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Read · Brownian motion · Explanation preview

Brownian motion: from wandering to a measurable law

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.

First Encounter · Zero Algebra Entrance

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).

Why do the cross terms vanish? Open the exact missing step →

Step 10 · The Bridge to the Argument
New Skill

keeping track of how far things went by squaring, so opposite directions stop cancelling.

Why Useful in the Paper

Section 5 says how far a particle typically wanders after a given time, and that statement is about the squared spread, not about a speed.

Continue With Your Choice of Guidance
More Guidance · Foundations

Review mean, variance, and root-mean-square displacement with worked algebraic examples in the Foundations library.

Open Mean, Variance & RMS Drawer →
Less Guidance · Laboratory & Paper

Test thousands of particles in the BM-01 tracer ensemble or jump straight to Einstein’s §5 displacement passage.

Open BM-01 Lab →Go to §5 Passage →

§4 · From random displacement to diffusion

definition · Within the stated model

Zero average is not no movement

What can we measure when left and right cancel?

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\langle x\rangle=0,\qquad\langle x^2\rangle>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.

Open the foundation: Mean, variance and RMS

Explore the equation · Modern model notation

Why the apparent speed depends on how you watch

vapp:=λxtv_{\mathrm{app}} := \frac{\lambda_x}{t}

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=2 D t\lambda_x = \sqrt{2\,D\,t}

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
  1. Choose the common starting point as zero and right as positive.
  2. Use four displacements, in micrometres: −3, −1, +1, +3.
  3. Their sum is zero. Divide by four: the signed mean is zero.
  4. Square each first: 9, 1, 1, 9 square micrometres.
  5. The squares add to 20. Divide by four: the mean square is 5 square micrometres.
  6. Take the square root: the RMS is approximately 2.236 micrometres.
  7. The mean distance is instead (3 + 1 + 1 + 3) / 4 = 2 micrometres. These are different questions, not inconsistent answers.
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.

    Open the explanation that addresses this

  • The physical reason for a step

    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.

    Open the explanation that addresses this

  • Simply too much at once

    This answer is not yet authored for this passage. No gloss, foundation, or invented explanation is offered in its place.

Source context: Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

derivation · Within the stated model

Why the square grows with time

What permits us to add the contributions of many random steps?

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=1nΔi)2⟩=∑i=1n⟨Δi2⟩=nl2\left\langle\left(\sum_{i=1}^n\Delta_i\right)^2\right\rangle=\sum_{i=1}^n\langle\Delta_i^2\rangle=nl^2

The mean square of the sum of independent centred increments is the sum of their mean squares.

t=nτ,D=l22τ,⟨x2⟩=2Dtt=n\tau,\qquad D=\frac{l^2}{2\tau},\qquad\langle x^2\rangle=2Dt

Elapsed time is n tau; defining D as ell squared over two tau gives mean-square displacement two D t.

Open the foundation: Probability and independence

Why the mean square is tractable

Choose a transition to open its explanation. Without JavaScript, open the worked bridge below; it contains the same four explanations.

  1. Expand the square
  2. Average each contribution
  3. Why do the cross terms vanish?
  4. Add equal mean squares
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⟩\htmlData{expression-id=totalMeanSquare}{\boxed{\left\langle \left(A + B\right)^{2}\right\rangle}}

After

⟨(A)2+2 A B+(B)2⟩\htmlData{expression-id=totalMeanSquare}{\boxed{\left\langle \left(A\right)^{2} + 2\,A\,B + \left(B\right)^{2}\right\rangle}}

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.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
1-1002-2
-11002-2
112422

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.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
112422

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.
ABA + B(A + B)²A² + B²2AB
-11002-2
1-1002-2

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.

Compare the paper’s step-distribution route (explanation preview) →

Average each contribution

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+2 A B+(B)2⟩\htmlData{expression-id=totalMeanSquare}{\boxed{\left\langle \left(A\right)^{2} + 2\,A\,B + \left(B\right)^{2}\right\rangle}}

After

⟨(A)2⟩+(⟨2 A B⟩)+⟨(B)2⟩\htmlData{expression-id=totalMeanSquare}{\boxed{\left\langle \left(A\right)^{2}\right\rangle + \left(\left\langle 2\,A\,B\right\rangle\right) + \left\langle \left(B\right)^{2}\right\rangle}}

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.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
1-1002-2
-11002-2
112422

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.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
112422

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.
ABA + B(A + B)²A² + B²2AB
-11002-2
1-1002-2

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.

Compare the paper’s step-distribution route (explanation preview) →

Why do the cross terms vanish?

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⟩+(⟨2 A B⟩)+⟨(B)2⟩\left\langle \left(A\right)^{2}\right\rangle + \htmlData{expression-id=crossTerm}{\boxed{\left(\left\langle 2\,A\,B\right\rangle\right)}} + \left\langle \left(B\right)^{2}\right\rangle

After

⟨(A)2⟩+(0)+⟨(B)2⟩\left\langle \left(A\right)^{2}\right\rangle + \htmlData{expression-id=crossTerm}{\boxed{\left(0\right)}} + \left\langle \left(B\right)^{2}\right\rangle

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.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
1-1002-2
-11002-2
112422

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.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
112422

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.
ABA + B(A + B)²A² + B²2AB
-11002-2
1-1002-2

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.

Compare the paper’s step-distribution route (explanation preview) →

Add equal mean squares

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⟩\htmlData{expression-id=totalMeanSquare}{\boxed{\left\langle \left(A\right)^{2}\right\rangle + \left(0\right) + \left\langle \left(B\right)^{2}\right\rangle}}

After

2 (l)2\htmlData{expression-id=totalMeanSquare}{\boxed{2\,\left(l\right)^{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.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
1-1002-2
-11002-2
112422

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.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
112422

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.
ABA + B(A + B)²A² + B²2AB
-11002-2
1-1002-2

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.

Compare the paper’s step-distribution route (explanation preview) →

Show every step here: Why the square grows with time
(A+B)2=A2+2AB+B2(A+B)^2=A^2+2AB+B^2

The square of A plus B contains two squares and twice the cross product.

  1. Let A and B be two signed increments. Both have mean zero.
  2. Independence gives average AB = average A times average B = 0.
  3. Averaging the expansion leaves average A² plus average B².
  4. For n increments, each squared increment contributes ℓ²; all pairwise products vanish in the expectation.
  5. Thus the total mean square is nℓ². Since t = nτ, replace n with t/τ.
  6. Name the coefficient ℓ²/(2τ) as D. The result becomes 2Dt.
  7. Four times t gives four times the mean square. Taking the square root gives twice the RMS.
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.

Earlier step: Zero average is not no movement

Source context: Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

derivation · Model approximation

From a step law to a density law

How can random individual steps produce a deterministic 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) dΔp(x,t+\tau)=\int_{-\infty}^{\infty}p(x-\Delta,t)\varphi(\Delta)\,d\Delta

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=12τ∫−∞∞Δ2φ(Δ) dΔD=\frac{1}{2\tau}\int_{-\infty}^{\infty}\Delta^2\varphi(\Delta)\,d\Delta

D is half the mean-square step divided by the step interval.

∂p∂t=D∂2p∂x2\frac{\partial p}{\partial t}=D\frac{\partial^2p}{\partial x^2}

The retained equation is the diffusion equation.

Open the foundation: A local expansion

Show every step here: From a step law to a density law
p(x,t+τ)≈p(x,t)+τ∂p∂tp(x,t+\tau)\approx p(x,t)+\tau\frac{\partial p}{\partial t}

Keep the density and its first time change on the left.

p(x−Δ,t)≈p−Δ∂p∂x+Δ22∂2p∂x2p(x-\Delta,t)\approx p-\Delta\frac{\partial p}{\partial x}+\frac{\Delta^2}{2}\frac{\partial^2p}{\partial x^2}

Keep value, slope and curvature of the spatial density on the right.

  1. The time derivative holds position fixed; the spatial derivatives hold time fixed.
  2. Insert the spatial expansion into the transition integral. The derivatives do not depend on the integration variable Δ.
  3. The integral of φ is one, so the first term integrates to p.
  4. For a symmetric φ, positive and negative Δ contributions cancel in the integral of Δφ.
  5. The remaining second-order contribution is half the second spatial derivative times the mean-square increment.
  6. Subtract p from both sides and divide by τ.
  7. Identify the mean-square increment divided by 2τ as D. This is the retained diffusion law, with the stated approximation conditions.
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.

Earlier step: Why the square grows with time

Source context: Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

derivation · Within the stated model

What the spreading curve predicts

How does the density law become a measurable displacement?

p(x,t)=e−x2/(4Dt)4πDt(t>0)p(x,t)=\frac{e^{-x^2/(4Dt)}}{\sqrt{4\pi Dt}}\quad(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\lambda_x=\sqrt{\langle x^2\rangle}=\sqrt{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.

Open the foundation: The Gaussian and its width

Show every step here: What the spreading curve predicts
  1. For positive t, change variable to u = x/√(4Dt). Then x = √(4Dt)u and dx = √(4Dt)du.
  2. The normalization factor cancels the factor in dx, leaving exp(−u²)/√π. Its integral is one.
  3. The first moment vanishes because x times the density is odd: the negative side cancels the positive side.
  4. For the second moment substitute x² = 4Dt u².
  5. The normalized integral of u² exp(−u²) is one half, by integration by parts.
  6. Multiplying 4Dt by one half yields 2Dt. Taking the nonnegative square root gives √(2Dt).
⟨x2⟩=4Dtπ∫−∞∞u2e−u2 du=2Dt\langle x^2\rangle=\frac{4Dt}{\sqrt\pi}\int_{-\infty}^{\infty}u^2e^{-u^2}\,du=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.

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.

Earlier step: From a step law to a density law

Source context: Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

§5 · From displacement to molecular scale

derivation · Model approximation

Why viscosity changes the spread

What fixes D for a small spherical tracer in a liquid?

The displacement law tells us what a given D predicts. A separate model connects D to a tracer’s physical surroundings. Let b be mobility, so a small force F produces mean drift bF. Stokes drag for a sphere gives b = 1/(6πηa).

Let c be number density. At isothermal balance the force density cF balances the osmotic-pressure gradient. With ideal osmotic pressure Π = cRT/N, the drift flux cbF becomes b(RT/N) times the density gradient. Equating it with the opposite diffusive flux gives D = bRT/N.

D=RT6πηaN=kBT6πηaD=\frac{RT}{6\pi\eta aN}=\frac{k_BT}{6\pi\eta a}

Stokes–Einstein diffusivity is R T over six pi viscosity radius N, or k B T over six pi viscosity radius.

The equality kB = R/N relates the two forms. For a prediction using modern constants it is convenient. For an inference of N, using a value of kB derived from that same N would defeat the point.

Open the foundation: Counting what crosses a boundary

Explore the equation · Modern model notation

Resistance to motion controls spreading

D=kB T6 π η aD = \frac{k_B\,T}{6\,\pi\,\eta\,a}

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

Symbolic equation

Symbolic here. Open the linked laboratory for a worked example and live values.

Diffusion coefficient
No accepted value is available here.
Boltzmann constant
No accepted value is available here.
Absolute temperature
No accepted value is available here.
Dynamic viscosity
No accepted value is available here.
Particle radius
No accepted value is available here.
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.

Show every step here: Why viscosity changes the spread
  1. Name the quantities: T is absolute temperature, η dynamic viscosity, a tracer radius, R molar gas constant, and N molecular number per mole.
  2. Mobility b is drift speed divided by force. Stokes drag supplies b = 1/(6πηa).
  3. The isothermal ideal osmotic-pressure relation is Π = cRT/N, where c counts tracers per volume.
  4. Differentiate with respect to position: the gradient of Π is (RT/N) times the gradient of c.
  5. Mechanical balance gives cF equal to that pressure gradient. Multiply by b to obtain the drift flux.
  6. Fick’s law gives the opposing diffusive flux as −D times the gradient of c.
  7. Zero total flux for this balance gives D = bRT/N. Substitute the Stokes mobility.
  8. Holding T, radius and constants fixed, doubling η halves D. The RMS displacement at a fixed time then falls by √2, not by two.
Assumptions and limits: Why viscosity changes the spread

Assumed here

  • Dilute spherical tracers in a homogeneous Newtonian liquid, Stokes mobility and ideal osmotic pressure.
  • R is a separately known molar gas constant; N denotes molecular number per mole.

What this does not establish

  • Slip, inertia, interactions, non-Newtonian response and gas corrections are not included.
  • The argument imports constitutive laws; conservation alone does not derive them.

Earlier step: What the spreading curve predicts

Source context: Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

derivation · Model approximation

What would let us count molecules?

Which additional measurements turn displacement into an estimate of N?

D=⟨x2⟩2t,N=RTt3πηa⟨x2⟩D=\frac{\langle x^2\rangle}{2t},\qquad N=\frac{RTt}{3\pi\eta a\langle x^2\rangle}

Diffusivity is mean-square displacement over twice time; molecular number is R T t over three pi viscosity radius mean-square displacement.

The first expression refers to the model mean square, or to an estimate obtained from an appropriate sample. The second is an inversion under the stated physical assumptions. An estimate needs uncertainty and checks of those assumptions; rearranging symbols does not remove experimental error.

Without an independent radius, the same D can result from many pairs of a and N. The data then select a compatible family rather than a unique molecular number. The existing synthetic laboratories explore this relationship but do not supply a historical measurement.

Open the foundation: Mean, variance and RMS

Open the foundation: Uncertainty, evidence and inference

Show every step here: What would let us count molecules?
  1. Begin with mean square = 2Dt. Divide both sides by 2t, for positive t.
  2. The result is D = mean square/(2t).
  3. The separate physical relation is D = RT/(6πηaN). Multiply both sides by 6πηaN.
  4. Divide by 6πηaD to obtain N = RT/(6πηaD).
  5. Insert mean square/(2t) for D. Dividing by that fraction multiplies by 2t/mean square.
  6. Cancel the factor two against six, giving N = RTt/(3πηa mean square).
  7. Notice that a and N occur as a product in the original relation. Without one, D alone cannot determine the other.
  8. A simulation using an assumed molecular number can test this inversion’s arithmetic, but cannot independently establish that number in nature.
Assumptions and limits: What would let us count molecules?

Assumed here

  • A suitable estimate of the model mean-square coordinate displacement or its time slope.
  • Independent values of R, T, η and tracer radius a, under the stated dilute-liquid model.

What this does not establish

  • A synthetic run generated from an assumed N is not independent evidence for N.
  • Diffusivity alone constrains the product aN when radius is unknown.
  • Measurement noise, drift, exposure and finite sampling require separate treatment.

Earlier step: Why viscosity changes the spread

Source context: Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

Explanation

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.

Open the tracer ensemble in this reading

BM-01 · A reproducible trial

Investigate the displacement argument

Static worked example

CurrentThese numbers match the current settings.

Model note
  • Primary output temperature: Host calculation (bm01.acceptedInputs). Owner bm01.acceptedInputs.
  • Primary output viscosity: Host calculation (bm01.acceptedInputs). Owner bm01.acceptedInputs.
  • Primary output particleRadius: Host calculation (bm01.acceptedInputs). Owner bm01.acceptedInputs.
  • Primary output observationInterval: Host calculation (bm01.acceptedInputs). Owner bm01.acceptedInputs.
  • Primary output boltzmannConstant: Host calculation (constants.modernSI2019). Owner constants.modernSI2019.
  • Primary output diffusionCoefficient: Host calculation (diffusion.stokesEinsteinD). Owner diffusion.stokesEinsteinD.
  • Primary output rmsDisplacement1d: Host calculation (diffusion.rmsDisplacement). Owner diffusion.rmsDisplacement.
  • Primary output modelSecondMoment: Host calculation (diffusion.moments). Owner diffusion.moments.
  • Primary output modelMeanNorm: Host calculation (diffusion.moments). Owner diffusion.moments.
  • Primary output modelRmsNorm: Host calculation (diffusion.moments). Owner diffusion.moments.
  • Primary output tracerPositions: Host reference calculation (recordTracers). Owner diffusion.recordTracers.
  • Primary output modelApparentSpeed: Host calculation (diffusion.apparentSpeed). Owner diffusion.apparentSpeed.
  • Primary output sampledApparentSpeed: Host calculation (diffusion.ensembleMoments). Owner diffusion.ensembleMoments.
  • Primary output traceCoordinates: Host calculation (diffusion.recordTracers). Owner diffusion.recordTracers.
  • Primary output traceTimes: Host calculation (bm01.measure). Owner bm01.measure.
  • Primary output histogramEdges: Host calculation (diffusion.displacementHistogram). Owner diffusion.displacementHistogram.
  • Primary output histogramCounts: Host calculation (diffusion.displacementHistogram). Owner diffusion.displacementHistogram.
  • Primary output histogramFrequencies: Host calculation (diffusion.displacementHistogram). Owner diffusion.displacementHistogram.
  • Primary output histogramModel: Host calculation (diffusion.intervalProbability). Owner diffusion.intervalProbability.
  • Primary output underflow: Host calculation (diffusion.displacementHistogram). Owner diffusion.displacementHistogram.
  • Primary output overflow: Host calculation (diffusion.displacementHistogram). Owner diffusion.displacementHistogram.
  • Primary output plotTimes: Host calculation (bm01.measure). Owner bm01.measure.
  • Primary output plotSampleMean: Host calculation (bm01.measure). Owner bm01.measure.
  • Primary output plotSampleMsd: Host calculation (bm01.measure). Owner bm01.measure.
  • Primary output plotSampleRms: Host calculation (bm01.measure). Owner bm01.measure.
  • Primary output plotSampleApparent: Host calculation (bm01.measure). Owner bm01.measure.
  • Primary output plotModelMean: Host calculation (bm01.measure). Owner bm01.measure.
  • Primary output plotModelMsd: Host calculation (bm01.measure). Owner bm01.measure.
  • Primary output plotModelRms: Host calculation (bm01.measure). Owner bm01.measure.
  • Primary output plotModelApparent: Host calculation (bm01.measure). Owner bm01.measure.
  • Primary output meanBand: Host calculation (diffusion.ensembleMomentBands). Owner diffusion.ensembleMomentBands.
  • Primary output secondMomentBand: Host calculation (diffusion.ensembleMomentBands). Owner diffusion.ensembleMomentBands.
  • Primary output recordingDraws: Host calculation (diffusion.recordTracers). Owner diffusion.recordTracers.
  • Primary output reusedRecording: Host calculation (bm01.measure). Owner bm01.measure.
  • Primary output ensembleSize: Host calculation (bm01.acceptedInputs). Owner bm01.acceptedInputs.
  • Primary output signedMean: Host calculation (diffusion.ensembleMoments). Owner diffusion.ensembleMoments.
  • Primary output meanSquare: Host calculation (diffusion.ensembleMoments). Owner diffusion.ensembleMoments.
  • Primary output lambdaX1s: Host calculation (diffusion.rmsDisplacement). Owner diffusion.rmsDisplacement.
  • Primary output lambdaX60s: Host calculation (diffusion.rmsDisplacement). Owner diffusion.rmsDisplacement.
  • Primary output signedMeanLowerBand: Host calculation (diffusion.ensembleMomentBands). Owner diffusion.ensembleMomentBands.
  • Primary output signedMeanUpperBand: Host calculation (diffusion.ensembleMomentBands). Owner diffusion.ensembleMomentBands.
  • Primary output meanSquareLowerBand: Host calculation (diffusion.ensembleMomentBands). Owner diffusion.ensembleMomentBands.
  • Primary output meanSquareUpperBand: Host calculation (diffusion.ensembleMomentBands). Owner diffusion.ensembleMomentBands.
  • Primary output kolmogorovDistance: Host calculation (diffusion.displacementHistogram). Owner diffusion.displacementHistogram.
  • Secondary output sampleMean: Host reduction (ensembleMoments). Owner diffusion.ensembleMoments.
  • Secondary output sampleMeanAbsolute: Host reduction (ensembleMoments). Owner diffusion.ensembleMoments.
  • Secondary output sampleMeanSquare: Host reduction (ensembleMoments). Owner diffusion.ensembleMoments.
  • Secondary output sampleRms: Host reduction (ensembleMoments). Owner diffusion.ensembleMoments.
  • Secondary output sampleMeanNorm: Host reduction (ensembleMoments). Owner diffusion.ensembleMoments.
  • Secondary output sampleMeanSquareNorm: Host reduction (ensembleMoments). Owner diffusion.ensembleMoments.
  • Secondary output sampleRmsNorm: Host reduction (ensembleMoments). Owner diffusion.ensembleMoments.
  • Seed 1905.
  • Accepted input revision 1.
  • Snapshot version 1.
  • Not modeled: Molecular collisions (no collision bath owns the displacement).
  • Show the code

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.

Record and observe

The preview records at most 8 MiB of latent paths. Its default is 400 tracers for 10 seconds at 0.02-second resolution. Larger requests are refused, never silently reduced.

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.

1 μm−5 μm+5 μmy displacement
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.
Fraction in each bin-4.6338 μm4.6338 μm
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 displacement0.77808 μm (modern-si-2019)
Mean-square coordinate displacement0.97223 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:=λxtv_{\mathrm{app}} := \frac{\lambda_x}{t}

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=kB T6 π η aD = \frac{k_B\,T}{6\,\pi\,\eta\,a}

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=2 D t\lambda_x = \sqrt{2\,D\,t}

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

μm · log scale0.02 s10 s
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)SampleModel
0.020.130380.13106
0.040.185560.18535
0.10.303680.29307
0.20.404160.41446
0.40.613880.58613
10.986020.92676
21.34951.3106
41.90351.8535
102.87232.9307

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.

Evaluator source digest: source:sha256:06f140ad57b373415f0056e72c0a61cee543139245eb0f242333675672ccd7b2

Show the code

Audited TypeScript reference evaluator: the owner on this device, or the host fallback for a FrankenSim capability.

stokesEinsteinD · src/physics/reference/diffusion/distributions.ts · revision workspace · sha256:a41fa1ff39026dbed9a701b368028012995d1835d9acebb2e32d5010f47404ec

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. */
export function stokesEinsteinD(
  {
    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))
    return outside(
      "diffusionCoefficient",
      "stokesEinsteinD",
      "T > 0, eta > 0, a > 0",
      "Enter positive finite temperature, viscosity, and particle radius.",
      "input",
      set,
    );
  if (medium !== "liquid")
    return outside(
      "diffusionCoefficient",
      "stokesEinsteinD",
      "stokes-gas-medium",
      "The liquid Stokes-drag model does not include the slip correction needed in a gas.",
      "model",
      set,
    );
  const k = thermalConstant(set);
  return number(
    "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).
StepExpressionValueUnit
The gas constant in SIR = 8.31e7 erg mol^{-1} K^{-1}8.31J mol^{-1} K^{-1}
Thermal energy per moleRT2411.1465J mol^{-1}
Drag on one sphere6 π η a1.2723450247038662e-8kg s^{-1}
Drag on a mole of spheresN 6 π η a7634070148223197kg s^{-1} mol^{-1}
Diffusion coefficientRT / (N 6 π η a)3.158402337396894e-13m^2 s^{-1}
Coordinate RMS at 1 sλ_x = √(2 D t) at t = 1 s7.947832833416785e-7m
Coordinate RMS at 60 sλ_x = √(2 D t) at t = 60 s0.000006156364840452743m

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rmsDisplacement · src/physics/reference/diffusion/distributions.ts · revision workspace · sha256:6d2d381222179a43397683f83d95eecf9dab253844f72dfcaa1eaab1302222eb

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

export function rmsDisplacement(D: number, t: number): Evaluation {
  if (!validDt(D, t))
    return outside(
      "rmsDisplacement1d",
      "rmsDisplacement",
      "D >= 0 and t >= 0",
      "Diffusivity and elapsed time must be finite and nonnegative.",
    );
  return number(
    "rmsDisplacement1d",
    "rmsDisplacement",
    scale(D, t),
    undefined,
    undefined,
    D > 0 && t > 0,
  );
}

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apparentSpeed · src/physics/reference/diffusion/distributions.ts · revision workspace · sha256:1f2137435fdc15b0a9b5c91cc4dafba680e8d52b0f5143fdbefdfa9945f6baec

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

export function apparentSpeed(D: number, tau: number): Evaluation {
  if (!validDt(D, tau) || tau === 0)
    return outside(
      "apparentSpeed",
      "apparentSpeed",
      "D >= 0 and tau > 0",
      "An apparent speed needs a positive observation interval.",
    );
  return number(
    "apparentSpeed",
    "apparentSpeed",
    (Math.SQRT2 * Math.sqrt(D)) / Math.sqrt(tau),
    undefined,
    undefined,
    D > 0,
  );
}

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ensembleMoments · src/physics/reference/diffusion/tracers.ts · revision workspace · sha256:1d87c8e69590b8871bcc7a6390089003f59d0b80bfc7eb1919a57dec661743e6

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. */
export function ensembleMoments({
  displacements,
  d,
  dt,
  viewport,
}: {
  displacements: Float64Array;
  d: number;
  dt?: number;
  viewport?: ViewportBounds;
}): Computation<EnsembleMoments> {
  if (
    !(displacements instanceof Float64Array) ||
    ![1, 2, 3].includes(d) ||
    displacements.length === 0 ||
    displacements.length % d !== 0 ||
    displacements.length > 30000 ||
    !displacements.every(Number.isFinite)
  )
    return invalid(
      ["displacements", "d"],
      "Provide finite, row-major displacements with one, two or three coordinates per tracer.",
    );
  const M = displacements.length / d,
    sums = new Float64Array(d * 3 + 1),
    corrections = new Float64Array(d * 3 + 1);
  function add(i: number, v: number) {
    const corr = corrections[i] ?? 0,
      sum = sums[i] ?? 0,
      y = v - corr,
      t = sum + y;
    corrections[i] = t - sum - y;
    sums[i] = t;
  }
  for (let i = 0; i < M; i++) {
    let norm = 0;
    for (let j = 0; j < d; j++) {
      const v = displacements[i * d + j];
      if (v === undefined) return failure("A required displacement coordinate was undefined.");
      if (v !== 0 && v * v === 0)
        return failure("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);
  }
  const axes = Array.from({ length: d }, (_, j) => {
    const mean = (sums[j * 3] ?? 0) / M;
    const meanAbsolute = (sums[j * 3 + 1] ?? 0) / M;
    const meanSquare = (sums[j * 3 + 2] ?? 0) / M;
    return {
      mean,
      meanAbsolute,
      meanSquare,
      rms: Math.sqrt(meanSquare),
    };
  });
  const meanSquareNorm = axes.reduce((s, a) => s + a.meanSquare, 0);
  if (![...sums, meanSquareNorm].every(Number.isFinite))
    return failure("The moment reduction exceeded the numerical range.");
  const rmsNorm = Math.sqrt(meanSquareNorm);
  let insideCount: number | undefined;
  let outsideCount: number | undefined;
  if (viewport) {
    let inc = 0;
    let outc = 0;
    for (let i = 0; i < M; i++) {
      let isInside = true;
      for (let j = 0; j < d; j++) {
        const v = displacements[i * d + j] ?? 0;
        const minVal = viewport.min[j] ?? -Infinity;
        const maxVal = viewport.max[j] ?? Infinity;
        if (v < minVal || v > maxVal) {
          isInside = false;
          break;
        }
      }
      if (isInside) inc++;
      else outc++;
    }
    insideCount = inc;
    outsideCount = outc;
  }
  const apparentSpeed =
    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.

displacementHistogram · src/physics/reference/diffusion/tracers.ts · revision workspace · sha256:746f24cee7b8573b8a986cd8e072b4786549eba47dffca106dbd38582d670c7b

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

export function displacementHistogram(
  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)) return false;
      if (i === 0) return true;
      const prev = edges[i - 1];
      return prev !== undefined && v > prev;
    })
  )
    return invalid(
      ["values", "edges"],
      "Use finite samples and strictly increasing histogram edges.",
    );
  const counts = new Float64Array(edges.length - 1);
  let underflow = 0,
    overflow = 0;
  const firstEdge = edges[0];
  const lastEdge = edges[edges.length - 1];
  if (firstEdge === undefined || lastEdge === undefined)
    return invalid(["edges"], "Histogram edges array must contain at least two finite bounds.");
  for (const value of values) {
    if (value < firstEdge) {
      underflow++;
      continue;
    }
    if (value > lastEdge) {
      overflow++;
      continue;
    }
    let lo = 0,
      hi = edges.length - 1;
    while (hi - lo > 1) {
      const mid = (lo + hi) >>> 1;
      const edgeMid = edges[mid];
      if (edgeMid !== undefined && value < edgeMid) hi = mid;
      else lo = mid;
    }
    const bin = Math.min(lo, counts.length - 1);
    const prevCount = 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
IntervalCount
Below the plotted range0
-4.6338 to -4.40210
-4.4021 to -4.17040
-4.1704 to -3.93870
-3.9387 to -3.7070
-3.707 to -3.47530
-3.4753 to -3.24370
-3.2437 to -3.0120
-3.012 to -2.78031
-2.7803 to -2.54861
-2.5486 to -2.31690
-2.3169 to -2.08527
-2.0852 to -1.85357
-1.8535 to -1.62186
-1.6218 to -1.390116
-1.3901 to -1.158412
-1.1584 to -0.9267615
-0.92676 to -0.6950728
-0.69507 to -0.4633824
-0.46338 to -0.2316939
-0.23169 to 039
0 to 0.2316943
0.23169 to 0.4633837
0.46338 to 0.6950726
0.69507 to 0.9267632
0.92676 to 1.158423
1.1584 to 1.390117
1.3901 to 1.62188
1.6218 to 1.85356
1.8535 to 2.08526
2.0852 to 2.31693
2.3169 to 2.54861
2.5486 to 2.78031
2.7803 to 3.0121
3.012 to 3.24371
3.2437 to 3.47530
3.4753 to 3.7070
3.707 to 3.93870
3.9387 to 4.17040
4.1704 to 4.40210
4.4021 to 4.63380
Above the plotted range0
Total, including both tails400
Notes and laboratory

Beside this passage

Margin notes, assumptions, and the laboratory stay here so the German argument keeps the main column.

Keep the tracer ensemble in view

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