R Parity: Exploring Its Significance and Benefits

[shakeel]

what is R Parity and why?

 

[EL]

what is R Parity and why?

R-parity is a symmetry often imposed in supersymmetric models.
Have a look here:
http://en.wikipedia.org/wiki/R-parity

 

[shakeel]

why there is a need for a new type of parity? is cpt voilated in supersymmetry

 

[mjsd]

this is to prevent unwanted terms in the theory so that experimental bounds on several phenomenologies are not violated.

 

[mormonator_rm]

why there is a need for a new type of parity? is cpt voilated in supersymmetry

R-parity is useful in the supersymmetric model of fundamental particles. You see, the standard fundamental particles, such as leptons, quarks, and the vector bosons all have positive R-parity, while the supersymmetric partners, such as sleptons, squarks, neutralinos and charginos have negative R-parity. This means that, in order to conserve the R-parity, only the following two-body decay transitions are allowed;

N --> NN
N --> SS
S --> NS

along with the following two-body scattering transitions;

NN --> NN
NN --> SS
SS --> SS
NS --> NS

where I have labeled "normal" (N) or standard fundamental particles and "supersymmetric" (S) fundamental particles. This means, for example, that a top quark could not decay into a charm quark and a neutralino, but a top quark could decay into a charm "squark" and a neutralino. Similarly, a "squark" cannot decay into another "squark" and a neutralino, but can decay into a quark and a neutralino. In all this, the standard decays which are well known in experimental physics remain allowed, such as quark transitions involving weak bosons. But it should be pointed out that a supersymmetric particle cannot decay into two new supersymmetric particles. All of these rules are a direct result of R-parity.

In order to prove the existence of R-parity, we now need to find some supersymmetric particles that exhibit these kinds of transitions. Experimentalists may be getting closer to this possibility, I hope.

 

[nrqed]

R-parity is useful in the supersymmetric model of fundamental particles. You see, the standard fundamental particles, such as leptons, quarks, and the vector bosons all have positive R-parity, while the supersymmetric partners, such as sleptons, squarks, neutralinos and charginos have negative R-parity. This means that, in order to conserve the R-parity, only the following two-body decay transitions are allowed;

N --> NN
N --> SS
S --> NS

along with the following two-body scattering transitions;

NN --> NN
NN --> SS
SS --> SS
NS --> NS

where I have labeled "normal" (N) or standard fundamental particles and "supersymmetric" (S) fundamental particles. This means, for example, that a top quark could not decay into a charm quark and a neutralino, but a top quark could decay into a charm "squark" and a neutralino. Similarly, a "squark" cannot decay into another "squark" and a neutralino, but can decay into a quark and a neutralino. In all this, the standard decays which are well known in experimental physics remain allowed, such as quark transitions involving weak bosons. But it should be pointed out that a supersymmetric particle cannot decay into two new supersymmetric particles. All of these rules are a direct result of R-parity.

In order to prove the existence of R-parity, we now need to find some supersymmetric particles that exhibit these kinds of transitions. Experimentalists may be getting closer to this possibility, I hope.

Thank you for a very informative post! I have tried to understand the exact consequences of R parity by reading articles on SUSY but it has never been clear like that.

So a consequence of this is that the lightest SUSY particle is absolutely stable, even if it is much more massive than non-susy particles, right?
What is the lightest susy particle in MSSM?

Thanks again for taking the time to post that.

 

[mormonator_rm]

Thank you for a very informative post! I have tried to understand the exact consequences of R parity by reading articles on SUSY but it has never been clear like that.

No problem. I know how it gets out there, they really don't take the time to break it down into plain English for us too often. Behind all that flowery technical language there is always a very simple and basic explanation.

So a consequence of this is that the lightest SUSY particle is absolutely stable, even if it is much more massive than non-susy particles, right?
What is the lightest susy particle in MSSM?

Yes, you are absolutely right! The lightest SUSY particle should be entirely stable, and if memory serves me right, it is the lightest neutralino that should be stable. This has broad implications for experimental physics in the SUSY sector.