# Why the photon can't produce electron and positron in space or in vacuum?

$$\frac{hc}{\lambda} = K_e + K_p + 2m_e c^2$$ could be the energy conservation equation for a photon of wavelength $\lambda$ decaying into a electron and positron with kinetic energies $K_e$ and $K_p$ and rest mass energy $m_e c^2$.

Why does this decay not occur in space or vacuum?

• h plank's constant ,N=c / λ,k(e-)kinetic energy of electron,kinetic energy of positron,m0c^2=.511 – Amr Hawk Nov 8 '14 at 17:05

You can't simultaneously conserve energy and linear momentum.

Let the photon have energy $E_{\gamma} = p_{\gamma} c$ and the electron have energy $E_{-}^{2} = p_{e}^{2}c^2 + m_{e}^{2}c^4$ and an analogous expression for the positron. Suppose the electron and positron depart from the interaction site with an angle $2\theta$ between them.

Conservation of energy.

$$p_{\gamma} c = \sqrt{(p_{e}^{2}c^2 +m_e^{2}c^4} + \sqrt{(p_{p}^{2}c^2 +m_e^{2}c^4},$$ but we know that $p_{p} = p_{e}$ from conservation of momentum perpendicular to the original photon direction. So $$p_{\gamma} = 2\sqrt{p_{e}^2 + m_e^{2}c^2}$$

Now conserving linear momentum in the original direction of the photon. $$p_{\gamma} = p_e \cos{\theta} + p_p \cos\theta = 2p_e \cos\theta$$

Equating these two expression for the photon momentum we have $$p_e \cos{\theta} = \sqrt{p_{e}^2 + m_e^{2}c^2}$$ $$\cos \theta = \sqrt{1 + m_e^{2}c^2/p_e^{2}}$$ As $\cos \theta$ cannot exceed 1 we see that this impossible.

Let us say that the photon has created pair of massive particles. There must exist a reference frame ("center of mass") in which the total momentum of the two particles is zero. So by momentum conservation, in the same frame the photon had to have momentum zero. But photon cannot have momentum zero, because then it would have zero energy. So momentum is not conserved in the center of mass frame, and, because of the Lorentz invariance, is not conserved in all frames. Therefore, the process is impossible.

In principle you could observe matter creation in a vacuum:

https://en.wikipedia.org/wiki/Schwinger_limit

And there are high power lasers being built to show this in experiments:

https://eli-laser.eu/science-applications/high-fields-physics/

The uncertainty principle is an inequality so the momentum times c times cosine which is energy may be made to equal the energy of the momentum and rest mass of the combined electron and positron. energy and momentum conservation is satisfied using the uncertainty principle.

This process in the vacuum cannot be directly measured as opposed to real measurements of particles on the mass shell, which your example shows. Show me that your process can be traced in the vacuum. As per your answer look at the final equation. Since processes in the vacuum are not traceable the equation should read $2\Delta Pc^2\cos^2$ greater than or equal to $[\hbar/2\Delta t]^2$. $2\Delta Pc^2 = (Energy)^2$ which for equality leads to $\cos^2 = 1. A$ photon leading to a electron-positron state may be bound with $l=0, s=1$ and has odd charge conjugation parity. The charge conjugation of a photon is -1 n photons with $n = 1, 3, 5,..... = -1$ is odd. Let $n = 1$ photon. This is a fundamental symmetry.

The uncertainty principle allows the energy and momentum to be compatible and the process may occur in the vacuum.