Originally Posted by

**PetTastic**
I can't pretend to understand the details of this, in particluar is it using SI units in anyway?

If it is, then it is the SI units that scale with the atom and the rest of physics, leaving the QCD Lagrangian is unchanged.

If it is not using SI units then no scale is defined anyway.

I am really not sure what you mean by this, since the symmetries of the Lagrangian don't have anything to do with units. Let's look at a very simple example, just to illustrate the concept; suppose we have a Lagrangian of the general form

which is just the derivative of some function A(r). This Lagrangian would be invariant under

** translations**, i.e. under transformations of the form

with a constant

*V*. This is easy to see, since the constant just drops out when the derivative is taken. However, the same Lagrangian is not invariant under

**rescalings** of the form

because the rescaling factor is not eliminated by the derivative operator, so the form of the Lagrangian changes under such a transformation.

This is not a physically meaningful example, it is intended only to illustrate the basic idea. Obviously, it is irrelevant what units A(r) is expressed in, the important aspect is only the symmetries of the system in question, i.e. what happens when one performs a transformation. Now, the QCD Lagrangian is

**very** much more complicated than this trivially simple example, but the upshot is that it isn't invariant under rescalings either, like our expression above.