Read the displacement argument, ask for its missing steps, and investigate the same relationships in three working laboratories.
This is newly authored explanatory text in modern notation, with editorial review pending. It is not the German source, an English translation, or a complete edition of the paper. The reviewed source faces and pinned facsimile remain in preparation. Section headings identify the argument being discussed, not a completed source inventory.
New explanatory text authored with AI assistance. Mathematical and editorial review remains pending; none of these passages is presented as Einstein’s wording.
Do particles that wander in all directions ever get anywhere?
Imagine placing a microscopic particle in a drop of water and marking where it is after a few moments. Pushed at random by invisible water molecules, it is just as likely to move left as right. In the symmetric arithmetic example below, signed displacements cancel even though all four endpoints differ from the starting point.
Authored arithmetic examples: Displacements of −3, −1, +1, and +3 units (three steps left, one left, one right, three right). These are an authored educational case, not a measured dataset.
Step 1 · Explore the particle displacements
Drag a marker or select it and use ← / → arrow keys to shift its position.
0 (start)
-8
-6
-4
-2
2
4
6
8
1
2
3
4
Particle 1-3 units(three steps left)
Particle 2-1 units(one step left)
Particle 3+1 units(one step right)
Particle 4+3 units(three steps right)
Signed Total (Sum)0 unitsMean: 0.0 units
Mean Absolute (|x|)2.00 unitsProposal A (ignore sign)
Mean Square (x²)5.00 sq unitsProposal B (square first)
Root Mean Square (RMS)2.236 units√(Mean Square)
Step 2 & 3 · What a signed sum tells us
When we add the displacements algebraically, opposite directions cancel out:
(-3) + (-1) + (+1) + (+3) = 0 units
A signed total of zero tells us that the average endpoint has not shifted. It does not mean every particle remained at rest. In the authored example all four particles have nonzero displacements; after your edits the displayed totals describe your chosen endpoints.
Step 4 & 5 · Two sensible proposals to keep information about distance
How do we keep track of how far particles wandered without opposite directions cancelling out? Both of the following proposals are completely sensible:
Proposal A (Ignore the direction): Take the absolute value of each displacement. For our authored example (−3, −1, +1, +3), the absolute values are 3, 1, 1, 3 units, giving a mean absolute displacement of 2 units.
Proposal B (Square each displacement): Multiplying any negative number by itself produces a positive number. The squared displacements are 9, 1, 1, 9 squared units, giving a mean square displacement of 5 squared units (and an RMS distance of √5 ≈ 2.236 units).
Step 6 & 7 · Scaling: What happens when displacements double?
Suppose after a longer interval every particle has wandered twice as far (−6, −2, +2, +6 units):
Mean Absolute Displacement:(|-6| + |-2| + |+2| + |+6|) / 4 = 4 units (doubles from 2)
Mean Square Displacement:(36 + 4 + 4 + 36) / 4 = 20 sq units (quadruples from 5)
Notice that when displacement distances double, the mean square quadruples (2² = 4 times larger), and its square root (RMS = √20 ≈ 4.472 units) doubles exactly in proportion to distance.
Step 8 · Why the mean square has a simple additive rule
Both proposals measure spread. The mean square has a useful property when independent, zero-mean displacements are added. This is a pedagogical bridge, not the paper’s printed calculation.
First expand the square of a sum. This algebra holds without an independence assumption:
(Δx₁ + Δx₂)² = Δx₁² + 2·Δx₁·Δx₂ + Δx₂²
Now assume the displacements over the chosen time intervals are independent and each has zero mean. Independence makes the average product equal the product of the averages, so the cross term vanishes on averaging—not in every outcome. Equal finite step mean squares then add in proportion to the number of intervals. This coarse-grained assumption is not a claim about molecular motion at arbitrarily short times.
Absolute values do not possess this mathematical linearity when steps are added together, which is why mean square has a particularly simple additive calculation.
Step 9 · Mean absolute displacement is not a wrong answer
Mean absolute displacement is not an incorrect calculation. It answers a slightly different question about the average absolute net displacement from the starting point and, in the ideal Gaussian distribution, it scales directly with the square root of time (equal to √(4Dt/π) along one coordinate).
Left and right can cancel in the average while every tracer moves. Squaring before averaging keeps track of the spread.
A signed mean answers where the ensemble’s centre has moved. It does not answer how far its members have wandered. For a centred distribution, rightward and leftward contributions balance even while the distribution broadens.
⟨x⟩=0,⟨x2⟩>0
The model’s mean displacement can be zero while its mean-square displacement is positive.
To retain the movement, square each displacement before averaging. Taking the square root of that mean square returns a length: the root-mean-square displacement, or RMS. A mean absolute displacement is a different observable, and neither is the length of a wandering trajectory.
Choose the common starting point as zero and right as positive.
Use four displacements, in micrometres: −3, −1, +1, +3.
Their sum is zero. Divide by four: the signed mean is zero.
Square each first: 9, 1, 1, 9 square micrometres.
The squares add to 20. Divide by four: the mean square is 5 square micrometres.
Take the square root: the RMS is approximately 2.236 micrometres.
The mean distance is instead (3 + 1 + 1 + 3) / 4 = 2 micrometres. These are different questions, not inconsistent answers.
Explore the equation · Modern model notation
Why the apparent speed depends on how you watch
vapp:=tλx
The apparent coordinate speed divides the typical coordinate distance by the chosen positive interval.
Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.
Input · Model result · Constant
Symbolic equation
Symbolic here. Open the linked laboratory for a worked example and live values.
Apparent coordinate speed
No accepted value is available here.
Coordinate RMS displacement
No accepted value is available here.
Observation interval
No accepted value is available here.
Read the equation aloud in words
Apparent coordinate speed equals the model coordinate root mean square displacement divided by the observation interval.
This quotient depends on the observation interval. It is not instantaneous physical velocity. At zero interval the quotient is undefined, even though the displacement is zero.
Model assumptions and every term’s meaning
The same one-coordinate ideal Brownian model as the RMS relation.
A positive observation interval is required for the quotient.
Lines joining recorded points are a rendering convention, not a velocity measurement.
Define the observable
This defines an interval-dependent comparison. It does not introduce a physical instantaneous Brownian velocity. Read the prerequisite
Apparent coordinate speed
A distance-per-interval statistic. It is not a molecular collision speed. Read the prerequisite
Divide by the same interval
The distance grows as the square root of time, while the denominator grows linearly. The quotient therefore decreases as the interval grows. Read the prerequisite
Model coordinate RMS
The same accepted one-coordinate model displacement used by the neighboring RMS equation. Read the prerequisite
Positive observation interval
A zero interval yields no apparent-speed value. The interface keeps the explanation instead of fabricating zero or infinity. Read the prerequisite
Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.
Explore the equation · Modern model notation
From spreading to a measurable distance
λx=2Dt
The typical coordinate distance is the positive square root of twice the diffusion coefficient times the observation interval.
Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.
Input · Model result · Constant
Symbolic equation
Symbolic here. Open the linked laboratory for a worked example and live values.
Coordinate RMS displacement
No accepted value is available here.
Diffusion coefficient
No accepted value is available here.
Observation interval
No accepted value is available here.
Read the equation aloud in words
The model coordinate root mean square displacement equals the square root of two times the diffusion coefficient times the observation interval.
Squaring measures spread without cancellation between directions. Taking the positive square root turns squared distance back into a distance. Four times the observation interval gives twice the model RMS, not four times.
Model assumptions and every term’s meaning
Independent, zero-mean Gaussian displacement increments in a homogeneous liquid.
One-coordinate model statistic, not a measured speed or a sample estimate.
The overdamped regime is assumed rather than established from additional particle and fluid measurements.
A model relation
This equality concerns the ideal model. A finite synthetic ensemble fluctuates around it; its sample RMS is displayed separately. Read the prerequisite
Coordinate RMS
One coordinate, not the total three-dimensional distance. Model RMS and sample RMS have different meanings. Read the prerequisite
Why a square root?
Two D t has units of squared length. Its positive square root has units of length and defines the typical displacement. Read the prerequisite
Build the mean square
Independent zero-mean increments add their variances. The definition of D makes the coordinate mean square equal to 2 D t. Read the prerequisite
Diffusion coefficient
This is the accepted model diffusivity, not a rate inferred from the synthetic data. Read the prerequisite
Observation interval
This is the interval used for the displayed displacement, not the simulation frame rate. Re-observing the trial preserves its paths. Read the prerequisite
Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.
Show every step here: Zero average is not no movement
Choose the common starting point as zero and right as positive.
Use four displacements, in micrometres: −3, −1, +1, +3.
Their sum is zero. Divide by four: the signed mean is zero.
Square each first: 9, 1, 1, 9 square micrometres.
The squares add to 20. Divide by four: the mean square is 5 square micrometres.
Take the square root: the RMS is approximately 2.236 micrometres.
The mean distance is instead (3 + 1 + 1 + 3) / 4 = 2 micrometres. These are different questions, not inconsistent answers.
Signed displacements can cancel. Their squares still record movement.
What this relies on
All displacements use the same origin, axis, units and observation interval.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: Zero average is not no movement
Assumed here
All displacements use the same origin, axis, units and observation interval.
What this does not establish
The model expectation need not equal the mean of one small sample.
Net displacement is not total path length.
What is getting in the way?
An unfamiliar word or symbol
This answer is not yet authored for this passage. No gloss, foundation, or invented explanation is offered in its place.
An algebraic move
Squaring before averaging is the move that keeps movement when signs cancel. The signed mean of −3, −1, +1, +3 is zero; the mean of those squares is not.
This answer is not yet authored for this passage. No gloss, foundation, or invented explanation is offered in its place.
The connection to the picture
This answer is not yet authored for this passage. No gloss, foundation, or invented explanation is offered in its place.
The purpose of the calculation
The calculation answers how far members of the ensemble have wandered, not where their centre has moved. That is why a zero signed mean is not a claim that nothing moved.
What permits us to add the contributions of many random steps?
Under the stated independent, unbiased, finite-variance step model, mean square grows in proportion to time. Four times the time gives twice the RMS displacement, not four times.
Write the displacement after n steps as the sum of their increments. Expanding its square exposes both the squared increments and their cross terms. Independence factors each expected cross term into the product of two means; centring makes that product zero.
⟨(i=1∑nΔi)2⟩=i=1∑n⟨Δi2⟩=nl2
The mean square of the sum of independent centred increments is the sum of their mean squares.
t=nτ,D=2τl2,⟨x2⟩=2Dt
Elapsed time is n tau; defining D as ell squared over two tau gives mean-square displacement two D t.
Read all four transitions here (also works without JavaScript)Expand the square
Pedagogical bridge, not the paper’s printed calculation
Expand the square
This is newly authored explanatory prose, not reviewed German text or a translation. The two-step calculation below illustrates the pairwise argument for any finite number of steps.
Changed subexpression: the total mean-square expression. The outline marks it on both sides.
Before
⟨(A+B)2⟩
After
⟨(A)2+2AB+(B)2⟩
Rule: Expand algebraic terms: use the stated equality without changing its assumptions.
Expand (A + B)² as A² + 2AB + B² before averaging. The 2AB term records how the two steps combine.
In fewer words
Two added steps produce three kinds of term, not just two squares.
Show every step in this transition
This algebra holds for every pair of step values, even correlated or biased ones. No probability assumption has been used yet.
What this step assumes
No new independence or centring premise is used at this step. The averages are assumed to exist.
Treat increments over the chosen intervals as independent, centred, and having the same finite second moment. In a Brownian model this is a coarse-grained assumption, not a claim about arbitrarily short times.
Two signed steps: see every possible outcome
Each step is +1 or −1 in arbitrary step units. Rows have equal probability within each selected model. These are exact finite teaching distributions, not observations or simulated Brownian paths.
Independent steps
All four pairs are equally likely. The cross products cancel only in the average, not in each outcome.
4 equally likely outcomes. Squared columns use squared step units.
A
B
A + B
(A + B)²
A² + B²
2AB
-1
-1
-2
4
2
2
1
-1
0
0
2
-2
-1
1
0
0
2
-2
1
1
2
4
2
2
Mean square: 2. Mean cross contribution: 0. Signed mean: 0. Mean absolute displacement: 1.
Always the same direction
Both individual steps still have zero mean and mean square one. Removing independence makes the cross contribution positive.
2 equally likely outcomes. Squared columns use squared step units.
A
B
A + B
(A + B)²
A² + B²
2AB
-1
-1
-2
4
2
2
1
1
2
4
2
2
Mean square: 4. Mean cross contribution: 2. Signed mean: 0. Mean absolute displacement: 2.
Always opposite directions
The same individual step averages now produce complete cancellation of the total displacement. Zero means alone are not enough.
2 equally likely outcomes. Squared columns use squared step units.
A
B
A + B
(A + B)²
A² + B²
2AB
-1
1
0
0
2
-2
1
-1
0
0
2
-2
Mean square: 0. Mean cross contribution: -2. Signed mean: 0. Mean absolute displacement: 0.
From two steps to many
For n steps, expand the square of their sum. There are n individual squared terms and cross terms for distinct pairs. Every cross average vanishes under independence and zero mean; equal step mean squares then give n times the one-step mean square. With t = nτ and D defined as the one-step mean square divided by 2τ, the result is 2Dt.
The mean absolute displacement is not a wrong answer. It also measures spreading, but absolute value does not distribute over addition, so it lacks this simple additive calculation. Neither absolute displacement nor RMS displacement is total path length.
Modern lens: where the independence model stops
Modern lens: inertia introduces a short-time regime in which successive motions need not be independent. Momentum relaxation time is a later dynamical interpretation, not a premise used in this teaching derivation.
Pedagogical bridge, not the paper’s printed calculation
Average each contribution
This is newly authored explanatory prose, not reviewed German text or a translation. The two-step calculation below illustrates the pairwise argument for any finite number of steps.
Changed subexpression: the total mean-square expression. The outline marks it on both sides.
Before
⟨(A)2+2AB+(B)2⟩
After
⟨(A)2⟩+(⟨2AB⟩)+⟨(B)2⟩
Rule: Linearity of averaging: average the sum by adding the averages. Independence is not required for this move.
Apply linearity of expectation to the two squared terms and the cross term. Linearity does not require independence.
In fewer words
Averaging a sum is the same as adding its averages.
Show every step in this transition
Finite second moments ensure these averages exist. Expectation distributes over addition and a fixed multiplier: ⟨A² + 2AB + B²⟩ = ⟨A²⟩ + 2⟨AB⟩ + ⟨B²⟩.
What this step assumes
No new independence or centring premise is used at this step. The averages are assumed to exist.
Treat increments over the chosen intervals as independent, centred, and having the same finite second moment. In a Brownian model this is a coarse-grained assumption, not a claim about arbitrarily short times.
Two signed steps: see every possible outcome
Each step is +1 or −1 in arbitrary step units. Rows have equal probability within each selected model. These are exact finite teaching distributions, not observations or simulated Brownian paths.
Independent steps
All four pairs are equally likely. The cross products cancel only in the average, not in each outcome.
4 equally likely outcomes. Squared columns use squared step units.
A
B
A + B
(A + B)²
A² + B²
2AB
-1
-1
-2
4
2
2
1
-1
0
0
2
-2
-1
1
0
0
2
-2
1
1
2
4
2
2
Mean square: 2. Mean cross contribution: 0. Signed mean: 0. Mean absolute displacement: 1.
Always the same direction
Both individual steps still have zero mean and mean square one. Removing independence makes the cross contribution positive.
2 equally likely outcomes. Squared columns use squared step units.
A
B
A + B
(A + B)²
A² + B²
2AB
-1
-1
-2
4
2
2
1
1
2
4
2
2
Mean square: 4. Mean cross contribution: 2. Signed mean: 0. Mean absolute displacement: 2.
Always opposite directions
The same individual step averages now produce complete cancellation of the total displacement. Zero means alone are not enough.
2 equally likely outcomes. Squared columns use squared step units.
A
B
A + B
(A + B)²
A² + B²
2AB
-1
1
0
0
2
-2
1
-1
0
0
2
-2
Mean square: 0. Mean cross contribution: -2. Signed mean: 0. Mean absolute displacement: 0.
From two steps to many
For n steps, expand the square of their sum. There are n individual squared terms and cross terms for distinct pairs. Every cross average vanishes under independence and zero mean; equal step mean squares then give n times the one-step mean square. With t = nτ and D defined as the one-step mean square divided by 2τ, the result is 2Dt.
The mean absolute displacement is not a wrong answer. It also measures spreading, but absolute value does not distribute over addition, so it lacks this simple additive calculation. Neither absolute displacement nor RMS displacement is total path length.
Modern lens: where the independence model stops
Modern lens: inertia introduces a short-time regime in which successive motions need not be independent. Momentum relaxation time is a later dynamical interpretation, not a premise used in this teaching derivation.
Pedagogical bridge, not the paper’s printed calculation
Why do the cross terms vanish?
This is newly authored explanatory prose, not reviewed German text or a translation. The two-step calculation below illustrates the pairwise argument for any finite number of steps.
The move: cross terms average away, not individual displacements.
Changed subexpression: the cross-term average. The outline marks it on both sides.
Before
⟨(A)2⟩+(⟨2AB⟩)+⟨(B)2⟩
After
⟨(A)2⟩+(0)+⟨(B)2⟩
Rule: For independent, zero-mean quantities, the mean of their product is the product of their means, hence zero.
Independence gives ⟨AB⟩ = ⟨A⟩⟨B⟩. The two means are zero, so the average cross term is zero. It is not zero in each individual outcome.
In fewer words
Independent, unbiased steps have no average cross contribution.
Show every step in this transition
These are two separate premises. Zero marginal means alone do not force ⟨AB⟩ to vanish: when B always equals A, both means are zero but AB is always positive. Pairwise uncorrelated zero-mean steps would also suffice; independence is the stronger premise used here.
What this step assumes
The increments A and B are independent.
Each increment has mean zero.
Treat increments over the chosen intervals as independent, centred, and having the same finite second moment. In a Brownian model this is a coarse-grained assumption, not a claim about arbitrarily short times.
Two signed steps: see every possible outcome
Each step is +1 or −1 in arbitrary step units. Rows have equal probability within each selected model. These are exact finite teaching distributions, not observations or simulated Brownian paths.
Independent steps
All four pairs are equally likely. The cross products cancel only in the average, not in each outcome.
4 equally likely outcomes. Squared columns use squared step units.
A
B
A + B
(A + B)²
A² + B²
2AB
-1
-1
-2
4
2
2
1
-1
0
0
2
-2
-1
1
0
0
2
-2
1
1
2
4
2
2
Mean square: 2. Mean cross contribution: 0. Signed mean: 0. Mean absolute displacement: 1.
Always the same direction
Both individual steps still have zero mean and mean square one. Removing independence makes the cross contribution positive.
2 equally likely outcomes. Squared columns use squared step units.
A
B
A + B
(A + B)²
A² + B²
2AB
-1
-1
-2
4
2
2
1
1
2
4
2
2
Mean square: 4. Mean cross contribution: 2. Signed mean: 0. Mean absolute displacement: 2.
Always opposite directions
The same individual step averages now produce complete cancellation of the total displacement. Zero means alone are not enough.
2 equally likely outcomes. Squared columns use squared step units.
A
B
A + B
(A + B)²
A² + B²
2AB
-1
1
0
0
2
-2
1
-1
0
0
2
-2
Mean square: 0. Mean cross contribution: -2. Signed mean: 0. Mean absolute displacement: 0.
From two steps to many
For n steps, expand the square of their sum. There are n individual squared terms and cross terms for distinct pairs. Every cross average vanishes under independence and zero mean; equal step mean squares then give n times the one-step mean square. With t = nτ and D defined as the one-step mean square divided by 2τ, the result is 2Dt.
The mean absolute displacement is not a wrong answer. It also measures spreading, but absolute value does not distribute over addition, so it lacks this simple additive calculation. Neither absolute displacement nor RMS displacement is total path length.
Modern lens: where the independence model stops
Modern lens: inertia introduces a short-time regime in which successive motions need not be independent. Momentum relaxation time is a later dynamical interpretation, not a premise used in this teaching derivation.
Pedagogical bridge, not the paper’s printed calculation
Add equal mean squares
This is newly authored explanatory prose, not reviewed German text or a translation. The two-step calculation below illustrates the pairwise argument for any finite number of steps.
Changed subexpression: the total mean-square expression. The outline marks it on both sides.
Before
⟨(A)2⟩+(0)+⟨(B)2⟩
After
2(l)2
Rule: Substitution: use the stated equality without changing its assumptions.
Here l² denotes the mean square of one step. The two remaining contributions are l² + l² = 2l².
In fewer words
Two equal contributions give twice one contribution.
Show every step in this transition
For n increments the expansion contains n square terms and one ordered cross term for every i ≠ j. The same independence and centring argument eliminates each cross average. If every step has mean square l², the result is nl².
What this step assumes
Each increment has the same finite mean square, called l² in this worked bridge.
Treat increments over the chosen intervals as independent, centred, and having the same finite second moment. In a Brownian model this is a coarse-grained assumption, not a claim about arbitrarily short times.
Two signed steps: see every possible outcome
Each step is +1 or −1 in arbitrary step units. Rows have equal probability within each selected model. These are exact finite teaching distributions, not observations or simulated Brownian paths.
Independent steps
All four pairs are equally likely. The cross products cancel only in the average, not in each outcome.
4 equally likely outcomes. Squared columns use squared step units.
A
B
A + B
(A + B)²
A² + B²
2AB
-1
-1
-2
4
2
2
1
-1
0
0
2
-2
-1
1
0
0
2
-2
1
1
2
4
2
2
Mean square: 2. Mean cross contribution: 0. Signed mean: 0. Mean absolute displacement: 1.
Always the same direction
Both individual steps still have zero mean and mean square one. Removing independence makes the cross contribution positive.
2 equally likely outcomes. Squared columns use squared step units.
A
B
A + B
(A + B)²
A² + B²
2AB
-1
-1
-2
4
2
2
1
1
2
4
2
2
Mean square: 4. Mean cross contribution: 2. Signed mean: 0. Mean absolute displacement: 2.
Always opposite directions
The same individual step averages now produce complete cancellation of the total displacement. Zero means alone are not enough.
2 equally likely outcomes. Squared columns use squared step units.
A
B
A + B
(A + B)²
A² + B²
2AB
-1
1
0
0
2
-2
1
-1
0
0
2
-2
Mean square: 0. Mean cross contribution: -2. Signed mean: 0. Mean absolute displacement: 0.
From two steps to many
For n steps, expand the square of their sum. There are n individual squared terms and cross terms for distinct pairs. Every cross average vanishes under independence and zero mean; equal step mean squares then give n times the one-step mean square. With t = nτ and D defined as the one-step mean square divided by 2τ, the result is 2Dt.
The mean absolute displacement is not a wrong answer. It also measures spreading, but absolute value does not distribute over addition, so it lacks this simple additive calculation. Neither absolute displacement nor RMS displacement is total path length.
Modern lens: where the independence model stops
Modern lens: inertia introduces a short-time regime in which successive motions need not be independent. Momentum relaxation time is a later dynamical interpretation, not a premise used in this teaching derivation.
Show every step here: Why the square grows with time
(A+B)2=A2+2AB+B2
The square of A plus B contains two squares and twice the cross product.
Let A and B be two signed increments. Both have mean zero.
Independence gives average AB = average A times average B = 0.
Averaging the expansion leaves average A² plus average B².
For n increments, each squared increment contributes ℓ²; all pairwise products vanish in the expectation.
Thus the total mean square is nℓ². Since t = nτ, replace n with t/τ.
Name the coefficient ℓ²/(2τ) as D. The result becomes 2Dt.
Four times t gives four times the mean square. Taking the square root gives twice the RMS.
Independent centred steps add mean squares; the root therefore grows as the square root of time.
What this relies on
Increments are independent, have zero mean, and share a finite mean square ℓ².
One step corresponds to a declared interval τ.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: Why the square grows with time
Assumed here
Increments are independent, have zero mean, and share a finite mean square ℓ².
One step corresponds to a declared interval τ.
What this does not establish
Correlated steps, bias or an infinite second moment change the argument.
This pedagogical walk does not describe fixed physical jumps in a liquid.
How can random individual steps produce a deterministic equation?
The exact step rule averages over possible displacements. With symmetry, finite variance and a justified smooth-scale approximation, its leading evolution is the diffusion equation.
Let φ(Δ) be the probability density for a displacement Δ during τ. To end at x, a tracer must start at x − Δ and then make that displacement. Adding over all possible increments gives the transition relation, written here in modern notation.
p(x,t+τ)=∫−∞∞p(x−Δ,t)φ(Δ)dΔ
The next density is the integral of the previous shifted density times the step density.
Expand to first order in time and second order in displacement. Normalization cancels the zeroth-order term. Symmetry removes the first moment. The second moment remains.
D=2τ1∫−∞∞Δ2φ(Δ)dΔ
D is half the mean-square step divided by the step interval.
Keep the density and its first time change on the left.
p(x−Δ,t)≈p−Δ∂x∂p+2Δ2∂x2∂2p
Keep value, slope and curvature of the spatial density on the right.
The time derivative holds position fixed; the spatial derivatives hold time fixed.
Insert the spatial expansion into the transition integral. The derivatives do not depend on the integration variable Δ.
The integral of φ is one, so the first term integrates to p.
For a symmetric φ, positive and negative Δ contributions cancel in the integral of Δφ.
The remaining second-order contribution is half the second spatial derivative times the mean-square increment.
Subtract p from both sides and divide by τ.
Identify the mean-square increment divided by 2τ as D. This is the retained diffusion law, with the stated approximation conditions.
Show every step here: From a step law to a density law
p(x,t+τ)≈p(x,t)+τ∂t∂p
Keep the density and its first time change on the left.
p(x−Δ,t)≈p−Δ∂x∂p+2Δ2∂x2∂2p
Keep value, slope and curvature of the spatial density on the right.
The time derivative holds position fixed; the spatial derivatives hold time fixed.
Insert the spatial expansion into the transition integral. The derivatives do not depend on the integration variable Δ.
The integral of φ is one, so the first term integrates to p.
For a symmetric φ, positive and negative Δ contributions cancel in the integral of Δφ.
The remaining second-order contribution is half the second spatial derivative times the mean-square increment.
Subtract p from both sides and divide by τ.
Identify the mean-square increment divided by 2τ as D. This is the retained diffusion law, with the stated approximation conditions.
Normalization preserves probability; symmetry removes drift; the second moment supplies diffusion.
What this relies on
The transition density is normalized, symmetric and has finite second moment.
The density is smooth on the step scale; the retained expansion has a controlled coarse-grained range.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: From a step law to a density law
Assumed here
The transition density is normalized, symmetric and has finite second moment.
The density is smooth on the step scale; the retained expansion has a controlled coarse-grained range.
What this does not establish
A Taylor truncation is an approximation unless justified by a suitable limiting procedure.
Physical independence breaks down at sufficiently short times; the limit is not a literal collision movie.
How does the density law become a measurable displacement?
For the unbounded, drift-free point-source model, spread is Gaussian at positive times and coordinate RMS displacement is √(2Dt). The starting distribution is a point mass.
p(x,t)=4πDte−x2/(4Dt)(t>0)
The point-source solution is the normalized Gaussian density for positive time.
Its symmetry gives zero mean. Its second moment is 2Dt. These are ensemble statements: the curve assigns probabilities to intervals, not destinations to individual particles.
λx=⟨x2⟩=2Dt
Coordinate RMS displacement is the square root of two D t.
The probability of finding a displacement between a and b is the integral of this density over that interval. At time zero, an interval containing the starting point has probability one; no finite bell represents that state.
For positive t, change variable to u = x/√(4Dt). Then x = √(4Dt)u and dx = √(4Dt)du.
The normalization factor cancels the factor in dx, leaving exp(−u²)/√π. Its integral is one.
The first moment vanishes because x times the density is odd: the negative side cancels the positive side.
For the second moment substitute x² = 4Dt u².
The normalized integral of u² exp(−u²) is one half, by integration by parts.
Multiplying 4Dt by one half yields 2Dt. Taking the nonnegative square root gives √(2Dt).
⟨x2⟩=π4Dt∫−∞∞u2e−u2du=2Dt
The second moment is four D t times the normalized Gaussian second-moment integral, giving two D t.
To check the density equation directly: the time derivative is p times (−1/(2t) + x²/(4Dt²)). The second spatial derivative is p times (−1/(2Dt) + x²/(4D²t²)). Multiplying the latter by D gives the former.
Show every step here: What the spreading curve predicts
For positive t, change variable to u = x/√(4Dt). Then x = √(4Dt)u and dx = √(4Dt)du.
The normalization factor cancels the factor in dx, leaving exp(−u²)/√π. Its integral is one.
The first moment vanishes because x times the density is odd: the negative side cancels the positive side.
For the second moment substitute x² = 4Dt u².
The normalized integral of u² exp(−u²) is one half, by integration by parts.
Multiplying 4Dt by one half yields 2Dt. Taking the nonnegative square root gives √(2Dt).
⟨x2⟩=π4Dt∫−∞∞u2e−u2du=2Dt
The second moment is four D t times the normalized Gaussian second-moment integral, giving two D t.
To check the density equation directly: the time derivative is p times (−1/(2t) + x²/(4Dt²)). The second spatial derivative is p times (−1/(2Dt) + x²/(4D²t²)). Multiplying the latter by D gives the former.
For an unbounded point source, the Gaussian has mean square 2Dt and coordinate RMS √(2Dt).
What this relies on
Constant positive D, an unbounded line, no drift, and a localized initial ensemble.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: What the spreading curve predicts
Assumed here
Constant positive D, an unbounded line, no drift, and a localized initial ensemble.
What this does not establish
A finite closed box has a different long-time distribution.
At t = 0 the distribution is a point mass, not an ordinary density.
Coordinate RMS is not the three-dimensional RMS distance.
Einstein's displacement argument: the typical distance grows with the square root of time, not with time itself. Every explanation stays on the page when reading-only is on.
Static worked case
For radius 0.5 μm, viscosity 1.35×10⁻³ Pa·s, and T = 290.15 K, the RMS displacement is about 0.8 μm in one second. This static worked case stays in the markup; loading the live ensemble does not replace it.
Keep the experiment beside the argument
This laboratory stays mounted while you change detail or open a foundation. Applying its controls explicitly starts a host calculation; opening an explanation never starts or restarts a trial.
Compare the movement of individual tracers with the statistics of the whole ensemble. Changing when or how you observe the trial does not generate different paths.
Same-seed viscosity comparisons use common random numbers, not independent trials.
Predict before comparing observation times
The independent-step model predicts four times the mean square, hence twice its square root. Your prediction never locks the explanation or controls.
These buttons use the accepted trial, not unsaved draft edits. The requested time must lie on its recording grid.
Accepted synthetic trial: 400 tracers, 1 seconds, coordinate mean 0.0029237 micrometres and coordinate RMS 0.98602 micrometres.
Accepted trial: seed 1905; 400 tracers; 293.15 K; viscosity 1 mPa·s; radius 0.5 μm. Observe at 1 s in a 10-second recording. Constants: modern SI 2019.
Displacements from a common origin, not a literal microscope image. Lines connect recorded positions and do not supply an instantaneous velocity. 0 / 400 endpoints lie outside this view; none are removed from the ensemble.
t = 1.00 s (true rate (1 s/s))
Natural rate: ~0.8 μm per second Brownian walk (scale bar: 1 μm)Simulation view: accelerated snapshot across 10 s
View only: no calculation or new draws.
Solid bars: the synthetic sample. Dashed line: probabilities of the same bins under the unbounded model, not a density curve. Counts beyond the plotted range: 0 left, 0 right.
Whole-ensemble statistics: signed coordinate x, total over 1 coordinate
Diffusion coefficient (model)
0.42944 μm²/s
Signed mean (sample)
0.0029237 μm (modern-si-2019)
Mean absolute coordinate displacement
0.77808 μm (modern-si-2019)
Mean-square coordinate displacement
0.97223 μm²
Coordinate RMS (sample / model)
0.98602 / 0.92676 μm(modern-si-2019)
Mean distance (sample / model)
0.77808 / 0.73945 μm (modern-si-2019)
Total mean square (sample / model)
0.97223 / 0.85888 μm²
Apparent coordinate speed (sample / model)
0.98602 / 0.92676 μm/s
Sampling bands under the model
99.9% per comparison, not a simultaneous guarantee across the table or repeated observations. These ranges are calculated from the model variance, not estimated from the observed sample, and are not uncertainties of the model itself.
Signed coordinate mean (μm): -0.15248 to 0.15248
Total mean square (μm²): 0.67299 to 1.0729
Inspect the model behind this trial
These equations describe the same accepted trial as the plots and table. Select a symbol or an operation to see what it means; model predictions are not sample estimates.
Explore the equation · Modern model notation
Why the apparent speed depends on how you watch
vapp:=tλx
The apparent coordinate speed divides the typical coordinate distance by the chosen positive interval.
Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.
Input · Model result · Constant
Static worked example
Values describe this accepted snapshot, not unsaved input edits.
Apparent coordinate speed
0.92676 μm/s
Coordinate RMS displacement
0.92676 μm
Observation interval
1 s
Read the equation aloud in words
Apparent coordinate speed equals the model coordinate root mean square displacement divided by the observation interval.
This quotient depends on the observation interval. It is not instantaneous physical velocity. At zero interval the quotient is undefined, even though the displacement is zero.
Model assumptions and every term’s meaning
The same one-coordinate ideal Brownian model as the RMS relation.
A positive observation interval is required for the quotient.
Lines joining recorded points are a rendering convention, not a velocity measurement.
Define the observable
This defines an interval-dependent comparison. It does not introduce a physical instantaneous Brownian velocity. Read the prerequisite
Apparent coordinate speed
A distance-per-interval statistic. It is not a molecular collision speed. Read the prerequisite
Divide by the same interval
The distance grows as the square root of time, while the denominator grows linearly. The quotient therefore decreases as the interval grows. Read the prerequisite
Model coordinate RMS
The same accepted one-coordinate model displacement used by the neighboring RMS equation. Read the prerequisite
Positive observation interval
A zero interval yields no apparent-speed value. The interface keeps the explanation instead of fabricating zero or infinity. Read the prerequisite
Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.
Explore the equation · Modern model notation
Resistance to motion controls spreading
D=6πηakBT
The diffusion coefficient is thermal energy divided by the viscous drag coefficient.
Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.
Input · Model result · Constant
Static worked example
Values describe this accepted snapshot, not unsaved input edits.
Diffusion coefficient
0.42944 μm²/s
Boltzmann constant
1.3806e-23 J/K
Absolute temperature
293.15 K
Dynamic viscosity
1 mPa·s
Particle radius
0.5 μm
Read the equation aloud in words
The diffusion coefficient equals the Boltzmann constant times the absolute temperature, divided by six times pi times the dynamic viscosity times the particle radius.
Thermal energy competes with viscous drag. Doubling viscosity halves the diffusion coefficient while reducing RMS displacement only by the square root of two. Radius means radius, not diameter.
Model assumptions and every term’s meaning
Dilute, approximately spherical tracers with no-slip Stokes drag in a homogeneous Newtonian liquid.
Wall corrections, interactions, inertia, and observation noise are not included.
The modern SI 2019 constant set is used; this is not a historical inversion exercise.
An ideal model
This equation assumes the dilute spherical tracer and Stokes-drag model; dimensional consistency alone does not establish those assumptions. Read the prerequisite
Diffusion coefficient
A squared-distance-per-time coefficient. This value comes from the accepted model calculation. Read the prerequisite
Why divide by drag?
Increasing the denominator reduces diffusivity at fixed thermal energy. This is a model dependence, not a rule that every larger quantity must reduce motion. Read the prerequisite
Thermal energy scale
Boltzmann constant times absolute temperature supplies energy per particle. Read the prerequisite
A known modern constant
The SI 2019 constant set supplies this value. It must not be treated as independent historical evidence when trying to infer molecular number. Read the prerequisite
Absolute temperature
Use kelvin. The laboratory holds viscosity as an independently supplied parameter. Read the prerequisite
The drag coefficient
Six pi times viscosity times radius is Stokes drag per speed for an isolated sphere in the assumed low-Reynolds regime. Read the prerequisite
Dynamic viscosity
Viscosity is in the denominator. Doubling it at fixed temperature and radius halves D, not the RMS displacement. Read the prerequisite
Particle radius
The input is a radius, not a diameter. Confusing them changes the predicted coefficient by a factor of two. Read the prerequisite
Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.
Explore the equation · Modern model notation
From spreading to a measurable distance
λx=2Dt
The typical coordinate distance is the positive square root of twice the diffusion coefficient times the observation interval.
Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.
Input · Model result · Constant
Static worked example
Values describe this accepted snapshot, not unsaved input edits.
Coordinate RMS displacement
0.92676 μm
Diffusion coefficient
0.42944 μm²/s
Observation interval
1 s
Read the equation aloud in words
The model coordinate root mean square displacement equals the square root of two times the diffusion coefficient times the observation interval.
Squaring measures spread without cancellation between directions. Taking the positive square root turns squared distance back into a distance. Four times the observation interval gives twice the model RMS, not four times.
Model assumptions and every term’s meaning
Independent, zero-mean Gaussian displacement increments in a homogeneous liquid.
One-coordinate model statistic, not a measured speed or a sample estimate.
The overdamped regime is assumed rather than established from additional particle and fluid measurements.
A model relation
This equality concerns the ideal model. A finite synthetic ensemble fluctuates around it; its sample RMS is displayed separately. Read the prerequisite
Coordinate RMS
One coordinate, not the total three-dimensional distance. Model RMS and sample RMS have different meanings. Read the prerequisite
Why a square root?
Two D t has units of squared length. Its positive square root has units of length and defines the typical displacement. Read the prerequisite
Build the mean square
Independent zero-mean increments add their variances. The definition of D makes the coordinate mean square equal to 2 D t. Read the prerequisite
Diffusion coefficient
This is the accepted model diffusivity, not a rate inferred from the synthetic data. Read the prerequisite
Observation interval
This is the interval used for the displayed displacement, not the simulation frame rate. Re-observing the trial preserves its paths. Read the prerequisite
Exact rational SI dimensions were checked at build time. This is a unit check, not a proof of the model. This teaching record is a draft, not a transcription of a printed equation.
RMS distance over recorded time
Solid: this sample. Dashed: the model. The time axis is logarithmic. Zero sample values, if any, remain in the table but cannot appear on a log scale. All times refer to the same recorded paths.Read the comparison as a table
RMS distance (μm)
Time (s)
Sample
Model
0.02
0.13038
0.13106
0.04
0.18556
0.18535
0.1
0.30368
0.29307
0.2
0.40416
0.41446
0.4
0.61388
0.58613
1
0.98602
0.92676
2
1.3495
1.3106
4
1.9035
1.8535
10
2.8723
2.9307
What is, and is not, being simulated
These synthetic paths are Gaussian independent increments for dilute spherical tracers in a homogeneous Newtonian liquid. They do not simulate individual molecular collisions, inertia, interactions, sedimentation, walls, localization error or motion blur.
Low Reynolds number and observation times long compared with momentum relaxation are assumed, not verified from fluid and particle data. Lines between samples are a drawing convention; they do not define an instantaneous Brownian speed.
The three latent coordinates are recorded together. Axis, dimension, observation and statistic changes reuse those coordinates. The histogram and statistics use every tracer, including those outside the view.
No FrankenSim WASM artifact is used. Integer random draws follow the pinned Philox mapping; Gaussian conversion uses this host’s math functions and is not claimed bitwise identical across browser engines.
Logical recording draws: 1200000. This result was assembled from a newly generated deterministic recording.
This is the function that produced the current snapshot.
In words
Take the temperature and the gas constant, divide by the number of molecules in a mole to get the energy scale of one molecule, then divide by the drag on a sphere of this radius in a liquid of this viscosity.
Mathematics
Implementation
/** SI: temperature K, viscosity Pa s, radius m, output m2/s. No ambient constants. */exportfunctionstokesEinsteinD({T,eta,a,medium="liquid",}:{T:number;eta:number;a:number;medium?:"liquid"|"gas"},set:ConstantSet,):Evaluation{if(![T,eta,a].every((v)=>Number.isFinite(v)&&v>0))returnoutside("diffusionCoefficient","stokesEinsteinD","T > 0, eta > 0, a > 0","Enter positive finite temperature, viscosity, and particle radius.","input",set,);if(medium!=="liquid")returnoutside("diffusionCoefficient","stokesEinsteinD","stokes-gas-medium","The liquid Stokes-drag model does not include the slip correction needed in a gas.","model",set,);constk=thermalConstant(set);returnnumber("diffusionCoefficient","stokesEinsteinD",(k.value*T)/(6*Math.PI*eta*a),set,undefined,true,);}
One worked example of this calculation. Constant set: Declared 1905-plan inputs (source review pending).
This function computes the listed outputs when it runs.
In words
The typical one-dimensional displacement is the square root of twice the diffusion coefficient times the observation interval.
Mathematics
Implementation
exportfunctionrmsDisplacement(D:number,t:number):Evaluation{if(!validDt(D,t))returnoutside("rmsDisplacement1d","rmsDisplacement","D >= 0 and t >= 0","Diffusivity and elapsed time must be finite and nonnegative.",);returnnumber("rmsDisplacement1d","rmsDisplacement",scale(D,t),undefined,undefined,D>0&&t>0,);}
Audited TypeScript reference evaluator: the owner on this device, or the host fallback for a FrankenSim capability.
This function computes the listed outputs when it runs.
In words
Divide the model RMS displacement by the observation interval. The quotient depends on how long you watch.
Mathematics
Implementation
exportfunctionapparentSpeed(D:number,tau:number):Evaluation{if(!validDt(D,tau)||tau===0)returnoutside("apparentSpeed","apparentSpeed","D >= 0 and tau > 0","An apparent speed needs a positive observation interval.",);returnnumber("apparentSpeed","apparentSpeed",(Math.SQRT2*Math.sqrt(D))/Math.sqrt(tau),undefined,undefined,D>0,);}
Audited TypeScript reference evaluator: the owner on this device, or the host fallback for a FrankenSim capability.
This function computes the listed outputs when it runs.
In words
Average the recorded displacements of every tracer, not a viewport subset.
Mathematics
Implementation
/** Compensated sums over ALL members, never a viewport-selected subset. Row-major M by d. */exportfunctionensembleMoments({displacements,d,dt,viewport,}:{displacements:Float64Array;d:number;dt?:number;viewport?:ViewportBounds;}):Computation<EnsembleMoments>{if(!(displacementsinstanceofFloat64Array)||![1,2,3].includes(d)||displacements.length===0||displacements.length%d!==0||displacements.length>30000||!displacements.every(Number.isFinite))returninvalid(["displacements","d"],"Provide finite, row-major displacements with one, two or three coordinates per tracer.",);constM=displacements.length/d,sums=newFloat64Array(d*3+1),corrections=newFloat64Array(d*3+1);functionadd(i:number,v:number){constcorr=corrections[i]??0,sum=sums[i]??0,y=v-corr,t=sum+y;corrections[i]=t-sum-y;sums[i]=t;}for(leti=0;i<M;i++){letnorm=0;for(letj=0;j<d;j++){constv=displacements[i*d+j];if(v===undefined)returnfailure("A required displacement coordinate was undefined.");if(v!==0&&v*v===0)returnfailure("A nonzero squared displacement is below the representable range.");add(j*3,v);add(j*3+1,Math.abs(v));add(j*3+2,v*v);norm=Math.hypot(norm,v);}add(d*3,norm);}constaxes=Array.from({length:d},(_,j)=>{constmean=(sums[j*3]??0)/M;constmeanAbsolute=(sums[j*3+1]??0)/M;constmeanSquare=(sums[j*3+2]??0)/M;return{mean,meanAbsolute,meanSquare,rms:Math.sqrt(meanSquare),};});constmeanSquareNorm=axes.reduce((s,a)=>s+a.meanSquare,0);if(![...sums,meanSquareNorm].every(Number.isFinite))returnfailure("The moment reduction exceeded the numerical range.");constrmsNorm=Math.sqrt(meanSquareNorm);letinsideCount:number|undefined;letoutsideCount:number|undefined;if(viewport){letinc=0;letoutc=0;for(leti=0;i<M;i++){letisInside=true;for(letj=0;j<d;j++){constv=displacements[i*d+j]??0;constminVal=viewport.min[j]??-Infinity;constmaxVal=viewport.max[j]??Infinity;if(v<minVal||v>maxVal){isInside=false;break;}}if(isInside)inc++;elseoutc++;}insideCount=inc;outsideCount=outc;}constapparentSpeed=dt!==undefined&&Number.isFinite(dt)&&dt>0?(d===1?(axes[0]?.rms??0):rmsNorm)/dt:undefined;return{kind:"accepted",data:{M,axes,meanNorm:(sums[d*3]??0)/M,meanSquareNorm,rmsNorm,...(apparentSpeed!==undefined?{apparentSpeed}:{}),...(insideCount!==undefined&&outsideCount!==undefined?{insideCount,outsideCount}:{}),},};}
Audited TypeScript reference evaluator: the owner on this device, or the host fallback for a FrankenSim capability.
This function computes the listed outputs when it runs.
In words
Count how many recorded displacements fall in each interval of the histogram.
Mathematics
Implementation
exportfunctiondisplacementHistogram(values:Float64Array,edges:Float64Array,):Computation<{counts:Float64Array;underflow:number;overflow:number;total:number}>{if(values.length===0||values.length>10000||edges.length<2||edges.length>1001||!values.every(Number.isFinite)||!edges.every((v,i)=>{if(!Number.isFinite(v))returnfalse;if(i===0)returntrue;constprev=edges[i-1];returnprev!==undefined&&v>prev;}))returninvalid(["values","edges"],"Use finite samples and strictly increasing histogram edges.",);constcounts=newFloat64Array(edges.length-1);letunderflow=0,overflow=0;constfirstEdge=edges[0];constlastEdge=edges[edges.length-1];if(firstEdge===undefined||lastEdge===undefined)returninvalid(["edges"],"Histogram edges array must contain at least two finite bounds.");for(constvalueofvalues){if(value<firstEdge){underflow++;continue;}if(value>lastEdge){overflow++;continue;}letlo=0,hi=edges.length-1;while(hi-lo>1){constmid=(lo+hi)>>>1;constedgeMid=edges[mid];if(edgeMid!==undefined&&value<edgeMid)hi=mid;elselo=mid;}constbin=Math.min(lo,counts.length-1);constprevCount=counts[bin];if(prevCount!==undefined){counts[bin]=prevCount+1;}}return{kind:"accepted",data:{counts,underflow,overflow,total:values.length}};}
Every histogram count, including tails
Coordinate intervals in μm; half-open bins, with the final endpoint included
Interval
Count
Below the plotted range
0
-4.6338 to -4.4021
0
-4.4021 to -4.1704
0
-4.1704 to -3.9387
0
-3.9387 to -3.707
0
-3.707 to -3.4753
0
-3.4753 to -3.2437
0
-3.2437 to -3.012
0
-3.012 to -2.7803
1
-2.7803 to -2.5486
1
-2.5486 to -2.3169
0
-2.3169 to -2.0852
7
-2.0852 to -1.8535
7
-1.8535 to -1.6218
6
-1.6218 to -1.3901
16
-1.3901 to -1.1584
12
-1.1584 to -0.92676
15
-0.92676 to -0.69507
28
-0.69507 to -0.46338
24
-0.46338 to -0.23169
39
-0.23169 to 0
39
0 to 0.23169
43
0.23169 to 0.46338
37
0.46338 to 0.69507
26
0.69507 to 0.92676
32
0.92676 to 1.1584
23
1.1584 to 1.3901
17
1.3901 to 1.6218
8
1.6218 to 1.8535
6
1.8535 to 2.0852
6
2.0852 to 2.3169
3
2.3169 to 2.5486
1
2.5486 to 2.7803
1
2.7803 to 3.012
1
3.012 to 3.2437
1
3.2437 to 3.4753
0
3.4753 to 3.707
0
3.707 to 3.9387
0
3.9387 to 4.1704
0
4.1704 to 4.4021
0
4.4021 to 4.6338
0
Above the plotted range
0
Total, including both tails
400
Notes and laboratory
Beside this passage
Margin notes, assumptions, and the laboratory stay here so the German argument keeps the main column.
Each HTML file contains the available explanation at every detail level, linked foundations, source references, and available build-time scalar results. It includes no private notes or running simulations. This remains an explanation preview, not a reviewed source edition.
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