Compute Limit of 4-Vectors: p and q

[parton]

I've the following problem. I have two four-vectors p and q where p is timelike (p^{2} > 0) and q is spacelike(q^{2} < 0).
Now I should consider the quantity

- \dfrac{2 (pq)^{2} + p^{2} q^{2}}{q^{2}}

and compute the limit q \to 0.

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 p=(p^{0}, \vec{0}) but it doesn't help.

 

[diazona]

Can you show some more detail of the work you did?

 

[parton]

My attempt so far was not successfully. I considered a special frame where p = \left( p^{0}, \vec{0} \right) which is possible, because p is timelike. Furthermore I defined q = (0, \epsilon, \epsilon, \epsilon). This will lead to:

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

and for arbitrary p we should have: - p^{2}.

But somehow I don't think that I can specify q in this way. Another choice of q, e.g. q = (\epsilon, \epsilon, \epsilon, \epsilon) would lead to a vanishing contribution = 0, 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?

 

[parton]

I've one further information, but I don't know if it helps: (p-q) \in V^{+}.

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

Which leads to \vec{p} = \vec{q} and therefore:

- \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}

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

But it appears questionable to do the computation like this.

Could anyone help me please?

 

[phsopher]

Pephaps the identity (p+q)² = p² + q² +2pq might be of help.

 

[parton]

... it does not really help