Timeline for Conformal transformation vs diffeomorphisms
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 28, 2019 at 23:52 | comment | added | MBolin | @knzhou I am a bit confused. Why diffeomorphisms are a symmetry of all resonable theories? | |
| Mar 28, 2019 at 23:50 | comment | added | MBolin | @NicolasFord, I think the reason your example is trivial because: 1) Rotations are an isometry of Euclidean metric and 2) Euclidean metric is independent of the point. So for your example, the definitions work. | |
| Mar 28, 2019 at 20:10 | comment | added | knzhou | @NicolasFord Yes, I agree 100%, the OP has a different issue. I'm just pointing out this other issue because it's a pet peeve of mine, and because otherwise OP would encounter a nasty surprise in a few pages and get confused all over again. | |
| Mar 28, 2019 at 20:09 | comment | added | Nicolas Ford | @knzhou: Thanks, that all makes sense, and it at least clears up everything for me. This really is not a good use of terminology! I have a feeling that this isn't exactly the crux of the original questioner's confusion, though; they seem to be stuck somewhere before this point. | |
| Mar 28, 2019 at 20:02 | comment | added | knzhou | @NicolasFord You're right that the formula as stated means what you said. However, this isn't the definition that is actually used later in the book. For example, even 5 pages later the book begins to speak of how to decide which classical field theories are conformally invariant. This doesn't make any sense if conformal transformations are a subset of diffeomorphisms, because just about all reasonable theories are diffeomorphism invariant, so knowing a theory is a CFT would be just about useless. | |
| Mar 28, 2019 at 20:00 | comment | added | knzhou | @NicolasFord A conformal transformation (in the sense of CFT) is a two-step process: first apply a diffeomorphism as you said, then apply an actual rescaling of the metric to cancel the $\Lambda$ factor. (This rescaling is allowed to also rescale the values of other fields.) At the end of the day, $g_{\mu\nu}'(x') = g_{\mu\nu}(x)$. | |
| Mar 28, 2019 at 19:46 | comment | added | Nicolas Ford | The definition I have in mind is that a conformal transformation is a diffeomorphism which preserves the metric up to scale, which I think is the same as the first sentence of your question. Rotations in the plane with the Euclidean metric is indeed a very particular example; it's just the example I understood you to be asking about. Certainly (a) not every conformal transformation is of this form, and (b) it's possible to put a metric on the plane for which rotations are not conformal. If this isn't what you're asking, then I'm afraid I still don't understand the question. | |
| Mar 28, 2019 at 19:39 | comment | added | MBolin | I think I don't know the mathematical definition. What I said about the Euclidean metric is that it is just a very particular example. From what I understand, conformal transformations should be a subset of diffeomorphisms, and you can see that this is not the case in general (for any metric) by taking the example of a rotation. | |
| Mar 28, 2019 at 19:34 | comment | added | Nicolas Ford | @MBolin: In the definition of "conformal" that I am familiar with from mathematics and the one that physicists use, which I have just learned in the last hour might be different, although I'm not sure how. But regardless, I believe that what you wrote in your question is what I thought "conformal" meant before this conversation. | |
| Mar 28, 2019 at 19:32 | comment | added | MBolin | @NicolasFord: sorry, there may be a difference in which definitions? | |
| Mar 28, 2019 at 19:31 | comment | added | Nicolas Ford | @MBolin: As I said in the last comment there may be a difference in the definitions that I'm not aware of, but it's worth pointing out that I think what you just said is not strictly true. There are certainly even linear transformations on the plane that are not conformal (and therefore not isometries). An example is multiplying the $x$ coordinate by 2, that is, $\begin{pmatrix}2&0\\0&1\end{pmatrix}$. | |
| Mar 28, 2019 at 19:28 | comment | added | Nicolas Ford | @knzhou: That's unfortunate! But it does seem like the definition I had in mind writing this is the one that the original question used in the first formula. Can you explain the discrepancy more clearly? I certainly don't want to contribute further to the confusion, so this answer ought to be edited if it's currently doing that. | |
| Mar 28, 2019 at 18:44 | comment | added | knzhou | This is definitely our (i.e. physicists’) fault. We stole a word from math and used it to mean several different things, and the notation conventionally used is so ambiguous one can never tell between them. | |
| Mar 28, 2019 at 18:43 | comment | added | knzhou | Your answer indeed gives one of the five definitions of the phrase “conformal transformation”. But it isn’t the one that CFT is actually about. Your definition makes conformal invariance just a subset of diffeomorphism invariance. | |
| Mar 28, 2019 at 18:42 | comment | added | MBolin | You are right, but I think that is not saying anything. Notice that for a simple metric like Euclidean metric, which doesn't depend on position, the formula for the transformation of the metric $g'_{\mu \nu}(x') = \frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} g_{\alpha \beta} (x)$ and the equation for an isometry $g'_{\mu \nu}(x') = g_{\mu \nu}(x') $ look the same. | |
| Mar 28, 2019 at 18:31 | history | answered | Nicolas Ford | CC BY-SA 4.0 |