I Particle collisions - a question on angles relative to beam

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1. Oct 13, 2016

"Don't panic!"

Consider the case in which an incoming particle collides with stationary target particle producing new particles through the interaction. For example, $$e^{-}+e^{+}\rightarrow X+\bar{X}$$ My question is, why in general do the particles produced in such an interaction propagate outwards are angles relative to the beam axis?

Can one use the classical analogy of collisions between billiard balls, in the sense that if the incoming ball strikes the stationary target ball the force acting on target ball acts in the direction normal to the contact point (along the line passing through the contact point and the centres of both billiard balls). If the incoming ball strikes the target ball off centre then this results in both balls moving off at angles relative to the original direction of propagation of the incoming billiard ball.

I'm completely unsure in the case where a particle and an anti-particle collide and annihilate, producing another particle anti-particle pair. Why are the momenta of the particle anti-particle pair at angles relative to the direction of propagation of the incident particle?!

2. Oct 13, 2016

Paul Colby

Consider the center of mass frame where the net momentum is zero. The $e^+$ and $e^-$ approach along the $+z$ and $-z$ axis. After they annihilate two particles will be ejected with opposite momentum but along an arbitrary direction. Transforming back to the lab frame yields two particles at some angle to the initial beam.

Well, by arbitrary I mean with some angle distribution relative to the incoming momentum determined by the differential cross section. The billiard ball example has a delta function cross section which is a special case.

3. Oct 14, 2016

"Don't panic!"

How does one show this mathematically?!

Why is there an angular distribution though? Is it simply because the momentum only needs to satisfy overall momentum conservation, but this doesn't mean that the different momenta of each of the produced particles can't be pointing along directions that are at angles relative to the beam axis, and so in general this will be the case?!

Heuristically, can one think of particle collisions in terms of colliding billiard balls, in the sense that when an incoming ball strikes a stationary target ball, then the resulting force acting on the target ball (and the reaction force acting on the incoming ball) acts along the direction normal to the point of contact (along the line connecting the centres of both balls that passes through the contact point). Hence, if the collision isn't exactly head-on, the forces acting on the two balls will cause both them to move off at angles relative to the original direction of the incoming ball?!

Last edited: Oct 14, 2016
4. Oct 14, 2016

Paul Colby

It is best shown using particle momentum as 4-vectors. Sorry I'm being lazy here. Look up and read an introductory high energy book that covers kinematics of collisions.

In the center of mass frame the net 3-momentum is zero. How many ways are there to add two vectors to get 0? Well, given any vector $p$ if the other one is $-p$ then the sum of the two is zero. There are on the face of it many possible $p$ to choose from. There are restrictions because $p^\mu p_\mu = m^2$ where $m$ is the particle rest mass.

Works for me. Picture the balls colliding in the center of mass. They need not collide on center on center. In fact they could miss each other entirely. The two momenta are along parallel directions. The perpendicular distance between these lines is called the impact parameter. As you point out two balls will recoil along a line through the point of contact (if there is one). The angle of this line is a function of the impact parameter.

5. Oct 14, 2016

"Don't panic!"

Is there any particular book that you would recommend?

This is my own working so far...

In the CoM frame the initial momenta satisfy $\mathbf{p}=-\mathbf{p}$, and so by momentum conservation, it must be that the final momenta satisfy $\mathbf{p}'=-\mathbf{p}'$, which are the momenta of the final particle $X$ and antiparticle $\bar{X}$.
Also, by energy conservation it follows that $E_{e}=E'_{X}$, where $E'_{X}$ is the energy of both the final particle $X$ and antiparticle $\bar{X}$.
Next, we can use that the mass is Lorentz invariant such that $$\left(\tilde{P}^{\mu}_{e^{+}}+\tilde{P}^{\mu}_{e^{-}}\right)^{2}=\left(P^{\mu}_{e^{+}}+P^{\mu}_{e^{-}}\right)^{2}$$ where $\tilde{P}^{\mu}$ are the 4-momenta of the particles in the CoM frame. We have then, that $$\left(E_{+}+E_{-}\right)^{2}-\left(\mathbf{p}_{e^{+}}+\mathbf{p}_{e^{-}}\right)^{2}=4m_{e}^{2}\\ \Rightarrow E_{+}E_{-}-p_{e^{+}}p_{e^{-}}\cos\phi=2m_{e}^{2}$$ I'm not quite sure how to proceed from this point however. I realise this might be the wrong approach though - I followed this approach originally to determine the threshold energy for particle creation in the lab frame. Would it be better to simply Lorentz transform the momenta?!

So is the point that there are many possible ways that the momenta of the outgoing particles can be oriented yet still satisfy the required constraints and so in general they will be oriented along directions at angles relative to the beam axis?!

This picture seems quite intuitive to me, so if it's a reasonable way to look at things heuristically then that's quite pleasing!

6. Oct 14, 2016

Paul Colby

The right hand side of the second set of equations you have are not quite right. The total "rest" mass is the sum of the rest energy of each particle plus the kinetic energy of each particle.

7. Oct 14, 2016

"Don't panic!"

Ah ok, so should it read
$$\left(E_{+}+E_{-}\right)^{2}-\left(\mathbf{p}_{e^{+}}+\mathbf{p}_{e^{-}}\right)^{2}=4E_{e}^{2}=4m_{e}^{2}+4\mathbf{p}^{2}$$ such that $$\mathbf{p}^{2}=\frac{1}{2}\left(E_{+}E_{-}-m_{e}^{2}-p_{e^{+}}p_{e^{-}}\cos\phi\right)$$ where $\mathbf{p}$ is the momentum of the electron (positron) in the CoM frame.