proton anti-proton annihilation.

malawi_glenn

1. The problem statement, all variables and given/known data

state why p + p_bar --> 1 photon

is forbidden


2. Relevant equations



3. The attempt at a solution

I have checked all quantum numbers and they are okay. I wondering altough if p_bar has intrinsic parity -1 (p has parity +1)??

According to the soloution, it is forbidden because of energy and momentum conservation. I have not been giving a single word about the production conditions and so forth. HOW can this (E and p) be violated if I choose whatever annihilation condition so that p and E is not violated??!!

George Jones

Which special frame of reference is often used in collision problems?

malawi_glenn

Center of mass. But that is not statet OFTEN CM frame is used. Sometimes I just want to **** my teachers.. :P

George Jones

Centre of momentum frame. Is there a center of momentum frame for a system consisting of of a proton and an anti-proton? Is a there center of momerntum frame for a system consisting of a single photon?

Good! This means that they're making you think!

malawi_glenn

And who are saying that the proton and antiproton has EXACTLY opposite momenta?

What if for example p has little more momentum than the p_bar? Is not one-photon creation possible then?

And if two photons are created, then parity is violated? So there must be a three photon creation in order to conserve all, IF the p and p_bar is travelling at exaclty the same momenta (but opposite). Is that right?

PsiPhi

I had a question similar to your one I had to solve. The question was "Show that a photon cannot spontaneously disintegrate into and electron-positron pair". The lecturer mentioned I could tackle this problem using four-momentum or classical conservation of energy and momentum.

I used the classical way, so I didn't have to deal with tensors
An assumption is made at the start, that the electron-positron pair don't move at relativistic speeds. Thus, a classical "Newtonian" means can be applied to the question.

So, all I did was work out the initial/final energy and momentum of the system. Using the conservation of energy equation I substituted it into the conservation of momentum equation and solved for the velocity v (This was the velocity of the electron and positron) . I found v to equal 2c, where c is the speed of light. This implies that the electron and positron was going at twice the speed of light which cannot happen.
This implies that the photon cannot spontaneously disintegrate.

I know this wasn't the same as your question, but I think the logic is the same but reversed and you may obtain a final velocity less than c. Not too sure though.

I hope this helps.

malawi_glenn

But protons are not elementary particle systems.

And if the photon has enough energy and is in the field of a nucleus, pair production is posssible.

The correct analogy would be "why can not an electron and positron annihilate and produce a single photon"
I know that two photons are produced, but that would violate parity if the electron and positron has relative angular momentum = zero. ?

I really love to think of the problems on a deeper level than just knowing the answer. I want to understad these things how they work.

If we look att this electron positron annihilation: e + posit = 2 fotons
parity_initial = (-1)(+1) (if zero angular momentum)
parity_final = (-1)(-1) i.e parity is violated??

I did the similar for the p + p_bar and either one or three photons are emitted, depending on how p and p_bar moves relative each other.

How would all the pro's out there solve this?;)

PsiPhi

Sorry mate, I don't seem to know enough about parities. I remember being it mentioned in quantum.

The question I got was from a relativistic dynamics unit.

Which topic of physics are you studying at the moment?

malawi_glenn

Iam studiyn nuclear and particle physics.

I wanna have all points on the exam tomorrow =( but exersices hwere no background information or descriptions of the collision and decays, I often want to give all possible solutions, because I am never sure what the teacher want me to answer...

as in this p + p_bar annihilation, why do they have to collide with exactly the same momenta (but reverse)? There is nothing in the text.. it really makes me angry :P

So my question for everyboy who know this stuff;
can the p and p_bar create a single photon if any of them has higher momentum than the other one? And if they collide with same but opposite momenta, do they create 3 photons to conserve parity?

George Jones

No.

If the total spatial momenta of the proton anti-proton system is non-zero, then you can always transform to a frame in which the total spatial momentum is zero. The spatial momentum of a single photon is non-zero in all frames.

Dick

If you don't like thinking about a special frame then think about a general frame and 4 vectors. Forgetting factors of 'c', we have E^2-p^2=m^2. So in these units the spatial momentum of a proton is always less in magnitude than its energy. Now ask yourself whether two such vectors can add together to yield a vector like the photon with E=p. Clearly not, it seems to me anyway.

malawi_glenn

okey, than I am fine =) Thanks!

We have not talked about this during the lecutres or lessons. And nothing in the book either..

So how would then the p + p_bar annihilate? 2 or 3 photons?

Dick

Either should be kinematically possible, though I'm really not sure. I think in general real world ppbar annihilation is pretty messy since they are composite particles. Ask the folks at CERN or the Tevatron.

malawi_glenn

okey, but is 2photon creation possible due to parity?
each photon has negative parity right? And no relative angular momentum between p and p_bar should give them negative parity. 2photons would have positive. am I totaly lost now? =)

Meir Achuz

You do seem lost with all those answers.
p and p_bar can annihilate to two photons.
p and pbar in the S=0, L=0 state have the same Q numbers as the pi_0.
The negative parity means that the photons will have polarization corresponding to the pseudoscalar E.B. This is how Yang suggested the measurement of the parity of the pi_0 fifty years ago.