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An expression for the mass ?

  1. Jun 9, 2006 #1
    An expression for the "mass"?..

    Let,s suppose we have a NOn-renormalizable theory, and that we wish to calculate the mass of a particle for that theory with lagrangian L, the "mass" will have an expression in terms of an infinite series:

    [tex] m \sim a(0)+ga(1)+g^{2}a(2)+g^{3}a(3)+......... [/tex]


    where every a(n) diverges in the form [tex] a(n)=\Lambda ^{n+1} [/tex] [tex] \Lambda \rightarrow \infty [/tex] (regulator)

    then my question is let,s suppose we truncate the series for a number N :

    [tex] m \sim a(0)+ga(1)+g^{2}a(2)+g^{3}a(3)+......g^{N}a(N) [/tex]

    where g is a finite constant and N say N=139 (for example), of course in that case we have 139 types of divergences, but let,s suppose we could express every divergence recursively in the form:

    .a(n)= c(n)+a(n-1)+a(n-2)+.............+a(0) where c(n) are finite constants and a(0) is the value of the regulator, being this formula valid for every n=1,2,3,4,..........139 so we could calculate our value for the mass in the form:

    [tex] m \sim C(0)a(0)+gc(1)b(1)+g^{2}b(2)c(2)+g^{3}b(3)c(3)+......... [/tex]

    so we have only the divergence a(0) left, then we apply the renormalization process to 2absorb2 this a(0)-->k (finite value) so our finite value for the mass would be:

    [tex] m \sim C(0)k+gc(1)b(1)+g^{2}b(2)c(2)+g^{3}b(3)c(3)+.........g^{N}c(N)b(N) [/tex] finite due to the fact that c(n),b(n) and k are finite constants.

    The question is if we could express always the ultraviolet divergences by a recursion formula...could we calculate any physical parameter truncating the series..even for Non-renormalizable theories?..thanks.
     
  2. jcsd
  3. Jun 11, 2006 #2
    Hmm; your statement,

    [tex]a(n)=c(n)+a(n-1)+a(n-2)+\ldots+a(0)[/tex]​

    doesn't sound quite right. Let us consider the case [itex]n=1[/itex] for your formula:

    [tex]a(1)=c(1)+a(0).[/tex]​

    You stated earlier that [itex]a(n)[/itex] diverges like [itex]\Lambda^{n+1}[/itex]. So, refering to the previous equation, since [itex]a(0)[/itex] diverges linearly and since you've fixed [itex]c(1)[/itex] to be a finite number, how could [itex]a(1)[/itex] possibly diverge quadratically?
     
    Last edited: Jun 12, 2006
  4. Jun 12, 2006 #3
    You can find the answer to your question in http://xxx.lanl.gov/abs/math-ph/0209025. It is the paper where to consider the case of non-Newtonians Physics where Lagrangian is depend of coordinates and momentums (velocity) only.
     
  5. Jun 26, 2006 #4
    although it can be very 2fuzzy2 mi intention was to express every UV divergence recursively..so if we knew how to give a "finite" meaning to

    [tex] \int_{0}^{\infty}dp [/tex]


    we could give a 2finite2 meaning to every integral of the form:


    [tex] \int_{0}^{\infty}dpp^{m} [/tex] m>0

    via a recursion formula..of course this would imply that we should "truncate" the series for a finite m in case there are infinitely many m.

    Another question, i have not been able to do that but...what would happen if we could obtain a finite value for the difference:

    [tex] \sum_{n=0}^{\infty}n^{m}-\int_{0}^{\infty}dxx^{m} [/tex] m>0

    Is always the difference above a finite quantity (let,s say C(m) where the constant C depends only on m a similar case is Euler Constant:

    [tex] \sum_{N=1}^{K}n^{-1}-ln(K)=\gamma [/tex] K-->oo
     
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