# What is transverse momentum?

1. Dec 17, 2009

### fisics101

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.

2. Dec 17, 2009

### mathman

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.

3. Dec 17, 2009

### fisics101

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?

4. Dec 17, 2009

### fisics101

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?

5. Dec 17, 2009

### Bob S

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

6. Dec 17, 2009

### fisics101

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?

7. Dec 17, 2009

### Bob S

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