z**h
发帖数: 224 | 1 and radius of gyration 的推导.
Gaussian chain. | s***e
发帖数: 911 | 2 The dynamics of polymer, by Edwards.
【在 z**h 的大作中提到】
: and radius of gyration 的推导.
: Gaussian chain.
| z**h
发帖数: 224 | 3 no higher moments derivation.
【在 s***e 的大作中提到】
: The dynamics of polymer, by Edwards.
| s***e
发帖数: 911 | 4 From page 11, (2.21), you can derive out the general distribution of
end-to-end distance, phi(|R|, N). All the higher moments can be derived from
here by integration over |R|^i * phi(|R|,N), d |R|. These integrations over
|R| should be simple.
【在 z**h 的大作中提到】
: no higher moments derivation.
| z**h
发帖数: 224 | 5 how about radius of gyration?
【在 s***e 的大作中提到】
: From page 11, (2.21), you can derive out the general distribution of
: end-to-end distance, phi(|R|, N). All the higher moments can be derived from
: here by integration over |R|^i * phi(|R|,N), d |R|. These integrations over
: |R| should be simple.
| s***e
发帖数: 911 | 6 Suppose the chain has N segments connecting N+1 points. The distance between
points i and j follows the distribution phi(r_ij, j-i). can be
calculated from the distribution. The radius of gyration is then:
R_G^2=Sum_{i,j} /(N+1)^2
【在 z**h 的大作中提到】
: how about radius of gyration?
| w********h
发帖数: 12367 | 7 Polymer Physics
by Rubinstein and Colby.
This book is the best one to sum up polymer dynamics and maybe thermodynamics
I have ever seen.
Although Doi and Edwards' book is classical,
I still cannot see it can benefit people outside of theoretical work on
Polymer dynamics a lot.
【在 z**h 的大作中提到】
: and radius of gyration 的推导.
: Gaussian chain.
| z**h
发帖数: 224 | 8 how about ? Thanks.
from
over
【在 s***e 的大作中提到】
: Suppose the chain has N segments connecting N+1 points. The distance between
: points i and j follows the distribution phi(r_ij, j-i). can be
: calculated from the distribution. The radius of gyration is then:
: R_G^2=Sum_{i,j} /(N+1)^2
| c*****e
发帖数: 238 | 9 Rg is just a sum of N independent random variables, thus its distribution is
known.
Knowing this we can calculate the mean value of its moment to arbitrary order
by the characteristic function, i.e., the Fourier transform of the probability
distribution, and taking derivatives.
Check any probability book, you should find the details.
【在 z**h 的大作中提到】
: how about ? Thanks.
:
: from
: over
| s***e
发帖数: 911 | 10 这个大概不是很简单. 需要计算这样的组合, 这样就
需要知道phi(r_{ij}, r_{kl})的联合分布. r_{ij}和r_{kl}有overlap时他们的
分布大概不是独立的, 所以讨论起来就比较复杂些.
【在 z**h 的大作中提到】
: how about ? Thanks.
:
: from
: over
| s***e
发帖数: 911 | 11 我大概想了一下. 如果走捷径利用r_{ij}的已知分布来就Rg^2的分布涉及到刚才说的
独立性问题, 可能不如直接从最基本的计算开始简单.
Rg^2 = Sum_{i,j} r_{ij}^2 /(N+1)^2, 其中的每一项:
r_{ij}^2 = (b_{i+1}+..+b_{j})^2,其中b_{i}是第i个segment的矢量.
如果计算, 交叉项都是零, = (j-i) b^2 (b是segment length).
这个正是freely-joint-chain的结果.
Rg^4涉及到Sum{i,j,k,l} r_{ij}^2*r_{kl}^2 /(N+1)^4, 其中的每项:
r_{ij}^2*r_{kl}^2
展开后是: Sum_{p,q=i+1,j; r,s=l+1,m} b_{p}*b_{q}*b_{r}*b_{s}
如果[i,j]和[p,q]没有overlap很简单, 平均后每项就是(delta_{pq}+delta_{rs})*b^4;
如果有overlap下面你得自己仔细讨论一下通项了,大概各项应该具有这样的形式:
b^4*(
【在 s***e 的大作中提到】
: 这个大概不是很简单. 需要计算这样的组合, 这样就
: 需要知道phi(r_{ij}, r_{kl})的联合分布. r_{ij}和r_{kl}有overlap时他们的
: 分布大概不是独立的, 所以讨论起来就比较复杂些.
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