Carrock

Perhaps one day in the dim and distant future there will be a very very very large baseline array microHertz telescope, with astronomers lobbying for a nanoHertz telescope. BTW I think astronomers could learn from chess, where they stopped at 'hypermodern' in the 1920s rather than continuing with ultrahypermodern, superhypermodern etc.

Would you expect your ancestors to be able to eat with a knife and fork? You claim that two million years of separate evolution is insignificant in a species that breeds at about one year compared to humans' 20ish+. Domestic cats have lived in Britain for about 2000 years. If your definition of species requires 'never interbreeds with other species' then there are far fewer species than generally accepted and the Scottish Wildcat has been extinct for at least 1500 years as has the British domestic moggy. Cro-magnons interbred with Neanderthals so perhaps H. Sap. isn't a species either... On the other hand. if you accept that Scottish Wildcats that have some domestic cat dna, but not enough to affect their aversion to humans cannot be tamed, then we seem to be in agreement.

From https://blogs.scientificamerican.com/running-ponies/a-cat-that-can-never-be-tamed/ I'd say the onus is on you to find a tamed wildcat. I recall reading some years ago ago of a couple of wildcat kittens being 'rescued' and reared as domestic cats. They were friendly as kittens but as they grew up they became impossible to handle.

The wild doesn't really exist in the world, except in places remote from people. Some Scottish domestic cats, both pets and 'semi-feral' cats which avoid humans, interact with Scottish wildcats and are at risk of being killed by eagles. A somewhat OTT description of the Scottish wildcat:

That would work as they all know Fake News is lying about the venue and indeed the announcement of his death is a conspiracy against them.

Genetic defects like that, caused by inbreeding, is a major point of naitche's post.

I didn't watch all the video but he appears to have taken a decrypted backup copy, fitted new bigger memory and restored from backup - no security issues. If you've forgotten the password how do you comply?

From https://www.apple.com/business/site/docs/iOS_Security_Guide.pdf Apple is deliberately a bit vague, but it seems the encrypted files cannot be read without using encrypted addressing which will be permanently disabled after about six hacking attempts. Physical hacking, using e.g. a TTM to measure the stored charge corresponding to each bit, is probably not practical.

Not at all. It was good to have a quick response.

You are assuming that they are either far ahead of us or that these things have solutions, I'm not sure either is a certainty, it could very well be that we are fast approaching as far as science can venture. I hope not btw... I'm assuming these things have solutions or don't have solutions or are unprovable like, probably, the continuum hypothesis or whether the proton is unstable. I'm not assuming 'they' are ahead of us, simply that more than 50% of the current number of civilizations were around a million (or 100 million) years ago. Not long in the context of 13 billion years since BB. It's then a reasonable assumption that at least one in a million of those millions of civilizations within, say, a million light years, was ahead of us in any given field subject to scientific method and have communicated their knowledge to anyone within a million light years who would rather look at the teacher's handbook, as it were, rather than work the problem. 'It could very well be that we are fast approaching as far as science can venture.' I hope not too. Back in 1900 physics' main big challenges were the photoelectric effect and the ultraviolet catastrophe. Happily our Rumsfeldian ignorance (known unknowns) has mostly been increasing faster than our knowledge ever since.

It could be a disaster for anyone working in fundamental physics, maths and some other sciences. There would be solutions for all the interesting current and future problems or proof there was no solution. If I were Galactic Dictator I would only only allow discussion of unsolvable problems like what is the best art or the best society, or discussion of largely solved problems (for us) like what is the best description of relativity. in so many areas, not just physics and maths, searching for knowledge is at least as valuable as what is discovered.

Both of these look good, not only for scienceforums. I don't use latex much on this forum, which is why it always seems like a major task to get [math] or \( etc right. Like almost everyone who didn't write the software, I feel the recent 'upgrade' was in effect a downgrade. I use gummi a fast wysiwyg latex editor on my computer. Not windows friendly, but can be easily installed on e.g. most linux Ubuntu OS using Synaptic Package Manager. (I use Linux Mint.)

Alternatively, the amount of americium in a typical new smoke detector is only about thirty times the maximum permissible body burden. From https://en.wikipedia.org/wiki/Americium-241 From https://en.wikipedia.org/wiki/Smoke_detector The danger in each home is negligible, but there are many million ionising smoke detectors in landfill, where containment fails and groundwater is contaminated. Optical detectors are much better for domestic fires and IIRC not much more expensive.

The Americium used in smoke detectors emits alpha particles which have very short range and can be stopped by a sheet of paper so not much hope detecting their radiation from a distance. Smoke detectors can be detected from a distance in the form of Americium contaminated runoff from landfill sites where old smoke detectors have been dumped. I have worked with these very cheap detectors and in practise the containment is far below nuclear industry standard.

[math]\int_a^\infty f (x)dx = \mathop {\lim }\limits_{l \to \infty } \int_a^l f (x)dx[/math] [math]\mathcal{F}_{x} [\sin(2\pi k_0 x)](k) = \int_{-\infty}^{\infty} e^{-2\pi ikx} \left( \frac{e^{2\pi ik_{0}x} - e^{-2\pi ik_{0}x}}{2i} \right)\, dx[/math] got it for now....

I'm really sorry about that late night rant, where I pressed 'submit' before engaging brain. It does, and I meant 'quoting only part of post', which is perfectly OK; I do that myself of course and sometimes don't quote a relevant part of the post, which I feel happened here - no bad intent by you. Inaccuracy by me again. You were, I thought rightly or wrongly, summarising my views in an inaccurate way and I should have said that or, better, nothing. Good we agree about tunnelling. Still apparently disagree here with you and others - precise terminology is a problem. I still maintain that per unit volume s orbital electrons are most likely to be found inside the nucleus. Since the orbital is much bigger than the nucleus that probability is still very low. Sorry again about my personal criticism, which was unjustified and should not have been posted. I've cancelled my downvote. Thank you for your excellent response.

Thanks for very useful reply to sandbox test which I've just read. I actually have latex on my computer but every time I try it here I feel I'm starting from scratch. I'll give it another go and probably give up again when I spend two minutes on a reply and twenty minutes finding an 'obvious' latex error such as the one you pointed out. 1) The OP posted in quantum theory and did not ask for a classical discussion. When I started reading about forbidden regions, barriers which cannot be tunnelled through etc I didn't want the OP to end up totally confused. 2) I just used a simple example for clarity and hinted 'More practical ways are available....' that there were a lot of other ways. I'm referring, probably imprecisely, to the square of the probability wave amplitude. On the small scale of an atomic orbital, you can only talk of probabilities unless you destroy its stability by localising it with a high energy probe. Your definition is essentially macroscopic which isn't useful here. At sufficiently small unit volume, e.g. part of an atom, you can only say something like 'there is a 30% probability that if I measure, there will be an electron in this region.' Some reasons I gave Hanke a downvote... This is true, but it is not what the OP has asked. The original question was why the electron does not fall into the nucleus, i.e. how is an atom different from a purely classical system of a charge in free fall towards another (opposite) charge, which of course is not a stable situation in the classical domain. So the OP wanted to know how this is possible, so I have attempted to answer the question. The spontaneous tunnelling through the nucleus - or any other classically forbidden region - is not the same as the electron “falling in”. I get rather irritated with ...this is true ... except it's not. It's perfectly reasonable to claim I've got it wrong, but to edit my post and then imply I've said meaningless rubbish like 'The spontaneous tunnelling through the nucleus - or any other classically forbidden region - is the same as the electron “falling in”' is not. The point I make that the electron doesn't significantly interact with the nucleus but does fall through it has been ignored. Claiming that the OP is correct in the idea that the electron is never in the nucleus requires justification by Heinke.

I would be interested in the mathematical detail you offer since I wonder if you are misplacing the use of electron density as an observable? Like any quantum effect electron density is not directly observable but it can be 'observed' by firing high energy electrons (accurate position, inaccurate momentum) at hydrogen atoms with an s1 orbital electron and observing the scattering positions. Also fire electrons at proton to calculate nuclear scattering and allow for it in result. More practical ways are available.... If you construct a 3d rectangular grid around the atom you will find the most scattering per unit volume at and near the nucleus. You will also find that the greatest number of scatterings is at the Bohr radius. I'll do the Swansont type calculations for an imaginary atom which has uniform electron probability up to the Bohr radius a0 and zero elsewhere. I'm not going to bother saying things which should be obvious but would lose marks in an exam. For that atom, the equation in http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydrng.html#c1 simplifies to P = k * (integral b to c of r^2 dr) where k is a constant. 1 = k * (integral 0 to a0 of r^2 dr) - probability of finding electron in this range is unity. 1 = k * (a0)3/3 i.e. k = 3/(a0)3 (integral 0 to 0.001a0 of r^2 dr) = 10-9 * (a0)3/3 * k = 10-9. (integral 0.1 to 0.101a0 of r^2 dr) = ( 0.001030301 * (a0)3/3 ) - ( 0.001 * (a0)3/3 )) * k = 3.0301 * 10-5 So for uniform electron probability density the radial probability for a 0.001a0 shell increases with distance from the nucleus. In the Bohr model as the radius decreases the volume of shell decreases faster than the probability per unit volume increases. Do you maintain that one of these is not true?

\int_a^{\infty} f(x)dx=\lim_{L\to\infty}\int_a^{L}f(x)dx

This is true, but it is not what the OP has asked. The original question was why the electron does not fall into the nucleus, i.e. how is an atom different from a purely classical system of a charge in free fall towards another (opposite) charge, which of course is not a stable situation in the classical domain. So the OP wanted to know how this is possible, so I have attempted to answer the question. The spontaneous tunnelling through the nucleus - or any other classically forbidden region - is not the same as the electron “falling in”. I don't know the meaning of 'classically forbidden' in this context. The nucleus is not a forbidden region for an s orbital electron. The electron does not tunnel through the nucleus - that would require some sort of barrier which the electron had insufficient energy to cross classically. In stable atoms, for quantum mechanical reasons there is simply no energetically favourable reaction the electron can have with the nucleus. That is an incomplete answer to the OP but accurate as far as it goes. I've answered that: nothing stops the electron going all the way into (and out of) the nucleus and it does exactly that. A classically analogous question would be: If I jump from an aircraft, why can't I go all the way into a cloud? The answer is that you can, but a cloud is not a solid object and there is no energetically favourable reaction which stops you leaving the cloud. ...and also the idea in your quote that "there is a lowest energy level, which corresponds to the minimum distance an electron can be with respect to the nucleus." Bound s electrons of any energy have a maximum probability, per unit volume, of being found inside the nucleus.

I'm honestly not clear what point you're making. If you don't agree with either of these I'll respond in detail to your last post. 1: [ Origin is centre of nuceus.] The probability of an electron in the hydrogen ground state being between 0 and 0.001a0 is 1.3 x 10^-9 The probability of an electron in the hydrogen ground state being between a0 and 1.001a0 is 0.54 x 10^-3 2: The probability of an s orbital electron in a hydrogen atom in any (very small) unit volume is a maximum at the antinode (i.e. centre of the nucleus) and is higher throughout the nucleus than anywhere else. (The quantitative calculation of probability at or near an antinode is not simple but measurements e.g. of the half life of an atom where an s electron is captured by a proton-rich nucleus can give an experimental value.)

The only point in choosing a different origin is to show that with any arbitrary origin the radial probability of finding the electron at r=0 is always 0. The solution is different of course but asymmetry does not cause the electron to have nonzero probability at r=0. I was trying to fight the idea that electrons in stable atoms do not interact with nucleons because the radial probability of them being inside nucleons is almost 0 but it seems that meme is too powerful for me.

It is possible for an electron in any s state (lowest energy or any higher bound state) in an atom to pass through the nucleus spontaneously, as I've said several times in this thread. Radial probability is not probability per unit volume. The radial probability density at r=0 is 0, wherever you choose the origin of the coordinates to be. Mathematically, r=0 is a point at the origin, here inside the nucleus, and the probability of finding an electron at r=0 is 0. You could choose the origin to be at any point in the nucleus or elsewhere, and the radial probability of finding the electron at r=0 is always 0. As calculus is involved, it would be somewhat more precise to say "As r approaches 0, the electron radial probability approaches 0, for any location of the centre of the radial coordinate system." As electron capture in metastable atoms shows, the concept of s electrons passing through the nucleus (and having a higher probability per unit volume of being in the nucleus than anywhere else) is not just an abstraction without physical meaning. If you scroll down to 's-type drum modes and wave functions' , https://en.wikipedia.org/wiki/Atomic_orbital#Qualitative_understanding_of_shapes really does have some good graphics of orbitals and its easy to see the maximum electron probability per unit volume for s orbitals is at the nucleus.

I agree the most probable radius of finding the electron is the Bohr radius. From http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydr.html#c1 So there is a factor of r^2 corresponding to the surface area at any radial distance. The surface area corresponding to the nuclear radius is tiny compared to e.g. the Bohr radius so the ratio of these surfaces is tiny. If the electron density was uniform throughout space, the probability of finding the electron at a given radius would be proportional to r^2. The probability per unit volume (for s orbitals) increases as the nucleus is approached and is higher inside the nucleus than anywhere else. Probability per unit volume is not used much since it doesn't provide much of use like e.g. the Bohr radius and the calculation of probability very near the antinode would be difficult and probably pointless. Probability per unit volume might be useful here.... The problem with radial probability, especially for s orbitals, is that it is often conflated with probability per unit volume.

Just to clarify, while electrons in a stable atom can't fall into the nucleus, s orbital electrons can pass through it. The nucleus is not inherently a forbidden space for electrons. From https://en.wikipedia.org/wiki/Atomic_orbital#Qualitative_understanding_of_shapes (scroll down for some good graphics)