# Homework Help: Probability of the polymer chain

1. Aug 3, 2017

### Kelly Lin

1. The problem statement, all variables and given/known data

2. Relevant equations
I want to check if I think it right!

3. The attempt at a solution
If
N=1: ← or → (2 configurations/ each length is l)
N=2: ← or → or ←← or ←→ or →← or →→
[COLOR=#black]------[/COLOR]→[COLOR=#black]----[/COLOR]←
(6 configurations/ folded polymer's length is l/2 andunfolded polymer's length is l)
Thus, the configuration written in the function of x:
$$2^{N} \hspace{1cm}\text{ for } x=l \\ (l/x)! \hspace{1cm}\textrm{ otherwise}$$

2. Aug 3, 2017

### Staff: Mentor

If N is the number of links of length l in the chain and x is the total length in a given configuration, how many links are in one direction and how many are in the other direction? How many ways are there are selecting the combinations of links that do this? (This is like a coin flipping problem).

3. Aug 3, 2017

### Kelly Lin

oh!So the length l doesn't mean the total length?

4. Aug 3, 2017

### Staff: Mentor

NO, it's the length of each link.

5. Aug 3, 2017

6. Aug 3, 2017

### Kelly Lin

Actually, I have another question. How come the second question asks about the total number of configurations? In my opinion, there are infinite configurations depend on unlimited x. Does the question ask about the average configuration number?
Thanks!

7. Aug 3, 2017

### Staff: Mentor

For a specified x, there are not an infinite number of configurations (at least not in 1D).

8. Aug 4, 2017

### Kelly Lin

But, I think the first question asks the total number of specific x. Or, can the polymer be folded? This really make the answers of two questions different.

9. Aug 4, 2017

### Staff: Mentor

It seems to me the implication is that it is folded along a straight line.

10. Aug 6, 2017

### Kelly Lin

Is it what the question means?

11. Aug 6, 2017

### Staff: Mentor

I don't think so. You need to figure out for a chain of identical joined links that can be folded along the x axis. So, for x = L, and N= 2, there are zero configurations. Leave off the arrows. For x = 2L and N = 2, there is only one configuration. For x = L and N = 1, there is only one configuration.

12. Aug 6, 2017

### Kelly Lin

But the question mentions that each segment can be orientated in positive or negative directions. Don't we consider the direction (arrow) in different cases?

13. Aug 6, 2017

### Kelly Lin

Also, why can't N=2 chain be folded? Thanks!!

14. Aug 6, 2017

### Staff: Mentor

No. It's just if there is a fold or not at each jumction. The focus should be in the junctions, not the links.

15. Aug 6, 2017

### Staff: Mentor

Yes If N=2, you can have x=1 with a fold.

16. Aug 6, 2017

### Kelly Lin

So you mean if N=2 there are 3 configurations?

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17. Aug 6, 2017

### Staff: Mentor

If N is 2 and x is 2, then there is 1 configuration. If N is 2 and x is 1, I would have to decide whether there are 2 configurations or 1. My inclination would be to count it as 1 configuration.

18. Aug 12, 2017

### Kelly Lin

I think this is more viable!

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19. Aug 12, 2017

### Kelly Lin

For the entropy in the system,
since
$$S=-k<lnP_{r}>=-k\sum_{r}{P_{r}lnP_{r}}$$
we get
$$S\approx -k\int{P(x)lnP(x)}dx=...=(-\frac{k}{2})(1-ln(\frac{2}{\pi N}))\approx -\frac{k}{2} \\ A=-\int S dT = \frac{1}{2}kT \\ U=A+TS=\frac{1}{2}kT-\frac{1}{2}kT=0$$
*A=free enegy; U=internal energy
So weird that U=0. Am I doing this wrong?? Thank you!!

20. Aug 12, 2017

### haruspex

Suppose r of the N are oriented one way and the remaining N-r the other way. What would the total length be?

By the way, the expression on the right in b) cannot correct. The argument to exp() must be dimensionless, but x is a length. I would rewrite it by replacing x by x/l everywhere.

21. Aug 15, 2017

### Kelly Lin

The result will be $$x=Nl-(N-r)l=rl$$
Oh! Then the configuration will be $$\frac{N!}{r!(N-r)!}=\frac{N!}{(\frac{x}{l})!(N-\frac{x}{l})!}$$
But, x can also be (N-r)l so the configuration above have to be multiplied by 2.
However, in this point of view, we view each section independently as an arrow. In the real situation, polymers will always turn their direction in the end. So, I think my table is more viable, right?

Although it is repalced by x/l, my result is still weird.(I got U=0)

Last edited: Aug 15, 2017
22. Aug 15, 2017

### haruspex

If you say so, but the simple hairpin view does lead to the target expression. If you are not getting that please post your working.

23. Aug 15, 2017

### Kelly Lin

I mean that I also cannot catch what the question wants. haha!
But my other questions are about internal energy! That's really weird, though…

24. Aug 15, 2017

### haruspex

You have the answer for qn a). For b), you need to divide by the total number of possible orientations of the N to get a probability. Are you familiar with Stirling's approximation for factorials?
For ease of algebra, I would work with r rather than x/l.

25. Aug 16, 2017

### Staff: Mentor

Shouldn't x be $x=|(N-2r)l|$?