I suspect that i'm missing some fundamental concept here, so please kindly help me figure it out (if any).

Suppose that superstring theory is a valid theory of nature. That is, every fundamental particle of nature, say $p$, is a vibrating string, with each vibrational mode corresponding to a different particle. Define $\Lambda$ to be the highest energy currently accessible at particle colliders and $p$ to be some fundamental particle that can only be experimentally detected at energies $>\Lambda$. Note that $p$ occurs with its supersymmetric (susy) partner, say $p^*$. Consider the vibrational mode of the string that corresponds to the particle $q$ that has exactly the same properties as $p$, except for the mass. More precisely, the mass of $q$ is sufficiently small such that $q$ can be experimentally detected at energies $\leq \Lambda$. Note that by supersymmetry, $q$ should also occur with its susy partner, say $q^*$. Observe that we now have a pair of susy particles $(q, q^*)$ that can be experimentally detected at energies $\leq \Lambda$. But we know that supersymmetry cannot be detected at energies currently accessible at particle accelerators, which is a contradiction ?

  • $\begingroup$ Indeed, any realistic ST must be such that SUSY is broken. This is not specific to ST but to any supersymmetric quantum theory that aspires to be phenomenologically realistic. But perhaps I didn't quite understand your question, because I'm not sure what role the particles $p,p^*$ play here. All you need is $q,q^*$, right? $\endgroup$ – AccidentalFourierTransform Aug 11 '18 at 21:43
  • $\begingroup$ @AccidentalFourierTransform, the $(p, p^*)$ corresponds to one vibrational mode of the string, whilst $(q, q^*)$ corresponds to another. So i wondered what would happen in the case of the vibrational mode that produces a susy pair with same properties as $(p, p^*)$ but differing only in mass (which means differing only in energy, by Einstein 's mass-energy relation). I'm not sure if this comment makes things clearer, but i just thought it could be helpful to explain my thought-process. $\endgroup$ – Isaac. Aug 11 '18 at 21:55
  • $\begingroup$ @AccidentalFourierTransform, On why i considered both $(p, p^*)$ and $(q, q^*)$: I was trying to come up with some low-energy predctions of string theory, based on its high-energy behaviour. Notice that since $p$ has a higher mass than $q$, it follows by Einstein's mass-energy relation that $p$ also has a higher energy than $q$. Since $p$ and $q$ have exactly the same properties except mass, it follows that $q$ can be viewed as a low-energy analogue of $p$. $\endgroup$ – Isaac. Aug 11 '18 at 23:48
  • $\begingroup$ Your statement that you also have a "vibrational mode of the string that corresponds to the particle q that has exactly the same properties as p, except for the mass" is wrong in bosonic/RNS string theory. The masses of particles are fixed by imposing constraints on the Hilbert space. Another thing is that in minimal supersymmetric standard model, massless particles get their mass by Higgs mechanism, not by being massive states of string theory. $\endgroup$ – Bruce Lee Aug 12 '18 at 3:39
  • $\begingroup$ @BruceLee, i thought the bosonic case was ruled out by the supersymmetricity of the string theory. $\endgroup$ – Isaac. Aug 13 '18 at 11:49

The question contains a peculiar argument along the lines "if there exists a pair of superpartners with the same mass at above experimental energies, then there must exist such a pair within experimental energies". I don't see what the exact reasoning is, but it is probably based on some misconception about how mass works in string theory.

In the standard model of elementary particles - which is what string theory has to reproduce in order to match reality - most particles are massless until the Higgs gets involved. It is the same in string theory. For example, quarks, electrons, and neutrinos all correspond to massless states of the string, which then acquire mass through interaction with a stringy counterpart of the Higgs mechanism. The higher vibratory modes of these string states are superheavy and not relevant to observable physics.

With respect to supersymmetry... If supersymmetry is unbroken, then yes, every state should have a superpartner of the same mass. But supersymmetry can be broken in many ways, and there are also string theory "vacua" which are not directly supersymmetric in the first place.

  • $\begingroup$ @Mitchelle, thank you very much for your detailed & enlightening response. $\endgroup$ – Isaac. Aug 12 '18 at 0:54
  • $\begingroup$ recommend you read Lee Smolin's book The Trouble With Physics $\endgroup$ – niels nielsen Aug 12 '18 at 2:51
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    $\begingroup$ @nielsnielsen That book is a massive misrepresentation of what string theory is, and what is string theorists' sociology. It is complete propaganda and that is not the way a scientist writes. $\endgroup$ – Bruce Lee Aug 12 '18 at 3:46
  • $\begingroup$ then you should discuss it with Smolin, not with me. $\endgroup$ – niels nielsen Aug 12 '18 at 4:36
  • $\begingroup$ @nielsnielsen You brought up the reference. I have the right to say why the reference is garbage, don't I? $\endgroup$ – Bruce Lee Aug 18 '18 at 7:54

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