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Fluence rate of gamma source - getting two different answers

  1. Aug 29, 2016 #1
    1. The problem statement, all variables and given/known data
    A 1 meter long uniform line source of high energy photons emits 10^7 photons /sec. Calculate the particle fluence rate (flux density) at a point P 1m away from the midpoint of the source.

    2. Relevant equations
    ## dN = number~of~particles ##
    ## da = differential~area##
    ##\Phi = fluence = \frac{dN}{da}##
    ##\phi = fluence~rate = \frac{d\Phi}{dt} = \frac{d}{dt} \frac{dN}{da} = \frac {\frac{dN}{dt}}{da}##

    3. The attempt at a solution

    I have tried to solve the problem two different ways, and I get 2 different answers.
    Define the origin (0,0) to be the location of the point P. Define the rod source such that it's endpoints lie at ##(1,-\frac{1}{2})## and ## (1, \frac{1}{2})##.

    1st method: I integrated using the pythagorean theorem and I got

    $$ r = \sqrt{y^{2} + 1^{2}}$$
    $$ \phi = \int {\frac{10^7}{4\pi r^{2}}} dr $$
    $$ = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{10^7}{4\pi (\sqrt{y^2+1})^2} dy\ $$
    $$ = \frac{10^7}{4\pi} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac {1}{y^2+1} dy\ $$
    $$ = \frac{10^7}{4 \pi} (arctan(y)) $$ from -0.5 to 0.5
    $$= \frac{10^7}{4 \pi} (arctan(0.5)-arctan(-0.5)) $$
    $$= 7.379 \times 10^5 \frac{photons}{m^2 s} (\frac{1m}{100cm})^2 = 73.79 \frac{photons}{cm^2 s}$$

    2nd method:
    Since ##rcos(\theta) = 1 ## always, I solved for ##r## in terms of ##\theta## and I got
    $$ r = \frac{1}{cos\theta}$$
    Also, since ##tan\theta_{max} = \frac{\frac{1}{2}}{1} ## we have ##\theta_{max} = tan^{-1} (0.5)##.

    Therefore
    $$ \phi = \int {\frac{10^7}{4\pi r^{2}}} dr $$
    $$= \int_{-tan^{-1}(0.5)}^{tan^{-1} (0.5)} \frac{10^7}{4\pi {\frac{1}{cos\theta}}^2} d\theta $$
    $$ = \frac{10^7}{4\pi} \int_{-tan^{-1}(0.5)}^{tan^{-1} (0.5)} cos^{2}(\theta) d\theta $$
    $$ = \frac{10^7}{4\pi} \frac{1}{2} (\theta + \frac{1}{2} sin(2\theta)) $$ from ##-tan^{-1}(0.5)## to ##tan^{-1}(0.5)##
    $$=\frac{10^7}{8\pi} (tan^{-1}(0.5)-(-tan^{-1}(0.5)) + \frac{1}{2} (sin(2tan^{-1}(0.5))-sin(-2tan^{-1}(0.5))))$$
    $$ = 6.87 \times 10^5 \frac {photons}{m^2 s}(\frac{1m}{100cm})^2 = 68.7\frac {photons}{cm^2 s} $$

    Why do the two methods give different answers? I am integrating over the same geometric object....only one way I am using the angle and the other way I am using the vertical component of r. Which method is correct?
     
    Last edited: Aug 29, 2016
  2. jcsd
  3. Aug 30, 2016 #2

    mfb

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    I think the substitution in the second method is missing something when you go from ##dr## to ##d\theta##.
     
  4. Aug 30, 2016 #3
    Thank you for catching that.

    Now that I have fixed that, I'm just getting 0.

    $$r= \frac 1 {cos\theta} = sec\theta$$
    $$dr = sec\theta tan\theta d\theta$$
    $$ \phi = \frac {10^7} {4\pi} \int_{-arctan(0.5)}^{arctan(0.5)} \frac {sec\theta tan\theta}{(\frac {1}{cos\theta})^2} d\theta $$
    $$=\frac {10^7} {4\pi} \int_{-arctan(0.5)}^{arctan(0.5)} \frac {tan\theta}{sec \theta} d\theta $$
    $$=\frac {10^7} {4\pi} \int_{-arctan(0.5)}^{arctan(0.5)} sin\theta d\theta $$
    $$=\frac {10^7} {4\pi} (0) = 0 $$

    If I divide the bounds of integration by 2 and multiply the integral by 2 i get:
    $$ \phi = \frac {10^7} {4\pi} 2\int_{0}^{arctan(0.5)} \frac {sec\theta tan\theta}{(\frac {1}{cos\theta})^2} d\theta $$
    $$=\frac {10^7} {2\pi} \int_{0}^{arctan(0.5)} \frac {tan\theta}{sec \theta} d\theta $$
    $$=\frac {10^7} {2\pi} \int_{0}^{arctan(0.5)} sin\theta d\theta $$
    $$=\frac {10^7} {2\pi} (-cos(arctan(0.5))-(-cos0)) = -1.7 \times 10^5 $$
    ?
     
  5. Aug 30, 2016 #4

    TSny

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    Should ##dr## actually be ##dy##?
     
  6. Aug 30, 2016 #5
    In the pythagorean theorem method, yes. But I'm trying to do it by integrating over the angle and get the same result as that.
     
  7. Aug 30, 2016 #6

    TSny

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    For the method of integrating over the angle, you will need to decide which of the following is the correct expression to start with:
    $$ \phi = \int {\frac{10^7}{4\pi r^{2}}} dr $$ $$ \phi = \int {\frac{10^7}{4\pi r^{2}}} dy $$
     
  8. Aug 30, 2016 #7
    ##r## is a function of ##y##

    $$y = rsin\theta$$
    $$r = \frac {y}{sin\theta}$$
    $$dr = d(\frac{y}{sin\theta})$$ = ?

    If I choose the second integral I just get a circular argument when I try to get the whole thing in terms of ##\theta##...
    $$ \phi = \int_{-.5}^{.5} \frac{10^7}{4\pi (\frac{y}{sin\theta})^2} dy$$
    $$=\frac{10^7}{4\pi} \int_{-.5}^{.5} (\frac{sin^2 \theta}{y^2}) dy$$

    but since ##y=rsin\theta##
    $$=\frac{10^7}{4\pi} \int_{-.5}^{.5} (\frac{sin^2 \theta}{(rsin\theta)^2}) dy$$
    $$=\frac{10^7}{4\pi} \int_{-.5}^{.5} (\frac{1}{r^2}) dy$$
     
  9. Aug 30, 2016 #8

    TSny

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    Have you decided which of the two integrals is the correct starting integral? ## \phi = \int {\frac{10^7}{4\pi r^{2}}} dr ## or ## \phi = \int {\frac{10^7}{4\pi r^{2}}} dy ##?

    If the second integral is the correct integral, then you need to express ##r## and ##dy## in terms of ##\theta## and ##d\theta## if you want to integrate over ##\theta##.

    To relate ##dy## to ##d\theta##, then ##y = rsin\theta## is not a good relation to use. That's because ##y = rsin\theta## relates three variables. Can you relate ##y## to ##\theta## without using a third variable. (Try a different trig function.)

    (You already found a nice way to relate ##r## to ##\theta## in your first attempt at integrating over ##\theta##.)
     
  10. Aug 30, 2016 #9
    I see what you mean now. Yes, I think the integral with dy is correct because the differential length we are integrating over is along the rod, not along the radius.

    $$\frac{y}{1} = tan\theta$$
    $$dy = sec^2 \theta d\theta$$
    $$\phi = \frac{10^7}{4\pi} \int_{-.5}^{.5} \frac{1}{r^2} dy$$
    $$= \frac{10^7}{4\pi} \int_{-arctan(.5)}^{arctan(.5)} \frac{sec^2\theta}{(\frac{1}{cos\theta})^2} d\theta$$
    $$=\frac{10^7}{4\pi} \int_{-arctan(.5)}^{arctan(.5)} cos^2 \theta sec^2 \theta d\theta$$

    wolfram alpha.....
    $$=7.3792 \times10^5 \frac{photons}{m^2 s} (\frac{1m}{100cm})^2 = 73.79 \frac{photons}{cm^2 s}$$
     
  11. Aug 30, 2016 #10

    TSny

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    Check your work in going from the first line above to the second line.
     
  12. Aug 30, 2016 #11
    Yes! Thank you! I edited the post.
     
  13. Aug 30, 2016 #12

    TSny

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    OK, good work. ##\; \; \cos^2\theta \sec^2\theta = 1##.

    Yes, as you say, it should be ##dy## instead of ##dr## in the original integral.
     
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