Timeline for Wrong explanation for why "electron can't exist in the nucleus"?
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Oct 31 at 17:24 | answer | added | ZeroTheHero | timeline score: 1 | |
| Oct 31 at 2:58 | comment | added | ZeroTheHero | @Andrew thank you for phrasing this in terms of Fourier analysis, as it is precisely what’s relevant here. The OP is using an extremely informal argument in which the important notion of simultaneous measurement is vacuous. | |
| Sep 15 at 7:28 | vote | accept | Amit Verma | ||
| Sep 17 at 11:24 | |||||
| Sep 10 at 5:58 | history | became hot network question | |||
| Sep 10 at 5:55 | history | reopened |
BioPhysicist benrg Vincent Thacker |
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| Sep 9 at 23:42 | review | Reopen votes | |||
| Sep 10 at 5:55 | |||||
| Sep 9 at 22:23 | history | closed |
naturallyInconsistent Jon Custer Miyase |
Not suitable for this site | |
| Sep 9 at 22:12 | comment | added | Andrew | If you don't have time to study Fourier transforms, you might consider trusting "us" (people who have) that you have not found a contradiction in the laws of physics, and it is both true that the electron wavefunction can't be fully localized in the nucleus for any significant amount of time and that there is no corresponding problem with the usual $1s$ wavefunction. | |
| Sep 9 at 22:10 | comment | added | Andrew | To compute the momentum, you need to do a Fourier transform of the position wavefunction. For a wavefunction localized within a nucleus, this integral will lead to a Fourier transform with a lot of support at super high wavenumbers (aka momenta). For a wavefunction like the $1s$ state which is spread out over the Bohr radius, the average momentum will be much smaller, and avoids the problem you are describing. These conclusions follow from a basic application of Fourier transforms. So if you want to understand the answer to this question in detail you should study Fourier transforms. | |
| Sep 9 at 21:34 | answer | added | benrg | timeline score: 8 | |
| Sep 9 at 17:05 | answer | added | JEB | timeline score: 5 | |
| Sep 9 at 16:55 | comment | added | controlgroup | Consider that momentum can be greater than $mv$. In the limit $v\to c$, momentum diverges towards infinity due to relativity. | |
| Sep 9 at 15:54 | history | edited | BioPhysicist | CC BY-SA 4.0 |
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| Sep 9 at 15:22 | review | Close votes | |||
| Sep 9 at 22:23 | |||||
| Sep 9 at 15:10 | vote | accept | Amit Verma | ||
| Sep 9 at 15:10 | |||||
| Sep 9 at 15:07 | comment | added | naturallyInconsistent | And again, if you understood the implication of "entire", you would have understood that you cannot just break up the wavefunction into bits and call it disproven. You can have as many nucleus-sized balls outside of the nucleus that you break the $1s$ orbital into pieces and claim that the electron cannot be inside them. That is just a silly misunderstanding on your part. Standard physics is talking about the entire wavefunction at once. That entire wavefunction cannot be localised into a ball as small as the nucleus, anywhere in the universe. | |
| Sep 9 at 15:06 | answer | added | BioPhysicist | timeline score: 8 | |
| Sep 9 at 15:04 | comment | added | Amit Verma | @naturallyInconsistent my second point is not about containing the $1s$ orbital in the nucleus it is that we can part by part say that electron also cant exist in the orbital outside the nucleus | |
| Sep 9 at 15:02 | comment | added | naturallyInconsistent | Sigh. Look, nobody is saying that the usual $1s$ orbital does not work. It is not that the electron's wavefunction cannot have bits that are inside the nucleus. It is that you cannot localise the entire electron's wavefunction inside the nucleus. Just because the person telling you this forgot to state the "entirely", or that you failed to read it, does not mean you found something wrong with standard physics. You either only found a sloppy wording, or you had a silly misunderstanding. | |
| Sep 9 at 14:56 | history | edited | Amit Verma | CC BY-SA 4.0 |
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| Sep 9 at 14:54 | comment | added | naturallyInconsistent | You are missing the forest for the trees. The wording may be sloppy, but the point is that the electron cannot be entirely within the nucleus. If the standard deviation of the electron's velocity is greater than the speed of light, then it also cannot be entirely contained within the speed of light, and that is also unphysical. You haven't really discovered anything interesting, just some sloppy wordings. | |
| Sep 9 at 14:44 | history | asked | Amit Verma | CC BY-SA 4.0 |