Solutions to Theory Problems

S1‒1
Theory Problem 1: Characterization of Soil Colloids (10 points)
Part A. Analysis of motions of colloidal particles (1.6 points)
A.1

The relation between the impulse and the momentum change is given by 𝑀 𝑣0 = 𝐼0 . Therefore,
𝑣0 =𝐼0.
𝑀(S1.1)For the situation considered here, the equation of motion reads(S1.2)

𝑀 𝑣 ̇ = −𝛾𝑣(𝑡).
Substituting the form of the solution given in the question sheet, 𝑣(𝑡) = 𝑣0 𝑒−(𝑡−𝑡0 )/𝜏 , we obtain
𝜏=A.1𝑀.
𝛾(S1.3)0.8 pt𝐼𝑣0 = 0𝑀𝑀
𝜏=𝛾

A.2
Thanks to the linearity of Eq. (S1.2), we can use the superposition principle, which tells us that𝑣(𝑡) is given by the sum of solutions for single collision events that occur before time 𝑡. This immediately
gives the solution as𝑣(𝑡) = ∑𝑖𝐼𝑖 −(𝑡−𝑡𝑖 )/𝜏𝑒,
𝑀(S1.4)

where the sum is taken in the range of 𝑖 that satisfies 0 < 𝑡𝑖 < 𝑡.
It is also not difficult to figure out this superposition principle, by considering the effect of a single collision as well as the velocity change between two consecutive collisions. From A.1, it is straightforward toshow that the velocity right after the 𝑖th collision is given by𝑣(𝑡𝑖 ) = 𝑣0 (𝑡𝑖 ) +𝐼𝑖,
𝑀

(S1.5)

where 𝑣0 (𝑡𝑖 ) is the velocity right before the collision. Also, since there is no collision during 𝑡𝑖 < 𝑡 < 𝑡𝑖+1 ,we have𝑣(𝑡) = (𝑣0 (𝑡𝑖 ) +𝐼𝑖) 𝑒−(𝑡−𝑡𝑖 )/𝜏 .
𝑀(S1.6)

In particular,𝑣0 (𝑡𝑖+1 ) = (𝑣0 (𝑡𝑖 ) +𝐼𝑖) 𝑒−(𝑡𝑖+1 −𝑡𝑖 )/𝜏 .
𝑀(S1.7)

Therefore, with 𝑣0 (𝑡1 ) = 0, we obtain𝑖−1

𝐼𝑗 −(𝑡 −𝑡 )/𝜏𝑒 𝑖 𝑗𝑀𝑗=1(S1.8)𝑣0 (𝑡𝑖 ) = ∑1

Solutions to Theory Problems

S1‒2
and, for 𝑡𝑖 < 𝑡 < 𝑡𝑖+1 ,𝑖𝐼𝑗 −(𝑡−𝑡 )/𝜏
𝑗𝑒.
𝑀
𝑗=1

(S1.9)

𝑣(𝑡) = ∑This is equivalent to Eq. (S1.4).
A.2

𝐼𝑣(𝑡) = ∑ 𝑖 𝑒−(𝑡−𝑡𝑖 )/𝜏𝑀𝑖
the inequality specifying the range of 𝑡𝑖 that needs to be considered:
0 < 𝑡𝑖 < 𝑡

0.8 ptPart B. Effective equation of motion (1.8 points)
B.1

From the definition of the model, we have𝑁(S1.10)

Δ𝑥(𝑡) = ∑ 𝑣𝑛 𝛿.
𝑛=1

Taking the average and using ⟨𝑣𝑛 ⟩ = 0, we obtain(S1.11)⟨Δ𝑥(𝑡)⟩ = 0.

For the mean square displacement, computing the square of Eq. (S1.10) and taking the average, weobtain𝑁𝑁⟨Δ𝑥(𝑡)2 ⟩ = ∑ ∑⟨𝑣𝑚 𝑣𝑛 ⟩𝛿 2 .

(S1.12)𝑚=1 𝑛=1Using ⟨𝑣𝑚 𝑣𝑛 ⟩ = 𝐶 for 𝑛 = 𝑚 and 0 otherwise, we find𝑁

⟨Δ𝑥(𝑡)2 ⟩ = ∑ 𝐶𝛿 2 = 𝑁 𝐶𝛿 2 .

(S1.13)𝑛=1

Since 𝑁 𝛿 = 𝑡, we obtain⟨Δ𝑥(𝑡)2 ⟩ = 𝐶𝛿𝑡.

(S1.14)B.1

1.0 pt⟨Δ𝑥(𝑡)⟩ = 0⟨Δ𝑥(𝑡)2 ⟩ = 𝐶𝛿𝑡

B.2
As described in the question sheet, the mean square displacement ⟨Δ𝑥(𝑡)2 ⟩ is a characteristic
observable of the Brownian motion, which of course takes a finite value for a given 𝑡. For the model
considered here, we have Eq. (S1.14), but we need to consider the limit 𝛿 → 0 to describe the Brownian
motion in this model. This requires that 𝐶𝛿 remains finite, so that 𝐶 ∝ 𝛿 −1 . It also follows that ⟨Δ𝑥(𝑡)2 ⟩ ∝ 𝑡.
▷ Note: The continuous time limit 𝛿 → 0 of the present model corresponds to what is called the overdamped Langevin equation. This reads, in the absence of external force as considered here,𝛾𝑑𝑥= 𝜉(𝑡)
𝑑𝑡(S1.15)2

Solutions to Theory Problems

S1‒3
with a Gaussian noise 𝜉(𝑡) that satisfies⟨𝜉(𝑡)⟩ = 0,

⟨𝜉(𝑡)𝜉(𝑡′ )⟩ = 2𝐷𝛿(𝑡 − 𝑡′ )(S1.16)

with the diffusion coefficient 𝐷. Here, 𝛿(𝑡) (not to confuse with 𝛿 in the problem) is called the delta𝑏function, which satisfies 𝛿(𝑡) = 0 for 𝑡 ≠ 0 and 𝛿(0) = ∞ but ∫𝑎 𝛿(𝑡)𝑑𝑡 = 1 for any 𝑎 < 0 and 𝑏 > 0.

B.20.8 pt
𝛼 = −1𝛽=1Part C. Electrophoresis (2.7 points)
C.1
For particles with velocity 𝑣 (> 0), only those in the range 𝑥0 − 𝑣𝛿 ≤ 𝑥 ≤ 𝑥0 pass the position 𝑥0
during a time interval 𝛿. Therefore, the number of such particles per unit cross-sectional area and perunit time is given by𝑥

𝑁+ (𝑥0 ) =011∫
𝑛(𝑥)𝑑𝑥𝛿 𝑥0 −𝑣𝛿 2(S1.17)

Using the Taylor expansion 𝑛(𝑥) ≃ 𝑛(𝑥0 ) + (𝑥 − 𝑥0 ) 𝑑𝑛𝑑𝑥 (𝑥0 ) and integrating, we obtain
𝑁+ (𝑥0 ) =

C.1

C.2

11 𝑑𝑛𝑛(𝑥0 )𝑣 −(𝑥 )𝑣2 𝛿.
24 𝑑𝑥 0(S1.18)

0.5 pt1 𝑑𝑛1(𝑥 )𝑣2 𝛿𝑁+ (𝑥0 ) = 𝑛(𝑥0 )𝑣 −24 𝑑𝑥 0

Let 𝑁− (𝑥0 ) be the counterpart of 𝑁+ (𝑥0 ) for particles with velocity −𝑣, then𝑁− (𝑥0 ) =11 𝑑𝑛𝑛(𝑥0 )𝑣 +(𝑥 )𝑣2 𝛿.
24 𝑑𝑥 0(S1.19)

With this equation, Eq. (S1.18), and 𝐽𝐷 (𝑥0 ) = ⟨𝑁+ (𝑥0 ) − 𝑁− (𝑥0 )⟩, we obtain
𝐽𝐷 (𝑥0 ) = −1 𝑑𝑛
1 𝑑𝑛(𝑥0 )⟨𝑣2 ⟩𝛿 = −(𝑥 )𝐶𝛿.
2 𝑑𝑥
2 𝑑𝑥 0(S1.20)

Comparing this with Eq. (4) in the question sheet for 𝑥 = 𝑥0 , 𝐽𝐷 (𝑥0 ) = −𝐷 𝑑𝑛𝑑𝑥 (𝑥0 ), we obtain
𝐷=1𝐶𝛿.
2(S1.21)

Plugging this into the result of B.1, we obtain⟨Δ𝑥(𝑡)2 ⟩ = 2𝐷𝑡.

(S1.22)

3

Solutions to Theory Problems

S1‒4
C.2

0.7 pt

1 𝑑𝑛
𝐽𝐷 (𝑥0 ) = −(𝑥 )𝐶𝛿
2 𝑑𝑥 01
𝐷 = 𝐶𝛿2⟨Δ𝑥(𝑡)2 ⟩ = 2𝐷𝑡

C.3

The force balance sketched in Fig. 2 is expressed by the following equation:(S1.23)

Π(𝑥)𝐴 + 𝑛(𝑥)𝐴Δ𝑥𝑄𝐸 = Π(𝑥 + Δ𝑥)𝐴.

Using the van 't Hoff equation for the osmotic pressure, Π(𝑥) = 𝑛(𝑥)𝑘𝑇 , and carrying out the Taylor
expansion of 𝑛(𝑥 + Δ𝑥), we obtain𝑑𝑛𝑛(𝑥)
=𝑄𝐸.
𝑑𝑥
𝑘𝑇

C.3

C.4

(S1.24)0.5 pt𝑑𝑛𝑛(𝑥)=𝑄𝐸
𝑑𝑥
𝑘𝑇

The equation of motion for ⟨𝑣(𝑡)⟩ is𝑀𝑑⟨𝑣(𝑡)⟩= −𝛾⟨𝑣(𝑡)⟩ + 𝑄𝐸.
𝑑𝑡(S1.25)

By solving this with the initial condition ⟨𝑣(0)⟩ = 0, we obtain⟨𝑣(𝑡)⟩ =𝑄𝐸(1 − 𝑒−𝑡/𝜏 ).
𝛾(S1.26)

Therefore,𝑢 = lim ⟨𝑣(𝑡)⟩ =𝑡→∞𝑄𝐸.
𝛾(S1.27)▷ Note: The student is expected to surmise that the solution to Eq. (S1.25) has a functional form
analogous to that to Eq. (S1.2), whose solution is given in the question sheet.

C.4

0.5 pt𝑄𝐸(1 − 𝑒−𝑡/𝜏 )⟨𝑣(𝑡)⟩ =𝛾𝑢 = 𝑄𝐸
𝛾

C.5

From the result of C.3 and Eq. (4) in the question sheet, we have𝐽𝐷 (𝑥) = −𝐷𝑄𝐸𝑛(𝑥).
𝑘𝑇(S1.28)

4

Solutions to Theory Problems

S1‒5
From the result of C.4 and Eq. (5) in the question sheet, we have
𝐽𝑄 (𝑥) =𝑄𝐸𝑛(𝑥).
𝛾(S1.29)

Plugging these into the flux balance condition, 𝐽𝐷 (𝑥) + 𝐽𝑄 (𝑥) = 0, we obtain
𝐷=C.5𝑘𝑇.
𝛾(S1.30)

0.5 pt𝑘𝑇
𝐷=𝛾Part D. Mean square displacement (2.4 points)
D.1 Combining the results of C.2 and C.5, 𝑘 = 𝑅/𝑁𝐴 , 𝛾 = 6𝜋𝑎𝜂, we obtain the following equation thatlinks the mean square displacement to 𝑁𝐴 :⟨Δ𝑥2 ⟩ =𝑅𝑇 Δ𝑡.
3𝜋𝑎𝜂𝑁𝐴(S1.31)

From the data given in the question sheet, the mean square displacement is estimated at ⟨Δ𝑥2 ⟩ =6.34 𝜇m2 . Plugging this and the values of the parameters given in the question sheet, we obtain
𝑁𝐴 = 5.6 × 1023 mol−1 .

(S1.32)▷ Note: In 1908, Jean Baptiste Perrin (1870-1942) carried out such an observation and obtained an estimate of 𝑁𝐴 , which turned out to be consistent with the values known at that time by other approaches.
This convinced the community of the fact that molecules and hence atoms do exist as constituents ofmatter. Perrin was awarded the Nobel Prize in Physics in 1926 for ``his work on the discontinuous structure of matter, and especially for his discovery of sedimentation equilibrium''. For more details, see, e.g.,
S. G. Brush, ``A History of Random Processes: I. Brownian Movement from Brown to Perrin'', Archive for
History of Exact Sciences, volume 5, pages 1−36 (1968).
▷ Note: On May 20, 2019, the definition of physical constants including the Avogadro constant 𝑁𝐴 waschanged. As a result, 𝑁𝐴 is now defined by a fixed value, not to be determined through measurements.

D.1D.2

1.0 pt𝑁𝐴 = 5.6 × 1023 mol−1𝑁Using Δ𝑥(𝑡) = ∑𝑛=1 (𝑢 + 𝑣𝑛 )𝛿 and Eq. (3) in the question sheet, we obtain⟨Δ𝑥2 ⟩ = (𝑢𝑡)2 + 2𝐷𝑡(S1.33)for general 𝑡. This can be rewritten as⟨Δ𝑥2 ⟩ = 𝑢2 𝑡 (𝑡 +2𝐷) = 𝑢2 𝑡(𝑡 + 𝑡∗ ),𝑢2(S1.34)

with 𝑡∗ = 2𝐷/𝑢2 . Therefore,𝑡⟨Δ𝑥2 ⟩ ∝ { 2𝑡for 𝑡 ≪ 𝑡∗
for 𝑡 ≫ 𝑡∗

(S1.35)5Solutions to Theory Problems

S1‒6
D.2

0.8 pt⟨Δ𝑥2 ⟩ = (𝑢𝑡)2 + 2𝐷𝑡 for general 𝑡𝑡⟨Δ𝑥2 ⟩ ∝ { 2𝑡𝑡∗ =for small 𝑡
for large 𝑡

2𝐷𝑢2

An example of the graph to answer:(log t*)D.1
Since the microbe does not change the swimming direction for 𝑡 ≪ 𝛿0 , we can use the result ofD.2 just by replacing 𝑢 by 𝑢0 . By contrast, for 𝑡 ≫ 𝛿0 , the motion of the microbe can be described by themodel considered in PART B, though its parameter 𝛿 is not an artificial parameter anymore but is now aquantity that characterizes the microbe's motion, 𝛿0 . The parameter 𝐶 is given by 𝐶 = 𝑢20 . Plugging thisinto Eq. (S1.14) and collecting all these results, we obtain⎧2𝐷𝑡for 𝑡 ≪ 2𝐷/𝑢20{⟨Δ𝑥2 ⟩ = ⎨𝑢20 𝑡2 for 2𝐷/𝑢20 ≪ 𝑡 ≪ 𝛿0{𝑢2 𝛿 𝑡 for 𝛿 ≪ 𝑡⎩ 0 00(S1.36)▷ Note: More precisely, one can show ⟨Δ𝑥2 ⟩ = (𝑢20 𝛿0 + 2𝐷)𝑡 for 𝑡 ≫ 𝛿0 . However, in order for the intermediate regime to exist, we have 2𝐷/𝑢20 ≪ 𝛿0 , from which it follows that 𝑢20 𝛿0 ≫ 2𝐷 and the expressionin Eq. (S1.36) is a good approximation.
▷ Note: The motion of the microbe described here is called the run-and-tumble motion, except that
it is usually assumed that the microbe changes the swimming direction (``tumbling'') at random timeintervals. Some bacteria including Escherichia coli is known to swim in this manner.

D.3

0.6 pt⎧2𝐷𝑡
for small 𝑡{⟨Δ𝑥2 ⟩ = ⎨𝑢20 𝑡2 for intermediate 𝑡{𝑢2 𝛿 𝑡 for large 𝑡⎩ 0 0Part E. Water purification (1.5 points)
E.1 The interaction energy 𝑈 (𝑑) has a barrier if 𝑐 is small enough, but the barrier disappears if 𝑐 exceeds
a threshold. This threshold is the critical concentration to derive in this question. The condition for thebarrier to disappear is given bymin 𝑈 ′ (𝑑) = 0.

(S1.37)
6

Solutions to Theory Problems

S1‒7This can be expressed by the following two equations:
𝐴𝐵𝜖(𝑘𝑇 )2 −𝑑/𝜆−𝑒= 0,𝑑2𝑞2 𝜆
2𝐴 𝐵𝜖(𝑘𝑇 )2 −𝑑/𝜆𝑈 ″ (𝑑) = − 3 +𝑒= 0.
𝑑𝑞 2 𝜆2𝑈 ′ (𝑑) =(S1.38)(S1.39)Solving these, we obtain𝑑 = 2𝜆 = √𝐴𝑞 2 𝜆𝐵𝜖(𝑘𝑇 )2(S1.40)

and therefore𝜆=𝑒2 𝐴𝑞 2.
4𝐵𝜖(𝑘𝑇 )2Plugging this into 𝑐 =𝑐=(S1.41)
𝜖𝑘𝑇 −2𝜆 , we obtain
2𝑁𝐴 𝑞 28𝐵2 𝜖3 (𝑘𝑇 )5.
𝑒4 𝑁𝐴 𝐴2 𝑞 6(S1.42)▷ Note: In the literature, it is also common to consider that the critical concentration is reached whenthe energy barrier becomes as low as the energy for 𝑑 → ∞, i.e., max 𝑈 (𝑑) = 0, although this does
not meet the requirements given in the question sheet. If this condition is used instead, we find 𝑐 =𝐵2 𝜖3 (𝑘𝑇 )5. This differs from Eq. (S1.42) only by a factor 𝑒2 /8 ≈ 0.92.
2𝑒2 𝑁 𝐴2 𝑞6𝐴

E.1

1.5 pt8𝐵2 𝜖3 (𝑘𝑇 )5𝑐= 4𝑒 𝑁𝐴 𝐴 2 𝑞 67