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

Homework Help: Limit computation

  1. Dec 19, 2009 #1
    I've the following problem. I have two four-vectors p and q where p is timelike ([tex] p^{2} > 0 [/tex]) and q is spacelike([tex] q^{2} < 0 [/tex]).
    Now I should consider the quantity

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

    and compute the limit [tex] q \to 0 [/tex].

    But I don't know how to perform the limit procedure. Could anyone help me please?

    I already tried to consider the problem in a special frame with [tex] p=(p^{0}, \vec{0}) [/tex] but it doesn't help.
  2. jcsd
  3. Dec 19, 2009 #2


    User Avatar
    Homework Helper

    Can you show some more detail of the work you did?
  4. Dec 20, 2009 #3
    My attempt so far was not successfully. I considered a special frame where [tex] p = \left( p^{0}, \vec{0} \right) [/tex] which is possible, because p is timelike. Furthermore I defined [tex] q = (0, \epsilon, \epsilon, \epsilon) [/tex]. This will lead to:

    [tex] - \dfrac{2 (pq)^{2} + p^{2} q^{2}}{q^{2}} = - p_{0}^{2} [/tex]

    and for arbitrary p we should have: [tex] - p^{2} [/tex].

    But somehow I don't think that I can specify q in this way. Another choice of q, e.g. [tex] q = (\epsilon, \epsilon, \epsilon, \epsilon) [/tex] would lead to a vanishing contribution [tex] = 0 [/tex], so I don't know how to compute the considered quantity. Obviously it depends on the choice of q.

    Any idea how to do that?
  5. Dec 21, 2009 #4
    I've one further information, but I don't know if it helps: [tex] (p-q) \in V^{+} [/tex].

    So, I also tried to consider a special frame where
    [tex] p-q = (p^{0} - q^{0}, \vec{0}) [/tex].

    Which leads to [tex] \vec{p} = \vec{q} [/tex] and therefore:

    [tex] - \dfrac{2 (pq)^{2} + p^{2} q^{2}}{q^{2}} = - \dfrac{2 (p^{0} q^{0} - \vec{p} \, ^{2})^{2} + p^{2} (q_{0}^{2} - \vec{p} \, ^{2})}{q_{0} - \vec{p} \, ^{2}} \simeq 2 \vec{p} \, ^{2} - p^{2} [/tex]

    Then I rewrite the last [tex] \vec{p} \, ^{2} [/tex] into [tex] \vec{p} \cdot \vec{q} [/tex] and finally obtain (again): [tex] -p^{2} [/tex].

    But it appears questionable to do the computation like this.

    Could anyone help me please?
  6. Dec 21, 2009 #5
    Pephaps the identity (p+q)2 = p2 + q2 +2pq might be of help.
  7. Dec 22, 2009 #6
    ...... it does not really help
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook