Matter and antimatter must interchangible

[Gamish]

If I have 1 KG of matter, and 1 KG of antimatter, and I combine the 2, then I get 2c^2 = 1.79751036 × 10^17 joules, then in turn, I convert the 1.79751036 × 10^17 joules back to 2 KG of mass. So, if antimatter is so anti, why can they be converted? :uhh:

 

[dextercioby]

If I have 1 KG of matter, and 1 KG of antimatter, and I combine the 2, then I get 2c^2 = 1.79751036 × 10^17 joules, then in turn, I convert the 1.79751036 × 10^17 joules back to 2 KG of mass. So, if antimatter is so anti, why can they be converted? :uhh:

If u knew a little bit more physics,u'd know that antiparticles don't differ from particles by anything,except electric charge.They have the same mass,the same spin,the same energy.Since mass is equivalent to energy,then pure energy (the 2 gamma photons) can be converted to mass (the particle and its antiparticle) and viceversa,mass (particles with rest mass) go into pure energy (gamma photons).
Actually the problem is a bit more tricky or subtle.The two photons should be one photon and one antiphoton,but since photons are chargeless (by their nature),the photon and the antiphoton are identical.

That's one great thing about relativity.Prediction of antiparticles.

Daniel.

 

[chroot]

When you convert the radiation back to matter, you will not produce just matter. You will, in fact, produce exactly one particle of antimatter for each particle of matter. There is no other way to conserve, for example, electric charge. Photons have no charge, but electrons do; you therefore cannot convert a photon into an electron without breaking the known laws of physics. You can, however, convert a photon into a electron and a positron, conserving the net zero charge.

- Warren

 

[Gamish]

Well, I'm sorry that I am not the best at physics, I have not even taken a class on physics, but I try to learn as much as posible.

And yes, I do know that the positron and the electron (same for all particles) have the exact same properties. But we still refer to them as anti-particles. If we combine matter with antimatter, why do they anniahlate, if they are the exact same thing?

 

[marlon]

Well, I'm sorry that I am not the best at physics, I have not even taken a class on physics, but I try to learn as much as posible.

And yes, I do know that the postrian and the electron (same for all particles) have the exact same properties. But we still refer to them as anti-particles. If we combine matter with antimatter, why do they anniahlate, if they are the exact same thing?

The "annihilate" because the interaction of a positron with some electron needs to respect certain conservation laws like charge, just like chroot mentionned. The reason why this is the case can be proven by QED...An electron and a positron have OPPOSITE charge so the net charge must be ZERO. For example a photon has zero charge and that is why an electron and a positron will be "transformed" into a photon. The energy of this photon is determined by other conservation laws like mass and energy...
This photon is the radiation in which these two particles are converted...

regards
marlon

 

[DB]

A couple of problems with what your doing and asking. Your are right to use E=mc^2 though I think you are multiplying Kg times m/s, instead of Kg tmes km/s. Also doing what you are doing will keep your running circles, which is pretty much the basis of anti-matter. As dextercioby said, anti-matter is just difference in charge of a particle; an electron being
$$e-$$,
its anti-matter particle is shown $$\bar{e}$$ a positron.
Same with any particle:

$$\bar{p},\bar\mu,\bar\tau$$

Particles and their anti particles combined to form pure energy, which happens often at microscopic levels. This can happen in just empty space with an energy fluctuation, over and over again. So your question is somewhat meaningless.

 

[DB]

P.S, sorry if I seem repetitive, by the time I had posted there were 3 more replies...!

 

[dextercioby]

The reason why this is the case can be proven by QED...An electron and a positron have OPPOSITE charge so the net charge must be ZERO.For example a photon has zero charge and that is why an electron and a positron will be "transformed" into a photon. The energy of this photon is determined by other conservation laws like mass and energy...
This photon is the radiation in which these two particles are converted...

You can, however, convert a photon into a electron and a positron, conserving the net zero charge.

I'm sorry,guys,BUT WRONG,WRONG!

I'm surprised at Marlon,i didn't expect that answer,especially having mentioned QED.

The Feynman diagram in QED is in the second order of pertubation theory and is topologically equivalent to the one describing the Compton effect ("switch propagators and change the direction of time).It has one electron,one positron and 2 PHOTONS,NOT ONE.I don't need QFT to prove to you guys that for free scattering electron and positron,u must have either the second photon (as i said,it should be an antiphoton,but massless+gauge invariance dictate zero electric charge for the photon,therefore photon=antiphoton),or another particle (boson for spin conservation) for having conserving 4 momentum in the scattering.This is 'classical' stuff,how could u f*** it up...? :yuck:

Daniel.

 

[chroot]

I'm aware that one photon cannot by itself turn into a pair as that reaction cannot conserve momentum; however, it wasn't really relevant to the discussion. In the presence of matter, however -- the usual case -- a single photon can turn into a pair.

- Warren

 

[Gamish]

A couple of problems with what your doing and asking. Your are right to use E=mc^2 though I think you are multiplying Kg times m/s, instead of Kg tmes km/s. Also doing what you are doing will keep your running circles, which is pretty much the basis of anti-matter. As dextercioby said, anti-matter is just difference in charge of a particle; an electron being
$$e-$$,
its anti-matter particle is shown $$\bar{e}$$ a positron.
Same with any particle:

$$\bar{p},\bar\mu,\bar\tau$$

Particles and their anti particles combined to form pure energy, which happens often at microscopic levels. This can happen in just empty space with an energy fluctuation, over and over again. So your question is somewhat meaningless.

1.I did use m/s in my calculations. I think you should re-check you math.

So your question is somewhat meaningless.

What is meaningless about it, read the question I asked.

Let me as my question in a different way.

If I had some antimatter (-e) , can I convert it to energy, the convert it to positive energy, the convert that energy to positive matter (+e)?

 

[chroot]

If I had some antimatter (-e) , can I convert it to energy, the convert it to positive energy, the convert that energy to positive matter (+e)?

You cannot convert antimatter by itself directly into energy (radiation) because doing so would violate the conservation of electric charge. You cannot directly turn energy (radiation) into only normal matter, again, because that would violate the conservation of electric charge. The only kinds of reactions that are permissible are those that destroy both a particle of matter and a particle of antimatter at the same time, or those that create a particle of matter and a particle of antimatter at the same time. What you propose is not possible, and has already been explained by several people in this thread. Please try to read and fully understand the responses you have been given before continuing to argue.

- Warren

 

[Astronuc]

What form of matter/anti-matter would be important, e.g. molecules of hydrogen (e^-, p) and anti-hydrogen (e⁺, p^-). The presence of anti-neutrons would be interesting.

The positron/electron pairs would annhiliate yielding 2 gamma-rays of approximately energy of 0.511 MeV or the rest mass of the electron/positron). The gamma-rays would scatter off elctrons (Compton scattering) and loose energy, and would not be able to reform electons or positrons. Electrons and positrons are formed in pairs (pair production) by the interaction of gamma-rays with a nucleus of an atom, but this requires a threshold of ~1.022 MeV or twice the electron rest mass.

The proton-antiproton annihilation would result in a whole lot of mesons, then these would decay into muons, electrons, neutrinos and more gammas, and I imagine that some the pions would interact with protons form K-mesons and resonance particles.

I just don't want to be nearby when someone annihilates 1 kg of antimatter.

Also, it is not said how the matter and anti-matter are brought together - instanteously, at rest or collided, solid, liquid or gaseous.

 

[DB]

1.I did use m/s in my calculations. I think you should re-check you math.

I know you used m/s in your calculations, I said that you would have had to convert into km/s because your anti/matter was being measured in kg. Though it's irrelevant now.

 

[DB]

What is meaningless about it, read the question I asked.

Let me as my question in a different way.

If I had some antimatter (-e) , can I convert it to energy, the convert it to positive energy, the convert that energy to positive matter (+e)?

Either way, Astronuc explains it all very well.

 

[chroot]

I know you used m/s in your calculations, I said that you would have had to convert into km/s because your anti/matter was being measured in kg. Though it's irrelevant now.

Actually, the fundamental SI unit of mass is the kilogram, not the gram.

- Warren

 

[marlon]

I'm sorry,guys,BUT WRONG,WRONG!

I'm surprised at Marlon,i didn't expect that answer,especially having mentioned QED.

The Feynman diagram in QED is in the second order of pertubation theory and is topologically equivalent to the one describing the Compton effect ("switch propagators and change the direction of time).It has one electron,one positron and 2 PHOTONS,NOT ONE.I don't need QFT to prove to you guys that for free scattering electron and positron,u must have either the second photon (as i said,it should be an antiphoton,but massless+gauge invariance dictate zero electric charge for the photon,therefore photon=antiphoton),or another particle (boson for spin conservation) for having conserving 4 momentum in the scattering.This is 'classical' stuff,how could u f*** it up...? :yuck:

Daniel.

Don't get so agitated man...What you are saying is true and both Chroot and me are well aware of that...check out previous posts if you will ...We were just making a sketch of the situation at hand by giving examples of why conservation laws are used. What you are saying isn't relevant to this discussion because we were not describing a complete annihilation reaction of a positron and an electron. Again we were exemplifying...


regards
marlon

 

[Gamish]

What form of matter/anti-matter would be important, e.g. molecules of hydrogen (e^-, p) and anti-hydrogen (e⁺, p^-). The presence of anti-neutrons would be interesting.

The positron/electron pairs would annhiliate yielding 2 gamma-rays of approximately energy of 0.511 MeV or the rest mass of the electron/positron). The gamma-rays would scatter off elctrons (Compton scattering) and loose energy, and would not be able to reform electons or positrons. Electrons and positrons are formed in pairs (pair production) by the interaction of gamma-rays with a nucleus of an atom, but this requires a threshold of ~1.022 MeV or twice the electron rest mass.

The proton-antiproton annihilation would result in a whole lot of mesons, then these would decay into muons, electrons, neutrinos and more gammas, and I imagine that some the pions would interact with protons form K-mesons and resonance particles.

I just don't want to be nearby when someone annihilates 1 kg of antimatter.

Also, it is not said how the matter and anti-matter are brought together - instanteously, at rest or collided, solid, liquid or gaseous.

OK, so, let me get this straight. If I annihilate 1 positron, and 1 electron, it will yield 1 anti-gamma-ray, and 1 gamma-ray, then, if we were to somehow convert this energy into matter again, it would yield 1 positron, and 1 electron?

 

[selfAdjoint]

OK, so, let me get this straight. If I annihilate 1 positron, and 1 electron, it will yield 1 anti-gamma-ray, and 1 gamma-ray, then, if we were to somehow convert this energy into matter again, it would yield 1 positron, and 1 electron?

The photon is its own anti-particle, so you only produce the one gamma ray. This ray will carry all the energy and momentum of the original particle and anti-particle.

 

[Gamish]

The photon is its own anti-particle, so you only produce the one gamma ray. This ray will carry all the energy and momentum of the original particle and anti-particle.

So, is it posible to convert this ray of energy back to either/and/or matter/antimatter?

 

[dextercioby]

So, is it posible to convert this ray of energy back to either/and/or matter/antimatter?

At elementary quantum level,the laws of physics are CPT invariant,and,in this case,and if I'm not mistaking in all cases,T invariant.So the answer is 'yes'.

Daniel.

PS.Sidenotes:
a)T-invariant:means the equations and all physically measurable quantities have the same form if:
$$t\rightarrow -t$$
b)CPT invariant:the...remain invariant if SIMULTANEOUSLY the following transformations are perfomed:
$$q\rightarrow -q;\vec{r}\rightarrow -\vec{r};t\rightarrow -t$$

 

[Gamish]

There is one thing I want to get clear, if I turn antimatter into energy, can it, in theory, be converted to regular matter? Or, when the conversion takes place, it automaticly becomes antimatter.

 

[dextercioby]

There is one thing I want to get clear, if I turn antimatter into energy, can it, in theory, be converted to regular matter? Or, when the conversion takes place, it automaticly becomes antimatter.

If 'there's one thing i want to get clear',is this one:Antimatter does not automatically get into energy.It is stable.It needs 'ordinary matter' to do that...Antimatter does never goes into matter.And viceversa.
Example:
A positron is a very stable particle and interacts with every other particle,just like the 'old-fashioned' electron does.If,accidentally,among those particles he interacts with,is one antipositron,vis.an electron,it will annihilate with it resulting 2 photons (i'm assuming 'free' interaction,in 'vacuum',no 'matter' in the surroundings).The same thing can be said about electrons as well.
An electron cannot be converted into a positron.The simplest QED process that might,in principle,allow such thing is the Compton effect.However,conservation of charge and lepton number won't allow it.

Daniel.
Daniel.

 

[DB]

I have a question. I pretty much understand anti-matter and all, but considering that in matter/anti-matter annihilations the rest mass of each particle is entirely converted into energy, and that this energy can then turn into matter and other particles, is the reason electrons are e- and protons are p+ in our universe today a coin flip?

 

[Gokul43201]

Not a coin flip but a simple matter of convention. If we called all positive particles 'negative' and negative particles 'positive', what difference would that make ? It's just a question of changing the names.

 

[Gamish]

Ok, I will re-word it again. If I have 1 electron and 1 positron, and they combine anbd annihilate, and somehow, they turn back into matter again, will I have 1 electron and 1 positron again, or is it posible that I can get 2 electrons, or 2 positrons?

 

[houserichichi]

No.

If you have an electron (charge = -1) and a positron (charge = +1), then their net charge is -1 + 1 = 0. When they annihilate they create a photon (which has charge of 0). So far so good. The reason the photon (two of them actually, but that's a moot point in this discussion) cannot be converted back into two electrons or two positrons is because the charge would be different. The charge of two electrons is (-1) + (-1) = -2 while the charge of two positrons is 1 + 1 = 2. You can't make charge magically appear from those photons so it can't happen.

To restate - charge of electron/positron together is 0. Charge of photons is 0. Charge of two electrons or two positrons is NOT 0 (not the same) so it can't happen.

Now my own question to the forum. Can an electron/positron annihilation that results in two photons come back and emit (is this the proper verb?) a muon and anti-muon or is this a violation of electron number conservation? If so I'm not entirely sure i understand the exact logic behind these kinds of conservation laws other than "they must be so". Any insight would be helpful. Thanks.

 

[Gamish]

Thanks houserichichi, and everyone else. So, if I am correct, when a e- and e+ annihilate, they become 2 photons with 0 charge, and it would be imposible to turn the, into matter again? There is no way at all, because it would violate conservation of energy laws?

As for you question houserichichi , it is believed that every entity of the universe has it's respective counterpart. We are not aware of all of them, such as the counterpart to space, but it may indeed be posible that a anti-muon exist, but I am not an expert, so I also would like to hear another answer.

 

[houserichichi]

Actually two photons can produce an electron/positron pair again since the charge is conserved. That is, charge of 0 goes in (electron/positron), charge of 0 is in the two photons, and charge of 0 comes out (electron/positron). Now I'm not sure how long that would last since wouldn't the positron and electron be attracted back to each other and annihilate again?

As for the antimuon it's a "real live" particle. The muon charge is -1 and the anti-muon charge is +1 (it's a muon with positive charge)...but now it makes sense since the electron number isn't conserved (which is why my previous query can't happen). Is electron number equivalent to electric charge in my previous argument or at least analogous?

I'd like to know how it is the two photons "know" to change back into an electron/positron pair rather than a muon/anti-muon pair...what physical interpretation does electron/muon/tau number have? I REALLY hope that's not a philosophical question and if it is please disregard.

 

[Astronuc]

Regarding positron-electron annihilation near rest, two gamma-rays are produced, each propagating in the opposite direction. Both photons will have an energy of approximately 0.511 MeV, which is equivalent to the rest mass of an electron. The gamma rays will more than likely scatter and degrade to lower energies in the EM spectrum. They most likely not form additional particles.

Pair production requires a gamma-ray of at least 1.022 MeV interacting with a nucleus (nuclear field) or an electron ( Electron-positron pair production by ultrarelativistic electrons in a soft photon field ), and presumably other subatomic particles. Thus the rest mass of a particle can be thought of as an energy threshold to its production.

Muons have rest mass of approximately 105.66 MeV (Emilio Segré, Nuclei and Particles) and so a 0.511 MeV gamma is not going to contribute to the formation of a muon.

Muons are normally found as decay products of higher mass particles such as pions and hadronic particles such as baryons like $\Sigma\, or\, \Lambda$particles, as well as K-mesons.

However, muons (and heavier particles) can also be created by positron-electron collisions, provided the total energy is sufficient. This would occur in an electron synchrotron for instance.

See http://www.desy.de/html/home/ (DESY home, aber auf Deutsch) or
http://www.desy.de/html/home/fastnavigator.html (English)

 

[chroot]

Is electron number equivalent to electric charge in my previous argument or at least analogous?

Lepton number (and electron number) is conserved all the way through. You start with an electron and a positron (net lepton number = 0, net electron number = 0) and create some photons. If the electron and positron had a large enough relative velocity when they collided, the photons could be energetic enough to create a muon-antimuon pair. The lepton number and electron number of that pair is, again, zero. The two numbers are conserved all the way through.

I'd like to know how it is the two photons "know" to change back into an electron/positron pair rather than a muon/anti-muon pair...what physical interpretation does electron/muon/tau number have? I REALLY hope that's not a philosophical question and if it is please disregard.

They don't know specifically to do anything. If you smash a gamma ray into a piece of metal, you can create a wide variety of pairs (e+, e- or u+, u- or p+, p-, etc.). Each has its own statistical probability of being created, but all valid reactions are possible, given enough energy.

- Warren

 

[houserichichi]

Chroot, thanks for the explanation. I had convinced myself muon-antimuon couldn't be produced but you've shed the light right on it for me! Thanks for the clarification folks!

 

[Invisible]

Does this sort of have something to do with a "Pair's Production"? I'm still in Physics 30, and we sort of learned something about this. I don't have my notes anywhere with me, but I'll explain what I remember by memory.

Annihilation: When matter and antimatter collide, matter is converted into energy. And then the next part, I seem to agree with Daniel about what he says that momentum is conserved.

Okay, say I have a high energy photon (ie. Gamma photon) and it strikes against a hard surface. Yes, both an electron and positron are created.

The Law Conservation of Charge in this case is that the existence of the negatively charged particle (an electron) from the collision of the gamma photon meant that there had to be another particle which was positively charged.

Law Conservation of Momentum (in this particle example):

- When matter and antimatter (electron and positron) collide, the total momentum before is zero.
- Yes, momentum has to be conserved, but in order for that to occur, two photons (agreeing with Daniel) must be produced which move in opposite directions, therefore, the total momentum of the photons is also zero!

Actually, I don't know if what I said has anything to do with it mainly because I'm still in high school... :-S

 

[houserichichi]

Sorry, this is back to something Chroot said:

If you smash a gamma ray into a piece of metal, you can create a wide variety of pairs (e+, e- or u+, u- or p+, p-, etc.). Each has its own statistical probability of being created, but all valid reactions are possible, given enough energy

So just to clarify, if we use a high enough energy in our experiment we could, in theory, create an e+/e- OR mu+/mu- OR tau+/tau- pair from two photons? If so could we ever be certain of the outcome or do we base our theory on the expected value of a probability, that is - we don't know for sure that an e+/e- pair will be created on each attempt if we so wish.

 

[shrumeo]

Howdy,

I'd like to know what happened at the beginning of the Universe then.
I'm under the impression that there was whole lots of matter and anti-matter at the time of the big bang, but there was a "slight" excess of what we call matter.

If there is such a thing as conservation of charge, then why was there an imbalance in the first place, and why can't we restore some of the imbalance?

 

[FZ+]

I'm under the impression that there was whole lots of matter and anti-matter at the time of the big bang, but there was a "slight" excess of what we call matter.

I don't think that is the general consensus. The general consensus - right now - is that there were equal amounts of antimatter and matter, but that symmetry is broken between the two - ie. antimatter is not exactly the same as matter, and for some reason, in our universe, decays a little more quickly. This is what leads to the imbalance. This, if I remember correctly, is borne out in a number of experiments which have illustrated broken symmetry, though results are not yet conclusive, since the effects observed do not fully account for the size of the imbalance.

Also, antimatter/matter imbalance is different from charge conservation. 1 proton + 1 antiproton -> 1 proton + 1 electron conserves charge, but still breaks the standard model.