Suppose a particle decays to three other particles. The masses of all particles are assumed to be known and we work in the rest frame of the parent particle. So there are 12 parameters for this because of the 4-momenta of the three daughter particles. Now the constraint of momentum conservation imposes 4 constraints and reduces the number of parameters to 8. Further, the energy-momentum relation for each particle imposes three more constraints and reduces the number of parameters to 5. Are there any other constraints that reduce the number of parameters to 2?


1 Answer 1


Well, yes and no. There is a set of Euler angles that describes the orientation of the product system in space, and the remaining two parameters are all that is needed to find the final $\left|\vec{p}\right|$s.

It's clearest to me when I am writing Monte Carlo generators. You do the physics in some conveniently defined coordinate system and then throw the dice to randomly assign a orientation for the system. Like this.

  • Choose one of the daughter particles, call the direction of it's three momentum the $\hat{z}$ axis.

    You can still describe all possible relationships between the three products, so there is no loss of generality.

  • Insist that the three momentum of the second daughter lies in the $\hat{x}$-$\hat{z}$ plane.

    Again, there is no loss of generality.

  • At this point you can use, say, $\theta_{1,2}$ and $\theta_{1,3}$ (the angles between the particles) to parameterize the physics. (That choice is not unique, of course, pick something that makes your math come out neatly.)

  • Choose $\phi \in \{0,2\pi\}$, $\cos(\theta) \in \{-1,1\}$, and $\psi \in \{0,2\pi\}$ (all uniformly) to describe the orientation of the whole system in space.

  • $\begingroup$ When you fix one direction, 1 parameter reduces. Next when you specify the plane as the xz plane, 4 more parameters reduce. This is because a plane is characterized by the normal and the angle(orientation). This normal has 3 components so the total comes to 4. Something seems to be wrong. $\endgroup$
    – venu
    Mar 22, 2013 at 4:43
  • $\begingroup$ I can see how you got there, but that's not the right way to think of it. We really are just chosing a set of Euler angles to describe the orientation of the system. Putting particle 1 on the positive z-axis fixes $\phi$ and $\theta$, and all that remains is a single rotation to fix $\psi$. $\endgroup$ Mar 22, 2013 at 4:49

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