Why do particles annhillate with their particles?

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In summary: What happens then?I don't understand. What prevents an electron and a positron that are not in proximity from annihilation?Perhaps this answer is too simple, but electrons and positrons have opposite charge, and in their bound state tey are in very close proximity to ano another. Therefore, the attraction between the two is very strong, and they get drawn together magnetically. Also, I think the question Friend mentions really is one of the big ones. The theory has been suggested that antimatter has a faster decay rate than matter, and this explains the predominance of matter in our universe, but it stills seems difficult to
  • #1
quantumfireball
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why do particles annhillate with their particles?

Why is an electron-positron bound state so short lived?
In other words why does an electron readily anhillate its corresponding antiparticle(positron)?
 
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  • #2
quantumfireball said:
Why is an electron-positron bound state so short lived?
In other words why does an electron readily anhillate its corresponding antiparticle(positron)?
Perhaps Murry Gell-Mann's law says it best. i.e. Whatever isn't forbidden is required. The annihilation of matter and antimatter does not violate any law of nature and so it can therefore happen and therefore "must" happen.

Pete
 
  • #3
quantumfireball said:
Why is an electron-positron bound state so short lived?
In other words why does an electron readily annihilate its corresponding antiparticle(positron)?

A better question might be why do they not always annihilate each other. Why does there exists at all free electron or positrons that do not annihilate each other. For it seems the more frequent occurrence is to have virtual electrons/positron spontaneously popping into existence and then immediately annihilating each other. And this begs the question as to why this isn't always the case?
 
  • #4
Perhaps this answer is too simple, but electrons and positrons have opposite charge, and in their bound state tey are in very close proximity to ano another. Therefore, the attraction between the two is very strong, and they get drawn together magnetically.

Also, I think the question Friend mentions really is one of the big ones. The theory has been suggested that antimatter has a faster decay rate than matter, and this explains the predominance of matter in our universe, but it stills seems difficult to imagine how a particle-antiparticle pair can form without immediately annihilating. I'm going to go look in CERN's website and see how they do it. I hear tell they've even made atoms, like anti-hydrogen which were stable (although I don't know how stable).
 
  • #5
quantumfireball said:
Why is an electron-positron bound state so short lived?
In other words why does an electron readily anhillate its corresponding antiparticle(positron)?

I don't understand. What prevents an electron and a positron that are not in proximity from annihilation?
 
  • #6
LURCH said:
Perhaps this answer is too simple, but electrons and positrons have opposite charge, and in their bound state tey are in very close proximity to ano another. Therefore, the attraction between the two is very strong, and they get drawn together magnetically.
Proton and electron in the hydrogen atom are as close as positron and electron in the positronium, but they don't annihilate.
 
  • #7
What doesn't annihilate in an electron-positron annihilation and what's conserved?
 
  • #8
Phrak said:
What doesn't annihilate in an electron-positron annihilation and what's conserved?

While rest mass itself is not conserved (since a photon has no rest mass) and energy is not conserved (a photon is created) the energy gained is equivalent to the mass lost through the formula [tex]E=mc^2[/tex].

Thus, while mass alone and energy alone are not conserved, the more general conservation of "mass-energy" is still valid.

Also, momentum is conserved in the process. For example, if the initial system had no linear momentum, two photons will result from the annihilation, each of equal energy going off in opposite directions. This system also has 0 net momentum. Thus, momentum is conserved.

Charge is conserved since the net charge before and after annihilation is zero.

Angular momentum is conserved as well. For instance, if needed, the process will result in three photons if the conservation of angular momentum is required. I should point out that theoretically any number of photons >=2 can be produced from this event. 1 will never be produced, since, as said before, linear momentum would not be conserved.
 
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  • #9
G01 said:
While rest mass itself is not conserved (since a photon has no rest mass) and energy is not conserved (a photon is created) the energy gained is equivalent to the mass lost through the formula [tex]E=mc^2[/tex].

Thus, while mass alone and energy alone are not conserved, the more general conservation of "mass-energy" is still valid.

Also, momentum is conserved in the process. For example, if the initial system had no linear momentum, two photons will result from the annihilation, each of equal energy going off in opposite directions. This system also has 0 net momentum. Thus, momentum is conserved.

Charge is conserved since the net charge before and after annihilation is zero.

Angular momentum is conserved as well. For instance, if needed, the process will result in three photons if the conservation of angular momentum is required. I should point out that theoretically any number of photons >=2 can be produced from this event. 1 will never be produced, since, as said before, linear momentum would not be conserved.

Yes, I've heard that either two or three photons result, depending upon whether the two particles are spin parallel or antiparallel. i didn't know there could be more.

I'm curious about the converse to the question that was asked by the OP. Say the two particles are not bound. Could it be that there's no combination of photons that will conserve mass, momentum and angular momentum?
 
  • #10
G01 said:
While rest mass itself is not conserved (since a photon has no rest mass) and energy is not conserved (a photon is created) the energy gained is equivalent to the mass lost through the formula [tex]E=mc^2[/tex].

Thus, while mass alone and energy alone are not conserved, the more general conservation of "mass-energy" is still valid.
Don't understand this. Isn't mc^2 energy (the rest energy)? When you compute a system's total energy, you also put mc^2 in the equation.
 
  • #11
G01 said:
While rest mass itself is not conserved (since a photon has no rest mass) and energy is not conserved (a photon is created) the energy gained is equivalent to the mass lost through the formula [tex]E=mc^2[/tex].
That is incorrect. In special relativity energy is always conserved. E.g. if an electron annihilates a positron which results in the production of two photons then the total energy before the annihilation is exactly the same as the energy before the annihilation. The initial energy is the sum of the rest energies of the two particles plus the kinetic energy of each particle. The final energy is the sum of the energies of each photon. The total inertial (aka relativistic) mass is conserved too since the total inertial energy is proportional to the total inertial mass.

Pete
 
  • #12
quantumfireball said:
Why is an electron-positron bound state so short lived?
In other words why does an electron readily anhillate its corresponding antiparticle(positron)?

Electrons and positrons are also both created from this interaction (and others, such as the electroweak coupling, though independent mechanisms for these phenomena are not guaranteed). Neutral kaons, which decay at different rates than their antiparticles, explain the ultimate accumulation of matter over antimatter (this is known as a violation of time reversal symmetry, or T violation, in CPT theory, and won the 1980 Nobel Prize in Physics; these particles are also the only ones known to disobey symmetric charge conjugation and parity, or CP violation, and the probability of divergence from CPT conservation corresponds to that measured for T violation though it occurs in an unassociated reaction).

http://physicsworld.com/cws/article/print/1327

Phrak said:
I don't understand. What prevents an electron and a positron that are not in proximity from annihilation?

Space?

lightarrow said:
Proton and electron in the hydrogen atom are as close as positron and electron in the positronium, but they don't annihilate.

Protons are hadrons and electrons are leptons.
 
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  • #13
shadowpuppet said:
Originally Posted by Phrak
"I don't understand. What prevents an electron and a positron that are not in proximity from annihilation? "

Space?

To that, you might add time. We might take Gell-Mann at face value, as quoted by Pete,

"Whatever isn't forbidden is required. The annihilation of matter and antimatter does not violate any law of nature and so it can therefore happen and therefore "must" happen,"

and ask, what forbids annihilation of matter and antimatter not in proximity?
 
  • #14
Phrak said:
To that, you might add time. We might take Gell-Mann at face value, as quoted by Pete,

"Whatever isn't forbidden is required. The annihilation of matter and antimatter does not violate any law of nature and so it can therefore happen and therefore "must" happen,"

and ask, what forbids annihilation of matter and antimatter not in proximity?

I some how doubt that...coupling antiparticles usually aren't stable enough to require time for decay. Perhaps the internal states of the positron and electron cannot respond to each other in time (at a distance) because the duration of electromagnetic transmission is finite and allows disordering entropic reactions within the leptons to occur (but this is just speculation as lepton sea partons have never been implicated in experimental results).
 
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  • #15
shadowpuppet said:
I some how doubt that...

What do you doubt?
 
  • #16
Phrak said:
What do you doubt?

I doubt that time prevents antiparticles from annihilating at a distance, unless you mean time as a constituent of space-time.
 
  • #17
Lifetime of positronium

A nice way to understand what happens in the electron-positron annihilation uses Feynman diagrams. Imagine a line representing an electron moving forward in time, then emitting a photon and being scattered. Then it is scattered backward in time with the emission of a second photon and now looks like a positron moving forward. The line segment between the two photon vertices is a "virtual" electron, since it doesn't conserve energy and momentum correctly, so it can only have a short life according to the uncertainty principle. So the two photons must be emitted when the two particles are close together--less than about the Compton wavelength apart. This is so short compared with the size of the positronium atom that we usually treat the interaction as effectively zero-range.
 
  • #18
shadowpuppet said:
Proton and electron in the hydrogen atom are as close as positron and electron in the positronium, but they don't annihilate.
Protons are hadrons and electrons are leptons.

Certainly, but I was answering to Lurch's comment:
Perhaps this answer is too simple, but electrons and positrons have opposite charge, and in their bound state tey are in very close proximity to ano another. Therefore, the attraction between the two is very strong, and they get drawn together magnetically
where it would seem that they annihilate just because they have opposite charge and are close to each other.
 
  • #19
lightarrow said:
Certainly, but I was answering to Lurch's comment:

where it would seem that they annihilate just because they have opposite charge and are close to each other.

Oh sorry, I guess it was directed at the wrong post.
 
  • #20
pmb_phy said:
That is incorrect. In special relativity energy is always conserved. E.g. if an electron annihilates a positron which results in the production of two photons then the total energy before the annihilation is exactly the same as the energy before the annihilation. The initial energy is the sum of the rest energies of the two particles plus the kinetic energy of each particle. The final energy is the sum of the energies of each photon. The total inertial (aka relativistic) mass is conserved too since the total inertial energy is proportional to the total inertial mass.

Pete
I see Pete. That makes perfect sense. Thank you for the correction. Sorry if my mistake mislead anybody.:redface:
 
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  • #21
G01 said:
I see Pete. That makes perfect sense. Thank you for the correction. Sorry if my mistake mislead anybody.:redface:

It so sucks to be wrong on this board, doesn't it? But no worries, I understood what you meant.
 
  • #22
Phrak said:
It so sucks to be wrong on this board, doesn't it? But no worries, I understood what you meant.

It depends on how you take being wrong. In my several years on this forum, Pete has shown himself to be an expert in this field and I know that. Also, I never expect to be correct all the time. I may have been incorrect with my explanation, but now I know more about this topic than I did before. It's a learning experience.
 
  • #23
Un fortunately the energy-momentum conservation does not tell you anything about dynamics -HOW does e+ e- anihilate? First it is good to realize that we have three different components here, e+. e- and the EM-field. Secondly that the lifetime of e+ and e- is approximately proportional to the overlap of the wavefunctions; close together large overlap -> small lifetime etc.

Compare with a ball rolling down a hill. Just because the energy is conserved, it does not give you any hint why the ball will roll down -right? This could be formulated within the principle of least action, where the lagrangian L=T-V is maximized (T=kinetic energy, V=potential energy), but T+V=E is conserved (i.e., action S is the time integral of L). Thus it is good to maximize T and minimize V.

Now the Lagrangian of two particles AND an EM-field contains a term including the magnetic vector potential, which is like a kinetic energy. Thus if it is possible (by means of charge conservation etc) to annihilate them and create a photon, i.e., increase the EM-field intensity, and hence the photon kinetic energy T_ph, then the action will increase -and that is what "nature" wants!

I hope it may explain a little bit...

Best whishes,
Per
 
  • #24
per.sundqvist said:
Un fortunately the energy-momentum conservation does not tell you anything about dynamics -HOW does e+ e- anihilate? First it is good to realize that we have three different components here, e+. e- and the EM-field. Secondly that the lifetime of e+ and e- is approximately proportional to the overlap of the wavefunctions; close together large overlap -> small lifetime etc.

Compare with a ball rolling down a hill. Just because the energy is conserved, it does not give you any hint why the ball will roll down -right? This could be formulated within the principle of least action, where the lagrangian L=T-V is maximized (T=kinetic energy, V=potential energy), but T+V=E is conserved (i.e., action S is the time integral of L). Thus it is good to maximize T and minimize V.

Now the Lagrangian of two particles AND an EM-field contains a term including the magnetic vector potential, which is like a kinetic energy. Thus if it is possible (by means of charge conservation etc) to annihilate them and create a photon, i.e., increase the EM-field intensity, and hence the photon kinetic energy T_ph, then the action will increase -and that is what "nature" wants!
And why the action have to increase? Don't understand the connection between this and the principle of least action (which state that, among all the possible trajectories with fixed initial and final points and times, the real one is that which makes the action stationary).
 
  • #25
per.sundqvist said:
Un fortunately the energy-momentum conservation does not tell you anything about dynamics -HOW does e+ e- anihilate? First it is good to realize that we have three different components here, e+. e- and the EM-field. Secondly that the lifetime of e+ and e- is approximately proportional to the overlap of the wavefunctions; close together large overlap -> small lifetime etc.

Compare with a ball rolling down a hill. Just because the energy is conserved, it does not give you any hint why the ball will roll down -right?

Thanks, Per. This is exactly what I needed. This is actually the very sort of notion but better, that inspires me to ask about annihilation; but in this classical setup, (replacing gravity with an accelerating reference frame) momentum is the converved quantity, as expressed in Newton's three laws, that tells you why the ball rolls down hill, it seems.
 
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  • #26
If we put a charge into a polarizable medium, we'll find lots of dipoles, very analogous to particle-antiparticle pairs. That is, an E&M field distorts the medium in which it exists, and thus distorts the medium's charge distribution. In QED, the medium is the real vacuum, and it is indeed polarizable. So we can imagine pair production, crudely to be sure, as another form of the photo-electric effect -- a photon zaps the vacuum, which, in turn, then emits a pair.

Why is the vacuum polarizable? The ultimate, fancy answer is gauge invariance. A more "nuts and bolts" answer is framed in terms of QED's interaction energy, which, of course, takes a lot of work to formulate. The interaction contains terms like b*d*a and b*d*a*, where * denotes transpose, and b* creates an electron, d* creates a positron, and a destroys a photon. So this first term is the classic case of a photon transforming into a pair. The second term, because of a*, creates a pair plus a photon from nothing. How's that again?

We can use a simple model-- everything is in a gravitational well. Just think of this as a substitute for the "potential" created by pair-pair interactions. We'll find bound states of pairs with wave like behavior, but they can't get very far from the well's boundary. That is, we have what are often called quasi-bound states; almost free. The something from nothing phenomena is pretty much like boiling or evaporation, that is thermal energy can, locally, give enough energy to free a quasi-pair along with a photon. (Seems like the something from nothing can be generated by a phonon, while the classic case is generated by a photon. ) When interactions are taken into account, the QED vacuum has a structure of pairs and photons. In reality, this structure can support thermal properties, that is, it can have a local temperature, which can have fluctuations, which can create pair-photon states.

No stuff, no creation from nothing. Stuff allows "creation of particles" in analogy to boiling bubbles in liquid, or surface evaporation. It just looks like something from nothing, but really isn't.

Regards,
Reilly Atkinson
 
  • #27


Interesting thread this.

Pardon me for taking a slightly different slant on the subject but I have always wondered what happens with regard to the gravitational field when matter and antimatter annihilate each other.
Consensus among physicists is that matter and antimatter behave the same as far as gravitational potential is concerned, i.e., both will attract matter and antimatter, so that a particle of antimatter would fall toward the Earth and not fly off to space.
According to the currently accepted cosmological theory, essentially equal amounts of matter and antimatter were initially created in the big bang.
Matter and antimatter promptly annihilated each other but there was an imbalance in favor of matter by about one part in a billion. This imbalance is what was left and now constitutes all the matter in our universe.
The radiation that resulted from the annihilation is now dissipated with the expanding universe, and is now seen as the cosmic microwave background (CMB) radiation.

But what about the gravity?
In the Big Bang, was there briefly the gravitational field of a billion additional universes before the annihilation took place?

The possibilities seem to be:
1. No there was not. This means that after annihilation there was also no residual gravitational field, in which case, contrary to what we currently believe, maybe matter and antimatter do in fact have opposite gravitational potentials.
2. Yes there was. This would mean that after the annihilation the massive naked gravitational field was radiated away as gravity waves. Maybe those gravitational waves, fields/potentials are still out there somewhere and can be detected or measured similar to the CMB
The question is, what would this residue look like today and how would we go about looking for it?

Any clarifications from you physicists or cosmologists?

Thanks and cheers,
Keijo
 
  • #29


According to quantum field theory, particles are just localized excitations of a field. And electrons and positrons are excitations of the same field. Roughly, you can think of an electron as a little bump in the field, and a positron as a little dip. From this viewpoint, it's no mystery that they can annihilate. Still, if all there was were electrons and positrons, this would never happen because the reaction would not conserve energy. It is because of the coupling of the electron field to the EM field that electrons and positrons can annihilate into photons.
 
  • #30


quantumfireball said:
Why is an electron-positron bound state so short lived?
In other words why does an electron readily anhillate its corresponding antiparticle(positron)?

Firstly they can annihilate because it is possible. Secondly, it is short lived since it is the overlap of their wave functions that determines the lifetime. Separate them enough and they will never annihilate. Or delay the time by applying a strong electric field, causing polarization, and hence reducing the overlap -> longer lifetime!
 

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