- #1
bananan
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Dear Lubos,
Heisenberg's uncertainty princple states that xp>=h bar/2
a string is smaller than the particles they make up, so x is smaller than the particles of the standard model.
The momentum of uncertainty p should be greater than the particles themselves.
How does string theory get around the mass-paradox:
The momentum uncertainty of a string confined in such a small location would be much larger than the particles they supposedly make up.
Lubos kindly replies:
Dear Dan,
the center-of-mass position X0 and the total momentum of a string, P0, behave just like for pointlike particles. They don't commute and follow the uncertainty principle. A certain X0 means uncertain P0 and vice versa, the product being above hbar/2.
Besides the zero modes (center-of-mass degrees of freedom), every string has infinitely many internal degrees of freedom. That's like an atom with many electrons, but you have infinitely many arranged along a string. The relative motion of pieces of string gives energy - expressed as the usual sum of the kinetic and potential contributions. And because they're relativistic strings, energy also means mass via E=mc^2.
The result is that the minimal size of the string - in the lowest-energy or lowest-mass states - is a compromise in which the kinetic terms from the internal degrees of freedom contribute the same as the potential terms, just like for X and P in the harmonic oscillator, attempting to minimize the energy while satisfying the uncertainty relation for the internal X,P degrees of freedom. This minimum occurs if the typical size of the string is comparable to a typical distance scale derivable from the tension of the string - the string length - that is conventionally believed to be close to the Planck scale 10^{-35} meters (a bit longer than that).
The actual numerical coefficient of the string is actually logarithmically divergent but this fact doesn't affect any finite-energy experiments.
Your question has the same answer in string theory just like for ordinary particles because it is the zero modes that matter here. The internal degrees of freedom are only relavant for experimental uncertainty considerations if your probe the internal structure of a string, and indeed, you will always find out that the "radius" of it seems to be of order the string length.
Best
Heisenberg's uncertainty princple states that xp>=h bar/2
a string is smaller than the particles they make up, so x is smaller than the particles of the standard model.
The momentum of uncertainty p should be greater than the particles themselves.
How does string theory get around the mass-paradox:
The momentum uncertainty of a string confined in such a small location would be much larger than the particles they supposedly make up.
Lubos kindly replies:
Dear Dan,
the center-of-mass position X0 and the total momentum of a string, P0, behave just like for pointlike particles. They don't commute and follow the uncertainty principle. A certain X0 means uncertain P0 and vice versa, the product being above hbar/2.
Besides the zero modes (center-of-mass degrees of freedom), every string has infinitely many internal degrees of freedom. That's like an atom with many electrons, but you have infinitely many arranged along a string. The relative motion of pieces of string gives energy - expressed as the usual sum of the kinetic and potential contributions. And because they're relativistic strings, energy also means mass via E=mc^2.
The result is that the minimal size of the string - in the lowest-energy or lowest-mass states - is a compromise in which the kinetic terms from the internal degrees of freedom contribute the same as the potential terms, just like for X and P in the harmonic oscillator, attempting to minimize the energy while satisfying the uncertainty relation for the internal X,P degrees of freedom. This minimum occurs if the typical size of the string is comparable to a typical distance scale derivable from the tension of the string - the string length - that is conventionally believed to be close to the Planck scale 10^{-35} meters (a bit longer than that).
The actual numerical coefficient of the string is actually logarithmically divergent but this fact doesn't affect any finite-energy experiments.
Your question has the same answer in string theory just like for ordinary particles because it is the zero modes that matter here. The internal degrees of freedom are only relavant for experimental uncertainty considerations if your probe the internal structure of a string, and indeed, you will always find out that the "radius" of it seems to be of order the string length.
Best