Colliding Particles at Right Angles
Most recent answer: 11/20/2012
Q:
I AM A CPA WITH NO PHYSICS BACKGROUND, BUT I HAD A DREAM THE OTHER NIGHT (ON THE QUANTUM LEVEL) THAT WAS SO VIVID I FEEL I NEED AN ANSWER. THE DREAM WAS ABOUT COLLIDING PROTONS AT A NINETY DEGREE ANGLE INSTEAD OF HEAD ON. IN MY DREAM IT REVEALED NEW SUB-ATOMIC PARTICLES. THE NINETY DGREE COLLISIONS OPENED A NEW DIMENSION. HAS THIS BEEN TRIED BEFORE? AM I CRAZY? PLEASE EMAIL ME WITH AN ANSWER THANK YOU!!
- robert bowles (age 53)
sarasota fl usa
- robert bowles (age 53)
sarasota fl usa
A:
It turns out that experiment is done all the time. Consider a standard particle collider, in which the particles collide head-on, as viewed in the lab. Say one of the particles is heading north at some speed u and the other is heading south at the same speed. View exactly the same event from the point of view of someone heading east according to the lab frame. One particle seems to him to be heading northwest, the other heading southwest. Adjust his speed right and those particles look like they're moving at right angles.
So the collisions we routinely see are right-angle collisions, as viewed from the right frame. The particles they produce are essentially the same as those produced by right-angle collisions.
You may wonder just what the right speed is for that viewer to move relative to the lab. It turns out to be u/(1+u2/c2)1/2. Since for interesting accelerators u is very close to c, this becomes nearly c/21/2, or ~0.71c.
Mike W.
So the collisions we routinely see are right-angle collisions, as viewed from the right frame. The particles they produce are essentially the same as those produced by right-angle collisions.
You may wonder just what the right speed is for that viewer to move relative to the lab. It turns out to be u/(1+u2/c2)1/2. Since for interesting accelerators u is very close to c, this becomes nearly c/21/2, or ~0.71c.
Mike W.
(published on 11/20/2012)