Four-vector of a proton as seen from the rest frame of another I have a question regarding the four-vector of two protons. My task is to determine the four-vector of one proton as seen from the rest frame of the other. 
I'm having a hard time understanding exactly what this is, and I'm wondering if I'm doing this right. Any and all help is greatly appreciated.

Two protons are moving towards each other at the same speed, each with a total energy of $1.5GeV$. 
I've calculated that their momentum are $6.24\cdot 10^{-19}kgm/s$, and the speed of each proton is $2.34\cdot10^8m/s^2$.
That means that the speed of one of the protons is $2.909\cdot10^8 m/s =-0.969c$ , if seen from the rest frame of the other one.
Now, my task is to calculate the proton's four-vector as seen from the rest frame of the other proton. 
We have
$p_\mu = (E, p_xc, p_yc, p_zc)\equiv(p_0,p_1c,p_2c,p_3c)$
and by definition
$p^\mu=\begin{pmatrix} p_0 \\-p_1c \\-p_2c \\-p_3c\end{pmatrix}$
I have
$p_0 = \gamma mc^2 = \frac{mc^2}{\sqrt{1-v^2/c^2}} = 1.0808\cdot10^{10}kg(m/s)^2$.
Where I used $v = -0.969c$.
And, 
$p_1c = \gamma mv_x'c = -5.90\cdot10^{10}kg(m/s)^2$
Where I used $v_x'=v=-0.969c$.
This leads to
$p^\mu=\begin{pmatrix} 1.0808\cdot10^{10}kg(m/s)^2 \\ 5.90\cdot10^{10}kg(m/s)^2 \\ 0 \\ 0\end{pmatrix}$
Now, my question is: Is that the four-vector of one proton as seem from the other, or am I wrong? Should I have used another speed (absolute instead of relative)?
 A: Let's use A and B to label the particles.
Method: first get the energy of proton B in the rest frame of A. From this you can then deduce the momentum. And to get the energy, although Lorentz transformation is one way to do it, there is a more elegant method involving the invariance of an inner product of vectors. Let $p$ be the 4-momentum of A and $q$ be the 4-momentum of B. Think about the quantity $p^\mu q_\mu$, evaluated in terms of components first in one frame and then another.
A: The Lorentz transform is a linear transform for transforming vector coordinates between inertial reference frames. So suppose we have energy E and momentum $p_x$ in the x direction in the lab frame. Then for some velocity dependent terms A and B, $E'=AE+Bp_x$ and $p_x'=BE+Ap_x$ where the $'$ represents the rest frame of the particle moving in the lab frame. 
In the rest frame of a particle, its momentum is zero and its energy is its rest energy. This gives us two equations in two unknowns. 
$$E'=m_pc^2=AE+Bp_x$$
$$p_x'=0=BE+Ap_x$$
You have two equations in two unknowns so this allows you to solve for A and B. 
Now the energies are equal for both protons, but the momentum is in the opposite direction. Let's call the energy and momentum for the other proton $E'_2$ and $p_{x2}'$. 
We use the same A and B because that's how you transform to the frame of the first proton you chose to pick. This means:
$$E'_2=AE-Bp_x$$
$$p_{x2}'=BE-Ap_x$$
A: As an alternative to calculations with invariants, algebraic calculations with components, or Lorentz Transformations, I'll suggest a geometric/trigonometric method (which could supplement the methods listed earlier by
helping to visualize and interpret what is being calculated).
Given the trigonometry problem below, how would you solve it?

*

*You are given an isosceles triangle $\stackrel{?}{\phi}=\stackrel{?}{\theta}$ (with the angles equal but unknown) but you know
the lengths $\stackrel{\checkmark}{m}$ and
the height $\stackrel{\checkmark}{h}=\stackrel{\checkmark}{m}\cos\stackrel{?}{\theta}$ . Compute vector OB's components parallel and perpendicular to vector OA
and compute the slope of OB with respect to OA.

Express all of your answers in terms of $m$ and $h$ and,
 in particular, [hint] an inverse-trig function of $(h/m)$.
Note that $B_{\textrm{parallelToA}}$ is determined by "dropping a perpendicular from B onto OA" (which is geometrically determined by a circle centered at O
with radius gradually-extending along OA until one finds that the corresponding tangent line meets B).



*

*Now repeat the logic of that problem with figure below (drawn on an "energy-momentum diagram"), which is constructed with the two proton 4-momenta OA and OB travelling in opposite directions with the same speed in the LAB frame. (It's useful to remember that the hypotenuse is the side opposite the right-angle and that the adjacent side goes with cosine.)


P.S.:
In spite of appearances, $OB$ and $BB_{parallelToAmink}$ are not
Minkowski-orthogonal. (For an isosceles triangle in the Lab frame with the height along the Lab's worldline, those rays do appear to be Euclidean-orthogonal.... which is exploited in
the Loedel-diagram approach. However, the Euclidean geometric relations
are limited to the timelines and spacelines for those two observers. Minkowski geometrical relations, however, are valid everywhere in this diagram, which are preserved when boosted into another inertial frame.)
