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Can anybody point me to an English translation of Heisenberg’s three part paper Über den Bau der Atomkerne?

Failing that, can you point me to a non-paywalled German version of part III? I have parts I and II in German, but not part III.

Thanks in advance. I shall award a 100-point bounty to the answer that best assists.

Edit 14/01/2017: and if nobody can answer that, here's a bonus question. But you have to answer both parts to qualify for the bounty: If the nuclear force is charge-independent then a) how come there are no diprotons, and b) what's responsible for the repulsion in the deuteron?

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Image courtesy of Dux college

John Duffield
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    You can bypass the paywall using sci-hub – lemon Jan 07 '17 at 16:46
  • Regarding Sci-Hub – rob Jan 09 '17 at 14:22
  • I suppose that with question (a) you mean that the answer should be "irrespective of electromagnetic interactions"? In any case, note that charge symmetry doesn't mean that the force doesn't depend on total isospin. The deuteron has isospin $I=0$, while a di-proton has $I=1$. – pppqqq Jan 14 '17 at 17:00
  • @pppqqq : no, I didn't mean irrespective of electromagnetic interactions. The nuclear force is supposed to be stronger than the electromagnetic force: "at short range the attractive nuclear force overcomes the repulsive electromagnetic force". I found the 1936 paper that is said to have established charge independence to be rather unconvincing myself. See Theory of Scattering of Protons by Protons by Gregory Breit et al. – John Duffield Jan 14 '17 at 17:13
  • That phrase on wiki refers to nuclei in general, it explains why they don't "fall apart". However, if you look at Segré's chart, you'll see that light nuclei's all tend to have $Z=N$, the neutrons being there to hold the nucleus together... which is why I find quite difficult to believe that electromagnetic force wouldn't play any role in a hypothetical di-proton nucleus. [...] – pppqqq Jan 14 '17 at 17:28
  • [...] You might want to give a look to Bertulani, "Nuclear Physics in a Nutshell". He asserts that "experimentally there is no $I=1, I_z=0$ bound state" (which would correspond to a neutron and proton with total isospin $I=1$). Charge symmetry then implies that there is no di-proton or di-neutron state (irrespective of electromagnetic forces, which would make the situation even worse for the di-proton). – pppqqq Jan 14 '17 at 17:31
  • @ pppqqq : all points noted. Thanks re Bertulani. Perhaps I should ask a separate question about isospin. – John Duffield Jan 14 '17 at 17:39
  • Isospin and the diproton/dineutron: http://physics.stackexchange.com/q/153422/44126, http://physics.stackexchange.com/q/78/44126. – rob Jan 15 '17 at 01:45
  • Meson spectrum and attractive vs. repulsive interactions: http://physics.stackexchange.com/q/221908/44126 and links therein – rob Jan 15 '17 at 01:53
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    Your "bonus questions" are different enough from your reference request that they might stand on their own rather than fit here as an edit. – rob Jan 15 '17 at 01:55
  • @rob : that so somebody can get the bonus even if there's no English translation. PS: I like to think that I know why the deuteron doesn't have spin 0, and why two neutrons repel. But any answers I gave would be unconventional, so I'll keep them to myself. – John Duffield Jan 15 '17 at 14:03

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