Neutron density of a beam

[dRic2]

Homework Statement

Show that the neutron density distribution function at any point in a monodirectional beam of monoenergetic neutrons moving along the x-axis is given by
$$n(x, \mathbf \omega) = \frac n {\pi} \delta( \mu -1)$$
where $n$ is the neutron density, $\delta( \mu -1)$ is the Dirac delta function, and $\mu$ is the cosine of the angle between $\mathbf \omega$ and the x-axis.

Homework Equations

$\int_{\Omega} n(x, \mathbf \omega) d \Omega = n$

The Attempt at a Solution

I simply checked that integrating over the solid angle gives the total neutron density:
$$\int_{\Omega} \frac n {\pi} \delta( cos \theta -1) sin \theta d \theta d \phi $$
$\mu = cos \theta \rightarrow d \mu = -sin\theta d \theta$ and by the properties of Dirac's delta function the above integral reduces to
$$\frac n {\pi} \frac 1 2 2 \pi = n$$

I think this is a valid proof, but I'm not very sure how to "derive" the expression in the first place. For example: why the Dirac's delta function is expressed in terms of $cos \theta$ instead of just $\theta$ ?

Thanks
Ric

 

[Charles Link]

Where did the $\frac{1}{2}$ come from? Otherwise what you did looks correct. I would have though the original expression needs an extra factor of $\frac{1}{2}$. Perhaps I'm mistaken. $\\$ Evaluating=i.e. integrating a delta function to the endpoint of integration (where the argument goes to zero) as opposed to integrating through it looks to be somewhat problematic.$\\$ Edit: Perhaps that is where the 1/2 comes from=I don't know how "sound" that is mathematically. I generally try to avoid that type of ambiguity. If it can be agreed upon that $\int\limits_{0}^{+\infty} \delta(x) \, dx=\frac{1}{2}$, then I guess you can say it works.

 

[dRic2]

Edit: Perhaps that is where the 1/2 comes from=I don't know how "sound" that is mathematically. I generally try to avoid that type of ambiguity. If it can be agreed upon that +∞∫0δ(x)dx=12∫0+∞δ(x)dx=12 \int\limits_{0}^{+\infty} \delta(x) \, dx=\frac{1}{2} , then I guess you can say it works.

Yes, sorry for not specifying. I'm still not very familiar with Dirac's delta function so I do not know how to state it with the proper mathematical rigor.

BTW I do not like my "proof" because it doesn't really tell you where that expression comes from, do you have any suggestions about it ?

 

[Charles Link]

This one reminds me of the scattering that occurs from a cone that scatters the incident beam into a ring at angle $\theta= \theta_o$. $\\$ By inspection $\frac{d \sigma}{d \Omega}= \frac{A_{base}}{2 \pi \sin{\theta}} \delta(\theta-\theta_o)$ in spherical coordinates, because integrating it over $d \Omega$ must give $\sigma_{total}= A_{base}$. $\\$ The $\frac{\delta(\theta-\theta_o)}{\sin{\theta}}$ can also be written as $\delta(\cos{\theta}-\cos{\theta_o})$. $\\$ See this post where this delta function with an angle in its argument just came up the other day=(see post 2): https://www.physicsforums.com/threa...a-rutherfords-experiment.965947/#post-6131309 $\\$ Basically $\delta(f(x)-f(a))=\frac{\delta(x-a)}{|f'(x)| } =\frac{\delta(x-a)}{|f'(a)|}$ if I'm not mistaken. (Calling it $f'(a)$ in the very last step might be incorrect. If $f'(a)=0$, it might need to remain as $f'(x)$).

 

[dRic2]

Sorry for the late replay. You gave me an excellent idea, but I still can't get the whole thing together. I have to think this through very carefully

 

[Charles Link]

Perhaps it is worth mentioning that with a delta function representation of the beam, you do not get any information on the $\theta$ dependence other than that it is located at one angle $\theta_o$. The delta function can be represented by very narrow Gaussians or other shapes, but these shapes essentially have zero width. $\\$ Once again, there is no information on the $\theta$ dependence other than that it all the power appears at one specific direction. $\\$ In this case, all the power is where $cos{\theta_o}=1$. The form that the delta function takes on for $n$, (or intensity $I$), is such that $P_{total}=\int I \, d \Omega$. The form of the delta function for the intensity $I=I(\theta, \phi)$ can normally be written down by simply looking at this integral $P_{total}=\int I \, d \Omega=\iint I(\theta, \phi) \, \sin{\theta} \, d \theta \, d \phi$ result. $\\$ The case you have is complicated by the fact that the intensity is at angle $\theta_o=0$ so that the integration could not take place on both sides of the delta function peak. Delta functions are normally considered to be symmetric about $x=0$ and thereby this one could be considered to have a 1/2 factor, i.e. $\frac{1}{2}=\int\limits_{0}^{+\infty} \delta(x) \, dx$, where $1=\int\limits_{-\infty}^{+\infty} \delta(x) \, dx$.

 

[dRic2]

Sorry if I'm not replying but I have been busy with exams in these days. I have to figure out by my self so I do not know what to add to your post. Thanks a lot. (Sorry for the errors but I'm with my phone setted to an other language and it is correcting everything)