# Why can an electron - antiproton pair not be created by a photon in free space?

sheelbe999

## Homework Statement

As in the title

Why can an electron - antiproton pair not be created by a photon in free space?

This question came up last term but our tutor told us that it is because there are "no stimuli", I'm sure there is a better answer than this. I seem to recall that it can be proved mathematically.

## The Attempt at a Solution

Staff Emeritus
Gold Member
Why can an electron - antiproton pair not be created by a photon in free space?

Did you really mean "electron - antiproton pair"?

Homework Helper
Assuming you meant electron-positron, my answer would be that since it is a 1 vertex feynman diagram, it is a virtual process. If you really meant electron-antiproton, then it violates all sorts of conservation laws.

AstroRoyale
That interaction cannot occur due to lepton and baryon number conservation. If you meant

photon -> electron + positron

This cannot happen in free space due to energy/momentum conservation. You need a heavy nucleus around for that interaction to occur to take some of the energy and momentum

sheelbe999
Sorry your right i ment electron-positron

so is the fact that there is only one vertex on the diagram a sufficiently strong answer?

Homework Helper
I think it is "sufficiently strong" but then again I think that E/p conservation violation is a "better" answer.

Staff Emeritus
Gold Member
Sorry your right i ment electron-positron

so is the fact that there is only one vertex on the diagram a sufficiently strong answer?

Is there always a frame in which the total spatial momentum of the electron-positron pair is zero? If there is, by conservation of spatial momentum, what is the energy of the photon in this frame?

sheelbe999
ah, that is a really good way of explaining it,

thank you very much to all of you

Tian WJ
Dear sheelbe999， your problem is very interesting, and I will do some mathematical formulation below.\\

Denote the rest mass of electron and positron as $m$, the critical (minimum) energy of the photon as $E_0$, and we turn to the Lorentz invariance $E^2-c^2p^2$. Throughout the reaction, the generated electron and positron couple are at rest with respect to the center of their momenta, hence
$$E_0^2-c^2p_0^2=(2mc^2)^2 （Eq1) ,$$
where $p_0$ is the momentum of the photon, and $p_0=E_0/c$.
$$0=4m^2c^4.$$
This is nothing more than ridiculous! So the process is impossible at all. This contradiction originating from $E^2-c^2p^2$ means that, conservation of energy and momentum don't hold at the same time. As a matter of fact, as is shown below, it is conservation of momentum that is violated.\\

Suppose the process below is permitted:
$$\gamma \to e^-+e^+.$$
We set the centroid of $e^-$ and $e^+$ at translation to be the reference frame, in which the momenta of the system $\{e^-,e^+\}$ is
vanishing. However, on the contrary, there's no reference in which a photon is at rest! Hence, this process fails to respect conservation of momentum and is therefore forbidden.\\

To set a photon as reference frame, its states and parameters of motion should be determined. The $x$ component of momentum $p_x$ reads $p_x=\frac{\hbar}{c} v_x$, where the constant $c$ is speed of light. Hence, $\Delta p_x=\frac{\hbar}{c}\Delta v_x$. The velocity of the photon reference is constant $v=c$, so does its $x$ exponent
$v_x$, hence $\Delta p_x=0$. However, according to uncertainty principle $\Delta p_x\cdot\Delta x\geq \hbar/2$, we have $\Delta x=\infty$. For $y$ and $z$ components, this phenomena hold, too. That is to say, when a photon acts as reference, whose momentum is determined, we canot locate the reference any more! Hence, a photon couldnot act as reference, nor could a flash of light where the