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