How Can p1 + p2 Exceed p in a Relativistic Elastic Collision?

[malawi_glenn]

Homework Statement

An relativistic proton collides with a proton at rest (in Lab-frame), the collision is elastic.

let incoming proton have momenta p, and the outgoing momenta = p1, p2.

The following is conserved:

\vec{p} = \vec{p}_1 + \vec{p}_2

\sqrt{m^2+p^2} + m = \sqrt{m^2+p_1^2} + \sqrt{m^2+p_2^2}

Gives for the angle between p_1 and p_2 (in lab frame). A minima occurs, which means that p1 = p2. One can show that this minima occurs so that: p1 + p2 > p. Explain why that is possible!

The Attempt at a Solution

MATLAB

m = 0.93828; % proton mass in GeV

p = 2; %GeV incomming proton

p1 = [0.01:0.01:2]; %range of outgoin proton #1s momenta.

p2 = sqrt((sqrt(m^2+p^2)+m-sqrt(m^2+p1.^2)).^2-m^2);

omega = acos((p^2-p1.^2-p2.^2)./(2*p1.*p2));
omega = 180/pi*omega;

plotting gives minima för p = 1.2GeV/c

I am very unsure about this, I think it is possible p1 + p2 > p science momenta is a vector quantity, so the magnitudes can change, but not the total (i.e the total vector after = total vector initial). More suggestions?

 

[haruspex]

I think it is possible p1+p2>pp1+p2>p p1 + p2 > p science momenta is a vector quantity, so the magnitudes can change, but not the total (i.e the total vector after = total vector initial).

Quite so. Just think of the triangle made by two added vectors and their resultant. Except in the degenerate case, the sum of the lengths of two sides of a triangle exceeds the length of the third.