9.5. Thermal, chemical, and mechanical switching[[Student version, January 17, 2003]] 325
higher for short chains. For example,N=4 6 gives a midpoint at around 35◦C(middle set of data
points in Figure 9.6). Since we have no more free parameters in the model, it had better be able
to predict that shift correctly.
Sullivan: Unfortunately, Equation 9.18 shows equally clearly that the midpoint is always atα=0,
which we found corresponds to one fixed temperature,Tm≡∆Ebond/∆Stotindependent ofN.It
looks like our model is no good.
Gilbert: How did you see that so quickly?
Sullivan: It’s a symmetry argument. Suppose I define ̃σi=−σi.Instead of summing overσi=±1,
Ican equally well sum over ̃σi=±1. In this way I show thatZ(−α)=Z(α); it’s an “even function.”
Then its derivative must be an “odd function,” and so must equal minus itself atα=0.
Gilbert: That may be good math, but it’s bad physics! Why should there be any symmetry
between being an alpha helix or being a random coil?
Putting Gilbert’s point slightly differently, suppose we have a tract of +1’s starting all the way
out at the end of the polymer and extending to some point in the middle. There is one junction
at the end of this tract, giving a penalty of e−^2 γ.Incontrast, for a tract of +1’s startingand
ending in the middle, there aretwojunctions, one at each end. But really, initiating a tract of
helix requires that we immobilize several bonds, regardless of whether it extends to the end or not.
So our partition function (Equation 9.18) underpenalizes those conformations with helical tracts
extending all the way out to one end (or both). Such “end effects” will be more prominent for short
chains.
Wecan readily cure this problem, and incorporate Gilbert’s insight. We introduce fictitious
monomers in positions 0 andN+1,but instead of summing over their values, we fixσ 0 =σN+1=
−1. Now a helical tract extending to the end (position 1 orN)will still have two “junctions” and
will get the same penalty as a tract in the middle of the polymer. Choosing− 1 instead of +1 at the
ends breaks the spurious symmetry between±1. That is, Sullivan’s discouraging result no longer
holds after this small change in the model. Let’s do the math properly.
Your Turn 9k
a. Repeat Your Turn 9g(a) on page 315 with the above modification, showing that the partition
function forN=2isZ 2 ′=r
[ 0
1]
·T^3
[ 0
1]
,whereTis the same as before, andris a quantity that
youare to find.
b. Adapt your answer to Your Turn 9g(b) (generalN)tothe present situation.
c. Adapt Equation 9.20 on page 315 to the present situation.SinceNis not infinite, we can no longer drop the second term of Equation 9.20, nor can we ignore
the constantpappearing in it. Thus we must find explicit eigenvectors ofT.First we introduce
the abbreviations
g±=eα−γλ±=eα
[
coshα±√
sinh^2 α+e−^4 γ]
.
Your Turn 9l
a. Show that we may write the eigenvectors ase±=[g
±−^1
e2(α−γ)]
.
b. Using (a), show that[ 0
1]
=w+e++w−e−,wherew±=±e2(γ−α)(1−g∓)/(g+−g−).
c. T 2 Find an expression for the full partition function,Z′N,interms ofg±.The rest of the derivation is familiar, if involved: We compute〈σav〉N=N−^1 ddαlnZN′ using your