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Momentum space measure -- change of variables

  1. Jan 2, 2015 #1
    I'm working through an article called "Cosmic abundances of stable particles -- improved analysis" (link -- viewable only in Firefox afaik), the result of which, equation (3.8), is cited quite a lot. I'm more interested in how they arrived there.
    Particularly, how come momentum space measure, [itex]\mathrm{d}^3\mathbf{p}_1\mathrm{d}^3\mathbf{p}_2[/itex], suddenly becomes [itex]4\pi p_1\mathrm{d} E_1 4\pi p_2\mathrm{d} E_2 \frac{1}{2}p_1 p_2 \mathrm{d}\cos\theta[/itex], where [itex]\theta=\angle(\mathbf{p}_1, \mathbf{p}_2)[/itex] and [itex]p_i=|\mathbf{p}_i|[/itex] (equation (3.2))? The dimensions don't match at all.. The closest I can come up with is [itex]\mathrm{d}^3\mathbf{p}_i = p_i^2 \sin\theta_i\mathrm{d}\theta_i\mathrm{d}\phi_i\mathrm{d}p_i \hat{\mathbf{p}}_i = [\mbox{integrate over angles}] = 4\pi p_i^2 \mathrm{d}p_i \hat{\mathbf{p}}_i [/itex], which after substituting with [itex]p_i \mathrm{d}p_i=E_i \mathrm{d}E_i[/itex] (because [itex]p_i^2 = E_i^2 - m_i^2[/itex]) gives [itex]\mathrm{d}^3\mathbf{p}_i = 4\pi p_i E_i \mathrm{d}E_i \hat{\mathbf{p}}_i [/itex]. So I think it should be something like [itex]\mathrm{d}^3\mathbf{p}_1\mathrm{d}^3\mathbf{p}_2 = 4\pi E_1 \mathrm{d}E_1 4\pi E_2 \mathrm{d}E_2 p_1 p_2 \cos\theta=16\pi^2 E_1 E_2 p_1 p_2\cos\theta\mathrm{d}E_1\mathrm{d}E_2[/itex]. With this I'm off by one diffenetial and 1/2...
    It also bugs me that they make change of variables, [itex]E_+ = E_1 + E_2[/itex], [itex]E_- = E_1 - E_2[/itex], [itex]s = (E_1 + E_2, \mathbf{p}_1 + \mathbf{p}_2)^2[/itex], which again leads to a peculiar result, [itex]\mathrm{d}^3 p_1\mathrm{d}^3p_2 = 2\pi^2 E_1 E_2 \mathrm{d}E_+\mathrm{d}E_-\mathrm{d}s[/itex] (equation 3.4). I mean, doing this rigorously doesn't work out, unless you assume that [itex]E_+ = E_+(E_2)[/itex], [itex]E_- = E_-(E_1)[/itex] and [itex]s = s(p_1 p_2 \cos\theta)[/itex] or something..

    Any ideas what went wrong in (3.2) and (3.4)?
    (Sorry for the question being too technical; if you feel this thread doesn't belong to this forum, move it away)
    Oh, and happy new year ..

    ---

    edit: ok, I have figured out that [itex]\mathrm{d}^3\mathbf{p}_2 = 2\pi p_2^2 \mathrm{d}p_2 \mathrm{d}\cos\theta[/itex] and evidently equation (3.2) is indeed missing a factor [itex]p_1 p_2[/itex]. Still, the change of variables into [itex]E_\pm[/itex] and [itex]s[/itex] puzzles me.
     
    Last edited: Jan 2, 2015
  2. jcsd
  3. Jan 7, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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