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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.

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1 Answer 1

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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 *)
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