How exactly will LHC detect superpartners?

  • Thread starter Thread starter bananan
  • Start date Start date
  • Tags Tags
    Lhc
bananan
Messages
173
Reaction score
0
How exactly will LHC detect superpartners? Presumably they will have higher mass, are there other ways to detect them?
 
Physics news on Phys.org
It really depends on the model. In models like the (Minimal Supersymmetric Standard Model) MSSM, there is a quantity called R-parity. Each Standard Model particle has R-parity +1 and each partner has R-parity -1. R-parity is conserved in interactions which means that the lightest partner (R-parity -1 state) will be stable (since it can only has R-parity +1 states that are lighter).

This Lightest Supersymmetric Partner (LSP) then leaves the detector without being seen. This is a real smoking gun because then the momentum in the plane transverse to the beam will not be conserved by the visible particles (the momentum parallel to the beam isn't either, but this is useless because one doesn't know the initail particle momenta, only their direction).
 
Severian said:
It really depends on the model. In models like the (Minimal Supersymmetric Standard Model) MSSM, there is a quantity called R-parity. Each Standard Model particle has R-parity +1 and each partner has R-parity -1. R-parity is conserved in interactions which means that the lightest partner (R-parity -1 state) will be stable (since it can only has R-parity +1 states that are lighter).

This Lightest Supersymmetric Partner (LSP) then leaves the detector without being seen. This is a real smoking gun because then the momentum in the plane transverse to the beam will not be conserved by the visible particles (the momentum parallel to the beam isn't either, but this is useless because one doesn't know the initail particle momenta, only their direction).

Thanks for the reply.

I guess while we are on subject, how will LHC see the higgs boson, and how could it see higher dimensions if they exist?
 
Severian said:
It really depends on the model. In models like the (Minimal Supersymmetric Standard Model) MSSM, there is a quantity called R-parity. Each Standard Model particle has R-parity +1 and each partner has R-parity -1. R-parity is conserved in interactions which means that the lightest partner (R-parity -1 state) will be stable (since it can only has R-parity +1 states that are lighter).

This Lightest Supersymmetric Partner (LSP) then leaves the detector without being seen. This is a real smoking gun because then the momentum in the plane transverse to the beam will not be conserved by the visible particles (the momentum parallel to the beam isn't either, but this is useless because one doesn't know the initail particle momenta, only their direction).

I wonder whether such a signature would prove SUSY, as opposed to a non-SUSY previously unknown particle with similar mass.
 
Yes, to prove it is SUSY you really need to do two things:

1. Prove the particles differ by spin 1/2.
2. Prove that the partners have the same couplings as the SM ones.

I have seen it claimed that 1 is only possible at a linear collider, but there have been some papers more recently on this, e.g. http://arxiv.org/abs/hep-ph/0605067 and http://arxiv.org/abs/hep-ph/0605286, which discuss measurements at the LHC.

As for the Higgs boson, there are various signals. For example, Higgs decays to two photons are nice because the diphoton invariant mass will give a peak at exactly the Higgs boson mass. This is a rather clean channel, and as soon as it is seen I am sure there will be a discovery announcement.

However, again, one has to be careful to make sure it is a Higgs and not something else. To do this, you really need to measure its spin (to see it is a scalar) and its couplings (to show they are proportional to the mass of the particle it couples to). Eventually if you can measure itself couplings you can actually reconstruct the shape of the mexican hat potential of the Higgs boson. Unfortunately this last thing won't be possible at the LHC.
 
Last edited by a moderator:
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Replies
4
Views
2K
Replies
23
Views
2K
Replies
9
Views
179
Replies
28
Views
4K
Replies
33
Views
7K
Replies
57
Views
15K
Back
Top