Physics of Soft and Biological Matter II: Problem Set 1

Physics of Soft and Biological Matter II: Problem Set 1

Owen A. Hickey May 13, 2014

Problem 1 Langevin Dynamics Using ESPResSo, 5 points

Use the provided Espresso script to simulate a single particle undergoing Langevin dynamics. Fit the short time and long time limits of the mean square displacement using a power law. How do the coefficient and exponent relate to the parameters of the Langevin equation in these two limits? Fit the velocity-velocity autocorrelation function. Relate the parameters in this fit to the variable found in the Langevin equation.

Solution: For the mean square displacement at short times the motion is ballistic and the velocity can be found via the equipartition theorem:

1

2 m < v

>= d

2 kT (1)

< x

/t

>= d 2

kT m

< x

>= d 2

kT m t

,

where d is the number of dimensions. At long times we have diffusive behaviour and the slope can be related to the definition of the diffusion coefficient:

D = x

2dt (2)

< x

>= 2dDt Dζ = Dγm = kT < x

>= 2d kT

γm t.

For the velocity autocorrelation function we again use the equipartition theorem to get the y-intercept:

1

2 m < v

>= d

2 kT (3)

< v

>= d kT

m ,

Problem 2 2

The decay can be understood by solving the Langevin eqnarray without noise:

d

x

dt

= −γ dx

dt (4)

< v(t)v(0) >= d kT

m exp (−γt) .

. . . . Problem 2 Diffusion of DNA, 5 points

a) Calculate the radius of gyration of a double-stranded λ-DNA strand with N

= 48490 base pairs. The Kuhn length is b ≈ 50nm and the distance between bases is l = 0.34nm such that the number of “steps” is N

l/b. Use the formula:

R

= 1 N

Z

0

Z

u

(r(u) − r(v))

dvdu, (5)

recalling that each subsection of a polymer is a random walk and thus

(r(u) − r(v))

= (u − v )b

. (6) Solution: First we must solve the integral

R

= 1 N

Z

0

Z

u

(r(u) − r(v))

dvdu R

= 1 N

Z

0

Z

u

(v − u)b

dvdu R

= b

N

Z

0

(N

/2 − u

/2 − N u + u

)du (7) R

= b

N

Z

0

(N

/2 − N u + u

/2)du R

= b

N

(N

/2 − N

/2 + N

/6) R

= 1 6 N b

,

Next the number of random “steps” needs to be expressed in terms of the number of bases:

R

= 1

6 N b

(8)

R

= 1 6

N

l b b

q

hR

i = r 1

6 N

lb q

hR

i = 371nm.

course name ^{PS #}

Problem 3 3

b) Calculate the hydrodynamic friction coefficient, ζ = 6πηR

,assuming R

= R

in water at 20

Solution: Note that η ≈ 0.001Pa·s:

ζ = 6π(0.001Pa · s)(371nm) (9)

ζ = 7.0 × 10

kg/s

c) Use the relation D = R

/τ to calculate the relaxation time of the DNA fragment.

Solution: Use the Nernst-Einstein relation to get D from ζ and solve:

D = R

τ kT

ζ = R

τ τ = R

ζ

kT τ = (371nm)

7.0 × 10

kg/s

4.11 × 10

J

τ = 0.23s (10)

. . . .

Problem 3 Metabolism of a Cell (Adopted from Biological Physics by Nelson), 5 points

a) Calculate the flux of a substance with concentration c

at infinity into a spherical cell of radius R with concentration 0 at the surface. Hint: Use Fick’s first law j = −D

combined with the fact that the flux through all spherical shells at a distance r is constant to get an expression for the flux.

Solution: first create a formula from the fact that flux is constant at all r:

I = A(r)j (r) = 4πr

j (r) j(r) = I

4πr

(11)

Now sub into Fick’s law and solve:

j = −D dc dr I

4πr

= −D dc dr c = A + B

r (12)

Using the boundary conditions c = 0 at r = R and c = c

as r goes to infinity we get c = c

1 − R

r

(13)

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Problem 3 4

Use Fick’s law to get the flux and sub in r = R:

j (r) = −D dc dr j(r) = −Dc

R r

j(R) = −Dc

R (14)

b) Find the flux for a cell of radius R = 1µm in an oxygen concentration c

= 0.2mole·m

.

Solution The problem is underspecified, lets take oxygen where D ≈ 2 × 10

m

/s:

I = A(R)j(R) I = 4πR

Dc

R I = 4πRDc

I = 4π(1µm)(2 × 10

m

/s)0.2mole · m

I = 5.0 × 10

mole (15)

c) Knowing that the metabolic rate of a cell is roughly 0.02mole kg

s

calculate the approximate maximum size of a cell.

Solution: Equate the total consumption of oxygen with the total maximum flux from part a and set the density to that of water:

A(R)j(R) = M V ρ 4πR

Dc

R = M 4 3 πR

ρ R =

s 3Dc

M ρ R =

s

3(2 × 10

m

/s)(0.2mole · m

) (0.02mole · kg

s

)(1000kg/m

)

R = 7.8µm (16)

. . . .

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