; the circles in the diagram are their sections, not an assumption of two-dimensional Coulomb's law. A uniform electric field has parallel plane equipotentials. For contours separated by the same small electric potential difference, smaller normal separation means a stronger electric field, approximately $|\mathbf{E}|\approx |\Delta \Phi |/\Delta n$. At a point where $\nabla \Phi =0$, the electric field is zero and has no direction to call normal.^[ref]

The superposition principle adds electric potentials as signed scalars. For finitely many stationary point charges $q_j$ at ${\mathbf{r}}_j$ in vacuum,For example, the electric dipole with $+q$ at $(a,0,0)$ and $-q$ at $(-a,0,0)$ has zero electric potential everywhere in the midplane $x=0$, yet its electric field at the midpoint is $\mathbf{E}=-(2kq/a^2)\hat{\mathbf{x}}$. With $q=2\mathrm{nC}$ and $a=0.10\mathrm{m}$, this is $-3596N/C\hat{\mathbf{x}}$. At $(2a,0,0)$, the same model gives $\Phi =2kq/(3a)\approx 119.867\mathrm{V}$ and $E_x=8kq/(9a^2)\approx 1598.222N/C$. Cancellation of electric potentials is quite different from cancellation of vector electric fields.

A motivating measurement is conductive-paper electric potential mapping. A documented teaching experiment uses slightly conductive paper on an insulating board, two circular electrodes in highly conducting ink, a steady supply, and a digital voltmeter. Connect the supply to the two electrodes, keep the voltmeter common lead on the negative electrode, and use its other lead as a movable probe. Adjust the supply using the measured electric potential difference between electrodes; the published procedure uses $10\mathrm{V}$. The observable is the voltmeter reading $\Phi (P)-\Phi (\mathrm{COM})$ at the probe position $P$. Locate several points reading $2\mathrm{V}$, record their positions on a separate map, and join them into an equipotential; repeat at other readings. Lift the probe between contacts so it does not damage the paper. This procedure makes the electrical reference potential concrete: every reading compares two contacts.^[ref]

From neighboring readings infer an electric field component by a finite difference, $E_x\approx -[\Phi (x+h)-\Phi (x)]/h$. Reconstruct two in-plane components with displacements along two independent directions. Alternatively, draw the electric field normal to mapped equipotentials, toward lower electric potential. The second diagram illustrates the same measurement principle with bar electrodes spanning a uniform sheet, insulating top and bottom edges, and a synthetic model $\Phi (x)=(50V/m)x$. The model readings $4.0\mathrm{V}$ at $x=0.08\mathrm{m}$ and $5.0\mathrm{V}$ at $x=0.10\mathrm{m}$ imply $E_x=-50V/m$. These values are calculated illustrations, not reported experimental data. The documented circular-electrode experiment similarly estimates the electric field from readings at successive centimetre marks.^[ref]

Useful controls include checking the voltmeter zero, checking that the electric potential difference along each painted electrode is negligible, and repeating reference readings to detect supply drift. Reversing the supply should reverse the signs relative to the same reference contact. Contact resistance, leakage electric currents, and touching the sheet can distort the map. A voltmeter has finite input electrical resistance and draws a small electric current; voltmeter loading changes the quantity being measured if that electrical resistance is insufficiently large relative to the sheet's effective source electrical resistance.^[ref][ref]

The paper carries a steady electric current, so this is a planar analogue rather than a direct measurement of isolated point charges in three-dimensional vacuum. For constant electrical conductivity, Ohm's law and steady charge conservation give $\mathbf{j}=-\sigma \nabla \Phi$ and $\nabla \cdot \mathbf{j}=0$, hence the two-dimensional Laplace equation away from electrodes. Insulating sheet edges impose zero normal electric current, while finite electrodes fix boundary values. This derivation explains both the analogy and its limits: inhomogeneous paper, finite boundaries, and finite electrodes can change the contours. One should not fit the three-dimensional point charge formula $kQ/r$ to a planar map without testing that model. Small probe spacing resolves spatial variation better, but amplifies uncertainty in the inferred finite difference. The published laboratory instructions explicitly ask students to assess whether their proposed field fit is appropriate.^[ref][ref]

The scalar description has a crucial limit. Faraday's law of induction for a stationary closed path $\Gamma$ sayswhere ${\Phi }_B$ is magnetic flux. Nonzero changing magnetic flux gives nonzero circulation, incompatible with a single-valued electric potential alone. More generally, a scalar electric potential and a magnetic vector potential together giveConsequently, in the presence of induction, two voltmeter lead routes between the same contacts can produce different readings if the enclosed magnetic flux changes. This follows by subtracting the two path line integrals; it is a deduction from Faraday's law of induction, not a claimed outcome of the paper experiment. The steady setup and negligible induction justify interpreting its readings as electric potential differences. Merely introducing a scalar electric potential that depends on time cannot supply a nonzero curl of the electric field.^[ref]

A short mathematical justification connects the statements above. Along a smooth path, the chain rule gives $d\Phi =\nabla \Phi \cdot d\mathbf{l}$. Integrating and using $\mathbf{E}=-\nabla \Phi$ yields the endpoint formula by the fundamental theorem for line integrals, proving path independence. Conversely, vanishing circulation around every closed path lets us define $\Phi (P)=-{\int }_{P_0}^P\mathbf{E}\cdot d\mathbf{l}$ independently of the route, with $\Phi (P_0)=0$. Zero curl is a local test; for a continuously differentiable curl-free vector field, a simply connected open region suffices for a global potential. The explicit finite-source formula already gives a global electric potential away from its source points.^[ref]

The dimensionless quantity relating induced electric polarization to the internal electric field through $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ in the static scalar model. It describes the material contribution; relative permittivity includes the vacuum term as well, ${ϵ}_r=1+{\chi }_e$. Zero electric susceptibility gives ${ϵ}_r=1$, not zero permittivity.

The scalar ${\chi }_e$ assumes a local response with linearity and isotropy, settled electric polarization, fixed material conditions, and no residual electric polarization at zero electric field. It relates $\mathbf{P}$ to the averaged total internal electric field, which already includes the field of the bound electric charges. It does not relate $\mathbf{P}$ to the field the external electrodes would produce in vacuum. The distinction is essential for a self-consistent calculation.^[ref]

Substitute the scalar relation into the definition of electric displacement field:Comparison with $\mathbf{D}=ϵ\mathbf{E}$ proves $ϵ={ϵ}_0(1+{\chi }_e)$ and ${ϵ}_r=1+{\chi }_e$. Both $\mathbf{D}$ and $\mathbf{P}$ have units $C/{\mathrm{m}}^2$, so ${\chi }_e$ is a dimensionless quantity. In a stipulated static model with ${ϵ}_r=3$, ${\chi }_e=2$: at a prescribed internal electric field, the material contribution $P$ is twice ${ϵ}_{0}E$. This is a hypothetical response, not a measured material constant. The electric susceptibility is a convenient bridge from microscopic dielectric polarization to a macroscopic capacitance measurement, but it is not a measure of electrical conductivity.

The electrostatic boundary conditions at a conductor relate the one-sided total electric fields at a smooth metal–vacuum interface. Choose a unit normal $\mathbf{n}$ from metal into the adjacent vacuum, let ${\mathbf{E}}_{\mathrm{metal}}$ and ${\mathbf{E}}_{\mathrm{vac}}$ be the two limiting electric fields, and let $\sigma$ be the signed surface charge density. A thin Gaussian pillbox with cap area $A$ has outward cap normals $+\mathbf{n}$ on the vacuum side and $-\mathbf{n}$ on the metal side. As its height tends to zero, its side electric flux tends to zero for bounded local electric fields, leavingThe minus sign comes from the second outward cap normal, not from the sign of the electric charge. A shrinking rectangular loop across the interface and the zero closed line integral in electrostatics give continuity of the tangential component:For a conductor in electrostatic equilibrium, ${\mathbf{E}}_{\mathrm{metal}}=0$, henceThus positive surface charge density gives an electric field away from metal and negative surface charge density gives an electric field toward metal. The derivative uses the same chosen unit normal, from the vacuum side. The ideal surface layer has no unique electric field value on the layer itself; these are limiting values. Ordinary surface electric charge preserves continuity of electric potential; an ideal electric dipole sheet would require a different model. ^[ref].

Here “exterior” means just outside the metal, including the vacuum within a cavity in a conductor. On a spherical inner wall of radius $a$, its unit normal from metal into the cavity in a conductor is $-\hat{\mathbf{r}}$. A central positive point charge $q$ gives ${\mathbf{E}}_{\mathrm{vac}}=q\hat{\mathbf{r}}/(4\pi {ϵ}_0{a}^2)$ there, soThe outward normal of the cavity in a conductor as a volume is instead $+\hat{\mathbf{r}}$; confusing these two unit normals reverses the inferred sign. For a stipulated spherical enclosure with inner radius $a$, outer radius $b$, no external sources and total isolated metal electric charge $Q_c$, spherical symmetry and Gauss's law give the analytical modelIts outer surface charge density is $(Q_c+q)/(4\pi b^2)$. Grounding to $V_c=0$ with zero electric potential at infinity instead gives $Q_c=-q$ and zero exterior electric field, while preserving the same nonzero electric field inside the charged cavity in a conductor.

As a numerical sign check, stipulate $q=1\mathrm{nC}$, $a=0.10\mathrm{m}$, $b=0.20\mathrm{m}$ and $1/(4\pi {ϵ}_0)\simeq 8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. The inner electric field is approximately $899N/C$ along $+\hat{\mathbf{r}}$, but ${\mathbf{E}}_{\mathrm{vac}}\cdot \mathbf{n}=-899N/C$ and ${\sigma }_{\mathrm{inner}}\simeq -7.96nC/{\mathrm{m}}^2$. An isolated neutral enclosure has $+1\mathrm{nC}$ on the outer wall, approximately $+1.99nC/{\mathrm{m}}^2$ and $225N/C$ just outside. Multiplying each surface charge density by its spherical surface area recovers the signed wall electric charges.

The local result uses the total electric field, including all induced and imposed sources. The factor $\sigma /{ϵ}_0$ follows from the zero metal-side electric field; a symmetric isolated sheet in vacuum instead has magnitude $|\sigma |/(2{ϵ}_0)$ on each side. ^[ref]. At sharp corners, limiting electric fields can be singular and the smooth local pillbox argument needs care. In material outside the metal, use the normal free-charge jump of the electric displacement field and the material's permittivity, as in electrostatic interface boundary conditions. ^[ref].

For two point electric charges in vacuum, Coulomb's law gives the interaction potential energywith zero at infinite separation. Here $r_{ij}$ is the distance between them and ${ϵ}_0$ is the vacuum permittivity. Opposite signs give negative potential energy and attraction; equal signs give positive potential energy and repulsion. For a finite collection of distinct point electric charges, sum over $i<j$ to count each pair once. A bulk ionic crystal requires the extended lattice sum, including both signs, rather than multiplying one attractive pair by the coordination number.^[ref]

At short distances, overlapping electron density makes the simple point-charge picture inadequate. Quantum mechanics, including the Pauli exclusion principle, matters to the steep repulsive contribution. Pure attraction proportional to $-1/r$ would collapse as $r\to 0$; it cannot set an equilibrium spacing on its own. A useful illustrative model isHere $r$ scales an assumed fixed arrangement, $A$ packages its attractive contribution, and $B/r^n$ represents short-range repulsion. The constants have units energy times length and energy times length to the power $n$, respectively. This is a model, not a fitted description of a named salt.^[ref]

Differentiating gives $r_0=(nB/A{)}^{1/(n-1)}$ and $U^{\prime \prime }(r_0)=(n-1)A/r_0^3>0$, so the stationary spacing is a stable minimum. For the diagram's stipulated $n=9$, put $x=r/r_0$ and $E_0=A(1-1/n)/r_0>0$. ThenThe dimensionless curves show how competing interactions select a spacing. Their energy scale and exponent are synthetic; neither is a measured lattice energy or a melting prediction.

Electrostatics studies time-independent electric charge distributions and their electric fields. In vacuum, its local equations in SI units areHere $\rho$ is charge density, ${ϵ}_0$ is vacuum permittivity, and $\Phi$ is electric potential. Gauss's law supplies the divergence equation; the static form of Faraday's law of induction supplies the zero curl. The Poisson equation becomes Laplace's equation where there is no electric charge. Prescribed electric charges give electric fields through Coulomb's law and the superposition principle; mobile electric charges on an electrical conductor instead require solving for their distribution together with the electric potential. ^[ref].

The useful distinction is between prescribed sources and prescribed boundary conditions. On a grounded electrical conductor, an external connection fixes its electric potential relative to an electrical reference potential and allows electric charge to enter or leave. On an isolated electrical conductor, total electric charge is fixed and its constant electric potential is an unknown. A neutral isolated electrical conductor can therefore have a nonzero local surface charge density through electrostatic induction. Equal electric potentials also do not imply equal electric charges: charge sharing between distant connected spheres gives a simple geometric example. ^[ref].

Electrostatics is useful because a spatial boundary value problem replaces the need to follow every moving charge carrier during charge relaxation. It describes the settled state of a capacitor and explains electrostatic shielding of sensitive equipment. Its assumptions must match the apparatus: stationary boundaries and sources, negligible time variation of the magnetic field, and enough time for charge relaxation. A metal carrying a maintained electric current through finite electrical resistance generally needs a nonzero internal electric field by Ohm's law; it is not a conductor in electrostatic equilibrium. Static macroscopic electric charge also does not mean that every Electron is motionless. ^[ref].

At normal incidence, with no optical dispersion and zero round-trip reflection wave phase modulo $2\pi$, the resonance condition in Fabry–Pérot transmission from multiple reflections isHere $\nu$ is frequency and $c$ is the speed of light in vacuum. A change of $\nu$ by one free spectral range advances the round-trip wave phase by $2\pi$. At fixed internal angle of incidence $\theta$, the same nondispersive model gives $c/(2nd\cos \theta )$. Holding the external angle of incidence fixed in a dispersive medium need not hold the internal $\theta$ fixed, because of Snell's law.

For stipulated $n=1$ and $d=10\mathrm{mm}$, the free spectral range is $14.9896229\mathrm{GHz}$ using $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. A frequency difference of $0.10\Delta {\nu }_{\mathrm{FSR}}$ is therefore $1.49896229\mathrm{GHz}$. Transmission repeats every free spectral range: one isolated scan cannot distinguish that separation from $1.10\Delta {\nu }_{\mathrm{FSR}}$. Independent knowledge of the optical spectrum, an additional filter, or a changed mirror separation is needed to assign orders.

Equal spacing in frequency does not imply equal spacing in wavelength. At normal incidence the exact adjacent resonance wavelengths obeyFor example, choosing $2nd=10{\lambda }_{10}$ makes the exact gap to order 11 equal to ${\lambda }_{10}/11$, while the high-order estimate gives ${\lambda }_{10}/10$. The approximation is already ten percent larger at order 10. With optical dispersion or frequency-dependent reflection wave phase, obtain local spacings from the total phase difference instead: $\Delta \nu \simeq 2\pi /(d\delta /d\nu )$ when its derivative varies little over one order.

Consider two identical, lossless parallel mirrors with intensity reflectivity $0\le R<1$, separation $d$, and a uniform transparent medium of refractive index $n$. Fix one polarization of an electromagnetic wave and an internal angle of incidence $\theta$ measured from the mirror normal vector. Illuminate with a monochromatic wave having planar wavefronts; the emerging contributions must overlap coherently in the detected direction. Define $T=I_{\mathrm{transmitted}}/I_{\mathrm{incident}}$ using the same incident and exit media. Each mirror transmits a fraction $1-R$ of incident power. Its reflected wave amplitude has magnitude $\sqrt{R}$, so two internal specular reflections multiply the circulating complex amplitude by magnitude $R$, not $R^2$.

Let $\lambda$ be the vacuum wavelength and choose the reflection wave phase convention in which the product of the two internal amplitude reflection coefficients is real and positive. The phase difference between successive transmitted contributions, evaluated on a common outgoing wavefront, isWith other mirror coatings or phase conventions, use the total $\delta =4\pi nd\cos \theta /\lambda +{\varphi }_{\mathrm{rt}}$, where ${\varphi }_{\mathrm{rt}}$ is the sum of the two internal reflection wave phases. The stated resonance positions presume ${\varphi }_{\mathrm{rt}}=0$ modulo $2\pi$; a constant nonzero ${\varphi }_{\mathrm{rt}}$ shifts them, while a frequency-dependent one can also change their spacing.

Why does the cosine appear rather than its reciprocal? Inside the gap, the extra two legs have optical path length $2nd/\cos \theta$. But the later emerging contribution starts $2d\tan \theta$ farther along the mirror. Comparing both on one outgoing wavefront subtracts $2nd\tan \theta \sin \theta$ from that path difference, leaving $2nd\cos \theta$. Equivalently, a plane wave's mirror-normal wavevector component is $k_z=(2\pi n/\lambda )\cos \theta$, so the round-trip wave phase is $2k_{z}d$. The diagram uses the same medium outside and inside to show the common-wavefront subtraction directly; at a refracting exit interface, tangential wavevector matching in Snell's law supplies the same result.

Absorb the common first-pass wave phase into the incident complex amplitude $E_{\mathrm{in}}$. Successive transmitted contributions are $(1-R)E_{\mathrm{in}}$, $(1-R)E_{\mathrm{in}}R{e}^{i\delta }$, and so on. The superposition principle and the convergent geometric series giveThe last step expands the squared complex modulus as $1+R^2-2R\cos \delta$. This transmission profile is often called the Airy transmission function; it is distinct from the Airy disk, a diffraction image of a circular aperture. The University of Washington's lab handout gives the same transmission profile and distinguishes mirror reflectivity from other sources of peak broadening.^[ref]

For an original example, set $R=3/4$. At $\delta =0$, the amplitudes add to $(1/4)/(1-3/4)=1$, so $T=1$. At $\delta =\pi$, they alternate in sign and add to $(1/4)/(1+3/4)=1/7$, so $T=1/49$. At $\delta =\pi /2$, the amplitude is $4/25+3i/25$ and $T=1/25$. Adding the individual intensities instead would give $(1-R{)}^2{\sum }_j{R}^{2j}=(1-R)/(1+R)=1/7$, independent of wave phase; that discards the wave interference responsible for the peaks.

In general the maximum is $T_{\max }=1$ at $\delta =2\pi m$, with integer $m$, and the minimum is $T_{\min }=[(1-R)/(1+R){]}^2$ halfway between peaks. Perfect resonant transmission requires identical lossless mirrors, coherent illumination, and sufficient time to establish the steady field. Absorption, unequal mirror transmission, finite beams that fail to overlap, and fluctuations of $d$ can reduce it. The exactly reflecting case $R=1$ is excluded: no incident power enters through a perfectly reflecting mirror, and the resonant limit $R\to 1^{-}$ is singular.

Far outside a localized steady source of size $a$, at distance $r\gg a$, its multipole expansion starts with the magnetic dipole moment if that moment is nonzero. In vacuum, using SI units,where ${\mu }_0$ is the vacuum permeability, $\hat{\mathbf{r}}=\mathbf{r}/r$, and the magnetic field is the curl of the magnetic vector potential. The magnetic dipole field falls as $r^{-3}$, and the magnetic vector potential as $r^{-2}$. The localized steady-current moment identities remove the putative $r^{-1}$ term in the magnetic vector potential. No magnetic monopole is needed. If different current loops cancel the total magnetic dipole moment, a higher term of the multipole expansion supplies the leading distant magnetic field. ^[ref].

The approximation is not valid at the center of a finite current loop. For a circular current loop of radius $R$ and electric current $I$, the exact on-axis magnetic field of a circular current loop and the distant approximation areTheir ratio is $B_z/B_{z,\mathrm{dip}}=(1+R^2/z^2{)}^{-3/2}$: at $|z|=10R$, the approximate magnetic field exceeds the exact magnetic field by about $1.50\%$ relative to the exact value. Near the source, retain its geometry rather than replacing it with a point source specified only by its magnetic dipole moment.

Free expansion occurs when a gas spreads into an initially evacuated region without driving an external load. The standard idealization uses a rigid, insulated vessel: initially a partition confines the gas to volume $V_1$, while the remaining compartment contains vacuum. Open a passage and wait for thermodynamic equilibrium; the same fixed amount of substance $n$ now occupies the larger accessible volume $V_2$. The outer vessel never changes size. This simple release separates three questions: what energy is conserved, whether temperature changes, and why entropy increases. ^[ref]

Choose the gas as the closed system, neglect heat exchange with the walls, and assume negligible changes in bulk kinetic energy and potential energy between the equilibrium endpoints. Insulation gives $Q_{\mathrm{in}}=0$; no moving outer boundary or external load receives work, so $W_{\mathrm{out}}=0$. The first law of thermodynamics therefore givesThe gas can develop internal motion during the release; this does not supply external work. The intermediate state need not have one equilibrium pressure or temperature, so there is no actual quasistatic process curve $p(V)$ to integrate. In particular, substituting the equilibrium ideal gas law into $\int pdV$ would describe a different process. ^[ref]

For an ideal gas, internal energy of an ideal gas depends only on temperature. With positive heat capacity at constant volume, $U(T_2)=U(T_1)$ implies $T_2=T_1$, and the ideal gas law gives $p_2/p_1=V_1/V_2$. Doubling the accessible volume thus halves the final pressure without changing the final temperature. This is an endpoint statement; it does not assert a spatially uniform, constant temperature during the rush. Insulation alone does not imply constant temperature: a reversible adiabatic expansion cools an ideal gas because it supplies work. ^[ref]

For a real gas, $U(T,V)$ can depend on volume as well as temperature, so unchanged internal energy need not mean unchanged temperature. For a fixed-composition simple compressible system in a single thermodynamic phase, equilibrium identities using a Maxwell relation giveHere $C_V$ is the whole sample's heat capacity at constant volume. The sign depends on the equation of state; for an ideal gas the numerator vanishes. For a finite free expansion, solve $U(T_2,V_2)=U(T_1,V_1)$, or integrate this slope along equilibrium states of equal internal energy; that mathematical curve does not describe the actual release. ^[ref]

Free expansion is useful because it tests how internal energy depends on number density while eliminating external expansion work. Its beautiful surprise is that an ideal gas can finish with the same temperature and internal energy but greater entropy: a thermometer alone cannot diagnose reversibility. See entropy change in free expansion for the calculation and temperature sensitivity of a free-expansion experiment for the experimental qualification. The continuous-flow throttling associated with the Joule-Thomson coefficient has a different balance: under its usual assumptions it preserves enthalpy, whereas the gas-only idealization here preserves internal energy. ^[ref]

Free fall is motion under gravity alone: there is no supporting normal force, engine thrust or appreciable drag force. In the Newtonian mechanics model, a test particle of mass $m$ obeys $m\ddot{\mathbf{r}}=m\mathbf{g}(\mathbf{r})$, hence $\ddot{\mathbf{r}}=\mathbf{g}(\mathbf{r})$. The mass cancels. Bodies released at the same position with the same velocity follow the same path in this model, even if their gravitational weights differ. ^[ref]

A dropped stone is one example; a coasting satellite is another. Free fall does not require motion straight down or decreasing distance to the attracting body. An ideal circular orbit has constant radius while its velocity continually turns under centripetal acceleration. The circular orbit as continuous free fall connects the familiar drop with orbital motion.

Near a surface, treating the gravitational acceleration as constant gives a useful first prediction. Over larger heights, the inverse-square law changes the acceleration during the fall. Radial free fall from rest uses conservation of mechanical energy to account for that change exactly, within the specified Newtonian gravitational field. The contrast matters: the simplest formula is an approximation to a physical model, rather than a law with the same numerical $g$ everywhere.

History also makes the idealization tangible. Galileo's Two New Sciences (1638), Third Day, Theorem II, derives distance proportional to the square of time for uniformly accelerated motion from rest; its first corollary gives successive distances in the ratios $1,3,5,7,\dots$ over equal time intervals. This is a kinematic statement, predating the gravitational inverse-square law. ^[ref]

On Apollo 15's final lunar surface excursion, David Scott released a hammer and a feather together. The near-vacuum removed the large difference in drag force familiar on Earth. NASA's report describes their arrival together within the accuracy of the release; this demonstration illustrates the free fall prediction without establishing infinite experimental precision. ^[ref]

The free spectral range is the wavelength or frequency spacing between adjacent orders or resonances of an optical instrument. It identifies when different orders can overlap. Always specify the coordinate: a uniform spacing in frequency need not be uniform in wavelength. For an ideal Fabry–Pérot interferometer at normal incidence, with mirror separation $d$ and constant refractive index $n$, Fabry–Pérot resonance spacing gives $\Delta {\nu }_{\mathrm{FSR}}=c/(2nd)$, where $c$ is the speed of light in vacuum. The free spectral range of an echelle grating follows a different order condition.

In vacuum, $\nu =c/\lambda$, where $c$ is the speed of light. Increasing wavelength decreases frequency, so the ordinary signed differentials satisfy $d\nu =-c{\lambda }^{-2}d\lambda$. A spectral slice's positive contribution to radiance must be unchanged. The change of variables formula therefore gives the absolute Jacobian determinantThese formulas use $L_{\lambda }$ per metre and $L_{\nu }$ per hertz. The factor $c/{\nu }^2$ has units $\mathrm{m}\mathrm{s}=\mathrm{m}\mathrm{H}{\mathrm{z}}^{-1}$, which converts the former density into the latter. If ${\ell }_{\mathrm{nm}}$ denotes the numerical density per nanometre in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{n}{\mathrm{m}}^{-1}$, its per-metre value is $10^9{\ell }_{\mathrm{nm}}$. Conversely, the per-nanometre value is $10^{-9}$ times the per-metre value. Likewise, a density per terahertz is $10^{12}$ times its per-hertz value because $1\mathrm{THz}=10^{12}\mathrm{Hz}$.

For a concrete stipulated spectrum, let $L_{\lambda }=C=10^9\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$ between ${\lambda }_1=500\mathrm{nm}$ and ${\lambda }_2=600\mathrm{nm}$, and zero outside. This is exactly $1\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{n}{\mathrm{m}}^{-1}$ throughout the band. Using the exact International System of Units value $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, the corresponding frequency bounds areThe transformed spectral radiance is $L_{\nu }=Cc/{\nu }^2$ on this interval. At ${\nu }_1$ it is approximately $1.200830743\times 10^{-12}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, and at ${\nu }_2$ it is $8.339102380\times 10^{-13}$ in the same units. These are respectively $1.200830743$ and $0.833910238$ per terahertz. Thus a flat wavelength density becomes a decreasing frequency density; relabelling the horizontal axis alone would miss both the shape and the units. ^[ref].

The equality of band radiances has a short exact proof:The reversed bounds and the Jacobian determinant preserve the same spectral measure. Each half-band, $500$–$550\mathrm{nm}$ and $550$–$600\mathrm{nm}$, carries $50\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, although the corresponding frequency widths are about $54.508$ and $45.423\mathrm{THz}$.

A spectrum's peak is consequently a statement about a particular density and coordinate. The frequency and wavelength peaks of Planck's law differ for precisely this reason. Integrated radiance over a specified physical band supplies a comparison that survives the change of variables.

Gas electron diffraction infers molecular geometry from the diffraction of Electrons by a gaseous sample. It makes a useful experimental test of shape: distances between both bonded and nonbonded atomic nuclei affect the signal, so a proposed straight arrangement and a bent arrangement can be distinguished through their pair separations.

A concrete modern apparatus is the Bielefeld instrument described by ^[ref]. A hot cathode supplies Electrons; apertures and magnetic lenses shape the beam. A temperature-controlled nozzle injects sample vapour across it inside a vacuum chamber. Pumps and a cold trap reduce unwanted residual gas. A central beam stop catches the strong unscattered beam; off-axis scattered Electrons expose imaging plates. A rotating sector can attenuate the intense small-angle region to accommodate detector range. The authors checked both the spatial scale of the plate scanner and its response to electron dose. These checks matter because an error in detector position or intensity calibration can masquerade as a structural difference.

What is measured is detector signal as a function of position, converted to scattering angle using the sample-to-detector distance. Randomly oriented molecules give radial information. After extracting the molecular intensity, compare it with predictions from a candidate arrangement and adjust the bond lengths and bond angles to improve the fit. The Wann Electron Diffraction Group's account of time-averaged GED, experiment and theory describes radial averaging and this refinement. The apparatus does not photograph one molecule with labelled atomic nuclei.

Why can distances survive random orientation? In a simplified rigid pair model with separation $r$, write $s$ for the magnitude of the difference between incident and scattered wave vectors. Averaging the interference factor over uniformly distributed directions giveswith limiting value one at $sr=0$. This follows by integrating the cosine; $t$ is the cosine of the angle between the pair and the scattering direction. The oscillations depend on $r$, although the orientation has disappeared. A real calculation weights many pairs by their scattering amplitudes and includes nuclear motion. The wave interference explanation and superposed pair-distance contributions are discussed in the same ^[ref].

Once a consistent structural model supplies two equal bond lengths $b$ and an outer-atom separation $d$, the law of cosines converts these to a bond angle. For example, the illustrative $d/b=1.5811$ corresponds to approximately $104.5^{\circ }$, while $d/b=2$ is straight. This geometric step explains what a shape measurement is trying to establish; it is not a new water measurement.

There are two essential limits. First, recovering labelled three-dimensional positions from a summed radial signal is an inverse problem: composition, candidate connectivity, and other structural information can be needed to distinguish competing fits. Karle's authored review Structure determination of gaseous and amorphous substances by diffraction methods (1977), abstract explains why too few independent observations make supplementary structural information indispensable. A good fit alone does not prove that the proposed arrangement is unique. Second, atomic nuclei are moving. Distances inferred by gas electron diffraction are motion-dependent averages; corrections are required before identifying them with a minimum-potential energy arrangement. For experimental distance conventions, see the IUPAC Gold Book: bond length. These limitations explain why model assumptions and the averaging convention must accompany a precise bond angle.