# Annihilation or Conversion

Buckeye
When electrons collide with positrons, we get gamma-rays and other particles. Keeping in mind E=mc^2, is the result better called "annihilation" or "conversion"?

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The process when a particle meets its antiparticle is called annihilation.

An example of "conversion" is "Internal conversion" , when an exited nucleus interacts with the atomic electrons which takes a way the excitation energy - instead of the nucleus emitting that energy in form of photons (gamma radiation)

Buckeye
The process when a particle meets its antiparticle is called annihilation.

An example of "conversion" is "Internal conversion" , when an exited nucleus interacts with the atomic electrons which takes a way the excitation energy - instead of the nucleus emitting that energy in form of photons (gamma radiation)

OK. Does the word "annihilation" mean the total destruction of the particles?

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Buckeye

Wikipedia is not necessarily a reliable source, but let me ask this. If annihilation means the total destruction of particles, then why do we get two photons? The production of something from the interaction of something else seems like a conversion. Why is this process called "annihilation"?

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Wikipedia is not necessarily a reliable source, but let me ask this. If annihilation means the total destruction of particles, then why do we get two photons? The production of something from the interaction of something else seems like a conversion. Why is this process called "annihilation"?

I know that article is a good one.

Energy can't be destroyed, only converted. So you can call it a conversion process of energy.

It is called annihilation since it refers to the particles in the initial state, they are the ones beeing totally destroyed. And the emphasis is on the particles perspective since we are dealing with particle physics.

This is consistent (the definition of annihilation vs. conversion) with the internal conversion process, where energy is converted to kinetic energy of an atomic electron instead of photon radiation (roughly speaking).

Fundamental, you gain only one photon in the electron-positron vertex, but that photon will be virtual! So in order to get a real photon, either the electron or the positron must radiate of a photon before annihilation (1st order feynman for real annihilation). http://www.shef.ac.uk/physics/people/cbooth/eegg.gif [Broken]
In this way, you combine two virtual diagrams, to constitute a real-non virtual-diagram.

And this is also in agreement of experiment, QED is the best tested physical theory there is. It is amazing i must say :-)

So a summary: Annihilation of particles, conversion of energy

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Buckeye
I know that article is a good one.

Energy can't be destroyed, only converted. So you can call it a conversion process of energy.

It is called annihilation since it refers to the particles in the initial state, they are the ones beeing totally destroyed. And the emphasis is on the particles perspective since we are dealing with particle physics.

This is consistent (the definition of annihilation vs. conversion) with the internal conversion process, where energy is converted to kinetic energy of an atomic electron instead of photon radiation (roughly speaking).

Fundamental, you gain only one photon in the electron-positron vertex, but that photon will be virtual! So in order to get a real photon, either the electron or the positron must radiate of a photon before annihilation (1st order feynman for real annihilation). http://www.shef.ac.uk/physics/people/cbooth/eegg.gif [Broken]
In this way, you combine two virtual diagrams, to constitute a real-non virtual-diagram.

And this is also in agreement of experiment, QED is the best tested physical theory there is. It is amazing i must say :-)

So a summary: Annihilation of particles, conversion of energy

OK.
What is the meaning of the equal sign (=) in the equation E=mc^2?

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OK.
What is the meaning of the equal sign (=) in the equation E=mc^2?

The "=" sign refers to that mass is a form of energy, just as there is kinetic energy, potential energy etc.

So the equal sign plays a role both in annihilation and conversion, annihilation is 'followed' by conversion. The initial energy of the positron and electron (both mass-energy and kinetic energy) is converted to photon(s) with the same energy.

The answer to the question "Does the equal sign indicate annihilation or conversion" is (again) that the question itself is wrong. The equal sign just tells you how much energy you can get from a mass $$m$$. Just as $$E_{kin} = mv^2/2$$, etc. In fact, the full formula for energy of a (free) particle (i.e no potential) is $$E^2 = (mc^2)^2 + (pc)^2$$.

Buckeye
Wikipedia reports the following:

Annihilation is defined as "total destruction" or "complete obliteration" of an object;[1] having its root in the Latin nihil (nothing). A literal translation is "to make into nothing". Annihilation is the opposite of exnihilation, which means "to create something out of nothing".

In physics, the word is used to denote the process that occurs when a subatomic particle collides with its respective antiparticle[2]. Since energy and momentum must be conserved, the particles are not actually made into nothing, but rather into new particles.

Based on the Physics definition, shown above, it seems that the word "annihilation" is extremely misleading. This Physics definition uses the word "made" in place of the word "convert". Do you think it would be better to change the Wikipedia definition to better match the overall conservation of energy and momentum?

The "=" sign refers to that mass is a form of energy, just as there is kinetic energy, potential energy etc.

So the equal sign plays a role both in annihilation and conversion, annihilation is 'followed' by conversion. The initial energy of the positron and electron (both mass-energy and kinetic energy) is converted to photon(s) with the same energy.

The answer to the question "Does the equal sign indicate annihilation or conversion" is (again) that the question itself is wrong. The equal sign just tells you how much energy you can get from a mass $$m$$. Just as $$E_{kin} = mv^2/2$$, etc. In fact, the full formula for energy of a (free) particle (i.e no potential) is $$E^2 = (mc^2)^2 + (pc)^2$$.

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I would not bother, as long as one knows what all is about and what is happening.

I would still stick that the energy is converted, from the form of real particles (electron,positron) into the form of photons (another kind of particle). The term annihilation refers to that the particles in the initial state stops to exist.

the term 'made' is a quaite amateurish word, i would write "the particles are not actually converted into nothing, but rather into new particles - c.f electron-positron colliders"

Buckeye
I would not bother, as long as one knows what all is about and what is happening.

I would still stick that the energy is converted, from the form of real particles (electron,positron) into the form of photons (another kind of particle). The term annihilation refers to that the particles in the initial state stops to exist.

the term 'made' is a quaite amateurish word, i would write "the particles are not actually converted into nothing, but rather into new particles - c.f electron-positron colliders"

I agree.

Unfortunately, not all textbooks are well written so many students will get the wrong idea. Government leaders who control funding are relying on Wikipedia for information and definitions when they justify the release of money to colliders, universities and the like. What happens when they get the wrong picture? Uh-oh, you misled me, so no more money for your project.

Unfortunately, Wikipedia is filled with poor writing which leads to poor education, misleading statements and wrong concepts. On top of that, many non-native English speakers read the English version and make translations into their native languages, quite easily carrying over the poor wording problem.

AS a member of an ISO subcommittee on surface chemical analysis (TC/201), I had the highly difficult job of proof-reading the vocabulary standard. We had battle after battle between the UK and the US on the use of very simple words such as "the", "a" and the next level up.

I like your description of annihilation. Maybe you should try to change the Wiki page on annihilation. Is suspect you will have problems to get that description accepted by the protectors of that page.

I would also love to have a decent definition of the term "particle" added to Wiki. Care to try?

As you know, I have another concept of what happens from so-called "annihilation". Unfortunately, no one has a detector that only measures momentum.

Ce la vie.

Buckeye
The "=" sign refers to that mass is a form of energy, just as there is kinetic energy, potential energy etc.

So the equal sign plays a role both in annihilation and conversion, annihilation is 'followed' by conversion. The initial energy of the positron and electron (both mass-energy and kinetic energy) is converted to photon(s) with the same energy.

The answer to the question "Does the equal sign indicate annihilation or conversion" is (again) that the question itself is wrong. The equal sign just tells you how much energy you can get from a mass $$m$$. Just as $$E_{kin} = mv^2/2$$, etc. In fact, the full formula for energy of a (free) particle (i.e no potential) is $$E^2 = (mc^2)^2 + (pc)^2$$.

Three related questions:

Why do the photons come out at right angles to the collision vector?

Is there a one-to-one relation between the number of destroyed pairs and the number of 0.511 MeV photons produced, i.e. conversion %, when using at threshold energies?

What is the probability or percentage of multi-level conversions? This means: How often do we get 4 photons from one annihilated pair when the collision energy is above 2 MeV or more?

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Unfortunately, not all textbooks are well written so many students will get the wrong idea. Government leaders who control funding are relying on Wikipedia for information and definitions when they justify the release of money to colliders, universities and the like.

:rofl: If government leaders already understood science on the level of Wiki, we would be living in a far more rational world !

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i) I agree that many textbooks and resources are quite unpedagocail.

ii) Wiki has many good articles on physics, I often refer to that place. But I would rather refer to textbooks, but wiki and hyperphysics and so on, everybody can reach.

iii) How to define a particle depends on what representation you use to describe a physical system. An elementary particle, I would say that it is an entity with no internal structure. It is not made up of anything else than it self and can not be divided into smaller pieces nor have excited modes nor undergo decay.

iv) What do you mean by "only measures momentum"? One wants to detect energy and ionization loss also, otherwise we don't know what particle we detected. Measuring only momentum is trivial, just have a track chamber and a magnetic field close to the interaction vertex. Measuring only momentum is useless.

v) The most probable annihilation mode of electron and positron is if they annihilate in (almost) rest, then the two photons will have an energy of 0.511MeV each, and emitted in right angles of each other.

vi) You can create higher energy photons from annihilation aswell, and also more than two photons from electron-positron annihilation aswell.

vi) The probablity for $p$ $p \leq 2[/tex] photon emission in one annihilation process are related to those of 2 photon emission as: $$\frac{\text{Rate}(e^+e^- \rightarrow p\gamma )}{\text{Rate}(e^+e^- \rightarrow \gamma \gamma )} = \mathcal{O}((p-2)\alpha)$$ Independent of energy, i.e 4 photon emission is (1/137)*(1/137) times the probability to get 2 photon emission. I thought you had 80 books about quantum physics, that you had read from cover to cover? Buckeye i) I agree that many textbooks and resources are quite unpedagocail. iii) How to define a particle depends on what representation you use to describe a physical system. An elementary particle, I would say that it is an entity with no internal structure. It is not made up of anything else than it self and can not be divided into smaller pieces nor have excited modes nor undergo decay. iv) What do you mean by "only measures momentum"? One wants to detect energy and ionization loss also, otherwise we don't know what particle we detected. Measuring only momentum is trivial, just have a track chamber and a magnetic field close to the interaction vertex. Measuring only momentum is useless. v) The most probable annihilation mode of electron and positron is if they annihilate in (almost) rest, then the two photons will have an energy of 0.511MeV each, and emitted in right angles of each other. vi) You can create higher energy photons from annihilation aswell, and also more than two photons from electron-positron annihilation aswell. vi) The probablity for [itex] p$ [itex] p \leq 2[/tex] photon emission in one annihilation process are related to those of 2 photon emission as:

$$\frac{\text{Rate}(e^+e^- \rightarrow p\gamma )}{\text{Rate}(e^+e^- \rightarrow \gamma \gamma )} = \mathcal{O}((p-2)\alpha)$$

Independent of energy, i.e 4 photon emission is (1/137)*(1/137) times the probability to get 2 photon emission.

iii) OK. Now define "entity".

iv) measure momentum only means you can not use magnetic or electric field properties. The guys who want to measure dark matter might disagree since, as far as we know, dark matter does not exhibit electromagnetic properties because if they did, we might have been able to detect some form of light (X, gamma, IR, radio...) from it. Is that a good enough reason to make a detector that truly measures only momentum?

v) What is the explanation for the right angle emission?

vi) I've not yet stumbled onto any experimental evidence (publication) of confirmed multi-photon emission from SLAC's compendia or others. Do you have a paper you can share?

My 80 books are currently resting on my shelves.

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iii) Separate and distinctive existence (reads in my philosophy book)

v) Energy and momentum conservation, here is a whole thread devoted to that process, where I try to teach a guy from eastern europe how it works. https://www.physicsforums.com/showthread.php?t=226986

vi) how about google: positronium - orthopositronium (which is 3 photon decay) ? which have been know for very long.

My class mates from Germany said that they might have seen 4 photon decay in positronium, can talk with them again when I see them on tuesday.

Buckeye
iii) Separate and distinctive existence (reads in my philosophy book)

v) Energy and momentum conservation, here is a whole thread devoted to that process, where I try to teach a guy from eastern europe how it works. https://www.physicsforums.com/showthread.php?t=226986

vi) how about google: positronium - orthopositronium (which is 3 photon decay) ? which have been know for very long.

My class mates from Germany said that they might have seen 4 photon decay in positronium, can talk with them again when I see them on tuesday.

How do you justify using a book on philosophy to explain a term used in physics?

Did I miss your answer on why photons come out at 90 deg angles to the collision axis?

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Scince the ultimate definition of what a particle is depends on what representation you have. Excitation of a field, a resonance of a string, a tiny tiny tiny ball with size smaller than our current resolution, etc. The definition I gave is a summarization of these, that an elementary particle is a thing/enity that can not be divided into smaller pieces. What that entity is, depends on what representation/model you have.

See my post #16: v) "Energy and momentum conservation". You might have mathematical skills enough to perform that simple calculation. ;-)

Buckeye
Your diatribe on momentum does not deal with particles that undergo conversion of mass to energy. Another suggestion?

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why don't you give it a try yourself? ;)

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besides, there is no collision axis if the electron and positron annihilates at rest.

Also, do you have reference on your statement "photons come out at 90 deg angles to the collision axis"

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"no one has a detector that only measures Momentum"

Can you elaborate? You are just goofing.

Have you studied orthopositronium yet?

Buckeye
besides, there is no collision axis if the electron and positron annihilates at rest.

Also, do you have reference on your statement "photons come out at 90 deg angles to the collision axis"

Have not read on the 90 deg angle aspect for a long time, but I found the following:

Check out NASA simulation at:
http://svs.gsfc.nasa.gov/vis/a000000/a000100/a000182/

Brazil JP
www.scielo.br/ scielo.php?pid=S0103-97332004000500085&script=sci_arttext - 24k

SLAC
http://www.slac.stanford.edu/pubs/slacpubs/0000/slac-pub-0211.pdf PAGE 2

Buckeye
"no one has a detector that only measures Momentum"

Can you elaborate? You are just goofing.

Have you studied orthopositronium yet?

Nope, not goofing. Thinking. Why can't we detect dark matter? Maybe because it has no EM properties which means a momentum only detector might be able to detect dark matter.

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Nope, not goofing. Thinking. Why can't we detect dark matter?
We haven't been able to detect dark matter yet since it does not interact under any of the forces other than gravity.
Buckeye said:
Maybe because it has no EM properties which means a momentum only detector might be able to detect dark matter.
Ok, could you please tell us what a "momentum only" detector is? Note that you are only permitted to talk about mainstream, or published, peer-reviewed material here. If what you are suggesting has not been peer reviewed, then the only place in which you can discuss it is the Independent Research forum.

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The 90 degree optimization just means that the differential cross section for 2photon production have its peak at theta = 90deg.

Compare with the differential cross section for 2muon production in an electron-positron annihilation: (E>>m_muoun)
$$\frac{d\sigma}{d\cos \theta}(e^+e^- \rightarrow \mu ^+ \mu ^-) = \pi\alpha ^2\frac{1+\cos ^2\theta}{8E^2}$$

$$\alpha \approx 1/137$$

The derivation can be found in Mandl and Shaw : Quantum field theory p 146.

If you do a similar calculation, but with the photons, you will get another diff cross section.

Dark matter interact via gravity, and you mean that we should have a device that measure only gravity-interacting particles by inducing a EM interaction discriminator?