How will our world be different if supersymmetry is an unbroken symmetry?

[petergreat]

With all kinds of low energy superpartner particles floating around, do we get the same types of atoms and molecules that build up our world? Will the periodic table of elements be larger or smaller? Is this world friendly to the evolution of intelligent life?

 

[arivero]

With all kinds of low energy superpartner particles floating around, do we get the same types of atoms and molecules that build up our world? Will the periodic table of elements be larger or smaller? Is this world friendly to the evolution of intelligent life?

More or less the same, at least for the partners of quarks and leptons. The spectrum is full of particles with the same mass charge and spin that the superpartners should have. It is only that they are QCD particles.

A different point is about the partners of gauge bosons, I can not imagine the world with a photino or gauginos.

Which are the interactions of the photino?

 

[haael]

The world would look rougly the same. The supersymmetric particles would still behave as dark matter. But if there are lighter, normal particles have more decay channels, so various particles are less stable. Probably most matter would break into the lightest supersymmetric particles, which would be fotino, sneutrino and selectron.

A different point is about the partners of gauge bosons, I can not imagine the world with a photino or gauginos.

It's not unlike ours :).

 

[Vanadium 50]

The world would look completely different. As one example, you could make atoms with selectrons rather than electrons, and the Pauli principle wouldn't apply. There would be no periodic table to speak of.

 

[arivero]

The world would look completely different. As one example, you could make atoms with selectrons rather than electrons, and the Pauli principle wouldn't apply. There would be no periodic table to speak of.

Ok, if selectrons were stable. But if they are not, it is not different from an atom with a pion.

True, with exact supersymmetry, the selectron would be stable. But even a small upwards breaking would do it very different.

But it is good you have grought the selectron example! The electron is the one that does not have a "pionic partner". This is, there are in nature exactly six spin zero, charge minus one, particles:

pion- at 139.5 MeV
kaon- at 493.6 MeV
D- at 1869 MeV
Ds- at 1968 MeV,
B-, at 5279 MeV
Bc-, at 6277 MeV

In the "unbroken supersymmety" scenary, we should get six spin zero, charge minus one, particles: 2 at 105 meV, 2 at 1776 MeV, 2 at 0.510 MeV. The two selectrons are the stable ones in your scenary, and yes it is the difference with the actual world.

It is interesting to note that if the electron were massless and the mass of the muon were exactly the mass of the pion-, the later would be stable at first order.

 

[Dmitry67]

massless charged particles would make EM field short-range.

 

[Vanadium 50]

What does that have to do with the OP's question?

 

[Physics Monkey]

Just for fun, I suppose the silliest answer would be that our world would be very different because there wouldn't be anything in it. Non-zero density or temperature breaks supersymmetry (basically because bosons are different from fermions). So truly unbroken supersymmetry would require a cold empty world.

 

[Prathyush]

Just for fun, I suppose the silliest answer would be that our world would be very different because there wouldn't be anything in it. Non-zero density or temperature breaks supersymmetry (basically because bosons are different from fermions). So truly unbroken supersymmetry would require a cold empty world.

Why does non zero density or temperature break super symmetry?

 

[Physics Monkey]

Why does non zero density or temperature break super symmetry?

There are lots of ways to see that something has to happen. At the algebraic level, finite temperature and density break Lorentz invariance. SUSY inherits this breaking. Physically, a finite density of weakly interacting bosons forms a superfluid while a finite density of weakling interacting fermions forms a Fermi liquid. These things have different heat capacities, for example. Or take the path integral point of view, bosons and fermions get different boundary conditions around the thermal circle leading to different spectra.

 

[haael]

At the algebraic level, finite temperature and density break Lorentz invariance.

Wait, how?

 

[Physics Monkey]

Wait, how?

Let's say either $$J^{\mu}$$ or $$T^{\mu \nu}$$ or both may have an expectation value that breaks Lorentz invariance.