# Renormalization and divergences

#### eljose

renormalization and divergences....

let suppose we have a formula for the mass in the form:

$$m=\int_{0}^{\infty}dxf(x)e^{-ax}$$ $$a=ln\epsilon$$

with epsilon tending to zero so a is divergent..but if we perform the integral numerically:

$$m=\sum_{j}w(x_{j})c_{j}f(x_{j})e^{-ax_{j})$$

so we could express the quantity a in terms of the mass m so $$a=g(m)$$ so we could put inside the integral to calculate the m:

$$m=\int_{0}^{\infty}dxf(x)e^{-xg(m)}$$ and from this equation obtain a value for the mass m.

I Know something similar is made for renormalizable theory..but why can not be made for non-renormalizable ones?...

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#### DarMM

Gold Member
What you have is a toy example of a single integral. A quantum field theory requires iterations of such substitutions across an infinite tower of integrals of progressively higher dimension as you move up in perturbative order.

In renormalizable theories the iterative substitution works. After performing the substitutions you worked out at the lower orders you're left with an unproblematic substitution at the current order. In non-renormalizable theories you are not. There will be an order of perturbation theory with an integral that even after the lower order substitutions will be left with $M > N$ divergences, with $N$ the number of parameters available to perform substitution with.

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