Solving Asymptotic Formula: Eq. 25 & 27

[JBD]

In the following equation,

$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2}
)cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$$if $$ 0<N_F(a) ≪ 1$$ and $$(x − aη)/aη ≫ 1/\sqrt{N_F (a)η}$$, how do you arrive at
$$P(x; a) ≃ \frac{2γ}{π^2η^2} (\frac{a^2}{(\frac {x^2}{η^2} − a^2)^2}
+\frac{1}{\frac{x^2}{η^2} − a^2} sin^2
(πN_F(a)\frac{x}{a}))$$Please see what I have done so far and check if I have errors in it.

Relevant Equations

$$\alpha (x; a) = \sqrt{N_F(a)\eta } (1 - \frac{x}{a\eta})$$Eq. 19

where $$\eta = 1 + L/D$$, $$N_F(a) = \frac{2a^2}{\lambda L}$$ and (additional definition) $$\gamma = \eta - 1$$

Starting with this:
Eq. 20
$$P(x; a)=\frac{1}{2\lambda(L+D)} ([C(α(x; a)) + C(α(x; −a))]^2 + [S(α(x; a)) + S(α(x; −a))]^2)$$
Eq. 24 (these two)
$$C[α(x; +a)] + C[α(x; −a)] ≃ \frac{1}{πα(x; a)} sin (\frac{πα(x; a)^2}{2}
) + \frac{1}{πα(x; -a)} sin (\frac{πα(x; -a)^2}{2}
) $$
and
$$S[α(x; +a)] + S[α(x; −a)] ≃ \frac{-1}{πα(x; a)} cos (\frac{πα(x; a)^2}{2}
) - \frac{1}{πα(x; -a)} cos (\frac{πα(x; -a)^2}{2}
) $$Here is what I have done so far:
$$(C[α(x; +a)] + C[α(x; −a)])^2 + (S[α(x; +a)] + S[α(x; −a)])^2 = \frac{1}{π^2α^2(x; a)} + \frac{1}{π^2α^2(x; -a)} + \frac{2}{π^2α(x; +a)α(x; −a)} [sin (\frac{πα(x; a)^2}{2})sin (\frac{πα(x; -a)^2}{2}) + cos (\frac{πα(x; a)^2}{2}
)cos (\frac{πα(x; -a)^2}{2})]$$

Using equation 19:

$$=\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2}
)cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]$$

And then in equation 20, the outer factor:
$$\frac{1}{2\lambda (L+D)} = \frac{\gamma}{2\lambda L \eta}$$

So the new equation for P is:
$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2}
)cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$$But this is where I am not sure what to do anymore with $$N_F (a) ≪ 1$$ and if $$(x − aη)/aη ≫ 1/\sqrt{N_F (a)η}$$ to arrive at equation 25.I'm trying to verify equations 25 and 27 in the paper linked below. I got confused on how to apply $$N_F(a) ≪ 1$$ to get the final result. I was able to do "Applying the Fresnel function asymptotic forms (24) to (20) and using the definition (19)" but then I got stuck here

"we deduce that if $$N_F (a) ≪ 1$$ and if $$(x − aη)/aη ≫ 1/\sqrt{N_F (a)η}$$ , we
get the following asymptotic formula:"

Eq. 25
$$P(x; a) ≃ \frac{2γ}{π^2η^2} (\frac{a^2}{(\frac {x^2}{η^2} − a^2)^2}
+\frac{1}{\frac{x^2}{η^2} − a^2} sin^2
(πN_F(a)\frac{x}{a}))$$

I could not reproduce this result (eq 25 in the paper) as well as eq 27.Page 14 Equations 25 and 27
Here is the link https://arxiv.org/pdf/1110.2346.pdf
There are occasional typographical errors in the paper. (from what I have verified so far, pages 1-13) [1]: https://i.stack.imgur.com/Qc5Ie.png

 

[Buzz Bloom]

But this is where I am not sure what to do anymore with NF(a)≪1

Hi JBD:

The only thing that occurs to me is to (1) replace sin(u) with u since u<<1, and (2) replace cos(u) with 1, since cos(u) ~= 1-u².

Hope this helps.

Regards,
Buzz

 

[JBD]

Hi JBD:

The only thing that occurs to me is to (1) replace sin(u) with u since u<<1, and (2) replace cos(u) with 1, since cos(u) ~= 1-u².

Hope this helps.

Regards,
Buzz

Thanks for helping. I did it differently and arrived at the final form but I had some assumptions that I don't know if allowed or not. Like for example, $$N_F(a) =\frac{2a^2}{\lambda L}$$ but since $$N_F(a) ≪1$$, I replaced it with $$\frac{a^2}{\lambda L}$$. Is this reasonable? If not, I would have arrived at a similar form but short of the factor 2. I also had to do it one more time in a fraction with large number in denominator. The numerator was one and I changed it to 2.

 

[Ray Vickson]

In the following equation,

$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2}
)cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$$if $$ 0<N_F(a) ≪ 1$$ and $$(x − aη)/aη ≫ 1/\sqrt{N_F (a)η}$$, how do you arrive at
$$P(x; a) ≃ \frac{2γ}{π^2η^2} (\frac{a^2}{(\frac {x^2}{η^2} − a^2)^2}
+\frac{1}{\frac{x^2}{η^2} − a^2} sin^2
(πN_F(a)\frac{x}{a}))$$

$P(x;a)$ has two parameters or arguments. When you go to the asymptotic form you are letting one of the arguments get large (or maybe small); would that be large (small?) $x$ or large (small?) $a$?

 

[Buzz Bloom]

but since

NF(a)≪1​

I replaced it with

a²λL.​

Is this reasonable?

I underlined "it" in the quote because I am not sure what the antecedent of this pronoun is. I am guessing you intend it to be what is equal to NF(a). I also do not understand what would make this assumptions reasonable.

Could you post the details of your work that leads to the correct answer except for a factor of 2?

Regards,
Buzz

 

[JBD]

$P(x;a)$ has two parameters or arguments. When you go to the asymptotic form you are letting one of the arguments get large (or maybe small); would that be large (small?) $x$ or large (small?) $a$?

$$x ≫ a$$ large x and small a
Sorry for the late reply.

 

[JBD]

I underlined "it" in the quote because I am not sure what the antecedent of this pronoun is. I am guessing you intend it to be what is equal to NF(a). I also do not understand what would make this assumptions reasonable.

Could you post the details of your work that leads to the correct answer except for a factor of 2?

Regards,
Buzz

I used cos (A-B) = cos A cos B + sin A sin B (I feel bad for not noticing this. I was too focused on applying the N_f(a) ≪1 and the other condition.)

$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2}
- \frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$$$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{2(1+\frac{x^2}{a^2\eta^2})}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [cos (2πN_F(a)\frac{x}{a})]]$$$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [\frac{1 + \frac{x^2}{a^2\eta^2}}{1 - \frac{x^2}{a^2\eta^2}}+cos (2πN_F(a)\frac{x}{a})]] $$$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{ \frac{x^2}{a^2\eta^2}}{ \frac{x^2}{a^2\eta^2}-1}+\frac{1 }{\frac{x^2}{a^2\eta^2}-1}-cos (2πN_F(a)\frac{x}{a})]] $$

$$\frac{ \frac{x^2}{a^2\eta^2}}{ \frac{x^2}{a^2\eta^2}-1} ≃ 1$$since$$ \frac{x}{a\eta}$$ is a large number$$P(x; a)≃ \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} +1-cos (2πN_F(a)\frac{x}{a})]] $$

I used 2 sin^2 A = 1 - cos 2A

$$P(x; a)≃ \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} +2 sin^2 (πN_F(a)\frac{x}{a})]] $$

$$N_F(a) = \frac{2a^2}{\lambda L}$$
so,
$$P(x; a)≃ \frac{\gamma}{ \eta} \frac{1}{π^22a^2\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} +2 sin^2 (πN_F(a)\frac{x}{a})] $$
$$P(x; a)≃ \frac{\gamma}{π^2 \eta^2} (\frac{1}{2a^2( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} + 2sin^2 (πN_F(a)\frac{x}{a})]) $$
$$P(x; a)≃ \frac{\gamma}{π^2 \eta^2} (\frac{1}{2( \frac{x^2}{\eta^2}-a^2)} [\frac{a^2 }{\frac{x^2}{\eta^2}-a^2} + 2sin^2 (πN_F(a)\frac{x}{a})]) $$
$$P(x; a)≃ \frac{\gamma}{π^2 \eta^2} ( \frac{a^2 }{2(\frac{x^2}{\eta^2}-a^2)^2} + \frac{1}{( \frac{x^2}{\eta^2}-a^2) }sin^2 (πN_F(a)\frac{x}{a})) $$

A factor of 2 missing in front and an extra 2 in the denominator.

 

[Buzz Bloom]

Hi JDB:

I now see how you derived the answer you got, and where it is different from Eq 25. What still puzzles me is where the assumption

N_F(a) = 2a²/λL​

comes from.

BTW, I now also see why my original suggestion is not helpful,

Regards,
Buzz

 

[JBD]

Hi JDB:

I now see how you derived the answer you got, and where it is different from Eq 25. What still puzzles me is where the assumption

N_F(a) = 2a²/λL​

comes from.

BTW, I now also see why my original suggestion is not helpful,

Regards,
Buzz

Fresnel Number is defined as $$N_f(a) = a^2 / \lambda L$$
where a is the size of slit, lambda is wavelength and L is distance from slit to screen
In the paper, the slit was size 2a and distance from slit to screen is L + D, but D = L so
$$N_f(a) = 4a^2 / \lambda 2L = 2a^2/ \lambda L$$

I don't know if eq 25 has typos (because I encountered some of them ) but I think it does not have typos. But then again, where did I get my math wrong? Thanks.

 

[Buzz Bloom]

Hi JDB:

My guess is that the text you were using has typos or just plain errors. If the problem was part of some work project, I think you should just accept that you got the right answer. If the problem was related to study for a course, I suggest you review your work with the professor - he may feel inclined to give you some extra credit for finding an error in the text.

Regards,
Buzz