IPhO 2023 · Theory problem 1

Characterization of Soil Colloids

Colloidal science is useful to characterize soil particles because many of them can be regarded as colloidal particles of micrometer size. For example, Brownian motion (random motion of colloidal particles) can be used to measure particle sizes.

Part A. Motions of colloidal particles (1.6 points)

We analyze the one-dimensional Brownian motion of a colloidal particle with mass \(M\). The equation of motion for its velocity \(v(t)\) reads:

\[ M\dot v=-\gamma v(t)+F(t)+F_{\mathrm{ext}}(t), \tag{1} \]

where \(\gamma\) is the friction coefficient, \(F(t)\) is a force due to random collisions with water molecules, and \(F_{\mathrm{ext}}(t)\) is an external force. In Part A, we assume \(F_{\mathrm{ext}}(t)=0\).

A.1 — 0.8pt

Consider that a water molecule collides with the particle at \(t=t_0\), giving impulse \(I_0\), and \(F(t)=0\) afterward. If \(v(t)=0\) before the collision, \(v(t)=v_0e^{-(t-t_0)/\tau}\) for \(t>t_0\). Determine \(v_0\) and \(\tau\), using \(I_0\) and necessary parameters in Eq.(1).

In the following, you may use \(\tau\) in your answers.

A.2 — 0.8pt

Actually, water molecules collide with the particle one after another. Suppose the \(i\)th collision gives the impulse \(I_i\) at time \(t_i\) and determine \(v(t)\) on condition that \(t>0\) and \(v(0)=0\). Also give the inequality specifying the range of \(t_i\) that needs to be considered for a given \(t\). In the answer sheet, it is not necessary to specify this range in the expression for \(v(t)\).

Part B. Effective equation of motion (1.8 points)

Results so far imply that particle velocities \(v(t)\) and \(v(t')\) may be regarded as uncorrelated random quantities if \(|t-t'|\gg\tau\). On this basis, we introduce a theoretical model to approximately describe the one-dimensional Brownian motion, where the velocity changes randomly at each time interval \(\delta\) \((\gg\tau)\), i.e.,

\[ v(t)=v_n\qquad(t_{n-1}<t\leq t_n), \tag{2} \]

with \(t_n=n\delta\) \((n=0,1,2,\cdots)\) and a random quantity \(v_n\). It satisfies

\[ \langle v_n\rangle=0,\qquad \langle v_nv_m\rangle= \begin{cases} C &(n=m),\\ 0 &(n\neq m), \end{cases} \tag{3} \]

with a parameter \(C\) depending on \(\delta\). Here \(\langle X\rangle\) indicates the expectation value of \(X\). That is, if you draw random numbers \(X\) infinite times, the mean will be \(\langle X\rangle\).

Now we consider the particle displacement \(\Delta x(t)=x(t)-x(0)\) for \(t=N\delta\) with an integer \(N\).

B.1 — 1.0pt

Determine \(\langle\Delta x(t)\rangle\) and \(\langle\Delta x(t)^2\rangle\) using \(C\), \(\delta\), and \(t\).

B.2 — 0.8pt

The quantity \(\langle\Delta x(t)^2\rangle\) is called the mean square displacement (MSD). It is a characteristic observable of the Brownian motion, which corresponds to the limiting case \(\delta\to0\). From this, we can show \(C\propto\delta^\alpha\) and \(\langle\Delta x(t)^2\rangle\propto t^\beta\). Determine the values of \(\alpha\) and \(\beta\).

Part C. Electrophoresis (2.7 points)

Here we discuss electrophoresis, i.e., transport of charged particles by an electric field. Suspension of colloidal particles with mass \(M\) and charge \(Q\) \((>0)\) is put in a narrow channel with a cross-section \(A\) (Fig.1(a)). We ignore the interaction between particles, effects of the wall, the fluid, the ions therein, and gravity.

Figure 1

Fig.1: Setting for Part C.

By applying a uniform electric field \(E\) in the \(x\)-direction, particles are transported and their concentration \(n(x)\) (particle number per unit volume) becomes non-uniform (Fig.1(b)). When \(E\) is removed, this non-uniformity gradually disappears. This is due to Brownian motion of particles. If \(n(x)\) is not uniform, the numbers of right-going and left-going particles may differ (Fig.1(c)). This generates a particle flux \(J_D(x)\), the mean number of particles flowing at \(x\) along the \(x\)-axis per unit cross-sectional area and unit time. This flux is known to satisfy

\[ J_D(x)=-D\frac{dn}{dx}(x), \tag{4} \]

where \(D\) is called the diffusion coefficient.

Now let's assume, for simplicity, that half of the particles have velocity \(+v\) and the other half have velocity \(-v\). Let \(N_+(x_0)\) be the number of particles with velocity \(+v\) that cross \(x_0\) from left to right per unit cross-sectional area and unit time. For particles with velocity \(+v\) to cross \(x_0\) in the time interval \(\delta\), they should be in the shaded region of Fig.1(c). Since \(\delta\) is small, we have \(n(x)\simeq n(x_0)+(x-x_0)\frac{dn}{dx}(x_0)\) in this region.

C.1 — 0.5pt

Express \(N_+(x_0)\) using necessary quantities from \(v\), \(\delta\), \(n(x_0)\), and \(\frac{dn}{dx}(x_0)\).

We define \(N_-(x_0)\) as the counterpart of \(N_+(x_0)\) for the velocity \(-v\). With this, we have \(J_D(x_0)=\langle N_+(x_0)-N_-(x_0)\rangle\). According to Eq.(3), we have \(\langle v^2\rangle=C\).

C.2 — 0.7pt

Determine \(J_D(x_0)\) using necessary quantities from \(C\), \(\delta\), \(n(x_0)\), and \(\frac{dn}{dx}(x_0)\). Using this and Eq.(4), express \(D\) in terms of \(C\) and \(\delta\), and \(\langle\Delta x(t)^2\rangle\) in terms of \(D\) and \(t\).

Now we discuss the effect of osmotic pressure \(\Pi\). It is given by \(\Pi=\frac{n}{N_A}RT=nkT\) with the Avogadro constant \(N_A\), the gas constant \(R\), temperature \(T\), and the Boltzmann constant \(k=\frac{R}{N_A}\). Let us consider the non-uniform concentration formed under the electric field \(E\) (Fig.1(b)). Since \(n(x)\) depends on \(x\), so does \(\Pi(x)\). Then the forces due to \(\Pi(x)\) and \(\Pi(x+\Delta x)\) must be balanced with the total force from the field \(E\) acting on the particles (Fig.2). Here we consider small \(\Delta x\), so that \(n(x)\) can be regarded as constant over this range, while \(n(x+\Delta x)-n(x)\simeq\Delta x\frac{dn}{dx}(x)\).

Figure 2

Fig.2: Force balance.

C.3 — 0.5pt

Express \(\frac{dn}{dx}(x)\) using \(n(x)\), \(T\), \(Q\), \(E\), and \(k\).

Let us discuss the balance of the flux now. Besides the flux \(J_D(x)\) due to the Brownian motion, there is also a flux due to the electric field, \(J_Q(x)\). It is given by

\[ J_Q(x)=n(x)u, \tag{5} \]

where \(u\) is the terminal velocity of particles driven by the field.

C.4 — 0.5pt

To determine \(u\), we use Eq.(1) with \(F_{\mathrm{ext}}(t)=QE\). Since \(v(t)\) is fluctuating, we consider \(\langle v(t)\rangle\). Assuming \(\langle v(0)\rangle=0\) and using \(\langle F(t)\rangle=0\), evaluate \(\langle v(t)\rangle\) and obtain \(u=\lim_{t\to\infty}\langle v(t)\rangle\).

C.5 — 0.5pt

The flux balance reads \(J_D(x)+J_Q(x)=0\). Express the diffusion coefficient \(D\) in terms of \(k\), \(\gamma\), and \(T\).

Part D. Mean square displacement (2.4 points)

Suppose we observed the Brownian motion of an isolated, spherical colloidal particle with radius \(a=5.0\,\mu\mathrm{m}\) in water. Figure 3 shows the histogram of displacements \(\Delta x\) measured in the \(x\)-direction at every interval \(\Delta t=60\,\mathrm{s}\). The friction coefficient is given by \(\gamma=6\pi a\eta\) with water viscosity \(\eta=8.9\times10^{-4}\,\mathrm{Pa\cdot s}\) and the temperature was \(T=25\,^\circ\mathrm{C}\).

Figure 3

Fig.3: Histogram of displacements.

\(\Delta x\;(\mu\mathrm{m})\) \(-10\) \(-9\) \(-8\) \(-7\) \(-6\) \(-5\) \(-4\)
\(N_{\mathrm{count}}\) 0 0 0 2 4 11 15
\(\Delta x\;(\mu\mathrm{m})\) \(-3\) \(-2\) \(-1\) \(0\) \(1\) \(2\) \(3\)
\(N_{\mathrm{count}}\) 35 44 58 68 59 50 26
\(\Delta x\;(\mu\mathrm{m})\) \(4\) \(5\) \(6\) \(7\) \(8\) \(9\) \(10\)
\(N_{\mathrm{count}}\) 15 5 4 3 0 1 0

D.1 — 1.0pt

Estimate the value of \(N_A\) without using the fact that it is the Avogadro constant, up to two significant digits from the data in Fig.3. The gas constant is \(R=8.31\,\mathrm{J/K\cdot mol}\). Do not use the value of the Boltzmann constant \(k\) given in General Instructions. As for the Avogadro constant, you might obtain a value different from that in General Instructions.

Now we extend the model in Part B to describe the motion of a particle with charge \(Q\) under an electric field \(E\). The particle velocity \(v(t)\) considered in Eq.(2) should be replaced by \(v(t)=u+v_n\) \((t_{n-1}<t\leq t_n)\) with \(v_n\) satisfying Eq. (3) and \(u\) being the terminal velocity considered in Eq.(5).

D.2 — 0.8pt

Express the MSD \(\langle\Delta x(t)^2\rangle\) in terms of \(u\), \(D\), and \(t\). Obtain approximate power laws for small \(t\) and large \(t\), as well as the characteristic time \(t_*\) where this change occurs. Draw a rough graph of MSD in a log-log plot, indicating the approximate location of \(t_*\).

Next, we consider swimming microbes (Fig.4(a)), in one dimension for simplicity (Fig.4(b)). These are spherical particles with radius \(a\). They swim at velocity either \(+u_0\) or \(-u_0\), the sign chosen randomly at every time interval \(\delta_0\) without correlation. The observed motion is a combination of displacements due to swimming and those due to the Brownian motion of a spherical particle.

Figure 4

Fig.4: (a) Motion of microbes. (b) Its one-dimensional version.

D.3 — 0.6pt

Figure 5 displays the MSD \(\langle\Delta x(t)^2\rangle\) of those microbes, showing different power laws for small, large, and intermidiate \(t\), as indicated by dashed lines. Obtain the power law for each time range and express it using necessary quantities from \(D\), \(u_0\), \(\delta_0\), and \(t\).

Figure 5

Fig.5: Mean square displacement of the microbes.

Part E. Water purification (1.5 points)

Here we discuss the purification of water including colloid-like soil particles, by adding electrolytes to coagulate them. Particles interact through van der Waals force and electrostatic force, the latter including effects of both surface charges and the surrounding layer of oppositely charged ions (such ions and their layer are called counter-ions and the electric double layer, respectively; see Fig.6(a)). As a result, the interaction potential for particle distance \(d\) (Fig.6(b)) is given by

\[ U(d)=-\frac{A}{d}+\frac{B\epsilon(kT)^2}{q^2}e^{-d/\lambda}, \tag{6} \]

where \(A\) and \(B\) are positive constants, \(\epsilon\) is the dielectric constant of water, and \(\lambda\) is the thickness of the electric double layer. Assuming that charges of ions are \(\pm q\), we have

\[ \lambda=\sqrt{\frac{\epsilon kT}{2N_Aq^2c}}, \tag{7} \]

where \(c\) is the molar concentration of ion.

Figure 6

Fig.6: (a) Surface charges of colloidal particles and counter-ions. (b) Definition of the distance \(d\).

E.1 — 1.5pt

Addition of sodium chloride (NaCl) to the suspension causes colloidal particles to coagulate. Determine the lowest concentration \(c\) of NaCl necessary for coagulation. It is sufficient to consider two particles without thermal fluctuations, i.e., \(F(t)=0\) in Eq.(1), and assume that the terminal velocity for the given potential force is reached instantaneously.

Official English text of the problem.

Answers

  • A.1 \[ \boxed{v_0=\frac{I_0}{M},\qquad \tau=\frac{M}{\gamma}.} \]

  • A.2 \[ \boxed{v(t)=\frac1M\sum_{0<t_i<t}I_i e^{-(t-t_i)/\tau},\qquad 0<t_i<t.} \] Immediately after a collision at the observation time, include \(t_i=t\).

  • B.1 \[ \boxed{\langle\Delta x(t)\rangle=0,\qquad \langle\Delta x(t)^2\rangle=C\delta t.} \]

  • B.2 \[ \boxed{\alpha=-1,\qquad\beta=1.} \]

  • C.1 \[ \boxed{N_+(x_0)\simeq \frac v2n(x_0)-\frac{v^2\delta}{4}\frac{dn}{dx}(x_0).} \]

  • C.2 \[ \boxed{J_D(x_0)=-\frac{C\delta}{2}\frac{dn}{dx}(x_0),\qquad D=\frac{C\delta}{2},\qquad \langle\Delta x(t)^2\rangle=2Dt.} \]

  • C.3 \[ \boxed{\frac{dn}{dx}(x)=\frac{QE}{kT}n(x).} \]

  • C.4 \[ \boxed{\langle v(t)\rangle=\frac{QE}{\gamma} \left(1-e^{-t/\tau}\right),\qquad u=\frac{QE}{\gamma}.} \]

  • C.5 \[ \boxed{D=\frac{kT}{\gamma}.} \]

  • D.1 \[ \boxed{N_A\simeq5.6\times10^{23}\,\mathrm{mol}^{-1}.} \]

  • D.2 \[ \boxed{\langle\Delta x(t)^2\rangle=2Dt+u^2t^2,\qquad t_*=\frac{2D}{u^2}.} \] For \(t\ll t_*\), the MSD is approximately \(2Dt\), with log-log slope \(1\). For \(t\gg t_*\), it is approximately \(u^2t^2\), with slope \(2\).

  • D.3 Define \(t_s=2D/u_0^2\). For the three distinct regimes shown, \(t_s\ll\delta_0\), and \[ \boxed{ \langle\Delta x(t)^2\rangle\simeq \begin{cases} 2Dt,&t\ll t_s,\\ u_0^2t^2,&t_s\ll t\ll\delta_0,\\ (2D+u_0^2\delta_0)t\simeq u_0^2\delta_0t,&t\gg\delta_0. \end{cases}} \] The respective log-log slopes are \(1,2,1\).

  • E.1 The critical concentration is \[ \boxed{c_{\mathrm{crit}}= \frac{8B^2\epsilon^3(kT)^5} {\exp(4)\,N_AA^2q^6},} \] where \(q\) is the elementary charge magnitude for NaCl. Coagulation from large separation occurs for \(c>c_{\mathrm{crit}}\). Exactly at \(c=c_{\mathrm{crit}}\), the stated noiseless, overdamped dynamics are marginal: the particles approach a finite separation asymptotically.

Solution

A.1

Integrating the equation of motion over the very short collision gives the impulse–momentum relation. The frictional impulse vanishes as the collision duration tends to zero, so

\[ M\bigl[v(t_0^+)-v(t_0^-)\bigr]=I_0. \]

Since the particle was initially at rest,

\[ v_0=v(t_0^+)=\frac{I_0}{M}. \]

After the collision, both forces \(F\) and \(F_{\mathrm{ext}}\) vanish. Thus

\[ M\frac{dv}{dt}=-\gamma v, \]

whose solution is

\[ v(t)=\frac{I_0}{M} \exp\left[-\frac{\gamma}{M}(t-t_0)\right]. \]

Comparison with the given form yields

\[ \boxed{v_0=\frac{I_0}{M},\qquad \tau=\frac{M}{\gamma}.} \]

A.2

The equation of motion is linear. Each collision produces an instantaneous velocity increment \(I_i/M\), which subsequently decays with relaxation time \(\tau\). Superposition therefore gives

\[ \boxed{v(t)=\frac1M\sum_{0<t_i<t}I_i \exp\left[-\frac{t-t_i}{\tau}\right].} \]

Only collisions after the initial time and before the observation contribute:

\[ \boxed{0<t_i<t.} \]

At a collision instant, the velocity has a jump \(I_i/M\). If \(v(t)\) denotes the velocity immediately after that collision, the upper endpoint is included.

B.1

During interval \(n\), the displacement is \(v_n\delta\). Hence, at \(t=N\delta\),

\[ \Delta x(t)=\delta\sum_{n=1}^{N}v_n. \]

Its expectation is

\[ \langle\Delta x(t)\rangle =\delta\sum_{n=1}^{N}\langle v_n\rangle=0. \]

For the second moment,

\[ \begin{aligned} \langle\Delta x(t)^2\rangle &=\delta^2\sum_{n=1}^{N}\sum_{m=1}^{N} \langle v_nv_m\rangle\\ &=\delta^2NC\\ &=C\delta t. \end{aligned} \]

The cross terms vanish because different intervals have uncorrelated velocities. Thus

\[ \boxed{\langle\Delta x(t)\rangle=0,\qquad \langle\Delta x(t)^2\rangle=C\delta t.} \]

B.2

At a fixed positive observation time, Brownian motion must have a finite, nonzero MSD in the diffusion limit. Since

\[ \langle\Delta x(t)^2\rangle=C\delta t, \]

the product \(C\delta\) must tend to a finite, nonzero constant as \(\delta\to0\). Therefore

\[ C\propto\delta^{-1}. \]

The remaining time dependence is linear:

\[ \langle\Delta x(t)^2\rangle\propto t. \]

Consequently,

\[ \boxed{\alpha=-1,\qquad\beta=1.} \]

This is the continuum limit of the effective model, describing motion on time scales much longer than the microscopic relaxation time \(\tau\).

C.1

Particles moving rightward cross \(x_0\) during the next interval \(\delta\) if they initially lie between \(x_0-v\delta\) and \(x_0\). Half of the particles in this region move rightward. Dividing their number per unit area by \(\delta\),

\[ N_+(x_0)=\frac{1}{2\delta} \int_{x_0-v\delta}^{x_0}n(x)\,dx. \]

Using the prescribed linear approximation,

\[ n(x)\simeq n(x_0)+(x-x_0)n'(x_0), \]

we obtain

\[ \begin{aligned} N_+(x_0) &\simeq\frac{1}{2\delta} \left[v\delta\,n(x_0) -\frac{(v\delta)^2}{2}n'(x_0)\right]. \end{aligned} \]

Therefore,

\[ \boxed{N_+(x_0)\simeq \frac v2n(x_0)-\frac{v^2\delta}{4}n'(x_0).} \]

This approximation requires the concentration to vary slowly over a distance \(v\delta\).

C.2

Similarly, particles moving leftward originate in \(x_0<x<x_0+v\delta\), giving

\[ \begin{aligned} N_-(x_0) &=\frac{1}{2\delta}\int_{x_0}^{x_0+v\delta}n(x)\,dx\\ &\simeq\frac v2n(x_0)+\frac{v^2\delta}{4}n'(x_0). \end{aligned} \]

Subtracting and averaging,

\[ \begin{aligned} J_D(x_0) &=\langle N_+(x_0)-N_-(x_0)\rangle\\ &=-\frac{\delta}{2}\langle v^2\rangle n'(x_0)\\ &=-\frac{C\delta}{2}n'(x_0). \end{aligned} \]

Comparison with Fick’s law, \(J_D=-Dn'\), gives

\[ \boxed{D=\frac{C\delta}{2}.} \]

Substituting into B.1,

\[ \boxed{\langle\Delta x(t)^2\rangle=2Dt.} \]

C.3

Consider the slab of cross-sectional area \(A\) and thickness \(\Delta x\). Its electric force in the positive \(x\)-direction is

\[ F_E=n(x)A\Delta x\,QE. \]

The net osmotic-pressure force is

\[ F_\Pi=A\Pi(x)-A\Pi(x+\Delta x) \simeq-A\Delta x\,\frac{d\Pi}{dx}. \]

Force balance gives

\[ n(x)QE=\frac{d\Pi}{dx}. \]

At uniform temperature, \(\Pi=nkT\), so

\[ \frac{d\Pi}{dx}=kT\frac{dn}{dx}. \]

Hence

\[ \boxed{\frac{dn}{dx}(x)=\frac{QE}{kT}n(x).} \]

The positive gradient is consistent with positively charged particles accumulating in the direction of the electric field.

C.4

Taking the expectation of Eq.(1), using the constant applied force and \(\langle F(t)\rangle=0\), yields

\[ M\frac{d\langle v\rangle}{dt} =-\gamma\langle v\rangle+QE. \]

The solution satisfying \(\langle v(0)\rangle=0\) is

\[ \boxed{\langle v(t)\rangle= \frac{QE}{\gamma}\left(1-e^{-\gamma t/M}\right) =\frac{QE}{\gamma}\left(1-e^{-t/\tau}\right).} \]

Therefore the terminal velocity is

\[ \boxed{u=\lim_{t\to\infty}\langle v(t)\rangle=\frac{QE}{\gamma}.} \]

C.5

At equilibrium, the diffusive flux balances the electric drift flux:

\[ -D\frac{dn}{dx}+nu=0. \]

Substituting the results of C.3 and C.4,

\[ -D\frac{nQE}{kT}+n\frac{QE}{\gamma}=0. \]

Thus

\[ \boxed{D=\frac{kT}{\gamma}.} \]

This is the Einstein relation between diffusion and friction.

D.1

Using the tabulated displacements as the histogram bin representatives, the total number of measurements is

\[ N_{\mathrm{tot}}=\sum_jN_j=400. \]

The weighted sum of squared displacements is

\[ \sum_jN_j(\Delta x_j)^2=2536\,\mu\mathrm{m}^2. \]

Therefore the estimated MSD over \(\Delta t=60\,\mathrm{s}\) is

\[ \widehat{\langle\Delta x^2\rangle} =\frac{2536}{400}\,\mu\mathrm{m}^2 =6.34\times10^{-12}\,\mathrm{m}^2. \]

From C.2,

\[ D=\frac{\widehat{\langle\Delta x^2\rangle}}{2\Delta t} =\frac{6.34\times10^{-12}}{120} =5.2833\times10^{-14}\,\mathrm{m}^2\,\mathrm{s}^{-1}. \]

The friction coefficient is

\[ \begin{aligned} \gamma &=6\pi a\eta\\ &=6\pi(5.0\times10^{-6})(8.9\times10^{-4})\\ &=8.3881\times10^{-8}\,\mathrm{kg\,s^{-1}}. \end{aligned} \]

Using \(k=R/N_A\) in the Einstein relation gives

\[ D=\frac{RT}{N_A\gamma}, \]

so

\[ N_A=\frac{RT}{\gamma D} =\frac{2RT\Delta t} {\gamma\,\widehat{\langle\Delta x^2\rangle}}. \]

With \(T=298.15\,\mathrm{K}\),

\[ N_A =\frac{8.31(298.15)} {(8.3881\times10^{-8})(5.2833\times10^{-14})} =5.5907\times10^{23}\,\mathrm{mol}^{-1}. \]

To two significant digits,

\[ \boxed{N_A\simeq5.6\times10^{23}\,\mathrm{mol}^{-1}.} \]

Neither an accepted value of \(N_A\) nor a numerical value of \(k\) was used.

D.2

With the terminal drift included, the displacement is

\[ \Delta x(t)=ut+\delta\sum_{n=1}^{N}v_n. \]

The random part has zero mean and second moment \(2Dt\). Thus the cross term between drift and random displacement averages to zero:

\[ \begin{aligned} \langle\Delta x(t)^2\rangle &=u^2t^2 +2ut\,\delta\sum_n\langle v_n\rangle +\delta^2\sum_{n,m}\langle v_nv_m\rangle\\ &=u^2t^2+2Dt. \end{aligned} \]

Therefore,

\[ \boxed{\langle\Delta x(t)^2\rangle=2Dt+u^2t^2.} \]

The crossover occurs when the two contributions are equal:

\[ 2Dt_*=u^2t_*^2, \]

giving

\[ \boxed{t_*=\frac{2D}{u^2}.} \]

The limiting power laws are

\[ \boxed{ \begin{aligned} t\ll t_*:&\qquad \langle\Delta x(t)^2\rangle\simeq2Dt,\\ t\gg t_*:&\qquad \langle\Delta x(t)^2\rangle\simeq u^2t^2. \end{aligned}} \]

Here “small \(t\)” refers to the effective diffusive model, with the microscopic relaxation already neglected and the terminal drift established.

For a precise description of the requested log-log sketch, define

\[ s=\frac{t}{t_*},\qquad Y=\frac{\langle\Delta x(t)^2\rangle}{2Dt_*}. \]

Then

\[ Y=s+s^2. \]

The graph follows the straight asymptote \(Y=s\), of slope \(1\), at small \(s\), bends smoothly upward near \(s=1\), and approaches \(Y=s^2\), of slope \(2\), at large \(s\). The two asymptotes intersect at \((s,Y)=(1,1)\), while the actual curve passes through \((1,2)\). Thus mark \(t=t_*\) at the bend. At that point,

\[ \langle\Delta x(t_*)^2\rangle =\frac{8D^2}{u^2}. \]

D.3

Write the displacement as the sum of thermal and swimming contributions:

\[ \Delta x(t)=\Delta x_B(t)+\Delta x_S(t). \]

The thermal motion and the swimming direction choices are taken to be independent, and both contributions have zero mean. Hence

\[ \langle\Delta x(t)^2\rangle =2Dt+\langle\Delta x_S(t)^2\rangle. \]

For \(t\ll\delta_0\), the swimming direction is effectively unchanged during the observation interval. Therefore

\[ \Delta x_S(t)\simeq\pm u_0t, \qquad \langle\Delta x_S(t)^2\rangle\simeq u_0^2t^2. \]

Thus

\[ \langle\Delta x(t)^2\rangle\simeq2Dt+u_0^2t^2 \qquad(t\ll\delta_0). \]

The two terms become comparable at

\[ t_s=\frac{2D}{u_0^2}. \]

Consequently, thermal diffusion dominates for \(t\ll t_s\), and swimming produces a ballistic regime for \(t_s\ll t\ll\delta_0\).

For long times, consider \(N\) complete swimming intervals, with \(t=N\delta_0\). Let \(s_j=\pm1\) denote the independently chosen sign in interval \(j\). Then

\[ \Delta x_S(t)=u_0\delta_0\sum_{j=1}^{N}s_j. \]

Because \(\langle s_j\rangle=0\) and \(\langle s_js_\ell\rangle=0\) for \(j\neq\ell\),

\[ \begin{aligned} \langle\Delta x_S(t)^2\rangle &=u_0^2\delta_0^2N\\ &=u_0^2\delta_0t. \end{aligned} \]

An incomplete final interval does not alter this leading behavior at \(t\gg\delta_0\). Therefore the long-time MSD is

\[ \langle\Delta x(t)^2\rangle \simeq(2D+u_0^2\delta_0)t. \]

Equivalently, the long-time diffusion coefficient is

\[ D_{\mathrm{eff}}=D+\frac{u_0^2\delta_0}{2}. \]

For a well-developed intermediate regime, as in the figure, the time scales must satisfy

\[ \frac{2D}{u_0^2}\ll\delta_0. \]

This also implies \(u_0^2\delta_0\gg2D\). The three observed asymptotes are therefore

\[ \boxed{ \begin{array}{c|c|c} \text{Time range}&\langle\Delta x(t)^2\rangle&\text{Log-log slope}\\ \hline t\ll 2D/u_0^2&2Dt&1\\ 2D/u_0^2\ll t\ll\delta_0&u_0^2t^2&2\\ t\gg\delta_0&u_0^2\delta_0t&1 \end{array}} \]

with the full long-time coefficient \(2D+u_0^2\delta_0\) retained when thermal diffusion is appreciable.

E.1

Define the positive energy scale

\[ K=\frac{B\epsilon(kT)^2}{q^2}. \]

Then

\[ U(d)=-\frac{A}{d}+Ke^{-d/\lambda}, \]

and

\[ U'(d)=\frac{A}{d^2}-\frac{K}{\lambda}e^{-d/\lambda}. \]

The force tending to increase the separation is \(-U'(d)\). Under the instantaneous-terminal-velocity assumption, each particle’s velocity is proportional to its force. For two identical particles with friction coefficient \(\gamma\),

\[ \dot d=-\frac{2}{\gamma}U'(d). \]

Thus the particles approach each other wherever \(U'(d)>0\).

For particles approaching from large separation to reach contact without thermal fluctuations, there must be no intervening repulsive region or stable finite-separation equilibrium. The critical condition is therefore the disappearance of the region where \(U'(d)<0\).

Rewrite the derivative as

\[ U'(d)=\frac{A}{d^2} \left[ 1-\frac{K\lambda}{A} \left(\frac d\lambda\right)^2e^{-d/\lambda} \right]. \]

Set \(z=d/\lambda>0\). The function controlling the sign is

\[ f(z)=z^2e^{-z}. \]

Its derivative is

\[ f'(z)=ze^{-z}(2-z), \]

so its global maximum occurs at \(z=2\), with

\[ f_{\max}=4e^{-2}. \]

Accordingly, \(U'(d)\geq0\) for every separation precisely when

\[ \frac{K\lambda}{A}\,4e^{-2}\leq1. \]

The critical screening length is

\[ \boxed{\lambda_{\mathrm{crit}} =\frac{\exp(2)A}{4K} =\frac{\exp(2)Aq^2}{4B\epsilon(kT)^2}.} \]

At this threshold, the two stationary points of the potential merge at

\[ d_{\mathrm{crit}}=2\lambda_{\mathrm{crit}}, \]

where both \(U'(d_{\mathrm{crit}})\) and \(U''(d_{\mathrm{crit}})\) vanish.

Increasing the salt concentration decreases \(\lambda\). Substituting \(\lambda_{\mathrm{crit}}\) into Eq.(7),

\[ \begin{aligned} c_{\mathrm{crit}} &=\frac{\epsilon kT} {2N_Aq^2\lambda_{\mathrm{crit}}^2}\\ &=\frac{\epsilon kT}{2N_Aq^2} \left[ \frac{4B\epsilon(kT)^2}{\exp(2)Aq^2} \right]^2. \end{aligned} \]

Therefore,

\[ \boxed{c_{\mathrm{crit}}= \frac{8B^2\epsilon^3(kT)^5} {\exp(4)\,N_AA^2q^6}.} \]

For NaCl, the ions are monovalent, so \(q\) is the elementary charge magnitude. The concentration of each ionic species equals the NaCl molar concentration; the factor \(2\) in Eq.(7) already accounts for both species. With SI parameters, the result has units \(\mathrm{mol\,m^{-3}}\).

For \(c>c_{\mathrm{crit}}\), \(U'(d)>0\) at every positive separation, and the deterministic motion brings an initially separated pair into contact. For \(c<c_{\mathrm{crit}}\), a pair approaching from sufficiently large separation becomes trapped at the outer local minimum of \(U\).

There is a precise qualification at equality. At \(c=c_{\mathrm{crit}}\),

\[ U'(d)\propto(d-d_{\mathrm{crit}})^2 \]

near \(d_{\mathrm{crit}}\). The overdamped approach from above therefore takes infinite time to reach that separation. Thus \(c_{\mathrm{crit}}\) is the critical threshold, or infimum of concentrations producing coagulation from large separation; the idealized dynamics require \(c>c_{\mathrm{crit}}\) for actual passage to contact.

The solution was checked against the official solution.