Carrock

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)

Sorry, not very rigorous like many in this thread but this one's easy to respond to. e.g. The concept of unlimited size means (by analogy with an unlimited but finite set of integers, for example) that however far you walk, you can always take one more step. i.o.w. the concept of a finite or countably infinite universe expanding for unlimited (but not infinite) time is perfectly OK. A universe which has expanded exponentially for infinite time is inconsistent with any current theory as its volume would be uncountably infinite. The 'I can't get to the infinite future from here' argument doesn't work very well since photons, which have infinite range, can get there from here. Models which predict unending expansion are therefor problematic.

Reference? Did you actually read my post? I, and perhaps you, stated the probability of the universe being as it is to be unity; but you say I am wrong because I believe the universe is almost certainly different from what it is(!) Where is an intelligent designer required in this (observationally refuted) universe or any of its more sophisticated successors? BTW I should clarify that Boltmann's assistant's universe and AFIK all similar universes are not fine tuned; they do not require an intelligent designer. I used the human concept of 'fine tuning' which was misleading; the universe is no more 'fine tuned' than a waterfall is 'fine tuned' to be beautiful. Claiming that the universe we exist in (not multiverse in many theories) necessitates the existence of an intelligent designer is just a creationist fantasy. No mention by you of eternal inflation Bayesian probability etc.

I'm guessing you're responding mainly to me. No. I don't know anyone who claims there is any possibility the universe is different from what it is. But given my limited knowledge, the odds against even the earth being the way I expect it to be next week are huge.... A sample of one where one person has won the lottery once (or twice) tells you nothing about fine tuning. IIRC there have been a few people who've won a major lottery in successive weeks. Why are these people special when single winners are not? Actually the chance of a single or double winner winning the lottery again, from just one ticket, is exactly the same as for someone who's never won. (In the real world, the chance of an innocent three times winner being convicted of fraud is significant.) In other words, tossing a fair coin and getting five heads in a row does not change the odds (50%) of getting a head on the next toss. Bayesian probability is calculated on current knowledge. i.e. here the probability of having five heads in a row is 100%; a sixth head is 50% not 1.5625%. As there is not perfect knowledge of the universe, it is possible and useful to predict aspects which have not yet been observed. For example: The original classical fine tuned universe theory (by Boltmann's assistant) postulated that a random low entropy fluctuation in a heat dead universe created the solar system and everything needed for human observers to evolve. It was noted, in a Bayesian way, that the probability of low entropy in so far unobserved parts of the universe must be less than one in a googolplex i.e. the universe must be minimally life friendly.. The discovery of low entropy in previously unobserved parts of the universe showed that fine tuning theories can be falsifiable. Various theories, such as eternal inflation, postulate the creation of an unlimited number of universes with different laws. It seems to me that the existence of observers in such theories is only really possible in intelligent life tolerant universes with tight parameters, much as human beings pretty much require a life tolerant part of earth, even though it's supposedly not impossible for them to exist briefly as Boltzmann brains. A key aspect of many/most anthropic theories is the prediction that the (unobserved) universe will be found to be minimally tolerant of life with human equivalent abstraction since even finer tuning is required for extra 'unnecessary' life friendliness. Note this may be one observable universe of a possible multiverse; already known facts are of limited or no value (Beyes) as in the disproof of Boltmann's assistant's theory. It may be be most other intelligent life communicates with yobba rays or whatever, but 'fine tuning' suggests intelligent life is so rare we may never find any despite improving search techniques. Even if this is the one and only universe, it is still worth producing theories which predict things which may exist but have not yet been observed.

If that was true, the chance of a creator existing without a creator to create it is, essentially, impossible. Plus another creator to create the creator of the creator..... BTW what is an essentially impossible chance?

"Shockingly inefficient design?" Only if you believe in a rather incompetent intelligent designer. I'll assume a weak anthropic principle i.e. that our universe is constrained by the necessity of having at least one instance of intelligent life capable of wondering if its universe is constrained by an anthropic principle. There's no obvious size or practical time constraint imposed by physics, and if the only 'necessity' is having at least one instance of intelligent life, then the laws are only constrained to allow a minimally intelligent-life-tolerant universe. Earth may have been extraordinarily lucky to have remained intelligent life tolerant for so long; the dinosaurs were around for 100 million years without evolving human equivalent abstraction and technology so that may also be extremely rare. I doubt any anthropic principle will ever become a truly falsifiable theory, but the continuing failure of SETI to find evidence of extraterrestrial life despite improving technology does fit with the idea, dating back to Boltzmann, that the universe is minimally intelligent life tolerant.

I glanced at lyincomey.com and noticed they were referencing FAKE NEWS sites like CNN etc. Surprisingly, neither Trump nor his supporters have realised that Comey must therefor be telling the truth...

If I look at an orange in daylight, or under artificial light with a different spectrum it always looks orange coloured to me because evolution has enabled me to see colours largely independent of the quality of the illuminating source. If I photograph the orange under both these conditions with no colour filtering, the orange in the photos looks to be different colours. Which if any of these colours is 'scientifically' correct?

pt and ps appear to be defined by you as constants. Therefor se is also constant. I presume this isn't what you intend. More clarity required.

Every Windows system I've ever used slows down substantially as 'updates' are added. Unless you're willing to run Windows without protection from recent viruses the only solutions are to buy a newer Windows version or run a free OS such as Linux. (Mint is good for beginners.)

There doesn't seem to be any generally accepted definition of the Copenhagen interpretation. This is as good as any. (From https://en.wikipedia.org/wiki/Heisenberg_cut ) In other words you could have an atom as the observed system and the rest of the universe as the measurement apparatus. Or you could have the solar system as the observed system and the rest of the universe as the measurement apparatus. (Voyager 2 would do as an observer.) This seems to me rubbish but it's the best version of the Copenhagen interpretation I'm aware of. Can anyone clearly describe a better one?

I don't get it. A cushy office job involves putting your jacket on a chair in your office in the morning and collecting it in the evening. Definitely no sitting down.