The tail with no slowest relaxation

Every settling argument about doublets so far has had a slowest member. The cancellation that leaves a tail had one relaxation and its settling was $\tau\ln(A/B)$. The slowest one decides it had eight spread over decades, all equally weighted, and found the slowest decided everything, its own share entering only through a logarithm. Its last open question was the one it could not answer with equal weights: what a real dielectric has is a distribution, and the settling depends on the product of the slowest time constant and the logarithm of its own share — so a spectrum weighted towards the fast end is much better than one weighted evenly, by an amount that is one measurement away.

This essay takes the measurement, and finds that “much better” undersells it. A spectrum weighted towards the fast end does not settle faster by a factor. It settles by a different law, one in which no relaxation is the slowest one that matters and the settling time is a power of the band.

A spectrum with a weighting #

The tail is one per cent of the step, divided among two hundred and one relaxations spaced geometrically over five decades of time constant, from 0.1 milliseconds to 10 seconds — dense enough that the sum behaves as a continuous spectrum, and wide enough that the middle of it is far from either end. Each relaxation’s amplitude is proportional to $\tau^{-m}$, normalised so that the whole tail is one per cent, and $m$ is the only parameter:

• $m = 0$ is the equal weighting the essay on the slowest doublet swept, now dense rather than eight lines;
• $m > 0$ puts more weight on the fast relaxations — at $m = 1$ each decade of time constant carries a tenth of the weight of the one below it;
• $m < 0$ puts it on the slow ones.

The two upper rows are the old law. Every band is reached at a time comparable to the slowest time constant, ten seconds, and each further decade of band adds a roughly constant increment — about twenty seconds on the even row — which is the signature of one exponential, the slowest, having taken over.

The bottom row is a different law. Each decade of band multiplies the time by ten. The increments grow, and the settling to 10⁻⁶ at 0.89 seconds is still an order of magnitude short of the slowest relaxation present: it is decided by relaxations in the middle of the spread, and the ten-second end of the spectrum never becomes the thing that matters.

Why a weighted spectrum gives a power law #

The error is a sum of exponentials, and for a dense spectrum the sum is an integral over the logarithm of the time constant. With the density $c\,\tau^{-m}$ per unit of $\ln\tau$,

$$e(t) = \int c\,\tau^{-m} e^{-t/\tau}\, d\ln\tau \;=\; c\,\Gamma(m)\,t^{-m}$$

when the integral runs over all $\tau$, which for a finite spread means for times well inside it. The substitution $u = t/\tau$ turns the integral into the Gamma function and leaves $t^{-m}$ outside. So between the two ends of the spread the tail is a power of time, and the power is the weighting’s exponent.

Normalising the density to a total amplitude $A$ over a spread that starts at $\tau_\mathrm{min}$ gives $c = mA\tau_\mathrm{min}^{m}$ when the far end is distant, and the settling time to a band $B$ follows in one line:

$$t_B \approx \tau_\mathrm{min}\left(\frac{\Gamma(m+1)\,A}{B}\right)^{1/m}$$

At $m = 1$ that is $\tau_\mathrm{min}A/B$: 0.1 ms times a hundred, ten milliseconds to reach 10⁻⁴, against the 9.71 measured. The formula contains the fastest time constant and not the slowest. A tighter band costs a factor of $10^{1/m}$ per decade — ten at $m = 1$, a hundred at $m = \tfrac12$ — and the slowest relaxation does not appear until the settling time reaches it.

The figure checks the closed form at the geometric centre of the spread, where for $m = 1$ the sum and $c\,\Gamma(m)\,t^{-m}$ agree to 0.3 per cent. At $m = \tfrac12$ the agreement is looser, and the reason is instructive: the density falls so slowly that the part of the integral beyond the ten-second cut is still worth $(t/\tau_\mathrm{max})^{m}/\bigl(m\,\Gamma(m)\bigr)$ of the whole — six per cent at the centre of five decades. The check carries that term as its tolerance rather than pretending the spread is infinite, and the slow end’s absence is a small, calculable help.

The boundary is at zero, and the even spectrum sits on it #

At $m = 0$ the integral no longer produces a power: the density is flat per decade and the error at a time inside the spread is the share of the decades still above it, $A\ln(\tau_\mathrm{max}/t)/\ln(\tau_\mathrm{max}/\tau_\mathrm{min})$, roughly. That falls by a fifth of the tail for every decade of time — so the one-per-cent tail is still at a fifth of a per cent after four decades — and only when the time reaches the slowest relaxation does the exponential take over and finish it. Settling at any band finer than the tail’s own share of a decade is therefore $\tau_\mathrm{max}$ times a logarithm, which is exactly the law the essay on the slowest doublet found with eight discrete lines. Its result was the $m = 0$ case, and $m = 0$ is the boundary.

Below it, with the weight on the slow end, the law is the same with a larger constant: the slowest decade carries most of the tail and decides the settling more completely.

Above it, the law changes character. Each weighting on the positive side has a settling time that is a power of the band and does not contain $\tau_\mathrm{max}$, provided the band is not so fine that the settling reaches the spread’s end. That is the precise sense in which the tail has no slowest relaxation: every relaxation slower than $t_B$ carries too little weight to matter at $t_B$, and the one that matters is whichever sits near $t_B$ itself. Which relaxation that is changes with the band.

The law figure widens the spread to eight decades so that each line has room to be straight before it meets the end. The measured slopes are −0.997 and −0.500 for $m = 1$ and 2, and −1.932 for $m = \tfrac12$ — three and a half per cent shallow, the same slow-falling density still feeling the cut at the far end. On these axes the old law is not a straight line at all; it is a curve that bends towards the vertical once the slowest relaxation takes over — the reverse of the straight lines and where they are not the curve, where a single pole’s asymptotes are straight on logarithmic axes and the curve is what joins them.

The two shapes, and what each says about a part #

This gives a settling measurement a diagnostic that needs no fit. How many relaxations a settling curve holds found that the individual amplitudes of a crowded spectrum cannot be recovered from a practical reading, and that the weight per decade can. The weighting exponent is a statement about the weight per decade, and it is read directly off the settling times:

• A constant difference between the times to successive decades of band — the ratio test of the essay on the slowest doublet — says one relaxation has taken over, and the difference divided by $\ln 10$ is its time constant.
• A constant ratio between them says a power law, and the ratio is $10^{1/m}$.

At $m = 1$ the successive times go 0.97, 9.71, 96.3 ms, ratios of 10.0 and 9.9; on the even row they go 2.23, 14.4, 31.8, 51.4 s, differences of 12, 17 and 20 seconds, converging towards the slowest relaxation’s $\tau\ln 10 = 23$. The two rows are distinguishable at the first three bands, from a reading anybody could take.

The same boundary in a noise spectrum #

The even weighting has a well-known relative in a different field, and the relationship is worth drawing because it explains why $m = 0$ keeps turning up in physical parts rather than being a mathematical curiosity.

A single relaxation process that switches at random with a time constant $\tau$ produces noise with a Lorentzian spectrum, flat below $1/(2\pi\tau)$ and falling as $1/f^2$ above it. Sum many such processes with their time constants spread evenly in $\ln\tau$ — the $m = 0$ weighting, each decade of time constant carrying the same weight — and the spectrum between the ends of the spread is $1/f$. That is the standard account of flicker noise, and the corner where averaging stops working is what its $1/f$ does to a measurement that averages longer and longer: the average stops improving because every decade of averaging time uncovers another decade of slow fluctuations carrying the same weight as the last.

That is the frequency-domain statement of the even row in the table above. Every decade of settling time uncovers another decade of slow relaxations carrying the same weight, so the error falls only by a fixed share per decade and the slow end has to be reached before it finishes. The dielectric that settles logarithmically and the resistor whose average never converges are the same weighting seen through two different measurements — which is also why neither can be summarised by one number, a fact the source resistance the flicker moves meets when a flicker corner shifts an optimum that white noise alone would have fixed.

On the fast side of the boundary the analogy continues. A spectrum weighted towards short time constants gives noise rising less steeply at low frequency than $1/f$, and it has a floor that averaging reaches; a tail weighted the same way settles as a power law that reaches any band in a finite, computable time. In both fields the exponent’s sign decides whether the measurement converges on its own or has to wait for the slowest process present.

The driven capacitor already has this weighting #

There is a circuit in which the weighting is not a hypothesis. The tail the source decides found that a capacitor driven through a small source resistance carries a tail of $R_s/R_i = R_sC_i/\tau_i$ from each relaxation. With the branch capacitances equal, that is an amplitude proportional to $1/\tau$ — exactly $m = 1$ — whatever the dielectric’s own spectrum looks like in the open circuit.

So the driven capacitor is on the power-law side of the boundary by construction. Its tail falls as one over time between the fastest and slowest branch, and its settling to a band is the fastest branch’s time constant times the tail over the band, $\tau_\mathrm{min}R_s/(R_{\parallel}B)$ in the dense limit, rather than any multiple of the slowest. That is the second, independent reason the driven capacitor on that page settled to 10⁻⁵ in 173 microseconds while a model with its absorption as an even-weighted tail wanted tens of milliseconds: the source changes the amplitude, and the way it changes it moves the spectrum across the boundary.

In the open circuit, and in the hold, the weighting is whatever the dielectric has, and that is where the next question is.

Where real dielectrics sit, which is near the boundary #

Measured discharge currents in many dielectrics fall as a power of time over several decades, $i \propto t^{-n}$ — the Curie–von Schweidler form — with exponents reported near one, and the continuous-spectrum picture used here is the standard way of reading that behaviour. If the charge still to be given back is the integral of such a current from $t$ onward, it falls as $t^{1-n}$, which in this essay’s terms is $m = n - 1$. An exponent just below one puts the part just on the slow side of the boundary; just above one, just on the fast side.

That mapping is a reading of a published form through this model, not something solved here, and it deserves the flag. What the arithmetic does establish is the consequence of where a part sits. A dielectric near $m = 0$ is at the worst possible place to be predicted from a single number, because on one side the settling is a power of the band that never sees the slow end and on the other it is a logarithm the slow end owns — and the coefficient that is about one reading is the same complaint made about a capacitor’s temperature coefficient: one printed number standing for a shape.

What the exponent means for a design #

On the fast side, the band is expensive and the spread is free. A tighter settling specification multiplies the time by $10^{1/m}$ a decade, so going from 10⁻⁴ to 10⁻⁶ at $m = 1$ costs a hundred times the time — but adding decades of slow relaxations at the far end costs nothing until the settling reaches them. A design on this side is limited by how fine it needs to settle, not by the dielectric’s slowest mechanism.

On the slow side, the band is cheap and the spread is everything. A tighter band adds an increment of about $\tau_\mathrm{max}\ln 10$, but that increment is proportional to the slowest relaxation. Here the slowest one decides it is the whole story, and the design lever is removing the slow decades.

At the boundary, both are true at once, which is why the even weighting took fourteen seconds to reach 10⁻⁴ from a tail whose fast-weighted cousin reached it in ten milliseconds. The same one per cent of absorption, differently distributed across the same five decades, spans three orders of magnitude of settling time.

The power law and the slope, held to account #

Inside the spread the sum is required to equal the power law $c\,\Gamma(m)\,t^{-m}$ at the geometric centre, to one per cent plus the calculable share of the density lying beyond the slow end. The Gamma function is evaluated by the Lanczos series, and a disagreement between the weighting’s exponent and the tail’s slope would stop the figure being drawn.

On the eight-decade spread the settling time is required to have slope −1/m against the band, to eight per cent, measured between bands whose settling times both lie at least thirty times inside each end. The three slopes are reported rather than assumed.

The settling times are found by bisection on a monotone sum — every amplitude is positive — so each is a single crossing and the table’s numbers are the curve’s, not an interpolation.

A spectrum with sharp ends and one exponent #

The spread is cut off sharply at both ends. A real spectrum tails away, and the power-law region is bounded by where the weighting stops being a power, which a measurement would have to find.

The weighting is one power across all five decades. A dielectric can have different exponents in different decades — one mechanism fast, another slow — and a settling curve would then change slope partway, which is detectable by the constant-ratio test applied band by band.

And the spectrum is dense by construction. How many relaxations a settling curve holds is why that is the honest choice: a spectrum finer than about two relaxations a decade cannot be told from a continuous one by any practical reading, so the continuous one is what a measurement can claim.

Still open: two exponents, the hold, and the exponent a measurement returns #

A spectrum with a break in it. Two weightings joined at a corner — fast-weighted below, slow-weighted above — would settle as a power law until the corner and as the slow end’s logarithm after it, and the band at which the law changes is a function of where the corner is. That is the realistic shape for a part with two absorption mechanisms, and the corner is the number a design would need.

The hold, with a weighted spectrum. A sample-and-hold gives back each relaxation’s unfinished charge in proportion to its capacitance, so the hold’s error weights the spectrum differently from the drive’s. Whether a dielectric on the fast side for the drive can be on the slow side for the hold — and so settle quickly and droop slowly, or the reverse — is the question that would decide which specification a hold capacitor should be bought on.

The exponent from a short measurement. The constant-ratio test needs three bands on the power-law line. How few bands, read how accurately, determine $m$ to a tenth — and so place a part on one side of the boundary or the other — is the fit that the inversion of a settling curve could not do for individual amplitudes and might do for one exponent.