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The diffeomorphism invariance of scalars is often written as: $$ \phi'(x') = \phi(x).\tag{1}$$ However, while scaling transformation is a type of diffeomorphism, in many places (say Di Francesco, Matthieu and Senechal page 38), you see the following for scalar fields: $$\phi'(\lambda x)=\lambda^{-\Delta} \phi(x).\tag{2.121}$$ This is taken to define the scaling dimension $\Delta$. Aren't these two definitions incongruent unless the scaling dimension is 0? I'm guessing no, I just have trouble seeing what is supposed to be happening here.

EDIT: To clarify a little bit, I don't fully understand why the prime appears in the second equation.

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  1. Eq. (1) is a passive general coordinate transformation of a scalar in GR, or is an active Lorentz transformation of a scalar in SR.

  2. In contrast, an active infinitesimal diffeomorphism generated by a vector field $X$ of a scalar $\phi$ in GR satisfies $${\cal L}_X\phi = X[\phi].$$

  3. Eq. (2.121) is an active scale transformation.

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    $\begingroup$ But where does the prime on the field in (2.121) come from? I would understand if the relation was $\phi(\lambda x)=\lambda^{-\Delta} \Phi(x)$. It would make sense to me because, well, you have a field defined on a point on the manifold, and then you see how its value there relates to some other, transformed point. In the first relation the prime refers to the components of the field in two different coordinate systems. But in the second one, if that is the interpretation, then I don't see what it is supposed to mean. $\endgroup$
    – Andreas Christophilopoulos
    Commented Sep 29 at 14:32
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I finally figured it out. The first one is just a passive coordinate change, as explained by @Qmechanic in their answer. But the other one is something completely different. It has nothing to do with how the field behaves in different points in spacetime, or with any coordinate change. Equation (2.121) is a symmetry of the action.

To be a bit clearer, DiFrancesco et al. are working in their book with a scale invariant action. What they want to say with this relation is that if you substitute $\phi'(x)$ for $\phi(x)$ in the action functional, it doesn't change:

$$ S[\phi']=S[\phi] $$

That's all there is to it really. In other words, classically one could say that if the field is a solution of the EOM, the transformed one also is.

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