# I Shift of momenta cures IR divergence?

1. Sep 26, 2017

### CAF123

Consider the following integral $$\int \frac{d^4k}{k^2}$$ It is UV divergent but is it IR finite or IR divergent? The integrand is singular as $k \rightarrow 0$ so this suggest an IR divergence but this is no longer the case if I make a shift of the loop momenta by say $p_1$ and write the same integral as $$\int \frac{d^4k}{(k+p_1)^2}$$

Usually we say an IR divergence can be cured by addition of a mass in the denominator (as then the integrand won't be singular as k goes to 0) but isn't the IR divergence also cured by simply making a lorentz transformation on the momenta (assuming $p_1^2 \neq 0$) ? I don't understand this result so where is the failure in the reasoning?

2. Sep 26, 2017

### Staff: Mentor

Don't you get a divergence at k=-p1 now?

3. Sep 26, 2017

### Orodruin

Staff Emeritus
Yes.

4. Sep 26, 2017

### CAF123

What if $p_1$ is large numerically? Then the divergence at $k=-p_1$ is not in the IR?

With a mass M, e.g in the integral $$\int \frac{d^4k}{k^2-M^2},$$ we have a divergence at $k=\pm M$. The mass is said to cure the IR divergence (because integrand no longer singular as $k\rightarrow 0$) but there is still this $k=\pm M$ divergence in place that is cured by the Feynman prescription.

So if $k=\pm M$ is not an IR divergence then why is $k=-p_1$ one?

5. Sep 26, 2017

### Staff: Mentor

It is still there, and renaming it won't help.
Substituting k -> k-p1 doesn't change the integral at all.
That is a different type of divergence (note the difference of squares instead of a squared difference), and there are ways to deal with it.

6. Sep 26, 2017

### CAF123

Ok,so if I took the IR limit (k->0) in the integral $\int d^4k/(k+p_1)^2$ then I lose all k dependence from the integrand and so the result is like $\int d^4k$ which is infinite. Is it in this sense that the IR divergence is still clear?

Are these 'ways' you mention the wick rotation to Euclidean 4 momentum etc?

7. Sep 29, 2017

### vanhees71

This integral is IR save since $\mathrm{d}^4 k \propto |k|^3$ (taking it as a Euclidean/Wick rotated QFT integral).