# Escape velocity of electron/positron pair

1. Jun 23, 2014

### xortdsc

Hi,

is there a way to compute the escape velocity/kinetic energy of a newly created electron/positron pair ? Or in other words: How much excess energy (beyond 2 times electron mass) has to be put into the creation such that they will escape each other into infinity ?

Thanks and cheers

2. Jun 23, 2014

### Bill_K

Two electron masses is already enough for them to escape to infinity. Creating them in a bound state would require slightly less energy than that.

The energy levels of positronium are half those of the hydrogen atom, so the ground state is at -6.8 eV.

3. Jun 23, 2014

### xortdsc

so that would mean that 2 electron masses will be just the amount of energy to be injected to produce the electron/positron and they would fly apart. During it's course through space they would slow down (due to attractive electro static forces) and come to rest at "infinity" ?

wouldn't that mean that their effective mass during flight would need to be less than their rest-mass as some of this energy is used for kinetic motion ? however special relativity states that the mass would increase when moving. how does this go together ?

4. Jun 23, 2014

### Staff: Mentor

Yes.

What do you mean with "effective mass"? The mass of the system in total is constant, two times the electron mass. The mass of the individual particles is constant as well and exactly the electron mass. Mass (sometimes called "rest mass") is not a velocity-dependent quantity. Energy is.

5. Jun 23, 2014

### ChrisVer

$E^2= p^2 + m^2$

mass doesn't change...
the kinetic energy changes by $p$ term

6. Jun 24, 2014

### xortdsc

are you saying the energy of the system increases as the particles move away from each other ? where is the conservation of energy gone ?

7. Jun 24, 2014

### xortdsc

when one considers conservation of energy of the whole system (which should be the case, i think), i'd have thought that the total energy is 2 times electron mass energy plus the kinetic energy needed to overcome the attractive coulomb potential between them. Then during flight this kinetic energy is drained by the coulomb force (so coulomb potential goes up while kinetic energy goes down as they move further and further apart) and both will be 0 at infinity.

Last edited: Jun 24, 2014
8. Jun 24, 2014

### ChrisVer

The coulomb potential falls the farther you go....
2m is enough for them to escape the attraction.

9. Jun 24, 2014

### xortdsc

no, it rises as particles of opposite charge increase their distance.
and "2m is enough for them to escape the attraction" must be wrong as the EM-force has infinite reach.

10. Jun 24, 2014

### nikkkom

Gravity has infinite reach too. It doesn't mean Voyagers will ever return to Solar system.

11. Jun 24, 2014

### xortdsc

sure, and in this case you can compute the amount of (kinetic) energy needed to overcome the gravitational potential. of course this does not mean that "after 2m it will escape". it's a matter of having or not having enough kinetic energy to overcome the potential. i wanna do the same for the electron/positron and you could if you don't start with the zero-distance initially. unfortunately the coulomb law breaks down when the distance becomes 0, so traditionally it is either not computable or it takes infinite energy.

12. Jun 24, 2014

### Staff: Mentor

as the two particles move apart, the potential energy increases as the kinetic energy decreases - total energy is conserved.

13. Jun 24, 2014

### Staff: Mentor

The Coulomb force falls off with distance. The potential, however, continues to increase, just more slowly.

Whether 2m is sufficient for escape or not depends on whether the kinetic energy at 2m is high enough to overcome the remaining potential energy barrier from 2m out to infinity. That's not a very high barrier at all (Don't take my word for this - calculate it yourself!) so in practice we usually think of 2m separation as complete escape.

14. Jun 24, 2014

### xortdsc

yes, that's what i thought. but how much potential (coulomb) energy does it have ? that was my initial question just with a different formulation.
the normal coulomb potential is insufficient as it would suggest that you can not even slightly separate two superimposed electron/positron without getting infinite energy in the calculation. there must be a more accurate way to compute it (or at least there should be, otherwise i'd consider this a serious flaw in the theory, no ?)

15. Jun 24, 2014

### Staff: Mentor

Plus the (negative) potential energy, which exactly cancels the kinetic energy all the time in the limiting case.
You can assume the initial separation to be small, but finite. Quantum mechanics makes sure this works - there is no way to localize particles at exactly one point in space.

@Nugatory: 2m means 2 times the electron mass, not 2 meters.

16. Jun 24, 2014

### xortdsc

the problem is that this won't help me to find out how much energy actually went into creating the particle pair which subsequently depart to infinity.
sure i can assume the initial separation to be small, but whatever distance i choose i will get different energies (the smaller the initial separation the higher the energy it would take, up to infinity).
But since the energy must be finite, at least for non-infinity-separations i was looking for a way to compute it, but the coulomb law won't help here :(

17. Jun 24, 2014

### xortdsc

I could turn the question around and ask: "Given an electron and a positron at rest separated by x meters, how much energy would they release upon annihilation ?"

18. Jun 24, 2014

### Bill_K

Metaquestion: Why do you keep asking the same question? It's been answered for you, multiple times.

Initially, the energy is 2mc2 + V(x), where V(x) = -e2/x is the Coulomb potential. After annihilation into photons, the energy is still 2mc2 - e2/x, by conservation of energy.

19. Jun 24, 2014

### xortdsc

I'm asking the same question, because the answers don't seem to fit. When you are saying that the initial energy is 2mc^2 + V(x), where V(x) = -e^2/x i can totally follow that. And sure you're right that by conservation of energy it still is on that level after they annihilated (possibly as radiation, possibly as a new pair that get's created).
However my initial question is how much energy is needed (has to be put into) to create the pair plus velocity from nothing (vacuum). In order to get that I'd need to know what's the energy difference between vacuum and the created pair as the energy is only a relative measure. And the energy level of the empty space cannot be zero. It's simply not possible to compute with the equations given so far or let's say they will always give unplausable results like infinity (still they get created in nature).

20. Jun 24, 2014

### ChrisVer

First remark, vacuum is not nothing...
Second remark, their creation needs 2m... that's more than enough to let them also escape.
afterall if they wouldn't escape they would have to be annihilated back, so you wouldn't observe them. So far we know that the energy needed to create observable pairs of electron+positron is ~2m (or ~1. MeV).

How do you do the calculations in classical mechanics?
in order to let a rocket for example leave the earth, you just need to give it some energy~ G Mm/R. This energy is enough and the rocket would leave the earth and fly to infinity.... it's kinetic energy would always drop, and its potential energy would always rise... what's the paradox of it at infinity?