(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# An expression for the mass ?

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