# Confusion about choice of renormalization scale in $\overline{\rm MS}$ mass

The $$\overline{\rm MS}$$ mass is function of the renormalization scale $$\mu$$. What does it mean to choose this scale $$\mu$$ as the $$\overline{\rm MS}$$ mass itself? I am giving some details below to make my question more concrete.

In M. Schwarz's book on QFT, the author discusses the effect of a fermion loop on the scalar mass. In Sec. 22.6.1 [Pg. 408], he considers a theory with scalar $$\phi$$ with mass $$m$$ and a Dirac fermion $$\psi$$ with mass $$M$$. This has the Lagrangian,

$$\mathcal{L}=-\frac{1}{2}\phi(\Box+m^2)\phi+\lambda\phi\bar{\psi}\psi+\bar{\psi}(i\gamma^\mu \partial_\mu-M)\psi \tag{22.41}$$

He then performs the renormalization exercise to calculate the counterterms in the on-shell scheme. He finally arrives at the difference between the pole mass and the $$\overline{\rm MS}$$ mass,

$$m_P^2-m^2_{\overline{\rm MS}}(\mu)=\frac{\lambda}{24\pi^2}(6M^2-m_P^2)-\frac{3\lambda^2}{4\pi^2}\int_0^1 dx[M^2-m_P^2 x(1-x)]\ln\frac{M^2-m_P^2 x(1-x)}{\mu^2} \tag{22.53}$$

This result is $$\mu$$ dependent, as expected. My confusion is when he considers the $$\phi$$ to be the Higgs boson and $$\psi$$ to be the top quark. With the choice of the parameters ($$\lambda$$, top mass, and $$m_P$$), he calculates that,

$$m_P^2-m^2_{\overline{\rm MS}}(m_{\overline{\rm MS}})=(18.6\,\mathrm{GeV})^2 \tag{22.54}$$

My question: How is it possible to put $$\mu=m_{\overline{\rm MS}}$$ in eq. (22.53) when it is $$m_{\overline{\rm MS}}$$ that we are calculating? Is there something simple that I am missing?

Thanks for any help.

There is nothing weird going on. Let $$\mu$$ denote the scale, and $$f=f(\mu)$$ the MS mass. We have an expression of the form $$f(\mu)= c\log \mu$$ for some constant $$c$$. We can choose the scale such that $$f(\mu)=\mu$$, which is around $$\mu=-c W(-1/c)$$, with $$W$$ the Lambert function. Your equation is a little bit more complicated, and so solving for $$\mu$$ explicitly is not as straightforward, but one can always do so, at least numerically. In any case, the point is that the MS mass is a function of $$\mu$$, and one can always choose $$\mu$$ so that the function is identical to $$\mu$$ itself — this is nothing but an equation for $$\mu$$, which is in principle solvable (although in practice it can become messy).

For example, if you have Mathematica, the following code'll do the trick:

λ = .5;
M = 10;
mp = 3;

eq[μ_?NumericQ] := mp^2 - μ^2 - (λ/(24 π^2) (6 M^2 - mp^2) - (3 λ^2)/(4 π^2) NIntegrate[(M^2 - mp^2 x (1 - x)) Log[(M^2 - mp^2 x (1 - x))/μ^2], {x, 0, 1}]);

NSolve[eq[μ], μ]
(* μ -> 3.42547 *)