# For dimensional regularization, why the arbitrary mass scale $\mu$ has the meaning of UV cutoff?

For sharp cut off regularization, we introduce the UV cutoff $$\Lambda$$. When we need to do momentum integral, we integrate the momentum ball with radius of $$\Lambda$$. This $$\Lambda$$ has the explicit physical meaning of UV cutoff.

For $$\phi^4$$ in $$4$$-dim, when we use dimensional regularization, we introduce an arbitrary mass scale $$\mu$$ $$S= \int d^Dx \frac{1}{2} (\partial_\mu \phi)^2 + \frac{m^2}{2} \phi^2+ \frac{\lambda \mu^\epsilon}{4!} \phi^4$$ with $$\epsilon = 4-D$$.

Up to now, the introduction of the arbitrary mass scale $$\mu$$ is just to keep parameter $$\lambda$$ dimensionless. It can be any number. But when we write the RG equation and beta function, we give the $$\mu$$ physical meaning of UV cutoff. However textbook doesn't explain why.

My question:

1. Why this arbitrary mass scale $$\mu$$ has the physical meaning of UV cutoff?
• This wont answer your question, but I believe you have to perform that integral before you use dimensional regularization, otherwise that integral doesnt make sense. – Craig Dec 19 '18 at 4:56
• Related questions here and here. – knzhou Dec 19 '18 at 20:40

Dude, you are confused! In the dimensional renormalization scheme (Feynman used to call the shell game of renormalization dippy Hocus-Pocus), it's the $$\epsilon = 4-d$$ which plays the role of UV cutoff.
The renormalization scale $$\mu$$ is the energy scale ($$p^2 \sim \mu^2$$) you anchor your renormalized parameters, such as coupling $$g_{renor}|_{p^2=\mu^2}$$, Usually, $$\mu$$ is chosen at the physical process scale in concern, rather than an UV cutoff scale $$\Lambda$$ which is assumed to be close to the Planck scale $$M_p$$.