Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Neutron density of a beam

  1. Feb 11, 2019 at 6:17 AM #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    ##\int_{\Omega} n(x, \mathbf \omega) d \Omega = n##

    3. 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
     
    Last edited: Feb 11, 2019 at 7:51 AM
  2. jcsd
  3. Feb 11, 2019 at 2:24 PM #2

    Charles Link

    User Avatar
    Homework Helper
    Gold Member
    2018 Award

    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.
     
    Last edited: Feb 11, 2019 at 3:51 PM
  4. Feb 11, 2019 at 4:10 PM #3
    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 ?
     
  5. Feb 11, 2019 at 4:31 PM #4

    Charles Link

    User Avatar
    Homework Helper
    Gold Member
    2018 Award

    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) ##).
     
    Last edited: Feb 11, 2019 at 4:59 PM
  6. Feb 12, 2019 at 10:59 AM #5
    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
     
  7. Feb 12, 2019 at 1:42 PM #6

    Charles Link

    User Avatar
    Homework Helper
    Gold Member
    2018 Award

    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 ##.
     
    Last edited: Feb 12, 2019 at 2:14 PM
  8. Feb 15, 2019 at 7:54 AM #7
    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)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?