How is transverse momentum defined and calculated?

AI Thread Summary
Transverse momentum is defined as the component of momentum perpendicular to the direction of a particle's motion, particularly in collision scenarios. When two objects collide and shatter, the total momentum of the resulting pieces should equal the momentum of the original objects. A 3D coordinate system is used to analyze momentum, with the center of mass at the origin and axes set up to account for perpendicular directions. The momentum of each shattered piece is calculated as a vector, with its magnitude determined by mass times speed. In relativistic contexts, the relationship between energy and momentum is more complex, requiring specific equations to account for the effects of high-energy collisions.
fisics101
Messages
5
Reaction score
0
What is transverse momentum?

I think I have a general gist but I may be wrong. I'm thinking that when two object collide and shatter, then all the shattered pieces' momentums should add to be the same as the momentum of the two objects just before colliding. But how would you find the momentum of a single shattered piece? Also, do perpendicular angles have anything to do with this?

Thanks.
 
Physics news on Phys.org
The general principal is to set up a 3-d coordinate system. The simplest would have the center of mass (which may be moving) as the origin and the line connecting the two objects as one axis. The other two axes would be any mutually perpendicular pair perpendicular to the first axis.
For each piece its momentum is a vector along its direction with magnitude mass times speed. In this system the total momentum should be 0.
 
Okay so I understand the first part about the 3-d coordinate system and the center of mass being at the origin. But I'm having difficulty picturing the perpendicular part.

So if I had a dangling orb, and I shot it, and it shattered, theoretically each piece would have a mutual piece traveling in the opposite direction with the same momentum?
 
Also, how would one set up an equation involving transverse momentum since it is not linear? Basically, how is the angle of the individual piece from the origin taken into account?
 
As you point out, when two particles collide, the vector sum of all the transverse momentum has to add up to zero. When it doesn't. there might be an invisible particle unaccounted for. In deep inelastic collisions, high-momentum-transfer collisions are sometimes referred to as high pT (for transverse momentum, or high p-perp (for perpendicular). Deep inelastic electron-proton collisions discovered quarks.
Bob S
 
Okay. That makes a lot of sense. So I get the whole concept of transverse momentum. But I still do not know how I would define it. Is transverse momentum still equal to mass*velocity? Or is it something different?
 
fisics101 said:
Okay. That makes a lot of sense. So I get the whole concept of transverse momentum. But I still do not know how I would define it. Is transverse momentum still equal to mass*velocity? Or is it something different?
For non-relativistic momentum, you are correct. For relativistic particles,

E2 = (pc)2 + (m0c2)2

where E = total energy, p = momentum, and (m0c2)2 is rest mass.

So p = sqrt(E2-(m0c2)2)/c

Bob S
 
Back
Top