An expression for the mass ?

[eljose]

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:

$$m \sim a(0)+ga(1)+g^{2}a(2)+g^{3}a(3)+...$$


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

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

$$m \sim a(0)+ga(1)+g^{2}a(2)+g^{3}a(3)+...g^{N}a(N)$$

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:

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

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:

$$m \sim C(0)k+gc(1)b(1)+g^{2}b(2)c(2)+g^{3}b(3)c(3)+...g^{N}c(N)b(N)$$ 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.

 

[TriTertButoxy]

Hmm; your statement,

$$a(n)=c(n)+a(n-1)+a(n-2)+\ldots+a(0)$$​

doesn't sound quite right. Let us consider the case $n=1$ for your formula:

$$a(1)=c(1)+a(0).$$​

You stated earlier that $a(n)$ diverges like $\Lambda^{n+1}$. So, referring to the previous equation, since $a(0)$ diverges linearly and since you've fixed $c(1)$ to be a finite number, how could $a(1)$ possibly diverge quadratically?

 

[cartuz]

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:

$$m \sim a(0)+ga(1)+g^{2}a(2)+g^{3}a(3)+...$$


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

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

$$m \sim a(0)+ga(1)+g^{2}a(2)+g^{3}a(3)+...g^{N}a(N)$$

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:

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

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:

$$m \sim C(0)k+gc(1)b(1)+g^{2}b(2)c(2)+g^{3}b(3)c(3)+...g^{N}c(N)b(N)$$ 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.

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.

 

[eljose]

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

$$\int_{0}^{\infty}dp$$


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


$$\int_{0}^{\infty}dpp^{m}$$ 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:

$$\sum_{n=0}^{\infty}n^{m}-\int_{0}^{\infty}dxx^{m}$$ 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:

$$\sum_{N=1}^{K}n^{-1}-ln(K)=\gamma$$ K-->oo