# What is the physical meaning of the renormalized (e.g. $\overline{\mathrm{MS}}$) mass?

In quantum field theory, a subtraction scheme is a way of splitting a Lagrangian $$\mathcal{L}$$ into a finite piece and a counterterm piece, $$\mathcal{L} = \mathcal{L}_f + \mathcal{L}_{ct}.$$ For example, in the on-shell scheme, $$\mathcal{L}_f$$ is set so its terms have the same coefficients as the corresponding terms in the 1PI effective action, evaluated at the pole masses, the real physical masses of the particles. By contrast, in the $$\overline{\mathrm{MS}}$$ scheme the counterterms are instead defined so that they have a simple canonical form.

As a result, the parameters in $$\mathcal{L}_f$$ depend upon the DR mass scale $$\mu$$. The coupling $$g(\mu)$$ has a nice physical interpretation: it is an estimate of the true interaction strength for a process at energy scale $$\mu$$ (see here). Quantitatively, if the 1PI effective action contains $$S^{\text{eff}} \supset f(q) \bar{\psi} \tilde{A}(q)\psi$$ then we have $$g(\mu) \approx f(\mu)$$.

However, I've never seen a physical interpretation of the $$\overline{\mathrm{MS}}$$ mass $$m^2(\mu)$$. It is not equal to the physical mass of the particle, i.e. the pole mass, as this quantity does not depend on $$\mu$$. In the case of the top quark, it is significantly different.

Does the $$\overline{\mathrm{MS}}$$ mass have a physical meaning? Can I conclude anything from the knowledge that it is, say, bigger than the pole mass? In general, what do renormalized masses mean?

• A duplicate that was not closed and instead answered is at physics.stackexchange.com/q/435266/50583 – ACuriousMind Mar 2 '19 at 8:55
• @ACuriousMind The top answer there is nice, but it seems to be answering the question "why is the pole mass different from the bare mass", and not "what is the meaning of the renormalized mass in a non on-shell scheme, where it isn't equal to the pole mass". – knzhou Mar 2 '19 at 12:26

This is not an entirely satisfying answer, but the mass can be interpreted in just the same way as a coupling. As stated above, the $$\overline{\text{MS}}$$ coupling at scale $$\mu$$ is the value so that computing at tree-level alone will give you approximately the all-order result when the energy scale of the process is $$\mu$$. One could think of a mass term in the same way: as a particular coupling constant that perturbs a theory with zero bare masses, affecting things like scattering cross-sections.