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Problem with four vectors

  1. Oct 6, 2009 #1
    I should calculate the limit of the following fraction

    [tex] \dfrac{- (pq) p^{2}}{-pq \pm \sqrt{(pq)^{2} - p^{2} q^{2}}} [/tex].

    with q --> 0, but I don't know how to do that.

    p and q are two four-vectors, so we have: [tex] pq = p_{\mu} q^{\mu} [/tex] and so on.

    Does anyone have an idea? Or at least: Is it somehow possible to "simplify" or rewrite the expression in the limit q --> 0 ?
  2. jcsd
  3. Oct 6, 2009 #2


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    Hi parton! :smile:

    If we're letting q --> 0, then we may as well put q = (a,0,0,0) and let a --> 0.

    Then it's -ap0p2/(-ap0 ± √(a2p02 - a2p2)), = -p0p2/(-p0 ± √(p02 - p2)),

    which I think is m2/(1 ± v).

    But I expect the result is different if q --> 0 along some other path.
  4. Oct 9, 2009 #3
    I would like to see a general solution to this as well. I poked at it but nothing immediately jumped out at me. The general solution must determine how [tex] -pq [/tex] compares to [tex]\sqrt{(pq)^{2} - p^{2} q^{2}}} [/tex] as q-->0. It's not obvious to me that the latter approaches zero any order "faster" so the ratio [tex] \dfrac{- (pq) }{ \sqrt{(pq)^{2} - p^{2} q^{2}}} [/tex] must approach some constant, but does that constant depend on q's path?

    I'm guessing there is either some nice identity that you can use if you expand it all out OR you may have to use some differentials to eventually end up with [tex] A p^{2}[/tex] where A is some constant.

    Although, if the answer really is path independent, then choosing your favorite path might be the way to go like tiny-tim suggested!
  5. Oct 9, 2009 #4
    I think you can do this without loss of generality by choosing a coordinate system where q = (a,0,0,0), like tiny-tim suggested. Then [tex](pq)^2 = (p_0a)^2[/tex], and [tex]p^2 q^2 = p^2 a^2[/tex], so then you have

    [tex]\frac{ -p_0 a p^2}{-p_0a \pm \sqrt{a^2(p_0^2 - p^2)}}[/tex]

    The a's can now be canceled out to give

    [tex]\frac{ -p_0 p^2}{-p_0 \pm \sqrt{p_0^2 - p^2}}[/tex]

    In another coordinate system, you would interpret [tex]p_0[/tex] as the component of p that points in the direction of q. I'm a bit surprised at the sign under the square root, since you should have [tex]p^2 \ge p_0^2[/tex], so maybe I've screwed something up; it's been a whiel since I worked with 4-vectors.
  6. Oct 9, 2009 #5
    Everything is OK: p2=m2 but p02>m2.
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