Probability of the polymer chain

[Kelly Lin]

Homework Statement

Homework Equations

I want to check if I think it right!

The Attempt at a Solution

If
N=1: ← or → (2 configurations/ each length is l)
N=2: ← or → or ←← or ←→ or →← or →→
------→----←
(6 configurations/ folded polymer's length is l/2 andunfolded polymer's length is l)
Thus, the configuration written in the function of x:
<br /> 2^{N} \hspace{1cm}\text{ for } x=l \\ <br /> (l/x)! \hspace{1cm}\textrm{ otherwise}<br /> <br />

 

[Chestermiller]

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).

 

[Kelly Lin]

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).

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

 

[Chestermiller]

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

NO, it's the length of each link.

 

[Kelly Lin]

NO, it's the length of each link.

Thanks! Reading mistake!

 

[Kelly Lin]

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).

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!

 

[Chestermiller]

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!

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

 

[Kelly Lin]

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

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.

 

[Chestermiller]

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.

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

 

[Kelly Lin]

Is it what the question means?

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

 

[Chestermiller]

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.

 

[Kelly Lin]

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

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.

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?

 

[Kelly Lin]

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

 

[Chestermiller]

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?

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

 

[Chestermiller]

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

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

 

[Kelly Lin]

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

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

 

[Chestermiller]

So you mean if N=2 there are 3 configurations?
View attachment 208439

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.

 

[Kelly Lin]

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.

I think this is more viable!

 

[Kelly Lin]

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

 

[haruspex]

View attachment 208865
I think this is more viable!

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.

 

[Kelly Lin]

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.

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)

 

[haruspex]

In the real situation, polymers will always turn their direction in the end

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.

 

[Kelly Lin]

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.

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…

 

[haruspex]

I mean that I also cannot catch what the question wants

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.

 

[Chestermiller]

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)

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

 

[haruspex]

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

Thanks, I forgot to check that.