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

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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.


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
George Jones
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Why can an electron - antiproton pair not be created by a photon in free space?
Did you really mean "electron - antiproton pair"?
 
  • #3
nicksauce
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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.
 
  • #4
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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
 
  • #5
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Sorry your right i ment electron-positron

so is the fact that there is only one vertex on the diagram a sufficiently strong answer?
 
  • #6
nicksauce
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I think it is "sufficiently strong" but then again I think that E/p conservation violation is a "better" answer.
 
  • #7
George Jones
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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?
 
  • #8
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ah, that is a really good way of explaining it,

thank you very much to all of you
 
  • #9
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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 $p_0=E_0/c$.
Therefore Eq1 leads to:
$$
0=4m^2c^4.
$$
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 donot 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.\\

Comments:\\
1. An extra particle whose rest mass in non-vanishing must be introduced to help a photon decay into electron-positron couple. Your tutor is correct.\\
2. The argument just above, \textsl{there's no reference in which a photon is at rest}, on the other hand, means a photon, or a flash of light, couldnot act as reference frame, not to say an inertial one, although light travels at constant rapidity in arbitrary directions. This is due to quantum effect.\\
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
numerous photons share a similarly regular motion.
 

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