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A phasor is a complex amplitude representing a sinusoid at one specified angular frequency: $a(t)=\mathrm{Re}(\hat{A}{e}^{\mathrm{i}\omega t})$. Its modulus gives the peak amplitude and its complex argument gives the wave phase in this convention. Both may instead use root mean square magnitudes if that normalization is stated.

Thorlabs' silicon FDS100 experiment sends a collimated, fibre-coupled laser beam through a beam splitter to the photodiode and a reference optical power meter. Varying reverse bias and electrical load tests linearity. The measured readout approaches saturation when its electric potential difference consumes the bias.^[ref] A separate low-signal setup uses a $637\mathrm{nm}$ laser, polarizers, an optical chopper, optical attenuators and a lock-in amplifier inside a light-tight enclosure. These are actual apparatus choices, not the parameters of the constructed example above.^[ref]

For an independent dark current check, Thorlabs enclosed a photodiode in nested metal boxes to exclude light and suppress interference, used a Keithley 6487 ammeter supplying reverse bias, and monitored temperature. Only one specimen per type was tested; the nearby temperature sensor did not directly contact the semiconductor. Such a dark control measures the baseline but cannot certify spectral responsivity.^[ref]

To infer spectral responsivity at a selected wavelength, measure the radiant flux actually reaching the test photodiode, rather than just the reference arm. A reference-arm measurement calibration must include the beam splitter ratio and path losses. With signed readouts in one fixed orientation, subtract a beam-blocked baseline and usefor direct electric current readout or a calibrated linear transimpedance amplifier, respectively. Repeating at several incident powers tests whether this ratio is constant. Blocking only the source estimates ambient light plus electronic baseline; covering the detector excludes ambient light too. Their difference distinguishes optical background from dark current and readout offset. For a broad optical spectrum, the ratio is a spectrum-weighted response, not generally $R$ at one wavelength.

For independent Poisson distribution collection, the root mean square shot noise in a one-sided rectangular frequency band of signal bandwidth $B>0$ isHere $I_d$ is reverse dark current; thermal electrical noise and amplifier noise also contribute.^[ref]

Subtracting a mean dark current removes its offset, not its fluctuations: subtracting a constant leaves the variance unchanged. Consequently $I_{\mathrm{ph}}\gg I_d$ makes dark current shot noise small relative to signal shot noise; it does not by itself establish high signal-to-noise ratio. With an independently measured baseline, the variance of that estimate must also be propagated. Increasing the measurement signal bandwidth raises the admitted noise. Calibration of wavelength, reference radiant flux, readout gain and collection geometry, together with drift and linearity, determines what measurement uncertainty can be claimed.

For free planar turning with constant $D_r$, no imposed torque and negligible inertia, model an unwrapped angle by the stochastic differential equationHere $W$ is standard one-dimensional Brownian motion, with increment variance $dt$, and brackets denote expected values over repeated realizations. The angular variance is conventionally reported in $\mathrm{ra}{\mathrm{d}}^2$. “Unwrapped” means recording accumulated turns continuously, rather than resetting the angle at each revolution. Directly subtracting wrapped camera angles creates artificial jumps; for an unmarked ellipsoid the wrapping period is $\pi$, and resolving successive turns additionally requires sufficiently frequent images.

For a planar arrow $\mathbf{u}=(\cos \varphi ,\sin \varphi )$, the orientational autocorrelation isFor an unoriented planar axis, use an autocorrelation unchanged by replacing either endpoint angle by that angle plus $\pi$:The same $D_r$ thus gives arrow and axis correlation times $1/D_r$ and $1/(4D_r)$, respectively, where a correlation time here is the lag at which the relevant curve reaches $1/e$. A short explanation is enough: $\Delta \varphi$ has a Gaussian distribution of variance $2D_{r}t$, so its characteristic function gives $⟨\cos (n\Delta \varphi )⟩=e^{-n^2{D}_{r}t}$. The integer $n$ selects what the observable can distinguish; it does not change the underlying angular spreading.

For an original numerical example, choose $D_r=0.25{\mathrm{s}}^{-1}$. After $2\mathrm{s}$, the angular variance is $1\mathrm{ra}{\mathrm{d}}^2$ and its root mean square displacement is $1\mathrm{rad}$, although the mean displacement is zero. The planar arrow autocorrelation is $e^{-0.5}\simeq 0.607$, while the planar axis autocorrelation is $e^{-2}\simeq 0.135$. Calling both “the orientation correlation” without specifying the observable would hide a factor of four in decay rate.

Using $u=r^2/(2{\sigma }^2)$ in the defining expected value integral givesThe last integral is the gamma function. Near zero, the integrand before substitution behaves as $r^{q+1}$, explaining the threshold $q>-2$; at infinity the Gaussian tail controls every power. In particular,^[ref]

For the chosen $\sigma =2\mathrm{cm}$, the mean radius is about $2.5066\mathrm{cm}$, its median is $2\sqrt{2\log 2}\simeq 2.3548\mathrm{cm}$, and its mode is $2\mathrm{cm}$. Its root mean square is $2\sqrt{2}\simeq 2.8284\mathrm{cm}$. Different summaries answer different questions; the component scale $\sigma$ is neither the radial mean nor the radial standard deviation.

The geometrical optics account of an optical trap adds the momentum flux changes of refracted and reflected light rays. It applies when the bead is large compared with the surrounding medium's wavelength. When the bead is much smaller, an induced electric dipole gives a different approximation. For a homogeneous dielectric sphere of radius $a$, define $\mathbf{p}(t)=\mathrm{Re}[\alpha {\mathbf{E}}_0{e}^{-i\omega t}]$, with ${\mathbf{E}}_0$ the peak complex electric field and $\alpha$ its SI polarizability. In a transparent, nonmagnetic, nondispersive host,where $n_p,n_m$ are particle and host refractive indices, $I$ is cycle-averaged light intensity, ${ϵ}_0$ is the vacuum permittivity and $c$ is the speed of light. The quasistatic electric dipole approximation requires $2\pi n_{m}a/{\lambda }_0\ll 1$, with ${\lambda }_0$ the vacuum wavelength, and a field nearly uniform across the bead.

For real quasistatic $\alpha$, building an induced electric dipole gives instantaneous potential energy $-\alpha E^2/2$. Averaging over one optical cycle givesThus $n_p>n_m$ gives positive polarizability and attraction toward larger light intensity; $n_p<n_m$ reverses this contribution. The factor $1/4$ uses a peak field: substituting a root mean square field without changing that factor would halve the force incorrectly. Dimensionally $\alpha |E_0{|}^2$ is energy, so its gradient is force.

This leading conservative term does not remove the optical scattering force. Even a nonabsorbing induced electric dipole reradiates light: at fixed wavelength and index contrast, its Rayleigh scattering cross-section scales as $a^6$, while $\alpha$ scales as $a^3$. A smaller bead can therefore improve the ratio of gradient to scattering forces while weakening the well relative to $k_{B}T$, where $k_B$ is the Boltzmann constant and $T$ the temperature. Optical force balance and resistance to thermal escape are separate questions. Intermediate bead sizes require a wave-scattering calculation; neither a ray calculation nor this small-dipole expression is uniformly valid there.

Use complex amplitudes only for signals at one common positive angular frequency $\omega$. Define the real terminal voltage and electric current byThe complex amplitudes here use peak magnitudes; both can instead use root mean square magnitudes without changing their ratio. Choose positive electric current entering the terminal assigned positive voltage, so $p=vi$ is absorbed power. For an ideal linear inductor and capacitor, their terminal laws $v_L=Ldi/dt$ and $i_C=Cd{v}_C/dt$ giveThe derivative multiplies a complex amplitude by $\mathrm{i}\omega$: this single algebraic step explains both signs. Taking $i(t)=I_0\cos \omega t$ as a reference gives $v_L(t)=-\omega L{I}_0\sin \omega t$ and $v_C(t)=I_0\sin \omega t/(\omega C)$ for the sinusoidal response. Thus inductor voltage leads electric current by $\pi /2$, while capacitor voltage lags electric current by $\pi /2$. A constant offset or a free transient is outside this single-frequency representation. ^[ref].

With the alternative $e^{-\mathrm{i}\omega t}$ convention, the same real signals use conjugated complex amplitudes and $Z_L=-\mathrm{i}\omega L$, $Z_C=+\mathrm{i}/(\omega C)$. Their physical leads and lags stay the same; the sign of $\mathrm{Im}Z$ changes. Also, some textbooks call the positive magnitude $1/(\omega C)$ the capacitive reactance and write the series expression as $X_L-|X_C|$. This article consistently uses signed $X_C=-1/(\omega C)$ and hence $X=X_L+X_C$. ^[ref].

Connect one physical resistor $R$ across an ideal lossless capacitor $C$, with no applied source and an ideal readout that draws no electric current. Represent the resistor by its noisy Thévenin equivalent. The resulting RC low-pass frequency response gives the stationary capacitor voltage power spectral densityIts mean square follows by the effective noise bandwidth calculation:This is also the equipartition theorem: the mean capacitor energy is $\frac{1}{2}C⟨v_C^2⟩=\frac{1}{2}{k}_{\mathrm{B}}T$. Raising $R$ makes the low-frequency spectrum taller and its width narrower in exactly compensating proportions. It changes the equilibration time $RC$, not the final variance.

The integral to infinity belongs to the ideal classical model. For a physical electrical circuit, assume thermal equilibrium, linear components with negligible extra dissipation, and a corner far below the frequencies where parasitic effects, frequency-dependent electrical resistance, or $hf\sim k_{\mathrm{B}}T$ invalidate the model; the omitted high-frequency tail must be negligible. A finite-bandwidth instrument measures only part of this variance. Its input capacitance contributes to the total $C$, and its finite input electrical resistance introduces both loading and its own thermal electrical noise.

For $T=300\mathrm{K}$ and $C=1\mathrm{nF}$, the predicted equilibrium variance is $4.141947\times 10^{-12}{\mathrm{V}}^2$ and the root mean square voltage is $2.0352\mu \mathrm{V}$. At $R=10\mathrm{k}\Omega$, $RC=10\mu \mathrm{s}$ and $B_{\mathrm{n}}=25\mathrm{kHz}$. At $R=100\mathrm{k}\Omega$, $RC=100\mu \mathrm{s}$ and $B_{\mathrm{n}}=2.5\mathrm{kHz}$; the equilibrium root mean square voltage is identical.

To obtain the growth curve, prepare $v_C(0)=0$ exactly and connect the resistor at $t=0$ while maintaining its temperature. The RC low-pass filter's causal impulse response is $h(t)=e^{-t/(RC)}/(RC)$ for $t\ge 0$. Classical ideal white noise with two-sided power spectral density $2k_{\mathrm{B}}TR$ givesThe factor $2k_{\mathrm{B}}TR$ here is the two-sided time-domain noise intensity; the equilibrium spectral integral above used the one-sided $4k_{\mathrm{B}}TR$. They give the same long-time limit. The variance approaches equilibrium with time constant $RC/2$, although a deterministic initial voltage decays with time constant $RC$. If the capacitor already starts in thermal equilibrium, its initial variance supplies the missing exponentially decaying term and the total variance stays constant.

The raw root mean square of a random variable is $\sqrt{E[X^2]}$, whereas its r.m.s. fluctuation about its mean is the standard deviation $\sqrt{E[(X-\mu {)}^2]}$. The former cannot be smaller than $|\mu |$. Experimental phrases such as r.m.s. noise or temperature spread usually refer to the latter and should state the reference level.

Consider a stipulated series circuit containing an ideal resistor of electrical resistance $R=100\Omega$ and an ideal capacitor of capacitance $C=1\mu \mathrm{F}$. The same electric current passes through both. Their terminal electric potential differences add by Kirchhoff's voltage law, soA quick derivation uses $i=Cd{v}_C/dt$: taking the derivative of the complex exponential multiplies the capacitor voltage phasor by $\mathrm{i}\omega$, hence ${\hat{V}}_C=\hat{I}/(\mathrm{i}\omega C)$. Dimensionally, $1/(\omega C)$ has units $s/F=V/A=\Omega$, matching $R$.

At frequency $f=1000\mathrm{Hz}$, the angular frequency is $\omega =2\pi f$. The signed capacitive electrical reactance, modulus and complex argument of the total electrical impedance areApply $V_{\mathrm{rms}}=1.00\mathrm{V}$ across the complete series circuit. Its root mean square electric current isWith voltage as the zero-phase reference, the current phasor has complex argument $+57.858092^{\circ }$: electric current leads. The phasor drops have moduli $V_{R,\mathrm{rms}}=0.532018\mathrm{V}$ and $V_{C,\mathrm{rms}}=0.846733\mathrm{V}$; these add as complex numbers, not as positive magnitudes. Their squared magnitudes sum to $(1.00\mathrm{V}{)}^2$.

The mean absorbed power uses the root mean square values:The ideal capacitor has zero cycle-averaged absorbed power; all the mean absorbed power becomes Joule heating in the resistor.^[ref] To see why the phase factor appears, average $p(t)=v(t)i(t)$ over a full cycle with $v(t)=\sqrt{2}{V}_{\mathrm{rms}}\cos \omega t$ and $i(t)=\sqrt{2}{I}_{\mathrm{rms}}\cos (\omega t-\varphi )$. The cosine addition formula reduces the average to $V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \varphi$; the term at $2\omega$ averages to zero. Independently, $p_C=d(C{v}_C^2/2)/dt$, and the capacitor energy returns to its starting value after each cycle.

At fixed applied $V_{\mathrm{rms}}$, the limit $\omega \to 0^{+}$ gives $|Z|\to \infty$, $\varphi \to -\pi /2$ and $I_{\mathrm{rms}}\to 0$: the ideal capacitor blocks steady electric current. The limit $\omega \to \infty$ gives $Z\to R$, $\varphi \to 0$ and $I_{\mathrm{rms}}\to V_{\mathrm{rms}}/R$. These are limits of the ideal lumped electrical circuit; unwanted lead inductance and component losses can invalidate that model at high frequency. At $\omega RC=1$, the equal legs of the impedance right triangle give $|Z|=\sqrt{2}R$ and $\varphi =-\pi /4$ without numerical evaluation. Measuring both the magnitude and the sign of the phase difference distinguishes this capacitor response from an equally large positive electrical reactance.

MIT's guide uses an illuminated vacuum photodiode, with collecting anode positive relative to the emitting cathode. A transimpedance amplifier gives $V_1=R_{F}I$; a multimeter reads its mean. The fluctuation path rejects below $100\mathrm{Hz}$, adds roughly tenfold voltage gain, then nominal hundredfold preamplification and a $1$–$50\mathrm{kHz}$ band-pass filter; the readout measures root mean square voltage. With diode bias off, a sinusoid at the test port calibrates $g(f)$. Restore bias, retain gain settings, vary lamp brightness, and repeat both readings. The guide warns about overload; it supplies a procedure, not measured results.^[ref]

The inference can be expressed without relying on nominal passband edges. Let $g(f)$ denote the calibrated voltage transfer after the $R_F$ conversion, so $K(f)=R_{F}g(f)$ and define $\mathcal{K}={\int }_0^{\infty }|K(f){|}^{2}df$. Under independent arrivals and an unchanged additive background,The slope of voltage variance against mean electric current can therefore estimate $q$. A free intercept accounts for a constant background; forcing the line through zero generally biases the slope. The root mean square readings must have their mean removed and then be squared. Subtracting two root mean square amplitudes does not subtract independent variances. Estimate the variance from successive records; averaging the waveforms first would suppress the fluctuation being measured.

Additional controls sharpen this interpretation. Block the light with detector bias, temperature, and gain unchanged; this preserves dark current and electronics while removing photocurrent. Repeat this dark comparison between illumination settings to check drift. If independent dark arrivals have mean magnitude $I_d$ and illuminated arrivals add $I_{\mathrm{ph}}$, their spectra add: $2q(I_d+I_{\mathrm{ph}})$. Subtracting the unchanged dark variance isolates $2q{I}_{\mathrm{ph}}\mathcal{K}$. This subtraction model assumes unchanged transfers and independence; switching off bias for gain calibration is a different control and need not preserve the dark background.

Vary measured $I_{\mathrm{ph}}$ rather than assuming lamp knob position is proportional to photocurrent. Check collection versus bias to identify a range where incomplete collection is negligible; a space charge limited tube can correlate transfers. Verify that the transimpedance amplifier remains linear and that its frequency response is unchanged over the illumination range. Changing signal bandwidth should change the excess variance through $\mathcal{K}$; compare inferred input spectra in several bands. Correlated fluctuations of the light source can add a contribution scaling approximately as $I_{\mathrm{ph}}^2$, while an electronics floor can hide the linear term at small electric currents. A fitted line over a short range alone does not establish independent transfers.

The gain calibration must include the same detector-to-readout path used for noise, or explicitly supply any omitted transfer. Detector capacitance, finite collection time, amplifier voltage and current noise, thermal electrical noise, pickup, clipping, and finite recording precision limit the inference. Current-dependent electrical impedance or heating can make a background current-dependent too. Since $\mathcal{K}$ contains squared gain, a small fractional error in gain produces approximately twice that fractional error in the predicted variance scale. Knowing a nominal gain and two cutoff settings is insufficient.

Write an ideal point-transfer model using the Dirac delta function:For this arrival stream with independent increments, distinct arrivals contribute the product of the means, which cancels when centering. The same-arrival term leaves the autocovarianceIts Fourier transform gives the two-sided power spectral density, while folding negative frequencies onto positive ones gives the one-sided convention:These spectra describe $\delta i$, excluding the spectral line from the mean direct current. The amplitude spectral density is $\sqrt{2qI}$ in $\mathrm{A}/\sqrt{\mathrm{Hz}}$; it is not a root mean square electric current until a measurement signal bandwidth is specified. Approximate white noise requires frequencies low enough that the single-transfer pulse shape and any correlations can be neglected. Neither the ideal impulses nor a flat spectrum extending to infinity describe a real detector at every timescale.

All spectra above are per hertz, with $f$ in cycles per second. One possible angular frequency convention defines a density ${\mathcal{S}}_I^{(2)}(\omega )$ integrated using $d\omega$ alone. Since $\omega =2\pi f$, it hasAnother common convention keeps the numerical density $qI$ but integrates it with $d\omega /(2\pi )$. State the integration measure as well as whether a spectrum is one-sided: a change of variable and a folding of signs are separate operations.

A signal-to-noise ratio compares a chosen measure of a signal with the corresponding noise measure. For a constant measured electric potential difference $V$ and a root mean square noise voltage $v_n$, an amplitude ratio is $|V|/v_n$; the corresponding power ratio under equal electrical resistance is its square. State the convention and measurement signal bandwidth, since changing either changes the numerical ratio.

Use a complex envelope $A(t)$ normalized so that $|A(t){|}^2=P(t)$; its units are ${\mathrm{W}}^{1/2}$. The optical carrier convention is $e^{-i{\omega }_{0}t}$: the physical electric field is proportional to the real part of $A(t)e^{-i{\omega }_{0}t}$ for a fixed transverse mode. With angular frequency offset $\Omega =\omega -{\omega }_0$, fix the Fourier transform conventionWrite $\tilde{A}=B{e}^{i\varphi }$ with real $B\ge 0$. A dispersive element adds spectral phaseConstant phase rotates the envelope; the linear term delays it by ${\varphi }_1$. The group delay is $d\varphi /d\Omega$. The quadratic coefficient ${\varphi }_2$ is group delay dispersion, abbreviated GDD, with units of time squared. Positive GDD makes higher frequencies arrive later with this convention. It is accumulated dispersion of a component; material group velocity dispersion is its corresponding coefficient per unit length.

To undo a quadratic stretcher spectral phase, the compressor must supply opposite GDD. Include the amplifier and all intervening optics in the sum:Setting this sum to zero does not cancel third-order dispersion, ${\varphi }_3$, or still higher orders. Residual spectral phase can produce asymmetry and satellite pulses even when the quadratic term vanishes. A diffraction grating system can make path length depend on frequency; matching the complete spectral phase over the occupied spectrum is more demanding than merely choosing opposite quadratic coefficients.

There is a useful proof of what phase correction can accomplish. By Parseval's identity,For a smooth decaying spectrum, the Fourier transform of $tA$ is $-id\tilde{A}/d\Omega$. Consequently the power-weighted mean arrival time and variance obeyAt fixed spectral amplitude, constant group delay minimizes this root mean square duration. Such a pulse is a transform-limited pulse, up to a time shift and constant phase. The nonnegative second term measures temporal broadening from unequal arrival times; the first term remains even after perfect phase correction.

MIT's laboratory guide mounts resistors on an aluminum box beneath a metal beaker shield, with short twisted leads. Its SR560 amplifier uses nominal gain $1000$ before a Krohn-Hite $1$–$50\mathrm{kHz}$ band-pass filter; an oscilloscope reads root mean square voltage. Calibration sweeps an attenuated sinusoid and measures the actual gain $g(f)$. Students alternate the resistor and a short circuit, subtracting squared readings $V_R^2-V_S^2$, and disconnect the ohmmeter during noise acquisition. Lead and input capacitance enter the gain-squared integral. With settings unchanged, varying electrical resistance or temperature tests the predicted mean square. The guide prescribes repeated readings and warns against averaging noise waveforms to zero. This is a documented teaching protocol, not a supplied dataset of measured outcomes.^[ref]

The subtraction requires care: shorting the input changes the source electrical impedance. An amplifier's input current noise can therefore produce different voltage noise in the two configurations; a shorted baseline alone does not remove that contribution. Gain drift, interference and insufficient observation time also limit the inference. These are limitations of interpreting the measurement, not corrections that can be inferred from the nominal signal bandwidth alone. In an independent analysis one would compare the measured difference withwhere $C$ represents the measured shunt capacitance and $g$ excludes that separately modeled shunt. If calibration already includes the same shunt response, applying the denominator again counts the attenuation twice. The thermal noise worked example and both original figures illustrate ideal models; they claim no measured agreement or measured value of the Boltzmann constant.

Model an ideal resistor of constant electrical resistance $R>0$ at uniform absolute temperature $T$ by its Thévenin equivalent: a zero-mean random voltage source in series with a noiseless $R$. In the classical range its one-sided, open-circuit voltage power spectral density, per hertz, iswhere $k_{\mathrm{B}}$ is the Boltzmann constant. Thus a positive-frequency interval of width $df$ contributes $4k_{\mathrm{B}}TRdf$ to the mean square voltage. This is approximately white noise only where the electrical resistance is effectively constant and $hf\ll k_{\mathrm{B}}T$, with $h$ the Planck constant. The ideal constant spectrum cannot describe arbitrarily high frequencies.^[ref]

For a real voltage, an equivalent two-sided power spectral density is $S_v^{(2)}(f)=2k_{\mathrm{B}}TR$ over signed frequencies. With a real linear time-invariant system of frequency response $H(f)$, the same output mean square isAngle brackets denote an expected value; for this zero-mean noise the mean square is its variance. Do not integrate $4k_{\mathrm{B}}TR$ over both signs of frequency: that counts the same fluctuations twice. All integrals here use frequency $f$ in hertz, not angular frequency $\omega =2\pi f$. The amplitude spectral density is $\sqrt{4k_{\mathrm{B}}TR}$ in $\mathrm{V}/\sqrt{\mathrm{Hz}}$, whereas the root mean square voltage requires a specified measurement signal bandwidth.

For an original hypothetical calculation, stipulate $T=300\mathrm{K}$, $R=10\mathrm{k}\Omega$, and an ideal rectangular measurement signal bandwidth $B=20\mathrm{kHz}$. Using $k_{\mathrm{B}}=1.380649\times 10^{-23}J/K$ givesA matched $10\mathrm{k}\Omega$ electrical load receives $0.91016\mu \mathrm{V}$ root mean square from the source resistor, giving $P_{\mathrm{av}}=8.283894\times 10^{-17}\mathrm{W}$. These are stipulated model predictions, not laboratory measurements. Increasing $R$ raises the open-circuit voltage noise but does not increase the matched available power. Restricting the measurement signal bandwidth reduces root mean square noise as its square root; it improves signal-to-noise ratio only if the desired signal is preserved.

For a nearly flat interface, the extra surface energy in the small-slope approximation is $(\gamma /2)\int |\nabla h{|}^{2}dA$. Each nonzero Fourier mode therefore has thermal variance $⟨|h_{\mathbf{q}}{|}^2⟩=k_{B}T/(A\gamma q^2)$ in a convention with height reconstructed as the sum of its mode amplitudes. Summing over a two-dimensional annulus of wavevectors givesThe smallest wavevector is set by the lateral size or pinning, and the largest by the breakdown of the continuum description, commonly a molecular length. A fixed mean height removes the zero mode. Gravity supplies an additional restoring term and changes the long-wavelength cutoff when it is retained. Roughness here means a height variance; its square root is the root mean square amplitude, even when the mean height fluctuation is zero.

A useful original model treats the thermopile absorber as one approximately uniform temperature $T_h(t)$ with heat capacity $C_{\mathrm{th}}$, connected to a fixed reference temperature $T_c$ by thermal conductance $G_{\mathrm{th}}$. Both parameters are positive constants here. Let $P_{\mathrm{abs}}(t)$ be an added absorbed power above the initial background. Linearize the outgoing heat transfer about that background. With $\Delta T=T_h-T_c$ measured from an initial state having $\Delta T=0$, conservation of energy gives the first-order linear differential equationThe left side is the rate of stored energy; the two terms on the right are added power and outgoing heat transfer. The effective thermal conductance can include several paths, rather than just heat conduction along the wires. The model assumes a stable reference, negligible electrical loading, a common absorber temperature, and changes small enough for constant parameters.

For a step from zero to constant $P_0$ at $t=0$, subtract the steady solution $\Delta T_{\infty }=P_0/G_{\mathrm{th}}$. The difference obeys exponential decay, so the initial condition fixesThe time constant has units $(\mathrm{J}{\mathrm{K}}^{-1})/(\mathrm{W}{\mathrm{K}}^{-1})=\mathrm{s}$. With the same constructed $N=20$ and $\alpha =40\mu \mathrm{V}{\mathrm{K}}^{-1}$, choose $C_{\mathrm{th}}=0.02\mathrm{J}{\mathrm{K}}^{-1}$, $G_{\mathrm{th}}=0.01\mathrm{W}{\mathrm{K}}^{-1}$ and $P_0=0.03\mathrm{W}$. ThenAt one time constant, the response is $1-e^{-1}\approx 0.632$ of its final value: $V_{\mathrm{oc}}(2\mathrm{s})\approx 1.52\mathrm{mV}$. Reaching $95\%$ requires $t=-\tau \ln (0.05)\approx 5.99\mathrm{s}$. A voltmeter read too soon therefore underestimates the steady signal even when its electrical calibration is perfect.

The steady ratio $V_{\infty }/P_0=N\alpha /G_{\mathrm{th}}$ is $0.080\mathrm{V}{\mathrm{W}}^{-1}$ in this constructed model. Halving thermal conductance doubles this ratio but also doubles the time constant if heat capacity stays fixed. Reducing heat capacity alone speeds the response without changing its steady value. Increasing $N$ gives a larger signal only at fixed individual temperature differences: adding legs can also change thermal conductance, heat capacity and electrical resistance. The algebra identifies coupled design choices, rather than predicting a particular manufactured detector.

A finite electrical resistance produces Thermal electrical noise. For nearly uniform absolute temperature $T$ and effective measurement signal bandwidth $B$, its root mean square open-circuit voltage isHere $k_{\mathrm{B}}$ is the Boltzmann constant. Thermal electrical noise in the sensing element is an important contribution, and the signal bandwidth enters its voltage variance.^[ref] If identical pairs have fixed electrical resistance $r$ and fixed $\Delta T$, then $R_{\mathrm{pile}}=Nr$: the signal grows as $N$ while this noise grows as $\sqrt{N}$. The resulting signal-to-noise ratio grows only as $\sqrt{N}$ under those assumptions. Reference temperature drift, environmental electromagnetic radiation, and meter noise remain separate limits; reducing signal bandwidth trades fast information for less integrated electrical noise.

An original teaching model exposes the roles of input loading, output loading and the supply. Choose a small-signal model with positive input electrical resistance $r_i$, transconductance $g_m>0$, and output electrical resistance $r_o>0$. Its controlled output current source draws $g_m{v}_i$ from the output node toward the reference node. This chosen sign gives an inverting voltage gain; it is not the historical contact circuit's sign convention.

Assume a fixed operating point, negligible reverse coupling, a frequency range where coupling capacitors pass the signal, and negligible other capacitances and transit delays. A collector resistor $R_D$ connects the output to a steady ideal supply $V_{DD}$; an external electrical load $R_L$ is coupled through a capacitor, so it draws no direct current. The input source has series electrical resistance $R_s$. All voltages below are small departures from the operating point. An ideal fixed supply has zero incremental voltage and therefore becomes the reference node in the small-signal model.

Define $R_O=R_D\parallel r_o$ and $R_P=R_O\parallel R_L$, where $a\parallel b=ab/(a+b)$. Input voltage division and output Kirchhoff's current law giveConsequently the port voltage gain, overall source-to-output voltage gain, and load current gain magnitudes areThis current gain uses load electric current, not all the controlled-source electric current. It is also a different port definition from the historical emitter-to-collector current gain. Larger $R_s$ attenuates the input; smaller $R_L$ attenuates the output.

For a sinusoid input signal, use root mean square voltages. The power gain from accepted input signal power $P_i=v_i^2/r_i$ to useful load power $P_L=v_o^2/R_L$ isUnlike $|A_i|$, this tends to zero as either $R_L\to 0$ or $R_L\to \infty$. Differentiation givesThus matching the output electrical load maximizes delivered power in this unilateral resistive model. For fixed source voltage, input loading multiplies $P_L$ by $[r_i/(R_s+r_i){]}^2$; the accepted-input power gain above excludes the loss in $R_s$.

Choose $r_i=R_s=1\mathrm{k\Omega }$, $g_m=2\mathrm{mS}$, $r_o=20\mathrm{k\Omega }$, and $R_D=2\mathrm{k\Omega }$. Then $R_O=20/11\mathrm{k\Omega }$. With $R_L=2\mathrm{k\Omega }$ and $v_s=20\mathrm{mV}$ root mean square, the input is $10\mathrm{mV}$ and the inverted output has magnitude $19.0476\mathrm{mV}$ root mean square. Hence $P_i=0.100\mu \mathrm{W}$ and $P_L=0.181406\mu \mathrm{W}$: $G_P=1.81406$. These are calculated values of the chosen model, not historical measurements.

With the same parameters but $R_L=100\Omega$, $|A_i|=1.89573$ while $G_P=0.359381<1$. At $R_L=10\mathrm{k\Omega }$, $|A_v|=3.07692$ while $G_P=0.946746<1$. Neither current gain nor voltage gain alone establishes power gain.

For the direct current load line in the figure, choose $V_{DD}=6\mathrm{V}$, $I_Q=1\mathrm{mA}$ and $V_Q=4\mathrm{V}$, consistent with $V_Q=V_{DD}-R_D{I}_Q$. Locally extend the model as $I=I_Q+g_m{v}_i+(V-V_Q)/r_o$. With the external electrical load isolated at direct current, its intersection with $I=(V_{DD}-V)/R_D$ is $V=V_Q-g_m{R}_O{v}_i$. Keeping $|v_o|$ small compared with the available voltage headroom is essential; extrapolation to arbitrarily large signals would be unphysical.

At the chosen operating point the collector supply delivers $V_{DD}{I}_Q=6\mathrm{mW}$: $2\mathrm{mW}$ is dissipated in $R_D$ and $4\mathrm{mW}$ in the output device. Under a small signal, that energy flow is redistributed to provide load power. The incremental controlled current source hides the supply's steady energy flow; it does not create energy. A real complete amplifier also needs its input biasing network and has further losses. This calculation establishes neither its total efficiency nor a universal law for point-contact transistors.

Consider the original two-electron example obtained by taking $N=2$ in the Laughlin wavefunction, with no additional polynomial for the centre of mass. Introduce $Z=(z_1+z_2)/2$, relative complex coordinate $w=z_1-z_2$, separation $r=|w|$, and dimensionless separation $u=r/{\ell }_B$. Since $|z_1{|}^2+|z_2{|}^2=2|Z{|}^2+|w{|}^2/2$, the probability density factorizes:Integrating out $Z$ leaves an isotropic relative probability density. Define $D_m(u)$ per unit dimensionless area and $p_m(u)$ per unit dimensionless separation:The physical areal probability density is $D_m(r/{\ell }_B)/{\ell }_B^2$. The probability of a separation between $r$ and $r+dr$ is $p_m(r/{\ell }_B)dr/{\ell }_B$. The extra $u$ comes from the Jacobian determinant in polar coordinates: a thin annulus has area $2\pi rdr$. Omitting that factor changes the question being answered.

Both wavefunction normalization and the peak shift can be checked exactly. The substitution $t=u^2/4$ gives the Gamma function integralFor $m>0$, logarithmic differentiation givesConsequently the areal maximum lies at $u=2\sqrt{m}$, while the most probable separation lies farther out, at $u=\sqrt{4m+2}$. For $m=1$ these are $2$ and $\sqrt{6}\simeq 2.449$; for $m=3$, $\sqrt{12}\simeq 3.464$ and $\sqrt{14}\simeq 3.742$. The same substitution gives $⟨u^2⟩=4(m+1)$, so the root mean square separations are $\sqrt{8}$ and $4$ respectively. The distinction between a probability density maximum and a radial maximum therefore has a directly calculable geometric origin.

This two-electron model explains exchange and pair avoidance. It does not establish a nonzero many-particle spectral gap, an extended quantum Hall plateau, or the topological properties of a fluid in the thermodynamic limit. Those require additional many-electron reasoning and experimental tests.