Anti-matter Creation Questions/ Discussion

In summary: I'm looking for. Thank you in advance!In summary, the expert summarizer asks a few questions about antimatter, beta decay, and how to explain reverse-beta decay. He is also looking for an explanation for a quantum effect that at high temperatures allows for the creation of Sphalerons.
  • #1
Matt-er
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Hello everyone!

I would first like to say that I am not exactly unfamiliar with these forums; I'm one of those people who is like a... "long -time thread watcher, first time poster." So yes, this is my first post. :smile: I have used these forums as a sort of reference library, looking up cool facts when I wanted to, and occasionally if I had questions of my own, I would usually find someone asking a similar question, with answers attached on the posts following. But this time, it's different (sounds like a movie, haha)... (also I hope this is the right section; I was unsure if maybe this should go in the quantum section. Thanks.)

Recently, I have been doing some research on antimatter, specifically about baryon-number violation (B + L), and no matter (no pun intended) where I looked, I only got answers that would confuse me further. So, I was hoping to clear up some general questions and most-likely some misconceptions I might have regarding antimatter as well as clarify exactly what's going on in different circumstances of antimatter creation. I would also like to discuss some of these questions in more detail, if and once they are answered, and perhaps turn this thread into a discussion (I hope that's not against the rules, as a discussion sometimes tends to go off-topic). Okay. Wow, that's a mouthful... anyway, so without further ado, here are my questions:
__________________

-1. Does antimatter and matter have to be both "free-floating" in order to annihilate? More specifically, if we shoot a bunch (say 30) of anti-protons into a nucleus of normal matter which contains a certain amount of protons (say 45 protons in the nucleus), will annihilation happen, and will there be 15 protons in the nucleus left? (after annihilation occurs; leave beta decay out of this specific question) OR is there something that prevents annihilation from happening inside of a nucleus? I think I know the answer, but I just want to be sure...​



-2. I have read that the only way that antimatter can convert into normal matter (or vice-versa) is only under a specific circumstance. However, this circumstance (which deals with a Sphaleron) is extremely rare and is technically still theoretical as it falls under one of the current UToE (unified theory of everything). I have a general question about this circumstance later, but here's my current question:
How does one then explain reverse-beta decay? I mean yes, I understand that there is clearly a different quantum effect going on here, but which one?
If you look at it from a general perspective, you have a nucleus actually take and gather energy from itself in order to transform the proton (lighter than a neutron) into a neutron and then also in order to conserve the charge of the proton, this energy creates an entirely new particle (the positron), right? But how can that happen? The positron is an anti-particle, in fact, it's an anti-lepton. Where did this anti- lepton come from? Sure, one can say that the nucleus demanded a neutron in order to be stable, and to conserve charge, it just created this particle, which by chance happens to be an antiparticle. But, energy doesn't just transform into matter/ antimatter instantly- this isn't Cinemax :biggrin:. There has to be some process that actually explains the formation of this antimatter from this energy. Is there some kind of quantum tunneling effect or something that I'm missing here?



-3. Stemming off from #2, the creation of an anti-leptons due to beta-decay, suggests that it might be possible to transform baryons into anti-leptons or anti-baryons into leptons, which is a theoretical concept of the GUT. Let's assume that it is the case that the GUT is true. Okay, but even in that case, the transformation of antimatter to matter (or vice-versa) only happens under the condition of (this is how I understand it) a saddle point of SU(2), which describes the motion of particles in Euclidean space time. And the chances of this Sphaleron appearing is virtually zero.
However, the temperatures of the early universe would have caused the rate of Sphaleron occurrence to be much more often due to (from what I have read) a different quantum effect. So, question is: What is this quantum effect that at very high temperatures allows the larger rate of creation of Sphalerons to occur? And what kind of high temperatures are we referring to (number-wise)?



-4. I'll be honest, I'm beyond confused with anything dealing with saddle points and SU(2) or SU(anything), haha. I don't know if I should start a new thread about this, but honestly, I would be grateful if anyone could explain exactly what this means and how one comes to such a conclusion of the possibility of a Sphaleron. To me, it looks like a mathematical description using matrices to describe particle movement and behavior, but how and in what way, I'm not exactly sure. Truth is, I'm not looking to get a semester- college course worth answer, just anything you can do to shed some light on this subject would be useful. Thanks.​
__________________

Okay. Sorry for dumping all that on you guys. I guess I've been so confused and filled with so many questions, that I just wanted to know immediately. I know it doesn't work that way, so please just do your best to answer any of these questions; you don't have to answer all of them- even a little piece of the puzzle is a big help. Who knows, maybe I'm not the only one with these questions, maybe others like me will benefit from this thread, so again, take your time and answer/ discuss away.

All questions/ comments/ critiques/ corrections are all welcome! But, please no flaming, name calling, other bad behavior. And truth is, I'm still in college; I haven't taken enough physics/ math courses to not be called "stupid," so go easy on me. :biggrin:

Thank you all; Awaiting your responses,
- Matt-er
 
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  • #2
1. Depending on the accuracy of the shooting, say you got all the protons, you could easily cause other types of creations/decays to take place. All you know for sure is that the left over particles plus their kinetic energy has to add up to the total energy you started with.

2/3. Beta decay, where a quark in a neutron (down-quark through a W boson), changes into an up-quark, making the neutron a proton, and off flies an electron and an anti-neutrino. This process is reversible in the sense that an electron with enough energy can cause an up to down-quark change, effectively turning a proton into a neutron.

4. There are a series of great lectures where you learn a great deal about using matrices as a mathematical description to describe particle movement and behavior (specifically polarization of photons and electrons) at

http://www.youtube.com/watch?v=pBh7Xqbh5JQ
 
  • #3
1. Annihilations release a lot of energy - I guess the first annihilation(s) would rip apart your nucleus long before the remaining antiprotons arrived. Yes, annihilations do happen in nuclei - and they look nearly the same as with free-floating particles.

2. Sphalerons are hypothetical objects, but within the Standard Model - no new unified theories are necessary. They have nothing to do with beta decay, independent of the direction.
Inverse beta decay is like the regular beta decay, just happening in the other direction.
There has to be some process that actually explains the formation of this antimatter from this energy
There is no (known) deeper reason for the interaction between different fields - it just happens, and the result are particles in this case. They are created in the process, and there is no process that does not produce them.
Note that beta decays conserve both baryon and lepton numbers separately.

What is this quantum effect that at very high temperatures allows the larger rate of creation of Sphalerons to occur?
High energy just means that high-energetic processes are more likely to occur.
 
  • #4
Matt-er said:
If you look at it from a general perspective, you have a nucleus actually take and gather energy from itself in order to transform the proton (lighter than a neutron) into a neutron and then also in order to conserve the charge of the proton, this energy creates an entirely new particle (the positron), right? But how can that happen? The positron is an anti-particle, in fact, it's an anti-lepton. Where did this anti- lepton come from? Sure, one can say that the nucleus demanded a neutron in order to be stable, and to conserve charge, it just created this particle, which by chance happens to be an antiparticle. But, energy doesn't just transform into matter/ antimatter instantly- this isn't Cinemax :biggrin:. There has to be some process that actually explains the formation of this antimatter from this energy. Is there some kind of quantum tunneling effect or something that I'm missing here?
In the first place, energy does not exist all by itself, it's a property of particles, including both kinetic energy and rest energy. In beta decay or any other interaction it's not a question of energy transforming into matter, rather you have one set of particles transforming into another set. Energy is conserved in the process, and some of the rest energy may transform into kinetic energy or vice versa, as required.

But there is no underlying process that "explains" it. And yes, in QFT it does happen instantaneously and at a single point. [Should point out that beta decay itself is actually a two-step process, in which a W meson is produced as an intermediate stage, which then decays.]

Where do the new particles come from? Particles are not fundamental either. After all, it is called Quantum FIELD Theory, and particles are just excitations of the fields. There is one electron field, for example, that exists always and everywhere. Normally it is in the vacuum state, with no particles (electrons and positrons) present. When an interaction "creates an electron", what it does is excite the field (that already existed) to a one-particle state.
 
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  • #5
Hello all! Thank you for replying.

It means a lot that someone takes the time to answer back or post/ reply, and I will do my best to "carry on the conversation." However, some of the answers that you have provided were not the ones I was looking for (mostly for #1); And, I think it has to do with the way my questions were formulated. Usually I like to add detail to my questions, but sometimes my lack of grammar (English is not my first language) mutates them into a giant mess. My apologies. Let me try to rephrase, then attend to the rest of your replies. Thanks.
__________

Regarding the first question- it was really a simple question of: Will sub-atomic particles annihilate if they are part of a nucleus? Ie: Can the annihilation of antiprotons (which are free-floating and Not part of a nucleus) and protons occur if the protons are part of a nucleus? (again; I think yes, but I might be unaware of some kind of effect that prevents this; that's all I want to know for this question).
__________

Now on to your replies:

edguy99 said:
4. There are a series of great lectures where you learn a great deal about using matrices as a mathematical description to describe particle movement and behavior (specifically polarization of photons and electrons) at



This is actually a tremendous help! Thank you. I will definitely look more into it later (haven't had much time to watch it, but it seems quite helpful so far; so thanks again).

mfb said:
2. Sphalerons are hypothetical objects, but within the Standard Model - no new unified theories are necessary. They have nothing to do with beta decay, independent of the direction.
Inverse beta decay is like the regular beta decay, just happening in the other direction.

Yes, you're right. Thank you for that correction- I never meant to say that Sphalerons were part of the beta-decay process; I was only wondering if there is a possible connection of baryon-number conservation violation of B+L and the event of reverse beta decay. And now that I think about it, I cannot see any real way that there could be, otherwise we would have detected other particles during a reverse beta decay event (and I mean other particles other than the W- boson).
Also I never suggested that a new unified ToE should be needed: I was merely saying that the idea of B+L violation was Part of a Current ToE.


mfb said:
There is no (known) deeper reason for the interaction between different fields - it just happens, and the result are particles in this case. They are created in the process, and there is no process that does not produce them.
Note that beta decays conserve both baryon and lepton numbers separately.

Bill_K said:
But there is no underlying process that "explains" it. And yes, in QFT it does happen instantaneously and at a single point. [Should point out that beta decay itself is actually a two-step process, in which a W meson is produced as an intermediate stage, which then decays.]

Where do the new particles come from? Particles are not fundamental either. After all, it is called Quantum FIELD Theory, and particles are just excitations of the fields. There is one electron field, for example, that exists always and everywhere. Normally it is in the vacuum state, with no particles (electrons and positrons) present. When an interaction "creates an electron", what it does is excite the field (that already existed) to a one-particle state.

I didn't realize it, but you know what? That makes sense, haha. I always thought that the particles give rise to the fields, not the other way around- learned it the opposite way in high school; shows what high school teaches us (nothing, haha). But yes, particle creation due to field excitation seems to make more sense now that I think about it. Thank you for this clarification; I will definitely have to dwell on this new knowledge to see what or if anything later "clicks" in my mind. Thanks again.

Bill_K said:
In the first place, energy does not exist all by itself, it's a property of particles, including both kinetic energy and rest energy. In beta decay or any other interaction it's not a question of energy transforming into matter, rather you have one set of particles transforming into another set. Energy is conserved in the process, and some of the rest energy may transform into kinetic energy or vice versa, as required.

Yes, I meant this- I meant the energy Of the Nucleus (in the Form of mass, etc) is used to help convert one of the protons into the other particles (during reverse beta decay). I probably tried to shove to much detail too quickly, and made it sound like some kind of crazy definition/ process. Thank you for this correction.

mfb said:
What is this quantum effect that at very high temperatures allows the larger rate of creation of Sphalerons to occur?
High energy just means that high-energetic processes are more likely to occur.

Well this is obvious, but I was referring to a special quantum effect that I read in an online physics paper regarding this issue. I can't remember where it was or which effect it was talking about... I'm sure I can find it again, and link it, or at least post screenshots of the information presented. If you give me a few days, I should be able to find it. Sorry for this inconvenience. Thank you for your time though.


Thanks again all; looking forward to some more replies,
- Matt-er
 
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  • #6
Matt-er said:
Regarding the first question- it was really a simple question of: Will sub-atomic particles annihilate if they are part of a nucleus? Ie: Can the annihilation of antiprotons (which are free-floating and Not part of a nucleus) and protons occur if the protons are part of a nucleus? (again; I think yes, but I might be unaware of some kind of effect that prevents this; that's all I want to know for this question).
I answered this:
mfb said:
Yes, annihilations do happen in nuclei - and they look nearly the same as with free-floating particles.
 
  • #7
@mfb: Yes, you did answer it. My fault for reading too quickly. My apologies. Thank you for your answer. But, now I have some more questions ^^ (and of course, anyone can answer them- thanks again):

mfb said:
Annihilations release a lot of energy - I guess the first annihilation(s) would rip apart your nucleus long before the remaining antiprotons arrived.

Why would the first annihilations do this? Why would that not just simply put all the remaining particles within the nucleus at just much higher excited states? Would it be because the remaining nucleus would be so unstable that it would undergo beta decay or even fission before the rest of anti-protons would arrive?
___________

Also I found the article (the one regarding the Sphaleron) again, as promised. And, you seem to be right, mfb, in the sense that higher temperatures mean simply "that high-energetic processes are more likely to occur." According to page 1, it says:

article from online said:
In a clever analysis, Kuzmin, Rubakov, and Shaposhni­ kov have argued that baryon-number violation in the standard model is unsuppressed at high temperature, specifically T >= 1 TeV. There is no suppression because the transition arises from classical thermal fluctuations rather than quantum tunneling. For example, consider the quantum mechanics of a particle in the one­ dimensional potential shown in Fig. 1.

At zero temperature the only connection between the two vacua is quan­tum tunneling, which is exponentially suppressed. At temperatures high compared to the potential barrier V0, the thermal distribution favors states with energy E >> V0 and the particle can move over the barrier classically; there is no suppression. At intermediate temperatures the particle has a certain probability of being thermally excit­ed over the barrier given by the Boltzmann distribution and proportional to exp( -f3Vo).

5wjb6e.png

Fig1.​

The article (which is actually a physics paper/ journal; now that I think about it, "article" sounds a bit misleading; sorry for this) can be located here: http://www.itp.kit.edu/~schreck/modern_cosmology_seminar/Sphalerons_small_fluctuations_and_baryon-number_violation_in_electroweak_theory.pdf


From my understanding, the sphaleron occurs at a saddle point, specifically at a point that is exactly in between two different vacuums of Euclidean Space-time. And it only occurs when the temperature is insanely hot. So I have a few questions:


1. Is my very brief understanding correct? (Even if it is, I would like to learn more information about this whole event; any extra information would be great. Thanks).

2. When it refers to the barrier, I assume it talks about the theoretical barrier that allows matter to convert to antimatter... is this correct or is it referring to something else?

3. When it says: "the thermal distribution favors states with energy E >> V0 and the particle can move over the barrier classically," What is the V0 represent (I assume velocity)? What does the W^2 in the picture represent? and also how does a particle move over the barrier "classically"?

4. What is the "Boltzmann distribution and proportional to exp( -f3Vo)" that the article refers to?


Thanks again in advance; looking forward to some more replies,
- Matt-er
 
  • #8
Matt-er said:
Why would the first annihilations do this? Why would that not just simply put all the remaining particles within the nucleus at just much higher excited states? Would it be because the remaining nucleus would be so unstable that it would undergo beta decay or even fission before the rest of anti-protons would arrive?
The annihilation of a single proton/antiproton pair releases 1880 MeV, mainly as fast pions. This is roughly the energy needed to completely disassemble an uranium nucleus into individual nucleons. Not the whole energy would be used to do that, but it is certainly more than enough energy to split most nuclei.


I'll leave the sphaleron questions for theoreticians.
 
  • #9
mfb said:
The annihilation of a single proton/antiproton pair releases 1880 MeV, mainly as fast pions. This is roughly the energy needed to completely disassemble an uranium nucleus into individual nucleons. Not the whole energy would be used to do that, but it is certainly more than enough energy to split most nuclei.


I'll leave the sphaleron questions for theoreticians.

Welp, in the mean time, while I wait for the answers to the sphaleron questions... I might as well ask this:

Do you think there would be some way to avoid the splitting of the nucleon after a reaction like that? Honestly, I did not think it would release that much energy.. I mean that is insanity, haha.

The reason I ask is that I had an idea a while ago that balanced on the idea that we could fire one antiparticle at a time at a nucleus, then after the annihilation, the nucleus (now without one of its sub-atomic particles) would become unstable, undergo beta decay (releasing energy), become stable, and then the process would repeat until theoretically the nucleus basically disappears (in a sense, this would provide tons and tons of "free" energy).
However, if as you point out, only one reaction has the potential to destroy the entire structure of the nucleus, then this idea is dead in the water. Unless of course, there is a way to prevent the split- ie: harnessing the force of the annihilation somehow, etc, which is what/ why I am asking.
 
  • #10
I don't know this sphaleron model, so I'll leave your question 1 for someone else.

2. When it refers to the barrier, I assume it talks about the theoretical barrier that allows matter to convert to antimatter... is this correct or is it referring to something else?

I assume they are talking about that bump in the potential energy plot. If the two vacua signify matter and antimatter, then that would indeed be the (theoretical) barrier between the two.

3. When it says: "the thermal distribution favors states with energy E >> V0 and the particle can move over the barrier classically," What is the V0 represent (I assume velocity)? What does the W^2 in the picture represent? and also how does a particle move over the barrier "classically"?

V0 seems to be the height of that bump in the potential, i.e. the energy barrier (see the figure again). Moving over a barrier classically basically means that you can jump or climb over an obstacle. That is clearly possible to do within the realms of classical physics.

The standard picture in physics is this: suppose you have a rolling ball in a landscape of valleys and hills (i.e. regions of low and high potentials, respectively). If it has a high enough kinetic energy it can climb out of a valley and pass over the hilltop. If it doesn't, it is bound to stay in the same valley. In the quantum case, particles can "cheat" and tunnel through the hill even though they haven't got enough energy to reach the top.

The ω's mainly seem to be used to show how the potential is shaped. I.e. that the minima and the maximum are shaped as parabolas, opening upwards and downwards, respectively.

4. What is the "Boltzmann distribution and proportional to exp( -f3Vo)" that the article refers to?


Thanks again in advance; looking forward to some more replies,
- Matt-er

In the paper it looks like they wrote [itex]\exp(-βVo)[/itex], where [itex]\beta=1/k_B T[/itex] is a measure of inverse temperature. That exponential factor is the main component of the Boltzmann distribution, which is a probability distribution used in statistical mechanics to tell how likely different states of a system are. As you can see, high energy states are suppressed at low temperatures and so on.
 
  • #11
Matt-er said:
Welp, in the mean time, while I wait for the answers to the sphaleron questions... I might as well ask this:

Do you think there would be some way to avoid the splitting of the nucleon after a reaction like that? Honestly, I did not think it would release that much energy.. I mean that is insanity, haha.

The reason I ask is that I had an idea a while ago that balanced on the idea that we could fire one antiparticle at a time at a nucleus, then after the annihilation, the nucleus (now without one of its sub-atomic particles) would become unstable, undergo beta decay (releasing energy), become stable, and then the process would repeat until theoretically the nucleus basically disappears (in a sense, this would provide tons and tons of "free" energy).
It takes a lot of energy to create antimatter - in our terrestrial environment - because there is no naturally occurring abundance of anti-protons (or anti-nucleons). So there is no opportunity for tons of "free energy".

The binding energy of nucleons in a nucleus is on the order of several MeV. An annihilation reaction of a proton-antiproton pair releases twice the rest mass (2 * 938.3 MeV = 1876 MeV), although some of that energy is in the form pions.

An interaction of an anti-proton with a nucleus may induce fission or spallation, depending on the nucleus, and how deep into the nucleus the anti-proton penertates before annihilation.
http://link.springer.com/article/10.1007/BF01281276#page-1
http://arxiv.org/ftp/nucl-ex/papers/0508/0508027.pdf

Clearly, some nuclei do not require much energy to fission. It is sufficient for U-235 or Pu-239 to absorb a thermal neutron (E ~ .024 eV) to fission. Others require neutrons in the MeV range.

One may wish to review - Handbook of Spallation Research
http://www.wiley.com/WileyCDA/WileyTitle/productCd-3527407146.html

One may also be interested in - Antiprotons from spallation of cosmic rays on interstellar matter
http://hal.in2p3.fr/docs/00/01/45/66/PDF/democrite-00010958.pdf [Broken]
 
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  • #12
Matt-er said:
The reason I ask is that I had an idea a while ago that balanced on the idea that we could fire one antiparticle at a time at a nucleus, then after the annihilation, the nucleus (now without one of its sub-atomic particles) would become unstable, undergo beta decay (releasing energy), become stable, and then the process would repeat until theoretically the nucleus basically disappears (in a sense, this would provide tons and tons of "free" energy).
However, if as you point out, only one reaction has the potential to destroy the entire structure of the nucleus, then this idea is dead in the water. Unless of course, there is a way to prevent the split- ie: harnessing the force of the annihilation somehow, etc, which is what/ why I am asking.
Even with a perfect source of antimatter/matter pair production, you need twice the rest energy of an proton to create one antiproton. The annihilation releases this energy again (but some parts are lost to neutrinos). "Antimatter-induced fission" could be possible, but why use antimatter if nuclear reactors can do the same with neutrons?

Realistic sources of antimatter need orders of magnitude more energy than the physical minimum, so this is completely unrealistic.
 

1. What is anti-matter?

Anti-matter is a type of matter that has the same properties as normal matter, but with opposite electrical charges. When anti-matter particles come into contact with normal matter particles, they annihilate each other, releasing a large amount of energy.

2. How is anti-matter created?

Anti-matter can be created through various processes, such as the decay of radioactive material or high energy collisions in particle accelerators. It can also be produced naturally in cosmic ray interactions in space.

3. Can anti-matter be harnessed as a source of energy?

Yes, anti-matter has the potential to be a highly efficient source of energy. When anti-matter particles annihilate with normal matter particles, they release a tremendous amount of energy. However, the process of creating and storing anti-matter is currently very expensive and difficult.

4. What are the potential uses of anti-matter?

Aside from energy production, anti-matter has potential applications in medical imaging and cancer treatment. It can also be used in precision measurements and in studying the fundamental forces of the universe.

5. Is anti-matter dangerous?

Anti-matter is not inherently dangerous, as it only becomes destructive when it comes into contact with normal matter. However, the process of creating and handling anti-matter is very complex and requires highly specialized equipment, making it potentially dangerous if not handled properly.

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