# Energy of electron + positron separated by distance

1. May 11, 2014

### xortdsc

Hi,

given the scenario of an electron and a positron (both assumed to be stationary) being separated by a given distance how could i compute the energy of that system ?
The total energy would certainly contain the rest-mass-energies of both particles, but i'm struggling to incorporate the electro-static energy (the usual electro static potential does not seem to be of help as it would suggest it requires infinite amounts of energy to separate the orignally superimposed electron/positron in the first place, or do i miss something here ?).

So mathematically:
E = 2mec2 + f(separation)
But what would f(separation) look like ?

Can somebody help me here ?

Last edited: May 11, 2014
2. May 12, 2014

### xortdsc

Hmm, is it that much of a tricky question ?

3. May 12, 2014

### Bill_K

Isn't it just E = 2mc2 - e2/r?

When the particles are infinitely far apart the energy reduces to 2mc2, since the rest energy includes the Coulomb self-energy.

4. May 12, 2014

### xortdsc

hmm, but that seems odd, as it would mean that at a certain distance E=0 and coming closer together E<0 up to E=-inf at r=0
Wouldn't make a whole lot of sense, or would it ?

5. May 12, 2014

### Bill_K

The total energy of the system can't fall below zero. At short distances you have to take into account the effects of QFT. And in particular, the Coulomb potential is modified by vacuum polarization.

Quoting this paper, for example,

They then give the first order modification of V(r).

6. May 12, 2014

### xortdsc

Ah thanks. That seems interesting. Gotta try out that formula and see what it spits out.

7. May 13, 2014

### xortdsc

Hi,

so I tried out the suggested corrected coulomb potential, but it doesn't seem proper and I cannot see what I'm doing wrong.

my definition goes like this (using cgs gauss units as in the paper, I'd presume, because of alpha=e^2/(hbar*c)):
Code (Text):
e = -4.80320451*10^-10 statC
m = 9.10938291*10^-28 g
hbar = 1.054571726*10^-27 erg*s
c = 29979245800 cm/s
alpha = e^2/(hbar*c) = ~1/137
qftCorrection[r] = 2*alpha/(3*Pi) * Integral[Exp[-2*(m*c/hbar)*x*r] * (1 + 1/(2*x^2)) * (x^2 - 1)^(1/2) / x^2, {x, 1, Infinity}]
coulombEnergy[r] = -e^2/r*(1 + qftCorrection[r])
pairEnergy[r] = 2*m*c^2 + coulombEnergy[r]
The main problem is that the energy for the electron/positron-pair (pairEnergy[r]) drops below 0 (which it shouldn't, right?) for distances smaller than around 2*10^-13 cm as can be seen in this plot.

Is there possibly some error in the units I'm using or did I misinterpret the formula somehow ?

#### Attached Files:

• ###### ModifiedCoulombPotential.png
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8. May 13, 2014

### The_Duck

The minimum radius of positronium (its radius in its lowest energy state) is of order the Bohr radius, very roughly $10^{-10}$ meters. Thus it always has positive total energy, because this is much greater than the radius you calculated at which the negative binding energy would exceed the positive rest mass energy of the constituents.

However, it seems to me that if you increased the charge of the electron enough, then you could make it so that the total energy of the positronium ground state would be negative. I'm unclear what this means--does the vacuum turn into a condensate of e+e- pairs or something?

9. May 13, 2014

### xortdsc

well, i'm not really concerned with positronium in ground state, but rather with electron/positron pair creation itself.

10. May 13, 2014

### The_Duck

OK, your worry seems to be "at the moment the e+e- pair is created, the potential energy is infinitely negative, therefore the total energy is infinitely negative."

If you want a classical picture of e+e- pair creation that resolves this worry, the e+ and the e- each start with infinite positive kinetic energy, canceling their infinite negative potential energy (and leaving a small finite positive remainder equal to the energy put into the e+e- pair by whatever process created it). This isn't too far from the correct quantum mechanical picture.

Last edited: May 13, 2014
11. May 13, 2014

### xortdsc

yes exactly.
but then how to get a actual number for the energy for a system in which a electron/positron pair is created which depart from each other, slowing down due to electro-static forces between them and at a distance of "d" actually stop (the "turning point") and start accelerate towards each other again ? That precisely what I wanna calculate. But how exactly ?

12. May 13, 2014

### The_Duck

Classically, the energy of such a system is V(Rmax), where Rmax is the maximum separation of the electron+positron.

Quantum mechanically, particles don't have definite trajectories.

13. May 13, 2014

### xortdsc

but the classical V will always be negative for positron/electron, right ? So it would only permit computing differences in energies and the difference to the zero separation is infinite.