Producing K+ from K0L via 'charge exchange'?

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In summary: I'll be looking into this a little more. In summary, a process called "charge exchange" can allow a K+ to be generated from a K0. However, this process is difficult to access and full details are not available.
  • #1
Anchovy
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I'm looking for ways that a [itex]K^{0}_{L}[/itex] can generate a [itex]K^{+}[/itex] and I'm aware that a so-called 'charge exchange' process can allow this. However, I can't access the well-cited publication that appears when googling, ("Kaon inelastic scattering and charge exchange on nuclei - Phys. Rev. C 19 1393 - CB Dover, 1979"), nor can I find a full downloadble version of any textbook that googling turns up, or even any powerpoint showing what happens.

Can anyone explain what such a process involves?
 
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  • #2
I guess it means that the Kaon interacts with a nucleon and changes...
for example in the paper you are asking they have:
[itex] ^{30}Si (K^- \bar{K}^0) ^{30}Al^{*}[/itex]
charge exchange..
So the Si's proton interacted with the negative charged Kaon and became a neutron for Al and a neutral Kaon...
 
  • #3
ChrisVer said:
I guess it means that the Kaon interacts with a nucleon and changes...
for example in the paper you are asking they have:
[itex] ^{30}Si (K^- \bar{K}^0) ^{30}Al^{*}[/itex]
charge exchange..
So the Si's proton interacted with the negative charged Kaon and became a neutron for Al and a neutral Kaon...

I'm not quite understanding... not sure why there's a negative kaon being mentioned now... is it illegal or against forum rules for you to post the pdf if you've been able to access it?
 
  • #4
I don't have access to the paper right now, but I can discuss generalities. First of all, ##K^0+X \rightarrow K^\pm +Y## implies that we are exchanging a ##d## or ##\bar{d}## for a ##u## or ##\bar{u}## quark, respectively. For the ##s\bar{d}## component, this is kind of easy, since we have ##d##s directly in ##p## or ##n## so an annihilation process can produce a ##u\bar{u}## and pair the ##\bar{u}## with the ##s## to get a ##K^-##. There are other processes available at higher orders.

From the abstract of the paper, they are talking about incident momenta of order ##300-800~\text{MeV}##, which is still of order the kaon masses, so one could imagine that an adequate description would be in terms of pion-intermediated strong interactions with the nucleus. This gives plenty of opportunities to match up ##u## and ##d## quark exchanges with the kaons.
 
  • #5
fzero said:
I don't have access to the paper right now, but I can discuss generalities. First of all, ##K^0+X \rightarrow K^\pm +Y## implies that we are exchanging a ##d## or ##\bar{d}## for a ##u## or ##\bar{u}## quark, respectively. For the ##s\bar{d}## component, this is kind of easy, since we have ##d##s directly in ##p## or ##n## so an annihilation process can produce a ##u\bar{u}## and pair the ##\bar{u}## with the ##s## to get a ##K^-##. There are other processes available at higher orders.

From the abstract of the paper, they are talking about incident momenta of order ##300-800~\text{MeV}##, which is still of order the kaon masses, so one could imagine that an adequate description would be in terms of pion-intermediated strong interactions with the nucleus. This gives plenty of opportunities to match up ##u## and ##d## quark exchanges with the kaons.

OK, many thanks.
 

1. How does charge exchange work in producing K+ from K0L?

Charge exchange is a process in which a neutral particle, such as the K0L, interacts with a charged particle, causing the neutral particle to gain or lose a charge. In the case of producing K+ from K0L, the K0L interacts with a proton, transferring a positive charge to the K0L and producing a K+ particle.

2. What is the purpose of producing K+ from K0L?

Producing K+ from K0L is important for studying the properties of the K0L particle and for understanding the fundamental forces and interactions of the subatomic world. It also has practical applications in medical imaging and cancer treatment.

3. How is charge exchange used in particle accelerators?

In particle accelerators, charge exchange is used to create new particles by colliding a beam of high-energy particles with a stationary target. This collision results in the transfer of charge between the particles, producing new particles that can be studied by scientists.

4. Is charge exchange the only way to produce K+ from K0L?

No, there are other methods for producing K+ from K0L, such as collisions with other particles or through nuclear reactions. However, charge exchange is a common and efficient method used in particle physics experiments.

5. What other particles can be produced through charge exchange?

Charge exchange can be used to produce a variety of particles, including pions, kaons, and neutrons. It is also used in experiments to create exotic particles, such as hyperons and antiprotons.

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