Ouestion About the Matter Antimatter Reaction

[JMS61]

When an electron and a positron encounter each other we get a matter antimatter reaction. If we added the rest mass of the electron and the rest mass of the positron together and then divided that number by the number of photons that the reaction generated, what would we get for an answer?

 

[zhermes]

Why don't you try it out?---wikipedia has the needed electron and positron masses, and the simplest case is that two photons are created.

 

[RocketSci5KN]

Hint: the mass of the electron and positron are the same. Just use E=Mc^2. Express
final answer in MeV or KeV. Hint2: Answer between 250KeV and 1.2MeV.

 

[JMS61]

Hint: the mass of the electron and positron are the same. Just use E=Mc^2. Express
final answer in MeV or KeV. Hint2: Answer between 250KeV and 1.2MeV.

Thank you, your patience with me is appreciated.

Is there a place on the internet where I can go to find out the wave frequency of the photons that are generated?

 

[RocketSci5KN]

E=h*c/lambda

http://en.wikipedia.org/wiki/Photon

 

[JMS61]

E=h*c/lambda

http://en.wikipedia.org/wiki/Photon

Thank you RocketSci5KN, I guess that the reaction gives off two gamma rays.

Now may I ask this question, "Is the local physics of a mass being accelerated by an outside source the same as the local physics of a mass being accelerated by an engine attached to the mass that is being accelerated?"

 

[cragar]

Now may I ask this question, "Is the local physics of a mass being accelerated by an outside source the same as the local physics of a mass being accelerated by an engine attached to the mass that is being accelerated?"

Yes isn't that the equivalence principle .

 

[K^2]

E-p annihilation usually produces 2 photons, but not always. There is a chance for 3+ photons produced, and when that happens, there is no guarantee that energies will be evenly split. However, you can never get a photon with energy greater than mass of electron/positron. (Assuming both particles were initially "at rest".)

 

[cragar]

you can produce 3 or more photons K^2 , interesting , How does conservation of spin work in that case? I'm just wondering

 

[JMS61]

Yes isn't that the equivalence principle .

Thank you cragar, and what I asked was a dumb question. I have been reading what wikipedia has to say about physics as was suggested by you guys and it is turning out that if one is not a physicist that one needs to do their home work before they as questions. It is also turning out that Experimental Physics is attempting to explore the questions that I was wondering about. One of the things that they are trying to do is to slow a photon down to see what happens when a photon drops out of light speed.

Again guys thank you for your patience and not laughing me off the message board. I do have one more question, it is about whether or not there are two different kinds of "black holes", but I think that probably belongs in astrophysics and I do have to do my homework first.

PS: Do you have any suggestions when it comes to homework on the subject of Black Hole Physics?

 

[K^2]

you can produce 3 or more photons K^2 , interesting , How does conservation of spin work in that case? I'm just wondering

Initial angular momentum isn't zero, obviously.

Basically, before annihilation, e-p pair briefly forms a bound state known as a positronium atom. Except for the fact that it can completely decay via annihilation, it behaves exactly like a hydrogen atom with reduced mass. If the annihilation happens from 1s state, only an even number of photons can be produced to conserve angular momentum. Rule of thumb in particle decay reactions is that the one with smallest number of products has the highest branching fraction. So you'd usually get 2 photons, but 4, 6, et cetera are also possible.

Of course, when the two random particles meet, their initial angular momentum is not generally zero. So when positronium forms, it is not in the 1s state. There are two things that can happen. Most common is that the positronium first decays to 1s state, emitting some low energy photons, just like hydrogen atom would. From there, above scenario takes place. Another possibility is that annihilation occurs before it decays to 1s. Say, it happens in 2p. Here, the angular momentum is one, so only an odd number of photons can be produced to conserve angular momentum. However, you can't produce just one, because that does not conserve translational momentum. So 3 is the lowest number you can get, and that's the most likely outcome. But again, 5, 7, and so on are also possible.

Notice that if we ignore angular momentum conservation, probability of 3 photon emission is much lower than probability of 2 photon emission. This means that a half-life of positronium in 2p is much higher than in 1s. And that's what usually gives it time to decay to 1s and annihilate into 2 photons.

 

[JMS61]

Initial angular momentum isn't zero, obviously.

Basically, before annihilation, e-p pair briefly forms a bound state known as a positronium atom. Except for the fact that it can completely decay via annihilation, it behaves exactly like a hydrogen atom with reduced mass. If the annihilation happens from 1s state, only an even number of photons can be produced to conserve angular momentum. Rule of thumb in particle decay reactions is that the one with smallest number of products has the highest branching fraction. So you'd usually get 2 photons, but 4, 6, et cetera are also possible.

Of course, when the two random particles meet, their initial angular momentum is not generally zero. So when positronium forms, it is not in the 1s state. There are two things that can happen. Most common is that the positronium first decays to 1s state, emitting some low energy photons, just like hydrogen atom would. From there, above scenario takes place. Another possibility is that annihilation occurs before it decays to 1s. Say, it happens in 2p. Here, the angular momentum is one, so only an odd number of photons can be produced to conserve angular momentum. However, you can't produce just one, because that does not conserve translational momentum. So 3 is the lowest number you can get, and that's the most likely outcome. But again, 5, 7, and so on are also possible.

Notice that if we ignore angular momentum conservation, probability of 3 photon emission is much lower than probability of 2 photon emission. This means that a half-life of positronium in 2p is much higher than in 1s. And that's what usually gives it time to decay to 1s and annihilate into 2 photons.

Either that or hypothetically there is such a thing as a heavy electron.

 

[K^2]

Hm? What do you mean? There are two more charged particles in the lepton family, namely the muon and tau particles, and you could think of these as heavy versions of electron, since pretty much all other properties are identical, but an e-p pair does not have enough energy to produce a muon, let alone tau. Muon, on the other hand, can and does decay into an electron and a couple of neutrinos. Muon-anti-muon reaction, therefore, can have far more possible outcomes than e-p annihilation.

 

[JMS61]

Hm? What do you mean? There are two more charged particles in the lepton family, namely the muon and tau particles, and you could think of these as heavy versions of electron, since pretty much all other properties are identical, but an e-p pair does not have enough energy to produce a muon, let alone tau. Muon, on the other hand, can and does decay into an electron and a couple of neutrinos. Muon-anti-muon reaction, therefore, can have far more possible outcomes than e-p annihilation.

Wikipedia:"leptons, comprising muons (μ−) and muon neutrinos (νμ); and the third is the tauonic leptons, comprising tauons (τ−) and tau neutrinos (ντ). Electrons have the least mass of all the charged leptons. The heavier muons and tauons will rapidly change into electrons through a process of particle decay: the transformation from a higher mass state to a lower mass state. Thus electrons are stable and the most common charged lepton in the universe, whereas muons and tauons can only be produced in high energy collisions (such as those involving cosmic rays and those occurring in particle accelerators)."

I guess that under certain conditions that an electron can take on mass and at a certain point that mass becomes a muon and is no longer an electron. And the muon becomes an unstable orderly group of particles that decay back into electrons. One then might also say that because an electron normally exists as a field, that a muon is a concentrated electron field that behaves as matter under certain conditions.

 

[K^2]

I am not aware of a single reaction where electron would become a heavier lepton. Muons and Tau are usually created in pair production reactions or in decay of mesons and baryons.

For example, a π⁺ meson may decay into a μ⁺ and ν_μ via emission of W⁺ bozon. In fact, that's almost always how it decays. That makes sense, since π⁺ has energy of 139.6 MeV, and μ⁺ is 105.7 MeV. The difference can be absorbed into kinetic energy of the muon and muon neutrino in proportion that would allow conservation of momentum.

Electrons and positrons, on the other hand, are only 512 keV. So there is not nearly enough energy to get a muon out of it. Of course, the muon wave function will still contribute due to tunneling effect and flavor oscillation, but with this energy gap, the contribution is so tiny as to be absolutely irrelevant.

 

[JMS61]

I am not aware of a single reaction where electron would become a heavier lepton. Muons and Tau are usually created in pair production reactions or in decay of mesons and baryons.

For example, a π⁺ meson may decay into a μ⁺ and ν_μ via emission of W⁺ bozon. In fact, that's almost always how it decays. That makes sense, since π⁺ has energy of 139.6 MeV, and μ⁺ is 105.7 MeV. The difference can be absorbed into kinetic energy of the muon and muon neutrino in proportion that would allow conservation of momentum.

Electrons and positrons, on the other hand, are only 512 keV. So there is not nearly enough energy to get a muon out of it. Of course, the muon wave function will still contribute due to tunneling effect and flavor oscillation, but with this energy gap, the contribution is so tiny as to be absolutely irrelevant.

Hypothetically if it was possible under certain circumstances to condense an electron field (cause it to take on mass), then there should be an example of it. A pion would be an example of it. The problem is that a pion is made up of quarks as is a proton and neutron. The pion eventually decays into versions of an electron and photon and a proton and a neutron does not decay into versions of an electron. Which would seem to indicate that not all quarks are the same, some are composed of condensed electron fields and some quarks are not, unless all mesons eventually decay into some version of an electron.

 

[K^2]

A neutron does decay into proton, electron, and a neutrino. A positron can decay into a neutron and a positron if the conditions are favorable for such decay. (E.g. in a heavy nucleus, resulting in beta-bar decay, or in a neutron star.)

But yes, there is a difference between quarks in the baryon (protons, neutrons, ...) and in the meson (pions, rho mesons, ...) The difference is color charge. Both baryons and mesons are color "neutral", but while the color charge of meson is completely canceled, the color charge of baryons is merely balanced.

Unlike electric charges, which there are just two of, there are 6 possible color charges. There are 3 colors and 3 anti-colors. Red, green, blue, anti-red, anti-green, anti-blue. The names have nothing to do with visible spectrum, but rather with concept of color neutrality.

Every quark has a color. Every anti-quark has an anti-color. Color is a conserved quantity. So when you create a red charge, you must also create an anti-red charge. Gluons, which are carriers of strong nuclear force carry a color and an anti-color. And that's what results in color-neutrality.

Suppose a red quark emits a gluon. That gluon must carry a color, and the only available option is red. So it takes a red charge away from the gluon. It also needs an anti-color, and the two available options are anti-green or anti-blue. If it takes anti-green, green charge must also be created and deposited on the quark. It becomes green. If it takes anti-blue, quark becomes blue.

Now, what quark can absorb this gluon? Since it caries red, it can be absorbed by anti-red. In that case, the color of anti-red gluon changes to anti- of whatever the emitting gluon has become. So red-anti-red becomes blue-anti-blue or green-anti-green. The other option is to have the same color that the emitting quark. So red-blue would become blue-red and red-green would become green-red.

It is the nature of strong nuclear interaction that if the possible emitted gluons don't have anything to interact with, it will result in pair-production for the interaction to occur. So you cannot have an isolated quark. Furthermore, you cannot have a group of quarks that are not color-neutral. If you have a red quark, you must either have an anti-red to go with it or both blue and green quarks.

Particles consisting of a quark with color and quark with anti-color are called mesons. Because they contain a quark and an anti-quark they can spontaneously annihilate, sometimes, producing electrons or positrons as a byproduct.

Particles consisting of three quarks, one of each color, are called baryons. Particles consisting of three anti-colors are their anti-particles. A baryon does not contain anti-quarks. It cannot annihilate spontaneously without violating color conservation. It must meet a particle containing 3 anti-colors, which would have to be an anti-particle baryon. It does not have to be its own anti-particle, however. A neutron will annihilate with an anti-proton just fine, resulting in production of an electron. This process is a little bit more complicated, however, since it involves weak interactions.

 

[JMS61]

A neutron does decay into proton, electron, and a neutrino. A positron can decay into a neutron and a positron if the conditions are favorable for such decay. (E.g. in a heavy nucleus, resulting in beta-bar decay, or in a neutron star.)

But yes, there is a difference between quarks in the baryon (protons, neutrons, ...) and in the meson (pions, rho mesons, ...) The difference is color charge. Both baryons and mesons are color "neutral", but while the color charge of meson is completely canceled, the color charge of baryons is merely balanced.

Unlike electric charges, which there are just two of, there are 6 possible color charges. There are 3 colors and 3 anti-colors. Red, green, blue, anti-red, anti-green, anti-blue. The names have nothing to do with visible spectrum, but rather with concept of color neutrality.

Every quark has a color. Every anti-quark has an anti-color. Color is a conserved quantity. So when you create a red charge, you must also create an anti-red charge. Gluons, which are carriers of strong nuclear force carry a color and an anti-color. And that's what results in color-neutrality.

Suppose a red quark emits a gluon. That gluon must carry a color, and the only available option is red. So it takes a red charge away from the gluon. It also needs an anti-color, and the two available options are anti-green or anti-blue. If it takes anti-green, green charge must also be created and deposited on the quark. It becomes green. If it takes anti-blue, quark becomes blue.

Now, what quark can absorb this gluon? Since it caries red, it can be absorbed by anti-red. In that case, the color of anti-red gluon changes to anti- of whatever the emitting gluon has become. So red-anti-red becomes blue-anti-blue or green-anti-green. The other option is to have the same color that the emitting quark. So red-blue would become blue-red and red-green would become green-red.

It is the nature of strong nuclear interaction that if the possible emitted gluons don't have anything to interact with, it will result in pair-production for the interaction to occur. So you cannot have an isolated quark. Furthermore, you cannot have a group of quarks that are not color-neutral. If you have a red quark, you must either have an anti-red to go with it or both blue and green quarks.

Particles consisting of a quark with color and quark with anti-color are called mesons. Because they contain a quark and an anti-quark they can spontaneously annihilate, sometimes, producing electrons or positrons as a byproduct.

Particles consisting of three quarks, one of each color, are called baryons. Particles consisting of three anti-colors are their anti-particles. A baryon does not contain anti-quarks. It cannot annihilate spontaneously without violating color conservation. It must meet a particle containing 3 anti-colors, which would have to be an anti-particle baryon. It does not have to be its own anti-particle, however. A neutron will annihilate with an anti-proton just fine, resulting in production of an electron. This process is a little bit more complicated, however, since it involves weak interactions.

Ok, so color creates stability. Thank you! I have found a website called "Hyper Physics". I am going to have study it before I can continue to ask intelligent questions. But because I have studied it a bit so far, I actually understand what you are saying.

Thank you again, you have given me an excitement for physics, which I have always loved.