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

Veneziano before the trialogue paper.

  1. Dec 3, 2007 #1


    User Avatar
    Gold Member

    Just noticed that http://xxx.lanl.gov/abs/physics/0110060 has an antecedent in http://www.slac.stanford.edu/spires/find/hep/www?j=EULEE,2,199 [Broken]

    It is a deep meditation. If one takes C and l (The Planck length, not the Planck mass) as fundamentals, then one can to build an angular momentum, mass.velocity*distance by introducing a mass. For instance the mass of the electron. Still, I have problems to see how does gravity -and Newton constant- work here. It seems that in a=G m /r^2 the G is born because of the m, and Veneziano seems to claim that both are unnecessary.

    Please read both papers, or the shortest one :-) and give your opinion.

    ah, it is free until december
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 3, 2007 #2


    User Avatar
    Science Advisor
    Gold Member
    Dearly Missed

    I see that the whole contents of EPL from July thru December 1986 is free to download until the end of the year. If you (or anyone else) sees any other paper that might be of special historical interest, please let us know.

    I cannot comment except that it seems like a wonderful, simple, and hopeful world back in 1986. One had a theory intended to explain aill of nature and it had NO free dimensionless constants and possibly only TWO dimensioned constants----the speed of light and a length. Everything else was to be derived geometrically from these two quantities.

    A beautiful moment, possbily even terrifying in retrospect

    I remember looking at the Trialogue on another occasion, especially the first 9 or 10 pages. I will take another look. The Lev Okun part is a bit reminiscent of his inspiring Physics Today (1990?) article about the meaning of mass.

    The string venture circa 1986 seems immeasurably remote from the present---the enterprise had an entirely different personality.
    Last edited: Dec 3, 2007
  4. Dec 3, 2007 #3


    User Avatar
    Gold Member

    Actually veneziano does a elementary but intriguing observation about quantum mechanics: he comes to say, it seems, that if an observable is a function of frequency then the Planck constant can be hidden under a fundamental length (sort of some Compton length) of the problem being solved. An he insinuates it is a general property, at least if the observables are wavelengths or frequencies (... but what if the observable is a combination of position and momentum?)
  5. Dec 3, 2007 #4


    User Avatar

    Since you have been doing a lot of work on the constants, I would be interested in your response. Did you get an idea from their paper?
    Would you be able to re-state everything as a fundamental/minimum length?
  6. Dec 3, 2007 #5


    User Avatar
    Gold Member

    Yes jal, I did not though about it first time I read the dialogue, but now I think Veneziano has a point. A lot of QM results can be presented using the Compton length instead of Planck constant. And well, position and time (c and L if you prefer) seem a lot more natural as basic constants that h, c, L. Still I would not tell they imply some minimum length.
  7. Dec 4, 2007 #6


    User Avatar
    Gold Member

    Related thoughts. Gravity in D dimensions has

    [G]= [c]^2 [m]^-1 [L]^(D-3).

    so that our usual D=4 Newton constant has dimensions [c]^2 [L] / [m] : it can be used to move between mass and length.

    Angular momentum, h, on the other side, has dimensions [c] [L] [m] so in some sense the role of Planck constant reciprocates with Newton constant.

    In 2+1 dimensions it is more intriguing, [G]= [c]^2 [m]^-1.

    In 1+1 dimensions [1/G]= [c]^[-2] [L] [m], and the constant [c^3/G] has the same dimensions than Planck constant.
  8. Dec 4, 2007 #7


    User Avatar
    Gold Member

    Consider now centrifugal force: [itex]a = V^2 / R [/itex]

    a stable circular orbit of a test particle around a mass M has

    [tex]V^2 / R = c^2 m_P^{-1} L_P^ {D-3} M / R^{D-2}. [/tex]

    [tex](V/c)^2 = (M/m_P) L_P^{D-3}/R^{D-3}
    = (M L_P^{D-3}) / (m_P R^{D-3})
    = (M/R^{D-3}) * (L_P^{D-3} / m_P). [/tex]

    and it shows how D=4 is interesting.

    Time ago I introduced Planck Area and Planck time, and asking
    [itex]R .V .T_p = A_p [/itex]
    gives [itex]V = c (L_p / R) [/itex]
    and above
    [itex] (L_p / R)^2= (M/R^{D-3}) * (L_P^{D-3} / m_P) [/itex]
    and thus
    [itex] 1 = (M/R^{D-5}) * (L_P^{D-5} / m_P) [/itex]
    So you can see, for D=4

    [itex]M * R = m_p * L_p = h/c [/itex]

    and we can emerge Planck constant: M is any mass originating the gravity force, [itex]R[/itex] is a radius, greater than [itex]L_p [/itex], such that a particle around M sweeps some unit of Planck area in one similar unit of Planck time. Then [itex]h [/itex] is [itex]M*R*c [/itex].
    Last edited: Dec 4, 2007
  9. Dec 4, 2007 #8


    User Avatar
    Gold Member

    Note that when [itex]h[/itex] emerges, gravity disappears. So I tend to agree with Veneziano, there are only two fundamental constants, and to produce explicitly the third one you must hide one of the others.

    It is a bit more obscure, of course. Not having [itex]h[/itex], you carry [itex]G[/itex] as a function of both [itex]m_P[/itex] and [itex]L_P[/itex], and you need an independent input of [itex]L_P[/itex] (the area condition) in order to emerge [itex]h[/itex] (and then BOTH Planck mass and length dissappear). The point is that for [itex]D=4[/itex] this extra input has at least some physical/geometrical meaning. For other dimensions, you can still use algebraically [itex]G m_p=c^2 L_P^{D-3}[/itex] to get [itex]m_P[/itex] from [itex]L_P[/itex], and then [itex]h[/itex] as the product, ie you always have [itex]h = c m_P L_P = c^3 G^{-1} L_P^{D-2}[/itex]. Perhaps it is just lucky that it has sense to speak of an area swept by some trajectory of a point particle, but not of a n-volume swept. Or perhaps we need branes :-D

    Veneziano seems more strict, it argues -it seems- that you do not need emerge [itex]h[/itex] at all, because measured quantities depend only on Compton lengths.
    Last edited: Dec 4, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook