# Producing K+ from K0L via 'charge exchange'?

1. Nov 16, 2015

### Anchovy

I'm looking for ways that a $K^{0}_{L}$ can generate a $K^{+}$ 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?

2. Nov 16, 2015

### ChrisVer

I guess it means that the Kaon interacts with a nucleon and changes...
for example in the paper you are asking they have:
$^{30}Si (K^- \bar{K}^0) ^{30}Al^{*}$
charge exchange..
So the Si's proton interacted with the negative charged Kaon and became a neutron for Al and a neutral Kaon....

3. Nov 16, 2015

### Anchovy

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. Nov 16, 2015

### fzero

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. Nov 24, 2015

### Anchovy

OK, many thanks.