# What makes General Relativity conformal variant?

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 ...

• I believe the absence of a mediating particle in the theory makes it conformaly invariant but I might be wrong. i.e since we are not talking about a quantum field with a quantum that can be identified as a particle then the fields could be scale invariant. – Constandinos Damalas Aug 16 '14 at 21:00
• GR isn't conformally invariant. It only becomes invariant if there are no massive particles. – Ben Crowell Aug 17 '14 at 0:15
• – Ben Crowell Aug 17 '14 at 0:20
• You may be interested in this beautiful compilation of papers: percacci.it/roberto/physics/conformal.html. I found in particular that the Flannagan paper (arXiv gr-qc/0403063) explained what it physically means to have a conformal transformation and explored the physics very well. – Arthur Suvorov Aug 17 '14 at 0:27
• @Amirpouyan: what he's saying is that the Weyl tensor is conformally invariant. What I"m saying is that almost no theories are conformally invariant. It's not conformally invarant, because it has a natural distance scale in the case where the mass of the system is nonzero. That's all to it. – Jerry Schirmer Aug 17 '14 at 12:57

A more physical attempt:

In general relativity, the metric tensor represents local clock and ruler measurments. 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 measurments (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.

• But in maxwell's theory a conformal transformation has nothing to do with vector potential, right? i mean even in maxwell's theory a conformal transformation changes the rates of clocks and the size of meters, then why maxwell's theory is invariant under this change of units while GR is not? , i guess the answer must be in the difference between gravitation interaction and EM interaction ... it maybe the kind of bosons that carry this two interactions ... or ...???!!! i'm seeking an answer based on the intrinsic properties of gravitation in GR ... sorry if i am arguing too much – user55867 Aug 17 '14 at 17:34
• @Amirpouyan: this is a classical field theory question. Any answer that involves particles is going to be wrong. General relativity is married the coordinate system in a way that Maxwell theory is not. In GR, a coordinate tranformation is a rescaling of $g_{ab}$, which takes the same role that $A_{a}$ does in Maxwell theory. That a change of scale makes $A_{a}$ transform differently than the way that $g_{ab}$ changes is really all there is to this. – Jerry Schirmer Aug 18 '14 at 15:11

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

• And I guess, one dimension, but every metric is trivially flat in one dimension. – Jerry Schirmer Aug 16 '14 at 23:42
• Hmm...so maybe it's an "or:" conformal invariance if $d=2$ or if there are no massive particles. – Ben Crowell Aug 17 '14 at 0:21
• Why maxwell's theory is conformal invariant in 4 dimensions but GR is not? what is the physical difference? massive particles? if yes, please explain it in a separate post. because i think it is the best fitted answer. explain what do you mean by massive particle in mathematical terms and why they are present in Gr? How does a gravitation theory look likne without massive particles? Thanks. – user55867 Aug 17 '14 at 8:29
• when you say on two dimensions, does this include coordinate transformations in Minkowski space (3+1) that leave two coordinates invariant? – diffeomorphism Feb 16 '16 at 20:27
• @diffeomorphism: conformal transformations in 3+1 are never equivalent to coordinate transformations of the type you describe. – Jerry Schirmer Feb 16 '16 at 21:13

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).

• Hi Truth, this seems better suited as a comment because it's too incomplete to serve as an answer. – Brandon Enright Aug 16 '14 at 23:27
• I would say this is the correct answer. Gravity is not conformal in 4D for the same reason that Yang Mills is not conformal in higher dimensions. The coupling specifies a preferred scale. – Dan Aug 20 '14 at 2:07