2
$\begingroup$

As you know

$\pi^-$ meson + proton ---> $K^+$ meson + $\Sigma^-$ particle.

(AntiUp,down) + (up up down) --> (up antistrange) + (down down strange)

I know that the quark number has be conserved in strong interactions. If that theory is correct, why doesn't the above equation have equal number of strange quarks on either sides? It clearly shows new types of quarks are made. Doesn't that mean that it is weak interaction?

What exchange particle transfers strangeness? If that question doesn't make sense or has a complex answer to it, please answer this question : what is the exchange particle in this interaction?

By the way, it clearly states the above equation is "Strong interaction"

I'm a high school kid. Please don't go too complex. Thanks.

Improve this question
$\endgroup$
  • 3
    $\begingroup$ It does have an equal number of strange quarks on both sides. There are net zero strange charges on each side of that equation. $\endgroup$ – dmckee --- ex-moderator kitten May 13 '14 at 0:47
5
$\begingroup$

You had

π− meson + proton ---> K+ meson + Σ− particle.

(AntiUp,Antidown) + (up up down) --> (up strange) + (down down strange)

since it has been correct.

(AntiUp,down) + (up up down) --> (up antistrange) + (down down strange)

each elementary particle composing the mesons carries quantum numbers that have to be conserved on the two sides of the interaction if it is a strong interaction. This is a scattering with the consequent production of two strange particles. It is strong, because the creation of the strange meson and the strange baryon was generated by the creation of a strange antistrange pair. All quantum numbers carried by the quarks/antiquarks are conserved in the right side of the interaction. That is indicative of the strong interaction, the conservation of all known quantum numbers allows the reaction to go fast and thus be characterized as strong.

The K+ meson created is stable enough to be seen in a bubble chamber picture, for example in this classic detection of the omega- all the decays are weak, so they can be seen in the chamber as separate vertices decaying into components, and then decaying further.

I know that the quark number has be conserved in strong interactions. If that theory is correct, why doesn't the above equation have equal number of strange quarks on either sides? It clearly shows new types of quarks are made. Doesn't that mean that it is weak interaction?

You had the quark content wrong. Yes, the correct one has the same strangeness left and right, 0.

What exchange particle transfers strangeness? If that question doesn't make sense or has a complex answer to it, please answer this question

what is the exchange particle in this interaction?

At the quark level it is a gluon can generates a strange-antistrange pair. The antiup from the K+ annihiates with an up of the proton in to a gluon which then creates a strange antistrange pair that couples up with the spectator quarks and changes the strangeness of the baryon and meson respectively.

By the way, it clearly states the above equation is "Strong interaction"

You do not give a link for this statement.

Look at the answer to a similar question here for the kind of Feynman diagram.

You might be interested in this link with teaching materials.

Improve this answer
$\endgroup$
2
$\begingroup$

You have the antimatter content of the mesons wrong (or did, prior to an edit). A meson is always a quark and an antiquark. (The antiquark wants to annihilate on the quark, which is why all the mesons are short-lived.)

The $\pi^-$ meson is made of an anti-up and a down, $\big\vert\bar ud\big>$.

The $K^+$ meson is made of an up and and anti-strange, $\big|u\bar s\big>$.

The antistrange in the $K^+$ and the strange in the $\Sigma^-$ add to give zero net strangeness on the right side, just as on the left.

Improve this answer
$\endgroup$
  • $\begingroup$ This is wrong. quarks have other quantum numbers that do not allow them to annihilate, even the up and down are separate quantum numbers ( baryon numbers) and have to match to annihilate as happens in the pi0 messon, when it goes to the next fastest, electromagnetic two gamma. $\endgroup$ – anna v May 13 '14 at 4:01
  • 1
    $\begingroup$ They want to annihilate, but sometimes they can't, which gives you the longer-lived mesons. It's not very wrong, and it's pedagogically useful for a beginning student. The crux of the confusion was the attempt to have all-matter and all-antimatter mesons. Your answer is also nice. $\endgroup$ – rob♦ May 13 '14 at 5:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.