# Pair-production derivation for h*nu_min

1. Oct 23, 2011

### fyziksdunce

γ+ M → e^- + e^+ + M

In pair-production, a photon of energy is absorbed in the neighborhood of a particle of mass M to make an e^-e^+ pair, each with mass m_e. The threshold photon energy for this process is known to be

h*nu_min = 2m_e*c^2 for M→∞
h*nu_min = 4m_e*c^2 for M→m_e

Derive a formula for h*nu_min that is valid for any M, and show that the result gives agreement in these two special cases.

Hint: Consider the simplest case of all three particles moving the Lab frame in the same direction with the same speed u.

From the hint it sounds like I should be using conservation of momentum or energy for the derivation. What I've come to for momentum is:

h*nu/c + M*u*c = 2m_e*u*c + M*u*c

which goes to

h*nu + M*u*c^2 = 2m_e*u*c^2 + M*u*c^2

Based upon my class work, the velocity of the particle with mass M is not changed in the pair production (it is an "essential spectator"). This gets me to something close to the first equation as the M*u*c drops from each side of the equation, but I have an extra u hanging around on the RHS.

So I tried energy:

h*nu + K_m + M*c^2 = 2(m_e*c^2 + K_m_e) + K_m

Where K_m is the kinetic energy of the particle with mass M and K_m_e is the kinetic energy of the electrons after the pair-production. I can sub in the usual 0.5mu^2 for the kinetic energies (with respectively correct masses, M and m_e, for m). But after canceling K_m from both sides, I don't have something that can satisfy conditions for M as above.

I feel like I'm missing something rather straightforward, but since I am a mathematician and not a physicist I feel like I don't have the intuition to make the leap.

Can someone provide a way ahead for this problem?

Thanks.

2. Oct 23, 2011

### vela

Staff Emeritus
The mass M is there to conserve momentum. If M is large, it just sits there, but since you're looking for a result valid for any M, I don't think you can assume it stays at rest. In fact, the hint suggests that you look at the case where the electron, positron, and M all end up moving at the same speed. (In the COM frame, all would be at rest, which is why you know you're finding the minimum energy.)

3. Oct 23, 2011

### fyziksdunce

By that argument, using COM:

h*nu = 2*m_e*u*c^2 + M*u*c^2, for M=>inf the M*u*c^2 goes to zero because the particle won't move, but with M => m_e, the required result isn't there (and I'm not sure what reasoning one would use to remove the u from the RHS).

Using COE:

h*nu + M*c^2 = 2*(m_e*c^2 + K_m_e) + K_M + M*c^2
then
h*nu = 2*(m_e*c^2 + K_m_e) + K_M

and K_M => K_m_e as M=>m_e, but how would I be able to translate KE to the problem at hand?

4. Oct 23, 2011

### vela

Staff Emeritus
By COM, I mean center of mass, not conservation of momentum. It's the frame where the total momentum of the system is 0.

Let's say the photon has momentum p. Its energy will be equal to pc. Similarly, let's say the mass M initially has momentum p' and energy E'. After the pair production, the two electrons and the mass M will be at rest. Conservation of energy and momentum gives you
\begin{align*}
p + p' &= 0 \\
pc + E' &= (2m+M)c^2
\end{align*}You can solve those equations for p, the momentum of the photon in the COM frame. Using the Lorentz transformation, you can find the momentum of the photon in the lab frame, where M is at rest. You could also try to solve the problem by writing the equations for the lab frame and solving them directly, but at first glance, it seems like that'll be more complicated.

A really useful relationship in relativity is $E^2-(pc)^2 = (mc^2)^2$. For example, for mass M, you have $E'^2-(p'c)^2=(Mc^2)^2$.

It's best to avoid using γ and velocities whenever you can. Sticking with energy and momentum makes the algebra a lot simpler.