helical ribbon

Dr. H.K. Sahu hks at igcar.ernet.in
Tue Nov 18 13:33:41 GMT 2003


Dear Dr. Ferrero
	I have amde an attemp to derive the form factor of a helical 
ribbon. I assume that the form factor is meant for X-ray scattering 
i.e. the electron distribution is what scatters; but it should be equally  
suitable for neutron scattering too. Further, it could be  taken that the  
charge is distributed unifromly on the ribbon or located only at the edges  
as in a DNA double helix. I give the form factor of the two cases separately, 
easier one first.
 
Case I.
=======
	It is a double helix with the charge only at the edges, the 
formula is simple. Assume that the charges extend from R1 to R2 from the 
mid axis of the double helix ( Clearly 2.R2 is the width of the ribbon) . 
Then 

P(q) = Integral(0 to 1) of 
[ F1(q,R1.sqrt(1-x^2),R2.sqrt(1-x^2)). F2(qLx/2) ] dx 

where L is the length of the helical ribbon, F1 & F2 are given as 
 
 F1(q,a,b) =  {2/(q.(a^2 - b^2))}.{a*J_sub_1(qa) - b*J_sub_1(qb)},
 F2(u) = sin(u)/u 
 
 with J_sub_1(x) is the Bessel function of order 1. J's are available in 
 almost all scientific libraries or can be evaluated as a series sum as

 J_sub_n (x) = (x/2)^n . sum(k=0 to inf) [{(-x^2/4)^k} / k! Gamma(n+k+1)}] 
 
 which in the case of n=1 becomes

 J_sub_1 (x) = (x/2)^n . sum(k=0 to inf) [[{(-x^2/4)^k) / (k+1)!].

	Note that this has already been averaged over all orientations of the 
helix about the axis. Thus, this is same as the form factor for the hollow 
cylinder of length L and radii R1 and R2.

Case II.
=======

	The charge is unifromly distributed over the ribbon. Here it is not 
proper to take an approach of the former case because the ribbon 
averaged over angles about the axis is not a solid cylinder. But the form 
factor without the above averaging can be derived from the 
geometric definition of the helical ribbon. Consider a flat ribbon of 
length L and width 2a. Twist it uniformly by a total angle of {2n.pi + 
beta}  over the length L. The define c = {2n.pi + b}/L. 

Thus, L = {2n.pi/c} + b/c, indicating n complete twists and a fraction 
b/2pi to constitute the helix
. 
The form factor is then given by

P(q_vector) = {2/(c.q_rho)} . [ 
{sin(q3.pi.n/c)/sin(q3.Pi/c)}.exp{(n-1).pi.q3/c}.M(q3/c,a.q_rho,2pi)} +
{exp(i.2pi.n.q3/c).M(q3/c,a.q_rho,b)} ]

where q_vector is the vector q (= (q1,q2,q3)), r_rho=sqrt(q1^2 + q2^2), 
i=sqrt(-1)  and the integral function M is defined as

M(p,q,r) = integral(0 to r) [ {exp(i.p.u).sin(q.cos(u))/cos(u)}du ] 
which on simplification, gives 
M(p,q,2pi)=(1 + exp(i.p.pi)).integral(0 to pi/2)[ {exp(i.p.u)+exp(i.p.(pi-u))}.
                                                 {sin(q.cos(u))/cos(u)} ]

	It may appear that the integrand diverges at pi/2 because of the  
cos(u) in the denominator, but actually in this limit 
{sin(q.cos(u))/cos(u)} is just q and there should not be any divergence.

	There may be some simplification possible but anyway these can be 
numerically evaluated. In case I can simplify further I shall inform.
Also if some one else has sent you other formulae would you please send 
those to me along with the e-mail address of the authors so that I can 
interact with them.
	Hope this mail is useful to you.
	With regards
	Sincerely yours
         H. K. Sahu






On Wed, 5 Nov 2003, Claudio Ferrero wrote:

> Dear All,
> 
> could somebody help me in finding out the expression of the form factor
> P(q) of an object with the shape of a helical ribbon ?
> 
> Many thanks in advance for your kind answers
> 
> Caudio Ferrero
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