Timeline for Why does this falsely lead to $g_{\mu \nu}$ always being the identity map?
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Jul 9, 2022 at 9:26 | vote | accept | Max | ||
| Jul 8, 2022 at 21:23 | answer | added | Níckolas Alves | timeline score: 3 | |
| Jul 7, 2022 at 13:26 | history | reopened |
John Rennie Kyle Oman Michael Seifert |
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| Jul 6, 2022 at 21:10 | comment | added | Max | @Qmechanic Sorry, I thought you were in charge of keeping the question closed, and that your reason was that I hadn’t fixed the indices. I now see that it’s still being reviewed, so please ignore my previous rant. Yeah, step (3) is where it went wrong, when replacing the basis with the dual after application with the metric. The equality simply doesn’t hold, as I’ve now come to understand. What I should’ve done was to expand the metric in the basis and then take the tensor product of everything to see what happens between basis vectors (even if doing so is a bit trivial at times) | |
| Jul 5, 2022 at 19:33 | comment | added | Qmechanic♦ | The upper & lower indices do not match in eq. (3). | |
| Jul 5, 2022 at 19:29 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Jul 5, 2022 at 15:19 | comment | added | Max | I figured out what I did wrong! I can’t post an answer since this is closed, but thank all of you for your help (especially Andrew!) | |
| Jul 5, 2022 at 14:21 | comment | added | Max | $\partial / \partial x^\mu$ is a vector, ie a map from the set of smooth, real functions defined on the manifold to real numbers. Inserting it into the metric gives a covector. I think the confusion here comes from the fact that $g_{ab} \left( \partial / \partial x^\mu \right)^a$ means that the vector is inserted into the first slot of the metric tensor. This yields a map $V \mapsto \mathbb{R}$ where $V$ is the space of vectors, ie a covector. | |
| Jul 5, 2022 at 14:09 | history | edited | Max | CC BY-SA 4.0 |
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| Jul 5, 2022 at 14:08 | comment | added | Andrew | Alternatively, perhaps what you intended was something like $$ g_{\mu\nu} = g_{ab} \frac{\partial x^a}{\partial x^\mu}\frac{\partial x^b}{\partial x^\nu} = g_{ab} \delta^a_\mu \delta^b_\nu = g_{\mu\nu} $$ which would represent a trivial coordinate transformation from one coordinate system to itself. (?) | |
| Jul 5, 2022 at 14:07 | comment | added | Andrew | You have to be really careful with partial derivatives. Usually you want to evaluate operator equations by acting on a test function. If you operate on a test scalar function $\phi$, then I would have written your last equation as \begin{equation} g_{\mu\nu} \frac{\partial \phi}{\partial x^\mu}\frac{\partial \phi}{\partial x^\nu} = \frac{\partial \phi}{\partial x_\mu} \frac{\partial \phi}{\partial x^\mu} \end{equation} I don't understand how you got from the partial derivative $\partial_\mu$ to the one form $dx^\mu$, but there must be an error there because the index heights don't line up. | |
| Jul 5, 2022 at 14:05 | review | Reopen votes | |||
| Jul 7, 2022 at 13:26 | |||||
| Jul 5, 2022 at 14:03 | history | edited | Max | CC BY-SA 4.0 |
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| Jul 5, 2022 at 13:53 | comment | added | Max | I know that it’s not always flat. I want to know why I get to that (false) conclusion for a general basis (hence the “why does X (falsely) lead to Y?” rather than “help me prove that X leads to Y”). Obviously something is wrong in this line of reasoning, but I can’t find out what that is. I think the confusion here comes from the use of abstract index notation. I added a paragraph to clarify that. Basically, Latin indices are not components in any basis, merely symbolic. | |
| Jul 5, 2022 at 13:52 | history | edited | Max | CC BY-SA 4.0 |
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| Jul 5, 2022 at 13:46 | comment | added | mmesser314 | Perhaps the confusion is the $g_{\mu \nu}$ is diagonal in flat space, and this was an attempt to prove that it is always flat? | |
| Jul 5, 2022 at 13:44 | comment | added | Qmechanic♦ | What is the title question (v2) supposed to mean? $g_{\mu \nu}$ is not always diagonal. | |
| Jul 5, 2022 at 13:42 | history | closed | Qmechanic♦ | Needs details or clarity | |
| Jul 5, 2022 at 13:40 | comment | added | Max | I’m using abstract index notation, so $(\partial/\partial x^\mu)^a = (\partial/\partial x^\mu)$. The index $a$ is only there to denote that it’s a tensor with one contravariant slot. In that sense, I’m using $g_{ab} T^b = T_a$ with basis vectors. I assumed that going from vector to covector using the metric (ie changing index from upper to lower or the other way around) was the same as switching between the basis vector and its corresponding dual, but maybe that is wrong? | |
| Jul 5, 2022 at 13:38 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Jul 5, 2022 at 13:34 | comment | added | Andrew | (a) All the equation $g_{ac}g^{cb}=\delta^b_a$ is saying is that $g g^{-1}=1$. (b) It is totally unclear to me what is going on in your last equation. What is $(\partial/\partial x^\mu)^a$? How did you go from $(\partial/\partial x^\mu)^a$ to $(dx^\mu)_a$? Of course this step is why you go from a lower $\mu$ index to an upper $\mu$ index, so that step must be incorrect, but I have no idea what logic you are using to write that down in the first place so I am not sure what error you are making. | |
| Jul 5, 2022 at 13:27 | history | asked | Max | CC BY-SA 4.0 |