A crystal can be grown from one well-mixed charge and still have a dopant concentration that changes along its growth history. The newly formed solid, the remaining liquid and a collected crop of solid are different objects. Their concentrations need not equal one another or the starting concentration.
The short answer is to follow the solute balance as the solid fraction changes. We will derive a constant-partition Scheil model, calculate original local and crop-average examples, and test what one composition reading can identify. The numbers are fictional teaching inputs. No partition coefficient is asserted for a particular yttrium-containing crystal, and no crystal-growth procedure or product acceptance limit is supplied.
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ROUTE 1 · CHAPTERS 1–6
Define the material account
Separate the host, dopant, mass basis and local partition law.
ROUTE 2 · CHAPTERS 7–13
Calculate local and crop values
Reconstruct conserved amounts and compare local concentrations with retained crops.
ROUTE 3 · CHAPTERS 14–20
Test inference assumptions
Check caps, competing mixing models and what concentration measurements can identify.
ROUTE 4 · CHAPTERS 21–27
Apply and challenge the model
Use sources, limits and independent crop calculations.
ROUTE 5 · CHAPTERS 28–32
Transfer the reasoning
Check sensitivity, averaging weights, growth coordinates and further reading.
Chapter contents
Define the material account · Chapters 1–6
1. Begin with the host and dopant distinction
2. Define the mass account and growth coordinate
3. State the constant-partition assumptions
Calculate local and crop values · Chapters 7–13
7. Calculate the total dopant already in solid
8. Obtain the average concentration of the whole solid crop
9. Read the local and average ledger
10. Average a selected interval rather than the entire solid
11. Convert the normalised account into an amount example
Test inference assumptions · Chapters 14–20
14. Apply a supplied concentration cap conditionally
15. Contrast complete solid mixing as a different model
16. Do not infer a growth mechanism from one local reading
17. Use two local readings under a common model
18. Check which calibration factors cancel
Apply and challenge the model · Chapters 21–27
21. Keep microscopic partitioning separate from bulk formulations
22. Ask what a real specimen would need beyond this model
23. Use the university source for the solidification law
24. Try an independent changed-crop exercise
25. Check the changed-crop solution
Transfer the reasoning · Chapters 28–32
28. Examine sensitivity to the partition parameter
29. Weight discrete crop records by their represented mass
30. Keep the growth coordinate distinct from time
CHAPTER 1 OF 32 · Define the material account
1. Begin with the host and dopant distinction
The Yttrium root explains why an optical-material name must distinguish a host from an added species. YAG identifies yttrium aluminium garnet; a doped specimen requires a fuller composition account. A starting formulation still does not by itself describe how an added species is distributed through a grown specimen.
The present article develops that spatial-composition question through a deliberately small solidification model. It does not treat the yttrium fraction in an ideal host formula as the same quantity as a dopant concentration. Nor does it use a vacancy-count conversion as a crystal-growth law.
For every number, name the constituent, material stage and denominator. A dopant mass fraction in a liquid charge, a local fraction in newly formed solid and a mean fraction in a crop may all be valid while differing numerically. The calculation becomes useful when the reader can trace how those objects are connected rather than shorten all three labels to concentration.
CHAPTER 2 OF 32 · Define the material account
2. Define the mass account and growth coordinate
Normalise the initial total mass to one. Let f be the fraction of that mass that has solidified, with zero ≤ f < one. The remaining liquid mass fraction is one − f. Let C0 be the initial dopant mass fraction and CL(f) the mass fraction in the remaining liquid.
Let CS(f) denote the dopant mass fraction in the infinitesimal solid increment forming at f. It is a local growth-history value. It is not automatically the mean concentration of all solid already collected.
Assume the system has no dopant loss, contamination or other material stream in this first account. The mass coordinate is specified directly. Length, elapsed time and pulled distance are different coordinates and would require relationships involving geometry, density and growth history before replacing f. A fraction of a crystal’s visible length is not automatically the fraction of initial charge mass that has solidified.
CHAPTER 3 OF 32 · Define the material account
3. State the constant-partition assumptions
Assume a well-mixed liquid at each stage, negligible redistribution of dopant within previously formed solid, and local interfacial partitioning CS = k CL. The partition coefficient k is positive and constant in the teaching model.
These assumptions allow newly formed solid to record the liquid concentration at its formation stage. They exclude complete homogenisation of the whole solid after every increment. They also exclude a time-dependent or composition-dependent k, multiple competing solid phases and an independently modelled liquid boundary layer.
The assumptions are not established by writing the word Scheil beside a material name. A real crystal’s applicable partitioning, diffusion and phase behaviour require their own evidence. We use a stipulated constant k so the consequences can be derived and checked exactly. The useful transfer is to recognise which observation or material condition would require changing the model, rather than import the fictional coefficient into a real YAG growth account.
The dopant amount remaining in liquid, per unit initial mass, is (one − f)CL. When an increment df solidifies, it removes CS df = k CL df from that liquid account. Therefore d[(one − f)CL] = −k CL df.
Expanding the derivative gives (one − f)dCL − CL df = −k CL df, so (one − f)dCL = (one − k)CL df. Dividing by the positive liquid amount and integrating from the initial state gives CL = C0(one − f)^(k − one).
The newly formed solid has CS = k C0(one − f)^(k − one). These are the constant-partition Scheil relationships under the stated account. The derivation makes the conservation law visible. A formula copied without its mass coordinate or phase definition can return correct arithmetic for a quantity other than the one the reader intended to compare.
CHAPTER 5 OF 32 · Define the material account
5. Interpret the partition coefficient before calculating
For k below one, the newly forming solid has less dopant than its adjacent liquid. The liquid becomes progressively enriched as solidification proceeds, so later solid increments can contain more dopant than earlier increments.
For k equal to one, the two phases have equal concentration at the interface, CL remains C0 and every solid increment has concentration C0 in this ideal account. For k above one, the early solid is enriched relative to the liquid, and the remaining liquid becomes depleted.
The same symbol k is sometimes replaced by an effective coefficient in a more detailed transport model. That is another relationship and should retain its definition. A fitted number need not be an independently measured equilibrium partition coefficient. Before interpreting k physically, establish whether it represents interfacial equilibrium, an effective growth response or merely a parameter used to summarise a particular concentration record.
CHAPTER 6 OF 32 · Define the material account
6. Work an original enriched-liquid example
Choose C0 = 0.01 and k = one-half. The initial charge therefore contains one percent dopant by mass in this fictional account. At f = zero, the first solid increment has CS = 0.005, or one-half percent, while the liquid begins at one percent.
At f = 0.75, the remaining liquid fraction is one-quarter. Its concentration is CL = 0.01 divided by the square root of one-quarter, giving 0.02. The newly forming solid has CS = 0.01.
The local solid concentration at that stage happens to equal the starting charge concentration, even though the liquid is twice as concentrated and earlier solid increments were leaner. That coincidence does not establish uniform growth or k = one. A local composition must be attached to its growth fraction and phase before it is used to interpret the distribution of the whole specimen.
CHAPTER 7 OF 32 · Calculate local and crop values
7. Calculate the total dopant already in solid
The remaining liquid contains (one − f)CL = C0(one − f)^k dopant per unit initial mass. Conservation therefore gives the total dopant amount in solid as MS(f) = C0[one − (one − f)^k].
For C0 = 0.01, k = one-half and f = 0.75, this is 0.01 times one minus one-half, giving 0.005. The liquid also contains 0.005, and the two amounts sum to the initial 0.01.
These are amounts per unit initial mass, not concentrations in their respective phases. The solid mass is 0.75 and the liquid mass is 0.25. Dividing the same dopant amount by those different phase masses gives different concentrations. Keeping amount and concentration separate is essential: the balance conserves dopant amount, while concentrations change because both the amount and the mass of each phase evolve.
CHAPTER 8 OF 32 · Calculate local and crop values
8. Obtain the average concentration of the whole solid crop
The mean dopant fraction in all solid formed up to f is CbarS = MS(f)/f = C0[one − (one − f)^k]/f for positive f. It averages the local solid increments over their actual mass coordinate.
At f = 0.75 in the original example, CbarS = 0.005/0.75 = approximately 0.006667. This is two-thirds of one percent, while the newly forming solid is at one percent and the liquid is at two percent.
All three values are consistent with one conserved account. A bulk assay of the collected solid would concern the crop mean under suitable sampling assumptions. A local assay near the latest growth stage would concern another quantity. Neither should be substituted for the other without the averaging relationship. The difference is not a calibration error manufactured by the model; it is the direct consequence of a varying composition history.
CHAPTER 9 OF 32 · Calculate local and crop values
9. Read the local and average ledger
The following values use the same fictional C0 = 0.01 and k = one-half. Concentrations are mass fractions; the solid dopant amount is expressed per unit initial charge mass.
| Solid fraction f | Liquid concentration | Newly formed solid concentration | Dopant amount in solid | Mean solid concentration |
|---|---|---|---|---|
| 0.36 | 0.0125 | 0.00625 | 0.002 | 0.005556 |
| 0.75 | 0.0200 | 0.01000 | 0.005 | 0.006667 |
| 0.96 | 0.0500 | 0.02500 | 0.008 | 0.008333 |
The final row is still only a mathematical output of the stipulated law. Whether its concentrations remain within an appropriate single-phase, dilute or constant-k regime must be checked separately. Increasing numerical enrichment does not prove that a real material continues to obey the same assumptions near the end of growth.
CHAPTER 10 OF 32 · Calculate local and crop values
10. Average a selected interval rather than the entire solid
Suppose a crop contains only material formed between f1 and f2, with f2 greater than f1. Its dopant amount per initial mass is C0[(one − f1)^k − (one − f2)^k]. Its mass fraction is f2 − f1.
The crop mean is therefore the amount divided by f2 − f1. For f1 = 0.36 and f2 = 0.75 in the original example, the residual powers are 0.8 and 0.5. The crop contains 0.003 dopant per initial mass and has mass fraction 0.39, giving mean concentration approximately 0.007692.
A crop mean depends on both selection boundaries. It is not simply the local concentration at the final boundary. Nor is it automatically the midpoint value. A real sampling statement should identify which material was included and whether the assay represents that selected interval. The model cannot supply a crop definition that the record never stated.
CHAPTER 11 OF 32 · Calculate local and crop values
11. Convert the normalised account into an amount example
For a separate paper ledger, let the initial mass be one kilogram while retaining C0 = one percent. The initial dopant amount is ten grams. At f = 0.75, the collected solid mass is 750 grams and contains five grams of dopant, while 250 grams of liquid contain the other five grams.
For the selected interval from 0.36 to 0.75, the crop mass is 390 grams and its dopant amount is three grams. Dividing gives approximately 0.7692 percent by mass. These numbers restate the same normalised balance in a physical mass unit.
They do not describe an actual growth run or imply a recovered product purity. The remaining liquid, earlier solid and chosen crop are all different material accounts. Scaling the initial mass changes the amounts proportionally while leaving the stipulated concentration profiles unchanged. Changing the material or growth conditions could change the model itself, so that is not merely the same scaling exercise.
CHAPTER 12 OF 32 · Calculate local and crop values
12. Check the early-growth limit
As f approaches zero, the whole-solid mean approaches k C0, the concentration of the first solid increment. The numerator and denominator both approach zero, but their ratio has a finite limit obtained by expansion or differentiation.
For the original k = one-half case, that limit is 0.005. It agrees with the first-increment calculation. This is a useful check because the whole crop is then dominated by material formed near the initial stage.
Do not evaluate the f = zero mean by direct division and label the zero-over-zero expression a physical concentration. At exactly zero solid mass, there is no collected solid crop whose bulk concentration can be measured. The limiting concentration describes progressively small positive crops under the model. A limit and an actually existing finite specimen are related ideas, but they should remain distinguished when interpreting endpoints.
CHAPTER 13 OF 32 · Calculate local and crop values
13. Examine the late-growth limit with its model boundary
For k positive, the total dopant amount in solid approaches C0 as f approaches one. The whole-solid mean approaches C0, consistent with putting all initial dopant into the complete final solid account if the idealisation is continued.
For k below one, however, the liquid concentration and local last-solid concentration diverge as f approaches one. A real mass fraction cannot grow without bound. This signals that extrapolating the dilute constant-partition single-phase law to that endpoint is not a physical completion of the process.
Additional phases, changing coefficients, finite mixing or other material effects may become important. The mathematical integrated amount can remain finite even when a local concentration law diverges, so conservation alone does not validate the local extrapolation. A responsible report states where the assumed law is being used and what independent phase or composition information limits it.
CHAPTER 14 OF 32 · Test inference assumptions
14. Apply a supplied concentration cap conditionally
Suppose a teaching problem independently restricts the liquid concentration to at most 0.04. This is a supplied model-domain cap, not a universal threshold for a named material. Retain C0 = 0.01 and k = one-half.
The condition 0.01/√(one − f) ≤ 0.04 gives √(one − f) at least one-quarter. Therefore f must be at most 0.9375. The earlier f = 0.96 row lies beyond this added cap and should not be used under the augmented contract.
The inequality does not predict which physical phenomenon occurs at that boundary. It simply locates where a stated concentration restriction becomes inconsistent with the extrapolated law. A real cap requires evidence appropriate to the particular material and phase behaviour. Keeping that distinction avoids turning an illustrative numerical limit into an unsupported operating recommendation or product criterion.
CHAPTER 15 OF 32 · Test inference assumptions
15. Contrast complete solid mixing as a different model
Now suppose all solid remains homogenised and equilibrated with the liquid throughout growth. With constant partitioning CS = k CL, conservation gives C0 = f k CL + (one − f)CL. Thus CL = C0/[one − f + kf] and the uniform solid concentration is k times that value.
For the original C0, k and f = 0.75, the denominator is 0.625. The liquid concentration is 0.016 and the uniform solid concentration is 0.008. The amounts are 0.004 in liquid and 0.006 in solid, again summing to 0.01.
This differs from the Scheil mean of approximately 0.006667 because the assumption about redistribution within the solid changed. Both calculations conserve mass. Conservation does not select between them by itself. The evidence must address mixing, diffusion and equilibrium assumptions for the particular specimen, rather than choose whichever formula produces a preferred concentration.
CHAPTER 16 OF 32 · Test inference assumptions
16. Do not infer a growth mechanism from one local reading
At f = 0.75, a local solid concentration equal to C0 is compatible with k = one-half in the Scheil model. It is also compatible with k = one, which gives uniform concentration C0 throughout.
The two parameter choices therefore reproduce that one local reading but predict different liquid and crop-average concentrations. For k = one-half, the liquid is at two C0 and the whole-solid mean is two-thirds C0. For k = one, both remain C0.
This exact counterexample shows that a local composition matching the starting charge does not prove the absence of segregation. The inference is not one-to-one over all possible k values at that stage. Additional growth-history readings, an independently known partition coefficient or another justified material relationship can help distinguish explanations. A more precise repeat of the same local value may leave the structural ambiguity unchanged.
CHAPTER 17 OF 32 · Test inference assumptions
17. Use two local readings under a common model
For two distinct growth fractions, the local solid ratio is CS(f2)/CS(f1) = [(one − f2)/(one − f1)]^(k − one). If both concentrations share a known basis and the same constant k applies, this gives a logarithmic relationship for k.
In the original example, the initial local solid concentration is 0.005 and the value at f = 0.75 is 0.01. Their ratio is two, while the remaining-liquid-fraction ratio is one-quarter. Therefore k = one + ln(two)/ln(one-quarter) = one-half.
With k identified under the stipulated law, C0 can be recovered from one of the local values. This is a conditional inverse calculation, not proof that the real growth follows Scheil assumptions. It also relies on identifying the growth fractions correctly. Concentrations measured at two arbitrary positions cannot be inserted as if those positions supplied f automatically.
CHAPTER 18 OF 32 · Test inference assumptions
18. Check which calibration factors cancel
A common positive multiplicative calibration factor applied to both local concentration readings cancels in their ratio. The inferred k from the two-point ratio is therefore unchanged under that particular shared-factor model.
The inferred C0 does not enjoy the same cancellation because it depends on the absolute concentration scale. If the factors differ between readings, even the ratio changes. An additive background generally does not cancel as a multiplicative ratio either.
This distinction helps separate profile-shape information from absolute composition information. Record the observation model and which errors are shared before claiming that a parameter estimate is calibration independent. A cancellation valid for k under one simple response model does not make the entire composition account independent of calibration, sample representativeness or the growth-coordinate assumptions used to assign each reading.
CHAPTER 19 OF 32 · Test inference assumptions
19. Recover a partition parameter from a crop mean
If C0, f and the whole-solid mean CbarS are supplied, the Scheil balance gives (one − f)^k = one − f CbarS/C0. Therefore k = ln[one − f CbarS/C0]/ln(one − f), provided the logarithms have admissible positive arguments.
For f = 0.75, C0 = 0.01 and CbarS = one/150, the right-hand residual is one-half. Dividing ln(one-half) by ln(one-quarter) returns k = one-half.
This inversion uses an average over the full specified solid crop, not a local final reading. Substituting CS(f) in place of CbarS changes the equation and can produce a plausible but incorrect fitted coefficient. The inverse should end by reconstructing the total solid amount, remaining liquid amount and mean. That forward substitution checks whether the inferred parameter actually reproduces the supplied object and denominator.
For a changed model, let k = two and retain C0 = 0.01. Then CL = C0(one − f), while newly formed solid has CS = two C0(one − f). The early solid is enriched and later increments become leaner.
At f = one-half, CL = 0.005 and the current solid increment has CS = 0.01. The total dopant amount in solid is 0.01[one − one-quarter] = 0.0075, so the whole-solid mean is 0.015.
Again the local current concentration equals the initial charge concentration while the crop mean differs. The direction of segregation changed, but the same local-versus-average distinction remains necessary. This is a useful transfer check because it prevents a learner from associating every segregation model with an enriching liquid tail. The sign of k minus one determines the trend within the stated law.
CHAPTER 21 OF 32 · Apply and challenge the model
21. Keep microscopic partitioning separate from bulk formulations
A partition coefficient compares concentrations in coexisting phases at an interface under its definition. A recipe percentage compares an added amount with a declared formulation denominator. A substitution fraction may compare a constituent count with a particular set of lattice sites.
Those quantities can be related in a complete material model, but they are not interchangeable. The existing yttria vacancy-fraction article works through constituent and charge ledgers in an ideal oxide formulation. It does not supply the partition law for a dopant in a growing optical crystal.
Before combining data, preserve the concentration basis: mass fraction, mole fraction, atomic fraction or site fraction. A constant k on one basis should not be copied to another without the relevant conversion and phase-composition relationships. The mathematical growth account is strongest when each symbol retains the same denominator throughout its derivation and when the actual material evidence supports that choice.
CHAPTER 22 OF 32 · Apply and challenge the model
22. Ask what a real specimen would need beyond this model
A physical interpretation requires the material identity, initial composition, growth-coordinate mapping, sampling locations and relevant phase behaviour. It also requires evidence about liquid mixing, solid redistribution and whether a constant coefficient is adequate over the range used.
A bulk assay can constrain an integrated amount while leaving local variation unresolved. A local assay can reveal a gradient while leaving the total collected crop uncertain. Combining them can be informative when their specimen relationship and units are established.
The purpose is not to demand every possible measurement before learning anything. It is to connect each proposed conclusion with the missing relationship that would make it specific. Our exact balances show which questions concern conservation and which concern transport or phase assumptions. That separation makes the next enquiry more useful than asking a single concentration number to describe a whole growth history.
CHAPTER 23 OF 32 · Apply and challenge the model
23. Use the university source for the solidification law
The University of Cambridge Materials Algorithms Project states the classical Scheil relationship and defines the local solid concentration, starting composition, fraction solidified and partition coefficient. Its implementation concerns alloys and is not a calibration for a yttrium-containing crystal.
Our charge composition, coefficients, crop boundaries and mass ledgers are original fictional examples. They illustrate how the stated law separates local, liquid and crop-average quantities. Applying it to a real garnet or another material requires independently justified phase and transport assumptions.
Read the Cambridge description of the Scheil model, checked on 25 October 2026. Its formula is used here as a model to derive and audit, not as evidence that every growth process or dopant has the same coefficient or profile.
CHAPTER 24 OF 32 · Apply and challenge the model
24. Try an independent changed-crop exercise
Use C0 = 0.02 and k = one-half. Find CL, the newly formed solid concentration, total dopant amount in solid and whole-solid mean at f = 0.84. Then find the mean concentration of the crop formed between f = 0.36 and f = 0.84.
As a second task, retain C0 = 0.02 but set k = two and f = one-half. Compare the local current concentration with the whole-solid mean. Explain why either local value alone is insufficient to describe the full crop.
Finally suppose the only record is a local concentration equal to the starting C0 at f = 0.75. Construct two constant-k explanations and state what additional observation would distinguish them. Work from the phase balance and crop definition before checking the answers. The changed numbers and coefficient direction test whether the method has transferred beyond the original table.
CHAPTER 25 OF 32 · Apply and challenge the model
25. Check the changed-crop solution
At f = 0.84, the remaining fraction is 0.16 and its square root is 0.4. Thus CL = 0.05, the new solid concentration is 0.025, and the solid dopant amount per initial mass is 0.02 times 0.6 = 0.012. The whole-solid mean is 0.012/0.84, approximately 0.014286.
Between f = 0.36 and 0.84, the residual square roots are 0.8 and 0.4. The crop dopant amount is 0.008 and crop mass fraction 0.48, so its mean is one/sixty, approximately 0.016667.
For k = two and f = one-half, CL = 0.01, current CS = 0.02 and whole-solid mean is 0.03. At f = 0.75, both k = one-half and k = one can give current CS equal to C0, but their liquid concentrations and whole-solid means differ. Each answer should be checked by reconstructing the appropriate phase amounts rather than comparing unlike concentrations alone.
CHAPTER 26 OF 32 · Apply and challenge the model
26. Simplify the half-partition crop mean exactly
For k = one-half, the selected-crop expression can be simplified by factoring a difference of squares. Let a = √(one − f1) and b = √(one − f2), with a greater than b. The crop dopant amount is C0(a − b), while its mass fraction is a² − b².
Their ratio is therefore Cbarcrop = C0/(a + b). For the original crop boundaries, a = 0.8 and b = 0.5, so the exact mean is 0.01/1.3. This gives the earlier approximately 0.007692 result without subtracting nearly equal quantities.
The simplified form also exposes which variables control the average. Moving either crop boundary changes the sum of residual square roots. It does not turn the average into a local endpoint reading. For the complete solid crop from zero to f, the mean is C0/[one + √(one − f)]. Comparing this with the local CS shows that their ratio is [one + √(one − f)]/[two √(one − f)]. At f = 0.75 that ratio is 1.5; at f = 0.96 it is three.
CHAPTER 27 OF 32 · Apply and challenge the model
27. Invert a crop target while preserving feasibility
As a paper inverse problem, suppose C0 = 0.01, k = one-half and the crop begins at f1 = 0.36. A supplied target crop mean of 0.01 implies a + b = one. Since a = 0.8, b = 0.2 and f2 = one − b² = 0.96.
That is an algebraic solution under the unaugmented model. If the separate liquid-concentration cap of 0.04 is also retained, it is infeasible because the model was restricted to f at most 0.9375. The two statements cannot both hold for that crop under the same assumptions.
At the cap, b is one-quarter and the largest available crop mean from the stated start is 0.01/(0.8 + 0.25), approximately 0.009524. A desired number does not override a model-domain constraint. This exercise concerns inverse consistency, not a recommendation to select a real growth endpoint. A real material decision would need independently supported coefficients and phase limits before the model could inform such a choice.
CHAPTER 28 OF 32 · Transfer the reasoning
28. Examine sensitivity to the partition parameter
At fixed C0 and f, the logarithmic sensitivity of CL to k is ln(one − f), which is negative after growth begins. Increasing k therefore reduces the model liquid concentration at that fixed stage.
For the local solid concentration, the sensitivity is one/k + ln(one − f). Its sign can change. At f = 0.75, it vanishes at k = one/ln(four), approximately 0.7213. A local concentration can therefore rise and then fall as k varies while the liquid concentration changes monotonically.
This explains why the earlier k = one-half and k = one examples can share a local solid value. The ambiguity is built into the forward relationship; it is not removed by assuming that a fitted coefficient must have one preferred numerical branch. If a local reading is used to infer k, specify the admissible parameter interval and any independent information selecting a branch. The inverse should be checked against a second phase or crop observation whenever the claimed uniqueness depends on that missing constraint.
CHAPTER 29 OF 32 · Transfer the reasoning
29. Weight discrete crop records by their represented mass
A set of concentration measurements does not define a crop mean until the represented material weights are supplied. Equal time intervals, equal distances and equal mass increments are different sampling conventions when the growth history is not uniform.
For a separate illustrative ledger, suppose one bin represents 0.1 of the initial mass with bin-average concentration 0.005, while another represents 0.4 with bin-average concentration 0.01. Their combined dopant amount is 0.0005 + 0.004 = 0.0045, and their combined mass is 0.5. The combined mean is 0.009.
The unweighted average of the two reported bin means would be 0.0075, which describes equal weighting of records rather than equal weighting of material. This discrete example is an averaging check, not another exact Scheil profile. It shows why a sampling ledger must state which mass each reported mean represents before those records are pooled. Increasing the number of equally spaced time readings cannot repair an absent mapping from time to represented crop mass by itself.
CHAPTER 30 OF 32 · Transfer the reasoning
30. Keep the growth coordinate distinct from time
The fraction f records how much of the initial charge has solidified. It is not automatically the elapsed fraction of a growth run. If a fictional run reaches f = 0.36 after one hour and f = 0.75 after three hours, the crop calculation still uses the solid-mass interval 0.39. Averaging the two concentrations with weights of one and two hours would answer a different question and would generally give the wrong crop composition.
To convert a concentration profile into a time trace, supply a separate measured or stipulated growth history f(t). To convert it into position along a crystal, supply the mass per unit length and the mapping from solidified mass to the retained specimen. A constant pulling speed alone does not establish either mapping if the cross-section changes. This distinction provides a practical evidence check: every integration weight should describe the physical material included in the sample. A time-weighted detector average becomes a mass-weighted crop average only after the relevant mass-flow relationship has been justified.
Is the final local concentration the crop average?
Not generally. The local value describes the increment forming at one growth fraction. The crop mean integrates the relevant solid increments and divides by their total mass. The two coincide only under additional conditions, such as a uniform profile in this model.
Does mass conservation prove the Scheil assumptions?
No. Complete solid mixing and negligible solid redistribution can both be formulated with conserved dopant amount while predicting different profiles. Conservation checks the ledger; material evidence selects the applicable redistribution and partitioning relationships.
Can the fictional coefficient be used for Nd:YAG?
No. No actual material coefficient is supplied here. The host, dopant, composition basis, phase behaviour and growth regime need their own evidence. The model teaches how a justified coefficient would enter the account.
Does a diverging tail predict an infinite real concentration?
No. It indicates that the extrapolated constant-partition law cannot remain a complete physical description at that endpoint. A finite integrated amount does not validate an unbounded local mass fraction.
CHAPTER 32 OF 32 · Transfer the reasoning
32. Continue through the relevant material routes
Return to Yttrium for host, dopant and material-evidence distinctions. For a different composition problem, read converting yttria mole fractions into oxygen-site vacancy fractions. Its charge and site denominators should remain distinct from this phase-partition model.
Continue through Learning That Transfers into Life and How X Works. A useful final explanation identifies the phase, growth coordinate and crop boundaries, then reconstructs the conserved amount before interpreting a concentration difference.