1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral from distribution function

  1. Dec 15, 2013 #1
    1. The problem statement, all variables and given/known data
    hi, so I've got this distribution function:
    [tex]f(z,p,t)=\frac{1}{2\pi\partial z\partial p}exp(-\frac{[z-v(p)t]^2}{2\partial z^2})exp(-\frac{[p-p_0]^2}{2\partial p^2})[/tex]

    where:
    [tex]v(p)=v_0+\alpha(p-p_0)[/tex]
    [tex]v_0=\frac{p_0}{m\gamma_0}[/tex]
    [tex]\alpha=\frac{1}{m\gamma_0^3}[/tex]

    I have to calculate the mean position as a function of time[tex]\langle z \rangle[/tex]

    2. Relevant equations

    I've got a hint too which doesn't help me at all to be honest ;] :
    All the integrals over z required to calculate the averages can be expressed in therms of [tex]I_\nu =\int_{-\infty}^{+\infty}\,ds \ s^\nu exp(-s^2)[/tex] with [itex]I_0=\sqrt{\pi}, I_1=0,\ and \ I_2=\sqrt{\pi}/2[/itex], by substitution and a suitable choice of the order of integration.



    3. The attempt at a solution
    Presented function is I think distribution of particles in a beam in longitudinal phase space, and in that case
    [itex]\int \,d^3p\ f(z,p,t)=n(z,t)[/itex] which is the particle density and [itex] \int \,d^3z\ n(r,t)=N[/itex] which is particles number.
    So I think that mean position should look like this:
    [tex] \langle z \rangle=\frac{\int\,d^3z\int\,d^3p \ z\ f(z,p,t)}{N}[/tex]
    so
    [tex] \langle z \rangle=\frac{\int\,d^3z\int\,d^3p \ z \ f(z,p,t)}{\int\,d^3z\int\,d^3p\ f(z,p,t)}[/tex]
    am I right? if yes, how to start solving this kind of integral. I mean something like even this :
    [tex] \int\,d^3z\frac{1}{2\pi\partial z\partial p} ...[/tex] those partials go before the integral .. or what?


    btw. I attached two files with the actual assignment, and a slide from lecture that is supposed to tell me everything ;)
     

    Attached Files:

  2. jcsd
  3. Dec 15, 2013 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    ##\delta## doesn't mean ##\partial##. ##\delta z## and ##\delta p## are just constants.
     
  4. Dec 15, 2013 #3
    oh... right :) thanks.

    what about 'the hint'? ;]
     
  5. Dec 15, 2013 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Besides what Vela has pointed out to you, there is also the important issue of whether ##\delta z^2## means ##\delta(z^2)## (an "error" parameter for ##z^2##) or whether it means ##(\delta z)^2## (the square of the "error" parameter for ##z##). Using simpler symbols like ##a## and ##b## instead of ##\delta z## and ##\delta p## would be a very great help, both to you and to us. (As as genera rule, choosing simpler notation helps when dealing with lengthy and complicated calculations; often, it is a good idea to replace the problem's notation by your own before starting on the problem. Then, in the final answer you can put back the original notation. Years of experience has taught me the value of doing that!)
     
  6. Dec 17, 2013 #5
    I'm not quite sure what is the [itex]\delta z[/itex] or [itex]\delta p[/itex]. It doesn't say, although I think that you guys are right that if [itex] f(z,p,t)[/itex] represents a distribution of particles in a beam then [itex]\delta z[/itex] and [itex]\delta p[/itex] could be the errors. Let suppose that it's acutally [itex](\delta z)^2[/itex] and [itex] (\delta p)^2[/itex] and that they're [itex]A[/itex] and [itex]B[/itex]

    I'm still not sure that for example in this integral [tex]\int\,d^3z\int\,d^3p\ f(z,p,t)[/tex]
    we've got [itex]\,d^3z [/itex] and [itex]\,d^3p[/itex] and if I assume that the beam goes in one direction (z) then can the integral be actually written like this? [tex]\int\,dz\int\,dp\ f(z,p,t)[/tex]

    if yes then is that right? :

    [tex]\int\,dz\int\,dp\ exp(\frac{-z^2+2zV(p)t-V(p)^2t^2}{2A})exp(\frac{-p^2+2pp_0-(p_0)^2}{2B})=\\ exp(-\frac{(p_0)^2}{2B})\int\,dz\int\,dp \exp(-\frac{z^2}{2A})exp(\frac{zV(p)t}{A})exp(-\frac{V(p)^2t^2}{2A})exp(-\frac{p^2}{2B})exp(\frac{pp_0}{B})[/tex]

    what about this second exp? it depends on both z and p? (others I can integrate over either z or p)
     
  7. Dec 17, 2013 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I suspect that p and z are one-dimensional, not 3-D (because you speak of a beam). If they were 3-dimensional your problem would be an uncomputable mess, because (at least for z) you would need geometric boundaries perpendicular to the beam's axis, etc.

    To make further progress you need to actually use the given form of v(p), in order to reduce everything to a couple of computable gaussian-type integrations. If you start by first doing the p-integration, you need to combine all the terms having p in them, whether they come from the first factor or the second one. In other words, your p-integration may also contain z as a "parameter", and that can modify the form of the later z-integration (assuming you integrate first over p and then over z).
     
    Last edited: Dec 17, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted