Electron-positron pair production requiring a nucleus

Question

I was told that a nucleus is required to be present to allow the conservation of energy and momentum in the production of an electron and positron from a high energy photon, during pair production.

However, if the nucleus is removed, how can it be shown that the energy and momentum can't be conserved?

What I tried to do is $$E_\zeta =h\nu=E_++E_-$$ $$h\nu=m_0c^2+K_++m_0c^2+K_-$$ and $$p_\zeta=p_++p_-$$

and using the relationship that $$p=\frac{E}{c}$$

to show that these statements are impossible but I don't know where to go from here.

Answer 1

A way to show that a single photon cannot decay into a massive particle-antiparticle pair can be wrote down by using the four-vector formalism. The decay is $$ k^\alpha=p_1^\alpha+p_2^\alpha $$ where $k$ is the four-vector of the photon while $p_1$ and $p_2$ are those of the particle and its antiparticle. Taking the square you have $$ 0 = m^2+p_1^\alpha p_{2\alpha} =m^2+E_1E_2-\vec{p_1}\cdot\vec{p_2} $$ where $E_i$ are the energies and $\vec{p_i}$ the spatial momenta with $i=\{1,2\}$. This is reference-independent, i.e. we didn't choose any particular reference system. Now a crucial observation: due to the symmetry of the system and the conservation of the spatial momentum, the particle and the anti-particle momenta must have the same module. Since the particles have also the same mass $m$, then also their energies must be equal $$ E_1=E_2=E \qquad |\vec{p_1}|=|\vec{p_2}|=p $$ and the mass-shell relation holds $E^2-p^2=m^2$. Now, substituting $$ 0 = m^2+E^2-p^2\cos\theta $$ where $\theta$ is the angle between the directions of the particle and the anti-particle (notice that the problem lies in just two dimensions, and also that the reference system in which the total momentum is zero corresponds to $\theta=\pi$, i.e. back-to-back particles). Now we just have to find a majorant of the equation above. Since $$ E^2-p^2\cos\theta \geq E^2-p^2 = m^2 \qquad \theta\in[0,\pi] $$ then $$ 0 \geq 2m^2 $$ which is wrong.