Timeline for Special relativity and diffeomorphism invariance
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 27, 2019 at 12:14 | answer | added | ACuriousMind♦ | timeline score: 7 | |
| Jan 27, 2019 at 3:56 | comment | added | Chiral Anomaly | I think that's right. I mean, I know the conclusion is right, and I think the proof really is that simple. The tensor $G$ is completely determined by the tensor $g$, so replacing $g\to \phi^*g$ must have the effect $G\to\phi^* G$. I'm sure this is spelled out in the literature, but I don't know of a specific reference right now. I'll bet Wald's General Relativity is a good resource, because it's exceptionally clear about the mathematical foundations. I don't have a copy handy to confirm this, but it's suggested by math.stackexchange.com/questions/1877969. | |
| Jan 26, 2019 at 21:03 | comment | added | Will | @DanYand To prove it, would it be sufficient to note that if $G(g)=0$, then applying a diffeomorphism to both sides of this equation we have $(\phi^{\ast}G)(g)=G(\phi^{\ast}g)=0$? Do you know of any good resources? | |
| Jan 26, 2019 at 19:51 | comment | added | Chiral Anomaly | That's true: if $G$ is a tensor and $G=0$, then applying a diffeomorphism to $G$ won't make it non-zero. And for the EFE, it's true that $G(g)=0$ implies $G(\phi^* g)=0$, but a concise rendition of the proof is eluding me right now. | |
| Jan 26, 2019 at 19:33 | comment | added | Will | @DanYand Is the point that as $G$ is a tensor, if $G=0$ then $\phi^{\ast}G=0$? | |
| Jan 26, 2019 at 19:18 | comment | added | Will | @DanYand Ok, cool. Thanks for your help. One thing I've always been slightly unsure about is what is meant by $G(g)=0$ and $\phi^{\ast}G(g')=0$. Surely if $g$ and $g'$ satsify the same equations of motion, one should have that $G(g)=G(g')=0$? | |
| Jan 26, 2019 at 19:13 | comment | added | Chiral Anomaly | It would take me some time to think through the subtleties of the compact notation, but the gist of it (and the words) seems to be correct. Given one solution of $G=0$, any other metric obtained from that one by a diffeomorphism (active or passive, doesn't matter) will be another solution. That's what physicists usually mean by diffeomorphism invariance. Of course, if other fields are involved (such as the EM field in the Einstein-Maxwell coupled system), then we need to apply the diffeomorphism to all of the fields in order to still have a solution. | |
| Jan 26, 2019 at 19:08 | comment | added | Will | @DanYand Cool. Also, is my calculation showing that GR is diffeomorphism invariant correct? Have I understood what is meant by diffeomorphism invariance correctly in the last of the display equations in my OP? | |
| Jan 26, 2019 at 19:06 | comment | added | Chiral Anomaly | Yeah, that's the way I'm thinking about it. | |
| Jan 26, 2019 at 19:05 | comment | added | Will | @DanYand Ah yes, you're on to something there. I wasn't really thinking about the fact that the background is fixed in SR, i.e. the metric is not dynamical, and therefore we can't vary it: this will generically change the metric such that it is no-longer Minkowski, and thus SR cannot be diffeomorphism invariant by construction. Would this be correct? | |
| Jan 26, 2019 at 19:02 | comment | added | Chiral Anomaly | Posting this as a comment because I'm not sure it addresses the issue. The EH action is understood to be a functional of the metric field $g_{ab}$, so we can vary the action with respect to $g_{ab}$. But in the EM action in SR, the metric is fixed: it's just a collection of coefficients in the integrand, not a collection of dynamic variables. Of course, we can generalize the EM action so that the metric is variable, and then the EM action is diffeomorphism-invariant (if we do it right). So maybe this isn't a problem with your calculations as much as clarifying what can be varied. | |
| Jan 26, 2019 at 18:42 | history | asked | Will | CC BY-SA 4.0 |