# Homework Help: Particle-photon interaction to create new particle

1. May 21, 2013

### thepopasmurf

1. The problem statement, all variables and given/known data
A particle with known rest mass energy, $m_{p} c^{2}$ pass through a cloud of monoenergetic photons with energy $E_{\gamma}$. The particle collides with a photon and a particle A, with mass $m_A$ is created. Show that the minimum energy of the particle required for the interaction is:

$\frac{E_{min}}{m_p c^2} = \frac{E_0}{m_p c^2} + \frac{m_p c^2}{4 E_0}$

where

$E_0 = \frac{(m_{A}^2 - m_{p}^2)c^4}{4 E_{\gamma}}$

2. Relevant equations

The relevant equations are the relativistic kinematic equations:

$\textbf{p} = (p_0, \vec{p})$

$E^2 = p^2c^2 + m^2c^4$

plus conservation of the four momentum (implying conservation of energy and momentum).

3. The attempt at a solution

So my first step was to consider the collision in the zero momentum frame since that gives the minimum energy to create the particle A. This also implies that the photon and particle are colinear, otherwise it would not be the zero momentum frame.

In this frame:

$E_p + E_{\gamma} = m_Ac^2$

considering the four momentum:

$p_p + p_{\gamma} = p_A$

$(E_p + E_{\gamma},0) = (m_Ac^2,0)$

The square of the four momentum of invariant, so square both sides:

$E_p^2 + E_{\gamma}^2 + 2E_pE_{\gamma} = m_A^2 c^4$

$(m_p^2c^4 + p_p^2c^2) + p_{\gamma}^2c^2 + 2E_pE_{\gamma}= m_A^2 c^4$

$m_p^2c^4 + 2p^2c^2 + 2E_pE_{\gamma} = m_A^2 c^4$

Third line comes from conservation of momentum.
Rearrange to give:

$E_p = \frac{(m_A^2 - m_p^2)c^4}{2E_{\gamma}} - \frac{2p^2c^2}{2E_{\gamma}}$

$E_p = 2E_0 - E_{\gamma}$

I feel that I'm very close with this result but I can't get to the required expression.

Any help would be appreciated. Thanks

2. May 21, 2013

### Staff: Mentor

Be careful with the notation. $E_\gamma$ in the rest frame of A does not have to be the same as $E_\gamma$ given in the problem statement.

3. May 21, 2013

### thepopasmurf

Ok, so here's my reworking of the problem:

In lab frame four-momentum is:

$p_p + p_{\gamma}$

The square of this is invariant

$(p_p + p_{\gamma})\cdot(p_p + p_{\gamma}) = \frac{E_p^2}{c^2} - p_p^2 + \frac{E_{\gamma}^2}{c^2} - p_{\gamma}^2 + \frac{2E_pE_{\gamma}}{c^2} - 2p_p\cdot p_{\gamma}$

In the zero momentum frame we have:

$(p'_p + p'_{\gamma}) \cdot (p'_p + p'_{\gamma}) = \frac{(E')_p^2}{c^2} + \frac{(E')_{\gamma}^2}{c^2} + \frac{2E'_pE'_{\gamma}}{c^2}$

Both of these values also equal $m_A^2c^4$

Sorry, but I'm not seeing where to go from here. I don't really see how to perform this transformation properly.

4. May 21, 2013

### Staff: Mentor

The left side has 4-vectors, the right side 3-vectors with the same symbol?

Let c=1.
Eγ=pγ=-pp

Therefore,
$m_a^2 = m_p^2+2E_{min}E_\gamma + 2E_\gamma^2$

Hmm, I don't think that leads to the target. Is that the minimal energy in the lab frame, or maybe some other frame?