# Fluence rate of gamma source - getting two different answers

1. Aug 29, 2016

### Hercuflea

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. Aug 30, 2016

### Staff: Mentor

I think the substitution in the second method is missing something when you go from $dr$ to $d\theta$.

3. Aug 30, 2016

### Hercuflea

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:
$$\phi = \frac {10^7} {4\pi} 2\int_{0}^{arctan(0.5)} \frac {sec\theta tan\theta}{(\frac {1}{cos\theta})^2} d\theta$$
$$=\frac {10^7} {2\pi} \int_{0}^{arctan(0.5)} \frac {tan\theta}{sec \theta} d\theta$$
$$=\frac {10^7} {2\pi} \int_{0}^{arctan(0.5)} sin\theta d\theta$$
$$=\frac {10^7} {2\pi} (-cos(arctan(0.5))-(-cos0)) = -1.7 \times 10^5$$
?

4. Aug 30, 2016

### TSny

Should $dr$ actually be $dy$?

5. Aug 30, 2016

### Hercuflea

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.

6. Aug 30, 2016

### TSny

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

7. Aug 30, 2016

### Hercuflea

$r$ 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 $y=rsin\theta$
$$=\frac{10^7}{4\pi} \int_{-.5}^{.5} (\frac{sin^2 \theta}{(rsin\theta)^2}) dy$$
$$=\frac{10^7}{4\pi} \int_{-.5}^{.5} (\frac{1}{r^2}) dy$$

8. Aug 30, 2016

### TSny

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 $\theta$.)

9. Aug 30, 2016

### Hercuflea

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.....
$$=7.3792 \times10^5 \frac{photons}{m^2 s} (\frac{1m}{100cm})^2 = 73.79 \frac{photons}{cm^2 s}$$

10. Aug 30, 2016

### TSny

Check your work in going from the first line above to the second line.

11. Aug 30, 2016

### Hercuflea

Yes! Thank you! I edited the post.

12. Aug 30, 2016

### TSny

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.