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Regge trajectories of hadrons

  1. Feb 2, 2006 #1
    OK the very origin of string theory is based on the observations of the regge trajectories pert. to hadrons. Funnily enough you get the relation that the ang. mom. is directly proporitional to the square of the energy (if i remember correctly) which would not be the case if we were dealing with, say, a rod.
    Well if you ignore the potential of the wave eq. for a moment (we set it to 0), we somewhat appr. get p^2~E where p is momentum and E is energy and ~ is the symbol for proportionality. That seems to be the reversed relation. Of course, this is the relation in classical physics, but classical physics is a apprx. of the more "correct" QM so let's stick to QM. And i'm igoring theory of rel. for the moment.

    Well, the above is a bit fuzzy, but if anyone knows why you have these opposite relations. I know there are lot of assymmetries here, like the first case concerns angular momentum, while the second case is not restricted in that manner.
  2. jcsd
  3. Feb 2, 2006 #2


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    Well, if you don't ignore relativity, the energy momentum relationship (c=1) is [tex]E^2 = p^2 - m^2[/tex]. The Regge trajectories DO have to be explained, i.e. they don't just fall out of obvious physics. That is why physicsts tried to explain them and, as you say, came up with the beginnings of string theory.
  4. Feb 2, 2006 #3
    well, they can relatively easily be explained if you consider the moment of inertia of a string and some simple relations. And of course, string theoretically, it can be explained, but that's really difficult. But even if the trajectories are explained one wonders about the relation i stated in my first message. I'm not sure if no one is able to explain it today of someone really slick with string theory can.
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