# Compute Limit of 4-Vectors: p and q

• parton
In summary, the problem is to compute the quantity -2(pq)^2/q^2 in the limit q->0, given that p is a timelike four-vector and q is a spacelike four-vector. The attempt to consider a special frame with p=(p^0,0) and q=(0,ε,ε,ε) did not lead to a successful solution. Another attempt to consider a special frame where p-q=(p^0-q^0,0) also did not provide a clear solution. The identity (p+q)^2=p^2+q^2+2pq was suggested, but it does not seem to be of much help in this case.
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.

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

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?

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.

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

... it does not really help

## What is a 4-vector?

A 4-vector is a mathematical concept used in physics and mathematics to describe quantities that have both magnitude and direction in four-dimensional spacetime.

## How do you compute the limit of 4-vectors p and q?

To compute the limit of 4-vectors p and q, you need to first determine the components of the vectors in four-dimensional spacetime. Then, you can use mathematical techniques such as vector addition and subtraction, dot product, and cross product to manipulate the vectors and eventually find the limit.

## What is the significance of computing the limit of 4-vectors p and q?

Computing the limit of 4-vectors p and q can provide valuable information about the behavior and properties of physical systems in four-dimensional spacetime. It can also help to predict and analyze the outcomes of various physical phenomena.

## What are some applications of computing the limit of 4-vectors p and q?

Some applications of computing the limit of 4-vectors p and q include studying the motion of particles in special relativity, analyzing the behavior of electromagnetic fields, and predicting the dynamics of quantum systems.

## Are there any limitations to computing the limit of 4-vectors p and q?

Yes, there are certain limitations to computing the limit of 4-vectors p and q. For example, the calculations can become very complex and difficult to solve for systems with a large number of interacting particles. Additionally, the accuracy of the results may be affected by experimental errors or uncertainties in the measurements of the 4-vectors.

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