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I have a question regarding the well known fact that General Relativity is not a conformal invariant theory or to put it in other words about the fact that it is conformal variant:

What are the physical assumptions in General Relativity about the gravitation that make the resulting formulation of gravity to be conformal variant?

I am not asking about a mathematical reasoning because that is obvious:

One performs a conformal transformation and he can check that the action and field equations are not invariant under this transformation. This mathematical part is obvious for me.

I am asking what is the physical assumption that make this happen? What physical property of gravity (which is absent in Electromagnetic interaction) makes it conformal variant (at least in GR)? is it weak equivalence principle? is it the requirement of the Corresponding principle? or what? there should be some physics behind this, it can't be a mere mathematical coincidence ...

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General relativity is only conformally invariant in two dimensions. This can be proven by making the transformation $g_{ab} \rightarrow \phi g_{ab}$, and seeing what transformation Einstein's equation${}^{1}$ makes. What you will find is that Einstein's equation will MOSTLY transform, but you will get terms proportional to $(d-2)(d-1)$ and derivatives of $\phi$. Hence, you only get conformal invariance in two dimensions.

If you want to prove this even more quickly, calculate the curvature of a conformally flat metric (i.e., one where $g_{ab} = \phi \eta_{ab}$.

${}^{1}$hint: $\Gamma_{ab}{}^{c} \rightarrow \Gamma_{ab}{}^{c} + \frac{1}{\phi}\left(\delta_{a}{}^{c}\nabla_{b}\phi + \delta_{b}{}^{c}\nabla_{a}\phi - g_{ab}\nabla^{c}\phi\right)$

Zo the Relativist
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A more physical attempt:

In general relativity, the metric tensor represents local clock and ruler measurements. If I multiply the metric tensor by a scalar constant, it should be obvious that this is inequivalent (in general) to a set of coordinate transformations, but, at the same time, I'm affecting local clock and ruler measurements (the ratio of the duration of an experiment at point a to the duration of an experiment at point b can change after the conformal transformation if $\phi(a) \neq \phi(b)$). Therefore, I'm representing a different physical state.

This is different from maxwell theory, where the vector potential has no direct physical meaning. Also, note that Maxwell theory has identity element $A_{a} = \vec{0}$, while GR has identity element $g_{ab} = \eta_{ab}$. Different behaviour under multiplication should be expected there, too.

Lagrangian
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Zo the Relativist
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Newton's constant is dimensionful. Hence the theory is NOT conformally invariant. In 2 dimensions, newtons constant is dimensionless. But then the apparent conformal symmetry is actually only a REDUNDANCY in the description (sometimes called weyl symmetry).

Truth
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