# Possibility of pair Production without interaction with matter?

1. Nov 6, 2007

### JamesJames

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

Is it possible for pair production to occur without interacting mwith matter?

2. Relevant equations

E+ + E- = E(gamma) - 2*m_e*c^2

3. The attempt at a solution

In the Coulomb field of nucleus (or electron), the incident photon energy can be converted to the creation of an electron-positron pair. A minimum of 2*m_e*c^2 is required for this purpose. Each of the annihilation photons can then interact with the surrounding medium via the photoelectric effect or, more likely, Compton scattering.

I've never been truly able to understand this question qualitatively. I keep imagining that the photon is not present but have always wondered how this would allow the interaction to occur physically?

Can this be shown explicitly using conserved quantities for this interaction?

2. Nov 6, 2007

### clem

For a photon, E^2-P^2=0.
For any number of massive particles, E^2-P^2>0.

3. Nov 7, 2007

### JamesJames

Are you saying that because E^2 + p^2*c^2 = m^2*c^4? This way since m^2*c^4 is always greater than zero, the left side must always be non zero. But , how does this show the connection to interaction?

Also, I should calrify my equation:
Conservation of Energy
E(photon) + m*c^2 = E(photon') + E(electron)

I'm still confused about where the conserved qualtities are related to the possibility of interaction.

4. Nov 7, 2007

### Dick

Not exactly, the reaction is photon->(e+)+(e-). So your four vector equation looks like (E(photon),p(photon))=(E(e+),p(e+))+(E(e-),p(e-)). So yes, the E's have to balance, but Meir's point is that |p(photon)|=E(photon) and |p(e+)|<E(e+) and |p(e-)|<E(e-) where I've put c=1. Can you balance the p's? Remember |a+b|<=|a|+|b|.

5. Nov 7, 2007

### JamesJames

Perhaps I;'m not correctly understanding what is physically going on. I can see that |p(photon)|=E(photon), so that part is fine. But I am confused about the other two equations:

|p(e+)|<E(e+)
|p(e-)|<E(e-)

How do they come about in relation to this problem? Also, why does |a+b|<=|a|+|b| come in if you consider them separate as you did above?

For normal pair production, in the lab system:

E(photon) + Mc^2 = 2m_e*c^2/sqrt(1-beta^2) + Mc^2/sqrt(1-beta^2)

p(photon) = 2m_ev/sqrt(1-beta^2) + Mv/sqrt(1-beta^2)

I can;t seem to relate this to "no interaction" though.

Last edited: Nov 8, 2007
6. Nov 8, 2007

### Dick

|p|<E because E^2=m^2+p^2. |a+b|<=|a|+|b| because that's the triangle inequality for vectors. Then you start losing me.

7. Nov 8, 2007

### JamesJames

For both energy and momentum, I have moving photon + Rest term = sum of 2 energies + rest term.

Perhaps, I need to back up: What does the earlier post mean physically when the following is said:
"For a photon, E^2-P^2=0.
For any number of massive particles, E^2-P^2>0."
What is the significance of the second line?

8. Nov 8, 2007

### Meir Achuz

E^2-p^2 for the initial (photon) state must equal E^2-p^2 for the final (pair) state.
If one is zero and the other greater than zero, they can't be equal.

9. Nov 8, 2007

### Dick

What is a 'rest term'? You keep writing an mc^2 on the photon side for no reason. Maybe that's part of the problem. There's no such thing.

10. Nov 8, 2007

### JamesJames

Thanks, that rest term should not be there. I think I understand some part of the Meir's post now. Here's the way I understand it...my question is at the end:

E^2 = p^2 c^2 + M^2 c^4, where M = m[0]

let c = 1

I can see that it must hold for the conservation to be valid. Just to clarify,

E^2-p^2 = M^2 for the initial (photon) state
E^2-p^2 = M'^2 for the final (pair) state

where M = M' = mass of electron or positron? Intuitively, I would have thought that M' = 2m(electron)? Basically, I understand the LHS of the two equations above, but not what the RHS should be. I've been of the impression that the RHS must be m[0]^2 c^4, but what is this m[0] for?

Question 1: For the final state, is it the combined masses of the electron and positron, which is twice the energy (not rest energy) of the electron which I called 2m(electron)?

Question 2: Why must E^2-p^2 for the initial (photon) state = E^2-p^2 for the final (pair) state. Can this be proved mathematically?

Question 3: Is Energy (electron) = Energy (positron)?
Is Momentum (electron) = Momentum (positron), in magnitude?

Last edited: Nov 8, 2007
11. Nov 8, 2007

### Dick

E^2-p^2 is the same for the initial state as the final state. So they are both zero. Just because the vector (E,p) for the initial state is equal to the vector (E,p) for the final state. That's actually the answer to question 2. They are the same vector. And that's the crux of the whole thread. The presence of an electric field implies the existence another another particle which should be taken into account. So you can't say the photon 4-vector is equal to the sum of the 4-vectors of the products. For question 1, no. Period. Question 3 would be true in a center of mass frame. But a photon has no center of mass frame. So no to that one too. In the absence of another particle the 4-vectors are conserved between initial and final states. Just concentrate on that.

12. Nov 8, 2007

### JamesJames

ok, I think I finally got what you are saying. Its making sense. The picture is forming correctly for me now. I tried balancing equations and one last question remains though:

When you guys let c = 1 in E^2 = p^2*c^2 + m^2*c^4, could I be able to show that E^2 is always larger than p^2*c^2 + m^2*c^4 WITHOUT letting c = 1? I can see that it would be the case if c = 1, but cannot wrap my head around the situation is c is not equal to 1.

[If you can helpt me to show this to be the case, then from my conservation equations(which I haven't posted here since they have become a bit long), I would be able to take it from there.]

13. Nov 8, 2007

### Dick

Good. If you throw the dimensions back in then you can still conclude E>c|p| if m is not zero. But all of the momenta are then multiplied by c. So none of the conclusions change. If a>b then ca>cb.

14. Nov 8, 2007

### JamesJames

I can rewrite that equation with brackets

(E)^2 = (pc)^2 + (mc^2)^2

tells me that E must always be greater than pc, where p = |p|. This is ONLY true if m is non-zero, because if m = 0, then E = pc.

So, in my final equation, have

2(E+)(E-) + C = 2(p+)(p-)cosx where C = (mc^2)^2

would I be correct in saying that
1. E+ in term 1 is always LARGER than p+ in term 3
2. E- in term 1 is always LARGER than p- in term 3
Therefore term 1 is always larger than term 3 and since term 1 is ADDED to a second term(term 2) on thr RHS, RHS will NEVER equal LHS and I could finish it from here. Let me know if I made sense with this and if I did, then I can finish it.

15. Nov 8, 2007

### Dick

I think so, if understand you correctly. On the photon side E=c|p|. On the electron side E<c|p| for both terms. So they can't balance.

16. Nov 9, 2007

### Meir Achuz

I will try to make this clearer.
Conservation of energy means $$E=E_1+E_2$$, where $$E$$ is photon energy, and $$E_1$$ and $$E_2$$ are the electron and positron energies
Conservation of momentum is
$${\vec p}={\vec p}_1+\vec p}_2$$.
It is just algebra to write
$$E^2-{\vec p}^2 =(E_1+E_2)^2-({\vec p}_1+\vec p}_2)^2$$.
The LHS is zero. Since E>p for any particle with mass, the RHS is greater than zero. Therefor, the process cannot occur.
If you are going to do relativity, you have to understand that whether c=c or c=1 can have no effect.

Last edited: Nov 9, 2007
17. Nov 9, 2007

### Count Iblis

You can do this problem without writing down any equations. Just consider conservation of energy. Suppose that it is possible for a photon to decay into an electron and a positron. Then the energy of the photon must be at least 2 m c^2. However, whatever the energy of a photon, there always exists a frame in which the photon's energy is lower than 2 m c^2....