Escape velocity of electron/positron pair

[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

 

[Bill_K]

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 ?

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.

 

[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 ?

 

[mfb]

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" ?

Yes.

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 ?

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.

 

[ChrisVer]

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

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

 

[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 ?

 

[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.

 

[ChrisVer]

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

 

[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.

 

[nikkkom]

"2m is enough for them to escape the attraction" must be wrong as the EM-force has infinite reach.

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

 

[xortdsc]

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

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 want to 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.

 

[Nugatory]

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

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

 

[Nugatory]

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

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.

 

[xortdsc]

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

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 ?)

 

[mfb]

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.

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.

 

[xortdsc]

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.

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 :(

 

[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 ?"

 

[Bill_K]

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

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 ?"

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.

 

[xortdsc]

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.

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).

 

[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?

 

[mfb]

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.

2m. This has been answered multiple times now. Those 2m already include the necessary kinetic energy to separate the particles.

 

[ChrisVer]

Also I think it's incorrect to have such a mechanical view on electron+positron pairs... (fundamental pointlike particles)
If they get created at one point, then their potential energy would be $-∞$. How could they escape that?

 

[xortdsc]

First remark, vacuum is not nothing...

yes, that's right. i meant vacuum. i said nothing as vacuum after all is as much nothing as it physically gets in our known universe. ;)

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).

so if you are saying that it only needs +2mc^2 of energy relative to the vacuum energy to create the pair and separate them to infinity, then where does the kinetic energy come from when they approach each other again (from infinity, or let's say "almost infinity", back to 0 and annihilation) ?
if what you say is true, the energy contained in each particle is only mc^2 so if that is all there is, it would need to convert this mass-energy into kinetic energy in order to conserve energy. i really don't see how this can possibly be correct.
so the question remains: "where does the kinetic energy come from ?"

if instead you do add the coulomb potential (which i think is correct) to the energy contained in the particles due to their opposite charge, then energy would be preserved, but on the other hand the energy level would drop below zero below a certain distance (so the vacuum is definitely something below 0 energy level), approaching -inf when the distance approaches 0 (so in this case the vacuum would have a -inf energy level).

can you see the problem in either approach ?

i think what is needed is a better formula for the coulomb potential which takes into account that if the 2 particles come very close they essentially become a dipole which vanishes as their separation approaches 0.

i'm sorry that i have to be that critical about it, but i think physics should be able to withstand critical questioning ;)

 

[ChrisVer]

From infinity to 0, the kinetic energy would come from the drop of the potential energy.
I don't understand why you say that the energy level would drop below zero at a certain distance.

+ don't try to think of a normal coulomb potential at very small distances...

 

[xortdsc]

From infinity to 0, the kinetic energy would come from the drop of the potential energy.

so you also have to add that potential energy it to the mass energy when the reference point is the vacuum energy.

I don't understand why you say that the energy level would drop below zero at a certain distance.

well, the energy drops below zero once the coulomb potential becomes smaller than -2mc^2 which it will below a certain distance.

+ don't try to think of a normal coulomb potential at very small distances...

yes exactly, I'm aware that you can not use the coulomb law for small distances. that's why I'm looking for a better way to compute it for the particular situation of close-to-zero separations (taking into account the vanishing dipole), because the REAL potential difference between 0 separation and let's say 1nm surely is finite. maybe qed came up with some integral or so ?

 

[Dadface]

so you also have to add that potential energy it to the mass energy when the reference point is the vacuum energy.



well, the energy drops below zero once the coulomb potential becomes smaller than -2mc^2 which it will below a certain distance.



yes exactly, I'm aware that you can not use the coulomb law for small distances. that's why I'm looking for a better way to compute it for the particular situation of close-to-zero separations (taking into account the vanishing dipole), because the REAL potential difference between 0 separation and let's say 1nm surely is finite. maybe qed came up with some integral or so ?

I think it might help if you do a search to find out whether there are any "known" limits of applicability of Coulombs law. I did a brief search and found the following which you might care to Google:

"Supersensitive Coulomb test proposed".

The article is from 97 and I haven't yet found out whether the proposed tests have been carried out. If you find anything interesting please report back.

 

[mfb]

@ChrisVer and xortdsc: To avoid "zero distance" we need quantum field theory, and I guess that is above the level of xortdsc's knowledge. Can we please ignore this point and start the analysis with a small, finite distance? Otherwise this thread leads to nothing but confusion.

 

[xortdsc]

well, just because it was beyond my knowledge i was asking this question in the first place ;)
is there nobody that has derived a modified coulomb law from quantum field theory to work out the energy involved at a certain separation in relation to vacuum energy ?
i can not start with a small finite distance, because what i need to know is how much energy does it take to create a electron/positron pair from vacuum which have enough kinetic energy so they can separate to distance x. so i defintely need to leave the classical physics and have to move into quantum field theory. the answer, if it has been found yet, must lie there somewhere.

 

[ChrisVer]

In QFT, the Coulomb potential is a non-relativistic limit of the QED (in the case you have a static particle- eg. the scattering of an electron from a proton, the proton will be "static" because it's very heavy).
In that case you can always check the Peskin & Schroeder

 

[mfb]

is there nobody that has derived a modified coulomb law from quantum field theory to work out the energy involved at a certain separation in relation to vacuum energy ?

There is no such thing as a certain "the separation of particles" in quantum mechanics.

i can not start with a small finite distance, because what i need to know is how much energy does it take to create a electron/positron pair from vacuum which have enough kinetic energy so they can separate to distance x.

You can, because you know the total energy needed to get to infinity. How often do we have to repeat that? If they can end up at a smaller separation, you can save the kinetic energy they would have (to escape to infinity) at this distance.

so i defintely need to leave the classical physics and have to move into quantum field theory. the answer, if it has been found yet, must lie there somewhere.

You can use classical mechanics if your final separation is sufficient to neglect quantum effects in the final state. And you can use classical mechanics as an approximation if the final separation is at the scale of the size of atoms. Everything smaller than that is not a proper final state anyway.

 

[nikkkom]

so if you are saying that it only needs +2mc^2 of energy relative to the vacuum energy to create the pair and separate them to infinity, then where does the kinetic energy come from when they approach each other again (from infinity, or let's say "almost infinity", back to 0 and annihilation)?

The total energy of the system of two e+ e- particles approaching each other does NOT increase as they come closer and closer. It stays exactly 2m.

You can think of it this way, if it makes the picture easier to digest: "kinetic" part of the total energy increases at the expence of decreasing rest mass of the particles, which decreases because bound system of e+e- close together (positronium) has less mass than two unbound particles (as all other bound systems do, from solar system to hydrogen atoms).

 

[xortdsc]

mass is either constant (invariant mass) or increases with kinetic energy (relativistic mass)
check this article: http://hepth.hanyang.ac.kr/~sjs/physics/mass.html

so how could it possibly pay the kinetic energy toll with its mass ? It makes no sense.
most people here don't seem to understand what i want to do. I don't care how much kinetic energy a electron needs to escape a positron in proximity, but how much energy is needed to create them. Okay, then you keep repeating it is 2 times the (invariant) rest-mass of the electron, but if mass is invariant, where does the kinetic energy come from ? from the coulomb potential (so there is additional energy involved) ! But the classical coulomb potential breaks down at close separations, because it doesn't consider quantum effects.
And saying that "it is not allowed to make the separation very small up to zero" is surely true for the classical coulomb potential formula, but that's just because the classical coulomb potential is a crude approximation which does not reflect nature when separations are small. In nature you can see particle-pairs spawning from vacuum with finite energy.
I really don't know how to make my problem any more clear... :/

 

[mfb]

most people here don't seem to understand what i want to do.

I think we do, but that is not possible, and you don't understand our answers.
To consider the region where the classical description does not work any more, you need quantum field theory. If you use this, and write down several pages of calculations, you arrive at the result that gets repeated over and over again: you need at least 2m to create an electron/positron pair that escapes to infinity, you need a bit less to create a pair that stays in a bound state (positronium).

 

[xortdsc]

and the argument...
"it is 2 times the (invariant) rest-mass of the electron, but if mass is invariant, where does the kinetic energy come from ?"
...does not raise any fundamental questions in you ?

and again, I'm aware that classical physics will not be sufficient and the "bug" is in the coulomb potential function (which didn't find much acknowledging here).

i have the impression people just keep repeating the textbook solution and don't even consider the arguments i give against it. if you're right there should be a plausable answer to my argument, right ?

 

[ChrisVer]

http://cds.cern.ch/record/404321/files/9905076.pdf
one example where you can see the energy you need and how it appears, outside of the main framework of QFT...