Biophysics II // Lecture 2 Julien Varennes January 18, 2015

1

Freely Rotating Chains, continued

Previous lecture we derived the following recursion relationship. < ~un · ~um >= b20 (cos β)|n−m| < uˆn · uˆm >= e−|n−m| ln(cos β) e−b0 |n−m|/ξ b0 ξ= ) ln(cos β)   ~ 2 >= b20 N + 2(N − 1) cos β + 2(N − 2)(cos β)2 + ... + 2(cos β)N −1 ⇒< R ~ 2 > by letting x = cos β. We can simplify < R      ~ 2 >= b2 N 2(1 + x + x2 + ... + xN −1 ) − 1 − 2 x + 2x2 + ... + (N − 1)xN −1 ⇒< R 0 ds is equivalent to the second series in the Let x = 1 + x + x2 + ... + xN −1 and notice that x dx above expression.   ds 2 2 ~ < R >= b0 N (2s − 1) − 2x dx Note that s looks like a Taylor Expansion.

1 − xN 1−x   1 + cos β cos β 2 2 N ~ ⇒< R >= b0 N − (1 − cos β) 1 − cos β (1 − cos β)2 ~ 2 >≈ N b2 with b2 ≡ b2 ( 1+cos β ). In the limit N → ∞, < R 0 1−cos β →s≈

~ 2 >. In the case of finite N , we need to add the following correction term to < R −2 cos β (1 − cosN β) (1 − cos β)2 −2 cos β −2 cos β (1 − cosN β) → 2 (1 − cos β) (1 − cos β)2 Note that if β is small then the chain looks stiff. N →∞⇒

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2

Thermodynamics

~ = T r (Ψ({~u})) δ(R ~− Partition function: Z = T r (Ψ({~u})) Define Z(R) we can calculate Free Energy and other thermo relationships.

P

~un ). Using this

~ = −kT ln Z(R) ~ = −kT ln PN (R) ~ F (R) ~ last lecture. Recall, we calculated PN (R) 3/2 −dR2 d ~ = PN (R) e 2 ~2 > 2π < R   R2 d ~ ⇒ F (R) = kT + ... ~2 > 2
∂F kT = fα ⇒ fα = dR ~2 > α ∂Rα d. R For the Freely Jointed Chain (FJC), the entropy is given as the following: ~ = S(R)

−kdR2 + ... ~2 > 2
~ which we aren’t interested in. Where ... indicates terms that are independent of R dQ = T dS =

−dkT R (dRα ) + ... ~2 > α
~ increases then dQin < 0 (heat is released). If R ~ decreases then dQin > 0 (feels cold). If R

3 3.1

Backbone Correlations Nearest Neighbor (nn) Correlations Only Ψ({~u}) =

N Y n=2

2

ψ2 (~un−1 , ~u)

The probability of the chain having some orientation is given by: RQ ~ − P ~u)Ψ d(~ u )δ( R ~ = RQ PN (R) d(~u)Ψ ~ = PN (R)

d~k i~kR~ e (2π)d

Z

RQ

~

~

d(~u)e−ik~u1 ψ2 (~u1 , ~u2 )e−ik~u2 ψ2 (~u1 , ~u2 )... RQ d(~u)ψ2 (~u1 , ~u2 )ψ2 (~u2 , ~u3 )...

We can simplify this expression by ignoring edge effects. ~ ~ Let M (1, 2) = e−ik~u1 /2 ψ2 (~u1 , ~u2 )e−ik~u2 /2 . We will use this equivalence and the Transfer Matrix Method for simplification. The Transfer Matrix Method states the following: Z Z d(1)...d(N )M (1, 2)M (2, 3)...M (N − 1, N )M (N, 1) = d(1) < 1|M N |1 > = T r(M N ) =

X

mN l and

l

P With {ml } being the set of eigenvalues of the matrix M . The solution to l mN l is strictly dominated by the largest eigenvalue when N → ∞. RQ Z ~k ~ ~ d(i)M (1, 2)M (2, 3)...M (N, 1) d i k R ~ = RQ e ⇒ PN (R) |k=0 (2π)d d(i)M (1, 2)|k=0 M (2, 3)|k=0 ...M (N, 1) Let m∗ be the largest eigenvalue in the set. In the integral m∗ will dominate over the terms with the other eigenvalues. ~ = PN (R) 3.1.1

Z

Z d~k i~kR~ mN d~k i~kR~ T r(M N (k)) ∗ (k) e ⇒ e d N 2 (2π) T r(M (0)) (2π) mN ∗ (0)

For the Freely Rotating Chain (FRC) Z Z=

d(ˆ u1 )d(ˆ u2 )...d(ˆ uN )δ(ˆ u1 uˆ2 − cos β)...δ(ˆ uN −1 uˆN − cos β)

In this case the transfer matrix will be M (ˆ u1 , uˆ2 ) = cδ(ˆ u1 uˆ2 − cos β), and c = 1/2π. In order to solve for the eigenvalues of M we will write M as a sum of spherical harmonics. M (ˆ u1 , uˆ2 ) =

X 2l + 1 l

M (ˆ u1 , uˆ2 ) =

X

4π

Pl (cos θ)Pl (ˆ u1 , uˆ2 )

∗ Pl (cos β)Ylm (ˆ u1 )Ylm (ˆ u2 )

l,m

3