FrontierScience / 16318af5-cfa6-4b2c-bf25-9eb3ac468b48 / We simulate a single polymer with the following model: As shown in the figure, a…

We simulate a single polymer with the following model: As shown in the figure, a polymer can be viewed as being composed of `n` rigid rods, each of length  `b`, connected end-to-end. Adjacent rods are connected by a pivot that allows for free rotation, so the total length of the polymer `L=nb`. However, the distance between the two ends of the polymer, denoted as

`\( r≠L \)`.

We will proceed with the following discussion under isothermal conditions.








We fix one end of the polymer (fixed end), and the other end (free end) is allowed to move freely. We define a spatial Cartesian coordinate system `xyz`, with the fixed end at the origin. Let the vector from the fixed end to the free end be `(r)→=(r_(x),r_(y),r_(z))`. 


Probability of the polymer's two ends being at a distance from `r` to `r+dr` is
`P(r)dr`



`\( P(r)dr=4\pi r^2\biggl(\frac{3}{2nb^2\pi}\biggr)^{\frac{3}{2}}e^{-\frac{3r^2}{2nb^2}}dr \)`




The formula for `P(r)` is similar to the Boltzmann distribution  `e^(−(3r²)/(2nb²))≈e^(−(U)/(kT))`, where `U` is the total internal energy of the polymer  and `k` is the Boltzmann constant.

Assuming that a fixed outward pulling force `f` is applied at both ends of the polymer, try to find the quantity 

 `S/r`.

where `S` is the total entropy of the polymer.


 Answer must be expressed in terms of  `f` and `T` and numerical constants.


 Let `S=0` when `r=0` for the polymer.

Think step by step and solve the problem below. At the end of your response, write your final answer on a new line starting with “FINAL ANSWER”. It should be an answer to the question such as providing a number, mathematical expression, formula, or entity name, without any extra commentary or providing multiple answer attempts.

We simulate a single polymer with the following model: As shown in the figure, a polymer can be viewed as being composed of `\( n \)` rigid rods, each of length  `\( b \)`, connected end-to-end. Adjacent rods are connected by a pivot that allows for free rotation, so the total length of the polymer `\( L=nb \)`. However, the distance between the two ends of the polymer, denoted as

`\( r≠L \)`.

We will proceed with the following discussion under isothermal conditions.








We fix one end of the polymer (fixed end), and the other end (free end) is allowed to move freely. We define a spatial Cartesian coordinate system `\( xyz \)`, with the fixed end at the origin. Let the vector from the fixed end to the free end be `\( \vec{r} = (r_x, r_y, r_z) \)`. 


Probability of the polymer's two ends being at a distance from `\( r \)` to `\( r + dr \)` is
`\( P(r)dr \)`



`\( P(r)dr=4\pi r^2\biggl(\frac{3}{2nb^2\pi}\biggr)^{\frac{3}{2}}e^{-\frac{3r^2}{2nb^2}}dr \)`




The formula for `\(  P(r)  \)` is similar to the Boltzmann distribution  `\( e^{-\frac{3r^{2}}{2nb^{2}}}\approx e^{-\frac{U}{kT}} \)`, where `\( U \)` is the total internal energy of the polymer  and `\( k \)` is the Boltzmann constant.

Assuming that a fixed outward pulling force `\( f \)` is applied at both ends of the polymer, try to find the quantity 

 `\( S/r \)`.

where `\( S \)` is the total entropy of the polymer.


 Answer must be expressed in terms of  `\( f \)` and `\( T \)` and numerical constants.


 Let `\( S = 0 \)` when `\( r = 0 \)` for the polymer.

Think step by step and solve the problem below. At the end of your response, write your final answer on a new line starting with “FINAL ANSWER”. It should be an answer to the question such as providing a number, mathematical expression, formula, or entity name, without any extra commentary or providing multiple answer attempts.