If we contract both sides of the Einstein field equation $$R_{\mu \nu }-{\frac {1}{2}}Rg_{\mu \nu }=-{8\pi G}T_{\mu \nu },$$ we will get $$\frac{2-n}{2}R=-8\pi G T_\mu^\mu,$$ where $n$ is the dimension of spacetime. Therefore in 2 dimensional spacetime we always have $T_\mu^\mu=0$, which implies any 2d field theory has conformal symmetry, at least classically. But this statement sounds unreasonable. What's wrong here?
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This may be helpful: Carlip, Lectures in (2+1)-Dimensional Gravity, http://arxiv.org/abs/gr-qc/9503024 which implies any 2d field theory has conformal symmetry, at least classically Do you mean any 2-dimensional field theory, or just any 2-dimensional spacetime? – Jul 20 '17 at 23:48
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It is true yes, unless you have a cosmological term – Slereah Jul 20 '17 at 23:54
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@BenCrowell Thanks, but is that paper for 3d rather than 2d? I mean field theory. Of course, the spacetime needs to have conformal symmetry in the first place so that we can say a field has such symmetry. – WunderNatur Jul 21 '17 at 00:45
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@Slereah I assume there is no cosmological term. I think I can add some terms in the Lagrangian with dimensional coupling to break the conformal symmetry. So why is that true? – WunderNatur Jul 21 '17 at 01:02
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Analogy: consider an infinitesimal symmetry transformation. Noether's theorem states that the current is conserved if the action doesn't change. But according to the least action principle the infinitesimal variation of the action should always be zero in classical theory! Conclusion: any current is conserved.
This is obviously wrong. But why?
Spoiler alert: if you like puzzles, you might want to try to figure this out on your own. Below is the answer, don't read it unless you want to :)
The answer to this is, of course, that the current is conserved if the symmetry doesn't change the action off-shell. That is, on any classical configuration, not necessarily satisfying the equations of motion.
A moment of reflection would convince you that the same thing happens here. A field is only conformal-invariant is $T_{\mu}^{\mu}$ vanishes off-shell. That is, without the implication of equations of motion. Einstein's equations have nothing to do with this.
Try checking that stress-energy trace vanishes off-shell for a massless scalar in 2d.
P.S. And yes, I believe Slereah's comment is wrong and misleading.
Prof. Legolasov
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Thank you very much. This is a point I missed before. I would like to paraphrase your ideas as following to check whether I got it correctly: Symmetry is defined to generate a horizontal motion in the action space (i.e. the action doesn't change) anywhere, not just at the lowest part (the Euler-Lagrange Equation). Therefore a symmetry of the action must be a symmetry of the Euler-Lagrange Equation, but the converse is not necessarily true, which happened in this question. The catch here is that usually the symmetry generator itself doesn't show up in the Euler-Lagrange Equation. – WunderNatur Jul 21 '17 at 05:04
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Now my question would be why a symmetry is defined to leave the action invariant, rather than, for example, leave the equation of motion invariant? – WunderNatur Jul 21 '17 at 05:10
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@WunderNatur no, the relevant thing is the symmetry of equations of motion. The symmetry of the action guarantees the symmetry of the equations of motion, however, only when it is valid off-shell. – Prof. Legolasov Jul 21 '17 at 05:19
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@WunderNatur my point is — equations of motion are only conformal-invariant when the stress-energy trace vanishes off-shell. – Prof. Legolasov Jul 21 '17 at 05:24
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I think the symmetry of an action is by definition valid off-shell. A symmetry of an action is sufficient but not necessary to a symmetry of equations of motion. That is, there can be a symmetry of equations of motion which is not a symmetry of the action. This answer provided lots of examples. – WunderNatur Jul 21 '17 at 05:32
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@WunderNatur I never said that there couldn't be one. Again, I only claim that equations of motion are guaranteed to be conformal-invariant when stress-energy trace vanishes off-shell, that is, is identically equal to zero, without implying eom for either the field itself or gravity. Which was your original mistake. – Prof. Legolasov Jul 21 '17 at 05:39
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Yes, and that's a good point, but which statement in my paraphrase do you think is incorrect or imprecise? Thanks. – WunderNatur Jul 21 '17 at 05:47
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@WunderNatur your paraphrase is correct. At first I got confused by your terminology, sorry. – Prof. Legolasov Jul 21 '17 at 05:50