Understanding 4-momentum and the Lorentz invariant product What I am trying to do is show that if $\textbf{P}_1$, $\textbf{P}_2$ are time-like, future point 4-vectors, then $\textbf{P}_1 .\textbf{P}_2\geq0$.
My understanding of the meaning of timelike is that $\textbf{P}_1 .\textbf{P}_1,\textbf{P}_2 .\textbf{P}_2>0$. I asked this question without detailing my thoughts so I thought I would ask again with more detail about what I've thought about.
What I was thinking was that if that the product is invariant under Lorentz transforms so if we consider the product in the rest frame of $\textbf{P}_1$ then the product looks like
$\begin{pmatrix}
m_1 c \\
\textbf{0}
\end{pmatrix}.
\begin{pmatrix}
m_2 c \gamma \\
m_2\textbf{v}\gamma
\end{pmatrix}
$
where $ m_1, m_2$ are the masses of particles, $\textbf{v}$ is the relative velocity, or alternatively I could have put the top entry  in terms of energy. So this evaluates to $m_1m_2c^2\gamma$ which is clearly $\geq0$. But there are a few points which I am uncertain about as a result of this topic being new to me. So I will formulate these into questions.
Is the argument valid?
Where have I used that the vectors are time-like? Is it when I assumed there was a Lorentz tranform taking us to the rest frame of $\textbf{P}_1$?
What if the vectors were space-like instead, do we get that $\textbf{P}_1 .\textbf{P}_2\leq0$, certainly we can't go to the rest frame?
Sorry if this is a bombardment of questions but essentially I am looking for some clarification of the ideas through the questions asked.
 A: The good thing about 4-vectors is that the inner product between two of them using the proper metric is always preserved, no matter from which reference frame you take it!
I don't know how familiar you are with the concept of metric, but an easy intuition of it is that it is the set of rules that define the mathematical structure of the inner product. It's easy to see in terms of matrices, since, as you know, in $\mathbb{R}^3$ we have that $\vec x_1 \cdot \vec x_2= \vec x_1 \vec x_2 ^T=\vec x_1 \mathbb{I} \vec x_2$, where $\mathbb{I}$ stands for the identity matrix. This is because in $\mathbb R^3$, the metric is represented by the identity.
The proper definition of inner product (scalar product) is thus $\vec x \cdot \vec y =\vec x G \vec y$, where $G$ is the metric of the space. The precise definition of this involves tensors, so I believe it's beyond the scope of the question, but I'll be happy to explain in another question if needed (though it requires a lot more of mathematical skills).
Now, the thing is that, when you perform a coordinate change, this is, you move to another reference system, not only the vectors change, but also the metric tensor does, and thus, the definition of the inner (scalar) product. Actually, it changes in a way so that it remains constant for every pair of vectors! (Note that this includes the inner product of a vector with itself, this is, the norm of the vector).
When translated to the 4-dimensional spacetime of relativity, this means that a time-like point vector will always remain such, and so will do the space-like and light-like 4-vectors. This also applies to the product between two vectors: It's always preserved between reference frames, since you also change the metric tensor, and thus the definition of inner product.
In your particular case, this statement is even stronger, since you're working in inertial reference frames, this is, flat spacetimes, in which the metric tensor can be represented by a diagonal matrix with all of its elements being $-1$ except for the time (first) component, which is $1$ (note that some authors represent it the other way arount, which is equivalent excep for the fact that the definitions of time-like and space-like vectors would be reversed). This is precisely the Minkovski metric. A Lorentz transformation is that which takes you from an inertial reference frame to another, and so, the metric remains flat (that of Minkovski), and that's why you never took it into account and still it worked.
Certainly, if you had space-like vectors instead, then the inner product would be <0.
Hope this helped!
