# IR and UV divergences

1. Apr 10, 2009

### zetafunction

perhaps is a dumb quetion but,

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

$$\int_{0}^{\infty} \frac{dx}{x^{3}}$$

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

$$\int_{0}^{\infty} udu$$ 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 ??

2. Apr 11, 2009

### 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.

3. Apr 11, 2009

### 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 $$\epsilon = 1/ \Lambda$$

with this epsilon tending to 0

4. Apr 11, 2009

### malawi_glenn

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

5. Apr 11, 2009

### zetafunction

of course is not the same taking the integral

$$\int_{0}^{\infty} d\lambda f( \lambda )$$

or taking the integral $$\int_{0}^{\infty} dp f(p)$$

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.

6. Apr 11, 2009