Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Massive string theory states

  1. May 14, 2018 at 1:48 AM #1
    Am I understanding it correctly that if you keep acting with an internal 2-dimensional creation operator on a string you will get heavier and heavier particle from the 11-dimensional (26 dimensional) point of view? If so, how come we haven't observed those particles in the real life? Is it because they are so heavy (say, 1 kilogram each) that their lifetime is very short?

    Also, there are two separate things going on: we can act with a creation operator as many times as we want, and we can also pick a wavelength corresponding to the creation operator of our choice. Are twe basically throwing *all* of this away on the name of it being "too heavy to observe"? Or am I missing something?
  2. jcsd
  3. May 14, 2018 at 9:38 PM #2
    Yes, in a realistic model they would usually be unobservable and only matter in the far UV, along with many other heavy states like Kaluza-Klein and winding states. However, in a braneworld model, the extra dimensions can be large and the string scale much lower. There are many papers on "TeV strings" which explore this possibility.
  4. May 14, 2018 at 9:44 PM #3
    So which of the states would actually be observed? Only the first excited states? But even then we have a problem: there are first excited states corresponding to different wavelengths. So how come we don't get different masses this way?
  5. May 14, 2018 at 10:12 PM #4
    The observed particles, like the elementary fermions and gauge bosons, correspond to massless string states. As in the standard model, some then acquire a small mass through an implementation of the Higgs mechanism (e.g. a gap between parallel branes playing the role of Higgs vev).
  6. May 14, 2018 at 10:22 PM #5
    By massless states, do you mean vacuum state from 1+1 point of view? If so, what is the point of string's ability to oscillate, if none of the excited states can be observed?
  7. May 14, 2018 at 11:26 PM #6
    The completely unexcited state of the string actually has "negative mass squared", it's unstable. You have to excite it just to reach the massless states.
  8. May 14, 2018 at 11:40 PM #7
    I thought that negative mass states is an excitation in x^0 direction, due to the -1 sign in the metric? If so, why would a vacuum state -- which is free of *all* excitations, *including* that one -- have negative mass?

    In any case, if the zero mass state would be the first excited state, that would imply that different first excitations provide the same amount of energy (just enough to cancel the negative energy). But that doesn't seem to be the case: I mean, the excitations with shorter wavelength would provide more energy? So how do you resolve this contradiction?
  9. May 15, 2018 at 12:56 AM #8
    The nonzero mass-squared comes from the zero-point energies of the string modes of oscillation. (Timelike excitations produce unphysical states with negative norm.)

    Excitation of modes with mode number > 1 produces massive states. The massless states only involve the first mode.
  10. May 15, 2018 at 1:11 AM #9
    But in QFT the bosonic vacuum has positive energy, fermionic vacuum has negative energy and supersymmetric vacuum has zero energy. Logically, this seem to imply that bosonic string would have positive vacuum mass and superstring would have zero vacuum mass. But, if I understand you correctly, you seem to be saying that vacuum has negative energy for a string? If so, that would suggest that string has fermionic degrees of freedom without bosonic ones, but thats not the case or is it?

    By first mode do you mean the mode with the largest wave length, or do you simply mean only one creation operator being used (regardless of wavelength)?
  11. May 15, 2018 at 1:38 AM #10
    The modes of the bosonic string do each have positive zero-point energy. But when you add them up, you get the divergent sum 1+2+3+..., which by various mathematical tricks (e.g. Ramanujan summation) can "equal" -1/12. Unfortunately I don't yet know a physical interpretation of this, but there seems to be a CFT argument (see Tong 4.4.1).

    And yes, by the first mode I do mean the mode with the largest wavelength.
  12. May 15, 2018 at 1:47 AM #11
    I remember that infinite sum leading to -1/12 in Polchinski. But right now I am trying to use Green Schwartz and Witten and it seems to avoid doing the infinite sum thing, since it uses commutators instead. Or are you saying Green Schwartz and Witten also has that sum, they just have it implicit rather than explicit? If so, where?
  13. May 15, 2018 at 1:51 AM #12
    Sorry, but my library sent its copy of GSW off to a warehouse! So I can't tell you what they are doing.
  14. May 15, 2018 at 2:07 AM #13
    Well, I don't have Polchinski with me at the moment. But I guess somehow they got d=26 from -1/12. The only thing I can think of at the moment that would produce d=26 is (d-2)/12=2. Was it the equation Polchinski used? If so, what did those 2 stand for, why was it 2 rather than 1?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted