# What does mathematical equivalence means here?

On Motls blog, http://motls.blogspot.com/2012/06/on-importance-of-conformal-field.html, while I was trying to understand what dimensional transmutation means, he said:

I said that by omitting the mass terms, we only get "classically" scale-invariant theories. What does "classically" mean here? Well, such theories aren't necessarily scale-invariant at the quantum level. The mechanism that breaks the scale invariance of classically invariant theories is known as the dimensional transmutation. It has a similar origin as the anomalous dimensions mentioned above. Roughly speaking, the Lagrangian density of QCD, $−Tr(F_{μν}F^{μν})/2g^2$, no longer has the units of $mass^4$ but slightly different units, so a dimensionful parameter $M^{δΔ}$ balancing the anomalous dimension has to be added in front of it. In this way, the previously dimensionless coefficient $1/2g^2$ that defined mathematically inequivalent theories is transformed into a dimensionful parameter M – which is the QCD scale in the QCD case – and the rescaling of the coefficient may be emulated by a change of the energy scale. So different values of the coefficient are mathematically equivalent after the dimensional transmutation because the modified physics may be "undone" by a change of the energy scale of the processes.

I did not understand how are the coefficients mathematically equivalent by this analysis?

• Not intending to be unwelcoming of your question, but does Lubos prohibit people from seeking clarifications on his blog? – 299792458 May 13 '15 at 12:15