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Absolute magnitude $M$, in the stellar convention used here, is the apparent magnitude a source would have at $10$ parsecs, with interstellar extinction removed. The hypothetical observation uses the specified photometric passband and magnitude zero point, preserving the source's emitted spectrum and luminosity. Moving every source to this common distance lets us compare intrinsic brightness without confusing it with distance. ^[ref].

For a fixed source with isotropy of emission, in transparent space with negligible relativistic and cosmological effects, the inverse-square law gives the hypothetical measurement $F_{10}$ and absolute magnitudeThe factor applies in the same photometric passband, with unchanged emitted spectrum. Thus absolute magnitude compares luminosity in that response, rather than automatically the total bolometric luminosity over all wavelengths.

Consider a toy source with $M=0$. Its apparent magnitude is $m=0$ at $10$ parsecs, $m=5$ at $100$ parsecs, and $m=10$ at $1000$ parsecs, neglecting interstellar extinction. Its absolute magnitude remains $M=0$ throughout: moving the observer changes received radiative flux, not emitted luminosity. Another toy source with $M=-5$ has $100$ times its passband output under the same isotropy and calibration assumptions. Negative absolute magnitudes are allowed too; "absolute" names the reference-distance convention, not an absolute value.

A calibrated standard candle reverses this reasoning: its known absolute magnitude and measured apparent magnitude give a distance modulus. This is why estimating intrinsic brightness is so useful for measuring astronomical distances. ^[ref].

The angle of incidence is the angle between the incident light ray and the line of the surface normal vector at the point of incidence. Use the acute angle on the incident side, with $0\le {\theta }_i<90^{\circ }$ for a nongrazing incident light ray. An angle measured from the surface itself is its complement $90^{\circ }-{\theta }_i$. The same convention applies to wavevectors in the transparent media with isotropy considered here.

A submerged object looks shallower because its emerging light rays bend away from the surface normal vector. Their backward extensions give an approximate virtual image nearer the surface. For an object at perpendicular depth $d$ in a medium of refractive index $n_{\mathrm{obj}}$, viewed through a flat boundary from a medium of refractive index $n_{\mathrm{obs}}$, the near-normal paraxial approximation gives $d_{\mathrm{par}}=(n_{\mathrm{obs}}/n_{\mathrm{obj}})d$. Thus the model water depth $12\mathrm{cm}$ appears approximately $9\mathrm{cm}$ from air. This is a bundle-of-light rays statement, not a deflection of the axial light ray.^[ref]

For general transmitted light rays, specify what “depth” means. Put the object at $(0,-d)$ below a horizontal boundary $z=0$. A light ray reaches the boundary at $(x,0)$, with object-side angle ${\theta }_{\mathrm{obj}}$ and observer-side angle ${\theta }_{\mathrm{obs}}$, both measured from the upward normal vector. For $x>0$, its backward extension meets the object's normal line $x=0$ at $z=-d_{\mathrm{axis}}$, whereThis axis-intercept depth is exact for that one transmitted light ray in the stated model. It tends to $d_{\mathrm{par}}$ as the angles tend to zero; at exactly zero the quotient is interpreted as a limit. These elementary tangent, sine and cosine relations show precisely where the small-angle approximation enters.

The original calculations below use $d=12\mathrm{cm}$, $n_{\mathrm{obj}}=4/3$ and $n_{\mathrm{obs}}=1$:

At appreciable obliquity these depths differ, so the full outgoing bundle has no single exact virtual image on the object's normal line. An oblique eye samples a narrow bundle; its local apparent position also depends on that bundle and cannot be inferred by declaring $d_{\mathrm{axis}}$ to be a universal image depth. A plane refracting boundary gives approximate image formation near normal incidence, with departures at larger angles. The construction assumes a flat surface, uniform transparent media with isotropy, and a propagating transmitted light ray; it does not extend to total internal reflection.

The Beer-Lambert law relates absorbance $A$ to an absorbing species' concentration $c$ and path length $l$ by $A=\epsilon lc$ in a homogeneous medium using sufficiently narrow-band light. The coefficient $\epsilon$ depends on the species, wavelength and chemical solution conditions. This makes Ellman's reagent useful for quantitative thiol assays once its released colored species has been calibrated. ^[ref]

Here $c$ is molar concentration, $l$ is the geometric length through the sample, and $\epsilon$ is the molar absorption coefficient. The ideal beam has parallel light rays, a single wavelength, and a medium with uniform composition and isotropy; finite spectral width must be small relative to features in the optical spectrum.^[ref]

For a spectrophotometer, distinguish the source's output from the reference that reaches the photodetector. Let $P_0$ be the received radiant power through a reference blank and $P$ the received radiant power through the sample, after subtracting the photodetector's dark response. Both refer to the same selected wavelength band and collection geometry. Define the blank-normalized transmittance and absorbance byThe reference blank includes the solvent and other assay ingredients except the absorbing target. Matching cuvettes and conditions makes their common losses divide out. Thus this $T$ is relative to the reference blank; it is not the absolute transmittance of the entire cuvette and its contents. Both radiant powers can be expressed in watts, while their ratio, $T$, and its logarithm, $A$, are dimensionless. An instrument displaying $40\%$ transmission means $T=0.40$, not $T=40$.

Why take a logarithm? Consider two consecutive identical absorbing layers. Each transmits a fraction $t$ of the radiant power entering it, so together they transmit $t^2$. Their absorbances add because $-{\log }_{10}(t^2)=2(-{\log }_{10}t)$. Doubling the sample length doubles absorbance while squaring transmittance. This turns multiplicative loss into an additive scale and makes a straight line useful for inferring molar concentration.

A short model derivation makes the assumption explicit. Suppose a thin layer of thickness $dx$ removes the fraction $kcdx$ of its incoming radiant power, where $k$ is a fixed constant of proportionality and $c$ is a uniform molar concentration. The corresponding ordinary differential equation, initial condition, and solution areTaking the base-ten logarithm gives $A=kcl/\ln 10$, so $\epsilon =k/\ln 10$. The exponential decay follows from equal layers removing equal fractions of what remains. Proportionality to molar concentration additionally assumes that the absorbing chemical species retains the same response as its molar concentration changes; the geometric argument alone does not establish that.

Black-body radiation is the electromagnetic radiation in thermal equilibrium with an ideal blackbody. It can be described as a photon gas: a cavity whose walls are all at the same temperature contains radiation with the universal Planck's law spectrum. A small opening in such a cavity approximates a blackbody emitter because incoming radiation undergoes repeated absorption inside. The radiation's equilibrium spectrum depends on temperature, rather than the wall material. ^[ref]

There are three useful quantities to keep distinct: directional spectral radiance $B_{\nu }$ or $B_{\lambda }$, total energy density $u$ in the cavity, and outward radiant exitance $j^{\ast }$ from its opening. Planck's law supplies the first; integrals over frequency and directions supply the others. For isotropy, every direction has the same $B_{\nu }$, so the full $4\pi$ solid angle givesHere $\hslash =h/(2\pi )$ is the reduced Planck constant, and the radiation constant is $a_{\mathrm{rad}}\approx 7.565733250\times 10^{-16}\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-4}$. The explicit Boltzmann constant makes this an SI formula; adopting natural units with $k_B=1$ recovers $u={\pi }^2{T}^4/(15{\hslash }^3{c}^3)$.

For a short integral outline, substitute $x=h\nu /(k_{B}T)$ and useThe expansion is a geometric series with nonnegative terms; repeated integration by parts evaluates each integral, and $\zeta (4)={\pi }^4/90$ is a value of the Riemann zeta function. This gives the coefficient above.

The photon gas also exerts radiation pressure, $p=u/3$. The factor $1/3$ comes from the angular average of the squared directional component of photon momentum, $⟨{\cos }^2\theta ⟩=1/3$. The thermodynamic derivation of blackbody radiation pressure gives the same result. In contrast, outgoing power involves one projected-direction factor over an outward hemisphere, yielding radiant exitanceThis is the Stefan–Boltzmann law with unit emissivity; $\sigma$ is the Stefan-Boltzmann constant. The factor $\pi$ follows from Lambert's cosine law, rather than the $4\pi$ full-sphere factor used for energy density.

For example, an ideal $6000\mathrm{K}$ cavity has $u\approx 0.980519\mathrm{J}{\mathrm{m}}^{-3}$, radiation pressure $p\approx 0.326840\mathrm{Pa}$, and outward radiant exitance $j^{\ast }\approx 7.34881\times 10^7\mathrm{W}{\mathrm{m}}^{-2}$. A stipulated $3000\mathrm{K}$ cavity has one-sixteenth each value because all three scale as $T^4$. These describe stored energy, mechanical pressure, and outgoing power per area, respectively. In a closed equilibrium cavity, equal opposing radiation balances the directional power flow even though $u$ and $p$ are nonzero.

The usefulness of black-body radiation is this common reference: Planck's law, Wien's displacement law, and the Stefan–Boltzmann law consistently connect spectral measurements, temperature, and total emission. The three illustrative curves in Planck's law are ideal equilibrium spectra; applying them to a material requires considering its emissivity and whether thermal equilibrium is a suitable approximation.

For a plane material-coordinate elastic wave with $u=(0,w(X_1,t),0)$, set $q=w_{,X_1}$. The longitudinal equation is $0={\partial }_{X_1}{P}_{11}(1,q,0)$. Arbitrary nonuniform pure transverse profiles thus require the displayed constitutive compatibility, which does not follow from isotropy alone. If it holds and ${\tau }^{\prime }(q)>0$, the shear equation closes as ${\rho }_0{w}_{tt}={\partial }_{X_1}\tau (q)$ with speed $c(q{)}^2={\tau }^{\prime }(q)/{\rho }_0$.

The distance modulus compares received brightness with intrinsic brightness through $\mu =m-M$, the difference between apparent magnitude and absolute magnitude in the same photometric passband and magnitude zero point. With no interstellar extinction, fixed emitted spectrum and luminosity, isotropy of emission, and the local inverse-square law, it determines Euclidean distance:The ratio inside the logarithm is dimensionless. To see the factor $5$, use the inverse-square law in the magnitude difference formula:For example, $m=5$ and $M=0$ give $\mu =5$, hence $d=10\mathrm{pc}\times 10=100\mathrm{pc}$. A tenfold increase in distance adds five magnitudes; an increase of one magnitude multiplies distance by $10^{1/5}\simeq 1.584893$. The distance modulus is a logarithmic scale for distance, not a distance in length units. ^[ref].

At $10$ parsecs, $\mu =0$ and $m=M$. Within $10$ parsecs, the distance modulus is negative; for example, $\mu =-5$ gives the positive distance $d=1\mathrm{pc}$. A negative distance modulus does not imply negative distance. At cosmological distances, the inference needs luminosity distance and a K correction; the simple local formula must not silently absorb these effects.

Interstellar extinction increases apparent magnitude. The observed distance modulus then includes both geometric dimming and attenuation; extinction bias in photometric distance explains how to separate them.

The Doppler effect is a change between emitted and received frequency caused by motion of the source or receiver. An approaching siren can send more wave crests per second to a microphone than the source emits per second; a receding siren sends fewer. The source need not change its own oscillation. The Doppler shift is $\Delta f=f_o-f_s$, where $f_s>0$ is emitted frequency and $f_o>0$ is received frequency, both measured in hertz.

For acoustic waves, motion must be specified relative to the material carrying the wave. The speed of sound is measured in that material's rest reference frame. Moving the source changes the spacing of successive wavefronts; moving the receiver changes how quickly it crosses those wavefronts. These two mechanisms give different exact factors in the classical Doppler effect for sound, even for the same closing relative velocity.

This makes a microphone a motion probe: a stable tone supplies a repeating clock, and changes in arrival spacing reveal the source's motion toward or away from the receiver. An acoustic Doppler measurement can test the prediction directly. A passing source also illustrates why geometry matters: its speed can stay constant while the received frequency changes continuously, as in Doppler frequency along a trajectory.

The following sound model assumes linear acoustics in uniform stationary material with isotropy, constant speed of sound and negligible wave dispersion. Source and receiver speeds are below the speed of sound, and there is a single direct propagation path. Air motion requires transforming source and receiver velocities into the air's rest reference frame; a nonuniform flow requires a more general propagation model. At the sonic threshold the simple periodic-arrival picture fails; moving-point acoustic wavefronts and the Mach cone describe the relevant geometry, while finite shock waves require further physics. Light in vacuum instead requires the relativistic Doppler effect: the acoustic formula cannot be converted into its exact optical counterpart just by replacing the speed of sound with the speed of light.

Electrical conductivity $\sigma$ measures the local response of current density to electric field. In the rest frame of an isotropic Ohmic conductor, the local form of Ohm's law isLarger positive electrical conductivity gives larger current density at the same electric field. It is a bulk material coefficient, distinct from the geometry-dependent electrical conductance of a whole component and from a sheet's surface conductivity. Electrical resistivity is ${\rho }_{\mathrm{el}}=1/\sigma$ in this scalar model. ^[ref].

The local and whole-wire forms fit together particularly neatly. For uniform current in a straight homogeneous wire, with length $L$, wire cross-sectional area $A$, constant electrical conductivity and terminal electric potential difference $V$, the axial electric field has magnitude $E=V/L$. ThusFor the defined electrical resistance example $\sigma =5\times 10^5(\Omega \mathrm{m}{)}^{-1}$, $L=1\mathrm{m}$, $A=10^{-6}{\mathrm{m}}^2$ and $V=5\mathrm{V}$, the local values are $E=5V/m$ and $j=2.5\times 10^{6}A/{\mathrm{m}}^2$; integrating gives $I=2.5\mathrm{A}$. One material law therefore supplies both a spatial description and a terminal relation.

The scalar form requires isotropy. In an anisotropic material the response can instead be a tensor, $J_i={\sum }_j{\sigma }_{ij}{E}_j$, and current density need not be parallel to electric field; uniaxial conductivity tensor is one existing example. These statements concern the material's rest frame. A moving electrical conductor instead uses the moving-conductor Ohm law.

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 Faraday effect is magnetic field-induced circular birefringence: opposite circular polarizations acquire different wave phases, rotating transmitted linear polarization. For a transparent medium with isotropy before applying the magnetic field, in the linear magnetic-response regime at fixed vacuum wavelength ${\lambda }_0$ and temperature $T$,Here $V$ is the Verdet constant, $L$ is the material path length, and $B_{\parallel }$ is the applied magnetic flux density component along the ray. This $\mathbf{B}$ is the external bias, distinct from the oscillating magnetic field of the light itself. Michael Faraday discovered the effect in 1845.^[ref]

Fix the sign before using the formula. On the forward path choose a right-handed Cartesian coordinate system, propagation along $+z$, and measure $\theta$ from $+x$ toward $+y$ for the electric field's oscillation line. Positive $\theta$ is counterclockwise in the $x$-right, $y$-up end view seen from $z>0$, looking toward the source. Then $B_{\parallel }=B_z$ and positive $V$ means positive rotation for positive $B_z$. This convention differs from the clockwise-positive convention in optical rotation. A linear polarization line is unchanged by $180^{\circ }$; a measured orientation determines accumulated rotation only modulo $\pi$, unless continuity selects a branch.

For uniform $B_{\parallel }$, the formula reduces to $\theta =VBL$. Its appeal is that a weak change of refractive index becomes a visible change of orientation. Magnetic reversal measurement of Faraday rotation turns this into an inference of $V$ or of the magnetic field path integral. Faraday rotation on a round trip explains the directional behavior used by an optical isolator. In an astronomical cold plasma, the high-frequency limit instead leads to the rotation measure law $\theta =\mathrm{RM}{\lambda }_0^2$; that wavelength dependence is not a universal law for transparent solids.

Fourier's law predicts the local conductive heat flux from the local temperature gradient. In a continuum approximation with isotropic thermal conductivity $k>0$, its vector form and one-dimensional form areThe heat flux $q$ has units $\mathrm{W}{\mathrm{m}}^{-2}$, the temperature gradient has units $\mathrm{K}{\mathrm{m}}^{-1}$ and the thermal conductivity has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. The rate of heat transfer $\dot{Q}$ has units $\mathrm{W}$. These units distinguish a local material response from the total power crossing a surface.

The minus sign makes heat flow toward lower temperature: for $k>0$, $\mathbf{q}\cdot \nabla T=-k|\nabla T{|}^2\le 0$. In the worked bar in thermal conduction, $dT/dx=-200\mathrm{K}{\mathrm{m}}^{-1}$ therefore gives $q=+400\mathrm{W}{\mathrm{m}}^{-2}$. Reversing the end temperatures reverses the heat flux. A zero temperature gradient gives zero conductive heat flux in this model.

Fourier's law is a constitutive relation expressing measured material behavior; it is not a consequence of conservation of energy alone. It applies locally and does not require a steady state. The straight-line bar solution additionally assumes constant thermal conductivity, constant cross-sectional area, no sources or side losses, and fixed end temperatures. Anisotropic materials require the directional response described by the thermal conductivity tensor. The scalar $k$ used in the examples expresses the stated isotropy assumption. ^[ref]

Historically, Joseph Fourier's Théorie analytique de la chaleur was published in 1822. In Chapter II, article 112, he writes the heat crossing a spherical surface as the product of thermal conductivity, elapsed time, surface area and the negative radial temperature gradient; article 113 balances the incoming and outgoing heat to obtain the heat equation for a sphere. The enduring idea is this connection between a local flow rule and a global prediction of changing temperature. ^{[ref] [ref]}

For constant particle speed and electric charge $q=ze$, in a transparent, linear, nonmagnetic medium with isotropy, the Frank-Tamm formula gives the produced photon density:Set it to zero below threshold. Here $e$ is the positive elementary charge, $x$ is track length, $\lambda =2\pi c/\omega$ is vacuum wavelength, and $\alpha$ is the fine-structure constant. With SI lengths in metres, the density has units ${\mathrm{m}}^{-2}$; multiplying by $dxd\lambda$ gives a dimensionless expected count. The explicit SI convention isThese conventions agree.^[ref]

To check the convention, start from $E=\hslash \omega =hc/\lambda$ and $d^{2}N/(dxdE)=\alpha z^2{\sin }^2\theta /(\hslash c)$. Changing variables uses the positive density Jacobian $|dE/d\lambda |=hc/{\lambda }^2$; since $h/\hslash =2\pi$, the coefficient becomes $2\pi \alpha z^2$. No factor of the material refractive index belongs in this Jacobian: $\lambda$ is not the in-medium wavelength. Multiplication by $\hslash \omega$ gives the corresponding energy loss density in this SI nonmagnetic convention:

For an original constant-index example, stipulate $n=1.50$, $\beta =0.90$, $z=1$ and $\alpha =1/137.036$. Integration from $400$ to $700\mathrm{nm}$ givesThus a stipulated $1\mathrm{cm}$ track produces about $222$ photons in this band in the ideal model. This is an emitted count before absorption, collection losses and detector response, not a brightness measurement or a measured water yield.

Consider a particle of mass $m$ in a rigid rectangular box of length $L$ along $x$ and wall surface area $A$, so $V=AL$ is the volume. Choose $+x$ toward the right wall. For an incoming normal velocity $v_x=v>0$, an elastic collision reverses the normal momentum:The impulse delivered to the wall points outward, toward $+x$, by Newton's third law. The particle's opposite momentum change is not the wall's impulse. A tangential velocity component is unchanged by the ideal smooth-wall collision.

Ignoring intervening particle-particle collisions in this counting model, the particle travels a normal distance $2L$ before hitting the same wall again. Its hit rate is $|v_x|/(2L)$; multiplying by the impulse per hit gives the time-averaged normal force and its contribution to pressure:For example, stipulate $m=4\times 10^{-26}\mathrm{kg}$, $|v_x|=500\mathrm{m}{\mathrm{s}}^{-1}$, $L=0.10\mathrm{m}$, and $A=0.010{\mathrm{m}}^2$. Then $V=0.0010{\mathrm{m}}^3$, the impulse per hit is $4\times 10^{-23}\mathrm{N}\mathrm{s}$, the hit rate is $2500{\mathrm{s}}^{-1}$, and $P_1=10^{-17}\mathrm{Pa}$. This is one particle's tiny time-averaged contribution to one wall's pressure, not an isotropic equilibrium gas by itself.

For $N$ particles of equal mass, add their force contributions and define $⟨v_x^2⟩=N^{-1}{\sum }_{i=1}^N{v}_{x,i}^2$. In the gas's rest frame, isotropy gives $⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩=⟨v^2⟩/3$, where $v^2=v_x^2+v_y^2+v_z^2$. HenceThe final equality uses the classical equipartition theorem. The kinetic pressure of an isotropic gas therefore recovers the ideal gas law from momentum transfers. Doubling the normal speed would double both the impulse per hit and the hit rate, giving four times the pressure contribution; pressure measures a mean squared speed. Interparticle elastic collisions redistribute velocities and help establish isotropy; the individual round-trip times need not remain $2L/|v_x|$, but the equilibrium momentum flux result survives. ^[ref]

For the Fourier transform convention ${\delta }_{\mathbf{k}}=\int d^{3}x\delta (\mathbf{x})e^{i\mathbf{k}\cdot \mathbf{x}}$ and covariance $⟨{\delta }_{\mathbf{k}}{\delta }_{{\mathbf{k}}^{\prime }}⟩=(2\pi {)}^{3}P(k){\delta }_D(\mathbf{k}+{\mathbf{k}}^{\prime })$, inversion gives $\xi (\mathbf{r})=\int d^{3}kP(k)e^{i\mathbf{k}\cdot \mathbf{r}}/(2\pi {)}^3$. Isotropy reduces the angular integral to $4\pi j_0(kr)$, where $j_0(x)=\sin x/x$ is the order-zero Spherical Bessel function. At zero separation this gives the variance if the spectrum integral converges. A nonnegative spectrum can produce negative correlations away from the origin.

The Mach angle of a supersonic flow with Mach number $M>1$ is $\mu =\arcsin (M^{-1})$. Infinitesimal pressure disturbances lie on the corresponding Mach cone, and a weak oblique shock approaches this angle as its strength tends to zero.

The Mach angle is the half-angle between a Mach cone generator and its backward motion axis for a steady point source in a uniform three-dimensional medium with isotropy and constant sound speed. It obeysThe relation concerns weak acoustic waves in linear acoustics. It is not the angle of a finite oblique shock produced by finite turning of the fluid flow; that angle also depends on the deflection and the gas properties.^[ref]

For a short proof, choose an old emission in the moving-point acoustic wavefronts construction and set $s=-\tau >0$. Its circle has center $(-Us,0)$ and radius $cs$. A tangent line from the present source at the origin meets the circle at a right angle to its radius. The resulting right triangle has hypotenuse $Us$ and opposite side $cs$, giving the displayed sine ratio. The emission age cancels: every wavefront has the same tangent directions, so a continuum of expanding spheres produces one simple Mach cone.

In an original example, stipulate uniform gas with sound speed $c=340\mathrm{m}{\mathrm{s}}^{-1}$ and a source moving through it at $U=680\mathrm{m}{\mathrm{s}}^{-1}$. Then $M=2$ and $\mu =\arcsin (1/2)=30^{\circ }$. An emission one second earlier is now a sphere centered $680\mathrm{m}$ behind the source with radius $340\mathrm{m}$. The selected sound speed defines this example; it is not a constant for air at every temperature. The Mach angle decreases as $M$ increases: its derivative is $-1/(M\sqrt{M^2-1})$ in radians, and its limits are $90^{\circ }$ as $M\downarrow 1$ and $0$ as $M\to \infty$.

A magnetic susceptibility tensor maps a small change of internal magnetic field intensity to a change of magnetization:This tensor reduces to $\chi {\delta }_{ij}$ for isotropy. With anisotropy, even a linear response need not point along the applied magnetic field intensity. As a stipulated example, $\mathbf{\chi }=\mathrm{diag}(1,1,3)\times 10^{-4}$ sends $(H,0,H)$ to $(H,0,3H)\times 10^{-4}$. State the axes and bias field when reporting its components. A constant tensor describes linear response; in a nonlinear material it is the Jacobian matrix of the response at the chosen operating point.

The Maxwell-Boltzmann distribution describes the random velocities and speeds of particles in a classical ideal gas at thermal equilibrium. Individual particles keep changing velocity through collisions, while the equilibrium probability distribution stays fixed. For one species of particle mass $m$ at positive temperature $T$, the characteristic scale iswhere $k_B$ is the Boltzmann constant. A signed velocity component follows a normal distribution; the nonnegative magnitude follows the Maxwell speed distribution. Their peaks differ because a speed measurement groups together all directions of motion.

Use the gas rest frame, three spatial dimensions, and nonrelativistic particles, with $k_{B}T\ll m{c}^2$ where $c$ is the speed of light. Assume a dilute gas whose interactions between encounters can be neglected in the ideal gas approximation and whose translational motion is classical. A sufficient condition for negligible quantum exchange effects is $n{\lambda }^3\ll 1$, where $n$ is the number density and $\lambda$ the thermal de Broglie wavelength. Diluteness alone does not guarantee thermal equilibrium; the gas must have equilibrated. These are the assumptions behind the classical Maxwell-Boltzmann velocity distribution. ^{[ref] [ref]}

The connection to Maxwell–Boltzmann statistics is the Boltzmann factor for translational kinetic energy, $E_{\mathrm{kin}}=m|\mathbf{v}{|}^2/2$. Normalizing $e^{-E_{\mathrm{kin}}/(k_{B}T)}$ over velocity gives a probability density; an occupation factor $e^{(\mu -E)/(k_{B}T)}$ instead specifies a mean occupation, with chemical potential $\mu$. Their normalizations serve different purposes. ^[ref]

This connects microscopic motion to measurements: a gas thermometer uses the ideal gas law, collisions and the mean free path govern transport, and measurements of spectral lines use thermal Doppler broadening to infer temperature. Each probes a different aspect of the same thermal motion.

In 1860, James Clerk Maxwell presented the component and speed laws in Part I, Proposition IV of Illustrations of the Dynamical Theory of Gases, pp. 22–23. His argument combined independence of the three velocity components with isotropy, and he explicitly compared the component law with the normal distribution of observational errors. Independence is an assumption of that argument; perpendicular coordinate axes alone do not establish it. ^[ref]

The Maxwell speed distribution is the probability density function for the magnitude $v=|\mathbf{v}|$. Unlike a signed velocity component, speed cannot be negative:For $0\le a<b$, the probability of a speed in $[a,b]$ is ${\int }_a^{b}g(v)dv$, and ${\int }_0^{\infty }g(v)dv=1$. The density has units of inverse speed, so its height alone is not a probability. ^[ref]

All velocities with magnitudes from $v$ to $v+dv$ occupy a thin spherical shell of volume $4\pi v^{2}dv$ in velocity space. The isotropy of the Maxwell-Boltzmann velocity distribution makes its density constant over each shell, giving $g(v)=4\pi v^{2}f(v{\mathbf{e}}_x)$ for any fixed unit direction ${\mathbf{e}}_x$. A larger radius offers more directions and more volume. The increasing shell factor initially wins over the decreasing Boltzmann factor, so $g(0)=0$ and the most probable speed is positive. This is the geometric reason that the peaks of the vector and speed densities disagree.

With the dimensionless quantity $u=v/s$, every such gas has the same rescaled speed density:For comparison, $w=v_x/s$ has the standard normal distribution density $\varphi (w)=e^{-w^2/2}/\sqrt{2\pi }$ on the whole real line. The Maxwell speed distribution is neither the positive half of this curve nor the distribution of $|v_x|$: it combines three components.

The first two moments of the Maxwell speed distribution give the mean and root mean square:Consequently the mean translational kinetic energy is $m⟨v^2⟩/2=3k_{B}T/2$, as required by the equipartition theorem. ^[ref] The mean speed weights fast particles more than the density peak does, and the root mean square weights them more strongly still.

Changing temperature or particle mass only changes $s\propto \sqrt{T/m}$. At fixed mass, four times the temperature doubles every characteristic speed; at fixed temperature, four times the mass halves them. A stretched density becomes lower so that its area stays one. For two species at the same temperature, equal mean translational kinetic energy therefore means different typical speeds.

For a steady point-like astronomical source with isotropy of emission, in transparent space with negligible relativistic or cosmological effects, conservation of energy spreads its luminosity over a sphere of surface area $4\pi d^2$. Its bolometric observed radiative flux isThis inverse-square law reflects geometric dilution: the same emitted power crosses increasingly large spheres. It does not require light to lose energy on the journey. For a fixed luminosity, doubling Euclidean distance quarters the observed radiative flux. ^[ref].

With $L_0=10^{26}\mathrm{W}$ and $d_0=20$ parsecs, the result is $F_0\simeq 2.09\times 10^{-11}\mathrm{W}{\mathrm{m}}^{-2}$. At $40$ parsecs, the same source gives $F_0/4\simeq 5.22\times 10^{-12}\mathrm{W}{\mathrm{m}}^{-2}$. An ideal collecting area of $1{\mathrm{m}}^2$ at $20$ parsecs intercepts approximately $2.09\times 10^{-11}\mathrm{W}$, before detector losses.