I'm conscious that this isn't necessarily a research level question, but I've asked this question on mathstackexchange, and received no answer. So I'm trying it here.


A usual mantra in field theories is the assertion that *only massless theories can be conformally invariant*. By a *theory* I mean an action $$ S = \int \mathcal{L} \, \mathrm{dVol}, $$ where $\mathcal{L}$ is the Lagrangian density, and the integral is taken over a 4-dimensional Lorentzian manifold with metric $g$. By *conformal invariance* I mean the statement that under the conformal rescaling of the metric $$ \hat{g} = \Omega^2g, $$ the Lagrangian transforms as $\hat{\mathcal{L}} = \Omega^{-4} \mathcal{L}$. Then, as the volume form transforms as $\widehat{\mathrm{dVol}} = \Omega^{4} \mathrm{dVol}$, the action $S$ is invariant, and the theory is said to be conformally invariant.


The usual physics explanation given is that "if a theory is supposed to be conformally invariant, then there cannot exist an intrinsic scale to it, such as mass or a Compton wavelength". Of course, this is a load of hand waving. I guess I don't strictly know what I mean by a *massless* theory. Maxwell's equations, for example, are a massless conformally invariant theory. My guess would have been that the mass of a theory is its ADM mass, but as has been pointed out in the comments, this is a property of a solution to a theory, not the theory itself. So, if $m$ is the mass of a theory, whatever it stands for exactly, and $m \neq0$, why must conformal invariance fail?



Why is conformal invariance only possible for massless theories?
A physicist would answer this question as follows. (Everything I'll say can be expressed in a way that the purest of mathematicians would understand, but that translation would take a lot of work, so I'll only do it on demand.)


In physics we have units of mass ($M$), length ($L$) and time ($T$). 


In special relativity we have a fundamental constant $c$, the speed of light, with units $L/T$. Thus, in special relativity, any length determines a time and vice versa. 


In quantum mechanics we have a fundamental constant $\hbar$, Planck's constant, with units $ML^2/T$. $\hbar / c$ has units $M L$. Thus, in theories involving both special relativity and quantum mechanics, any mass determines an inverse length, and vice versa.


Relativistic quantum field theory involves both special relativity and quantum mechanics. Thus, in relativistic quantum field theory, if we have a particle of some mass $m$, it determines a length, namely $\hbar / c m$. This is called the 
**[Compton wavelength](https://en.wikipedia.org/wiki/Compton_wavelength)** of that particle. 


Some physicists may prefer to use $h = 2 \pi \hbar$ in the definition of the Compton wavelength, but this isn't important here: what really matters is that when you have a relativistic quantum field theory with a particle of some given mass, it will have a preferred length scale and will thus not be invariant under scale transformations, hence not under conformal transformations... *unless that mass is zero*, in which case the Compton wavelength is undefined.


What is the meaning of the Compton wavelength? Here's a rough explanation. In quantum mechanics, to measure the position of a particle, you can shoot light (or some other kind of wave) at it. To measure the position accurately, you need to use light with a short wavelength, and thus a lot of energy. If you try to measure the position of a particle with mass $m$ more accurately than its Compton wavelength, you need to use an energy that exceeds $mc^2$. This means that the collision is energetic enough to create another particle of the kind whose position you're trying to measure. So, in relativistic quantum field theory, we should imagine any particle as part of a 'cloud' of 'virtual' particles which can become 'real' if you try to measure its position too accurately. The size of this cloud is not sharply defined, but it's roughly the Compton wavelength. So, the theory of this particle is not invariant under conformal transformations.


