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Asymptotic Formula

  1. Jun 18, 2017 #1


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    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

    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)
    $$C[α(x; +a)] + C[α(x; −a)] ≃ \frac{1}{πα(x; a)} sin (\frac{πα(x; a)^2}{2}
    ) + \frac{1}{πα(x; -a)} sin (\frac{πα(x; -a)^2}{2}
    ) $$
    $$S[α(x; +a)] + S[α(x; −a)] ≃ \frac{-1}{πα(x; a)} cos (\frac{πα(x; a)^2}{2}
    ) - \frac{1}{πα(x; -a)} cos (\frac{πα(x; -a)^2}{2}
    ) $$

    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

    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
  2. jcsd
  3. Jun 18, 2017 #2
    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.

  4. Jun 19, 2017 #3


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    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.
  5. Jun 19, 2017 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    ##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##?
    Last edited: Jun 19, 2017
  6. Jun 19, 2017 #5
    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?

  7. Jun 20, 2017 #6


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    $$x ≫ a$$ large x and small a
    Sorry for the late reply.
    Last edited: Jun 20, 2017
  8. Jun 20, 2017 #7


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    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

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

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

    A factor of 2 missing in front and an extra 2 in the denominator.
  9. Jun 20, 2017 #8
    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
    NF(a) = 2a2/λL​
    comes from.

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

  10. Jun 20, 2017 #9


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    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.
  11. Jun 20, 2017 #10
    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.

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