Particle-photon interaction to create new particle

[thepopasmurf]

Homework Statement

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}}

Homework 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).

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

 

[mfb]

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.

 

[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&#039;_p + p&#039;_{\gamma}) \cdot (p&#039;_p + p&#039;_{\gamma}) =<br /> \frac{(E&#039;)_p^2}{c^2} + \frac{(E&#039;)_{\gamma}^2}{c^2} + \frac{2E&#039;_pE&#039;_{\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.

 

[mfb]

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

Let c=1.
E_γ=p_γ=-p_p

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?