Timeline for Why can't General Relativity be written in terms of physical variables?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 16, 2012 at 7:23 | comment | added | Joss L | Diffeomorphism is a symmetry by this definition, but $R$ is not diffeo invariant if "invariant" is defined as "unchanged". Yang mills is "special" because the exact form that the Lagrangian changes by is zero (so the Lagrangian is both SU(N) invariant and symmetric). In GR under a diffeo the Lagrangian changes by an exact form that is not zero (under an infinitesimal diffeo generated by a vector field $\xi$, the change in the Lagrangian N-form $L$ is $d(\xi \cdot L)$, where $\cdot$ denotes contraction into the first index). | |
| Jan 16, 2012 at 6:35 | comment | added | GuSuku | Lagrangian changing only by an exact term is a fair definition of symmetry - the same definition that is used in Yang-Mills to say that $F_{\mu \nu}F^{\mu \nu}$ is $SU(N)$ gauge symmetric (upto exact terms). So, by this definition, why is diffeomorphism not a symmetry (and only a covariance) and why is $R$ not diffeomorphism invariant (while SU(N) is a symmetry of Yang-Mills and $F_{\mu \nu}F^{\mu \nu}$ is $SU(N)$ invariant)? | |
| Jan 16, 2012 at 5:28 | comment | added | Joss L | How do you define symmetry? The gauge transformations of GR are the diffeomorphisms, since under a diffeo the Lagrangian changes by an exact form. | |
| Jan 16, 2012 at 5:09 | comment | added | GuSuku | Since covariance is not the same as symmetry (or invariance), what then is the gauge symmetry of GR? | |
| Jan 16, 2012 at 3:56 | comment | added | Joss L | I'm not sure why they are called curvature invariants; I've more often heard them called curvature scalars which sounds like a better name to me. The Lagrangian is "diffeomorphism covariant" (i.e. there are no "background fields") which I would distinguish from "diffeomorphism invariant" - although I think people use the terms in different ways (confusingly). The Lagrangian $L(g)$, thought of as a n-form on spacetime, certainly changes under a diffeo $\phi$, since $L(g)\neq \phi^*L(g)$, but it is "covariant" in that $L(\phi^* g)=\phi^*L(g)$. | |
| Jan 16, 2012 at 3:15 | comment | added | GuSuku | May be my understanding is flawed but, under what kind of transformations are the curvature invariants (say, Ricci scalar) an invariant (a scalar) under? Also, do you agree that the gauge symmetry of GR is diffeomorphism invariance? Then, shouldn't the Lagrangian $R$ (and the measure of the action, $\sqrt{-g} d^4x$) be invariant under the gauge symmetry? | |
| Jan 15, 2012 at 20:33 | comment | added | Joss L | Curvature invariant $\neq$ diffeomorphism invariant. Under the infinitesimal diffeomorphism generated by the vector field $\xi$, $R$ transforms by $R \rightarrow R + \mathcal{L}_{\xi}R$. If $\xi$ is not a Killing vector then this diffeo is not an isometry (by defn of Killing vector). In any case, this is all irrelevant to my original question, which was about gauge invariants, not curvature invariants. | |
| Jan 15, 2012 at 12:30 | comment | added | GuSuku | Some other scalars/invariants: en.wikipedia.org/wiki/… . Looks like Kretschmann scalar corresponds to $F_{\mu \nu}F^{\mu \nu}$ of Yang-Mills and Chern-Pontryagin scalar to $F_{\mu \nu} \tilde{F}^{\mu \nu}$. I dont know what Euler scalar would correspond to. | |
| Jan 15, 2012 at 11:47 | comment | added | GuSuku | You are referring to an isometry, and not invariance under GCT. $R$ is called a scalar precisely because it is invariant under GCT/diffeomorphism. | |
| Jan 15, 2012 at 8:44 | comment | added | Joss L | How is $R$ gauge invariant? The only way that you could have $R=R+\mathcal{L}_{\xi}R$ (for any vector field $\xi$) is if $R$ is a constant, which it isn't in general. | |
| Jan 15, 2012 at 6:09 | history | edited | GuSuku | CC BY-SA 3.0 |
Struck off a wrong comment.
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| Jan 15, 2012 at 5:17 | comment | added | GuSuku | A better way to draw parallels between the two worlds is to consider matter spinors coupled to GR in vielbein formalism. Here, each term has to be invariant under both gauge symmetries. | |
| Jan 15, 2012 at 4:57 | comment | added | GuSuku | Sorry for the incorrect remark. My mistake! $E$ and $B$ are indeed invariant, but only because QED is abelian gauge. In non-abelian Yang-Mills, the only lowest order invariants under gauge symmetry are the ones I mentioned. Varying Yang-Mills w.r.t. $A$ gives its field equations (which is not put-together using those invariants). In GR, the gauge symmetry is GCT. Under GCT, $R$ is an invariant. Varying Einstein-Hilbert action w.r.t. metric $g$ gives its field equations. | |
| Jan 14, 2012 at 18:01 | comment | added | Joss L | I don't think I understand this. In E&M, under a gauge transformation $A \rightarrow A + d\xi$ , $E$ and $B$ are unchanged. In GR, under a gauge transformation $g \rightarrow g + \mathcal{L}_{\xi}g$, $R$ is not unchanged unless it is constant. | |
| Jan 14, 2012 at 7:27 | history | edited | GuSuku | CC BY-SA 3.0 |
added 314 characters in body
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| Jan 14, 2012 at 7:18 | history | answered | GuSuku | CC BY-SA 3.0 |