Why Are IR and UV Divergences the Same?

[zetafunction]

perhaps is a dumb quetion but,

given a IR divergent integral (diverges whenever x tends to 0)

[tex] \int_{0}^{\infty} \frac{dx}{x^{3}} [/tex]

then using a simple change of variables x=1/u the IR integral becomes an UV divergent integral


[tex] \int_{0}^{\infty} udu [/tex] which is an UV divergent integral (it diverges whenever x tends to infinity)

then why we call IR or UV divergences if they are essentially the same thing ??

 

[Avodyne]

They are not the same thing. An IR divergence is one that arises from integrals over low momentum or energy, and a UV divergence is one that arises from integrals over high momentum or energy.

 

[zetafunction]

yes of course, but from the mathematical point of view a change of variable would turn an IR divergence into a UV one, for a mathematician both functions or divergences would be the same since from the cut-off we can define a function [tex] \epsilon = 1/ \Lambda [/tex]

with this epsilon tending to 0

 

[malawi_glenn]

But the physical variable is not the same, that is why we call them different

 

[zetafunction]

of course is not the same taking the integral

[tex] \int_{0}^{\infty} d\lambda f( \lambda ) [/tex]

or taking the integral [tex] \int_{0}^{\infty} dp f(p) [/tex]

in the first integral we integrate over the wavelength (meters) whereas in the second we integrate over moment (kg.m/second) but for a mathematician both singularities would seem the same.

 

[Vanadium 50]

But it's physicists that are using these terms, and they do distinguish between divergent in E and in 1/E.