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April 13

Feynman Lectures. Lecture 52. Ch.52-7

...
There are a lot of other features that were predicted. For example, it turns out that the spin, the angular momentum, of the cobalt nucleus before disintegration is 5 units of ℏ, and after disintegration it is 4 units. The electron carries spin angular momentum, and there is also a neutrino involved. It is easy to see from this that the electron must carry its spin angular momentum aligned along its direction of motion, the neutrino likewise. So it looks as though the electron is spinning to the left, and that was also checked. In fact, it was checked right here at Caltech by Boehm and Wapstra, that the electrons spin mostly to the left. (There were some other experiments that gave the opposite answer, but they were wrong!)
The next problem, of course, was to find the law of the failure of parity conservation. What is the rule that tells us how strong the failure is going to be? The rule is this: it occurs only in these very slow reactions, called weak decays, and when it occurs, the rule is that the particles which carry spin, like the electron, neutrino, and so on, come out with a spin tending to the left. That is a lopsided rule; it connects a polar vector velocity and an axial vector angular momentum, and says that the angular momentum is more likely to be opposite to the velocity than along it.

— Feynman • Leighton • Sands, The Feynman Lectures on Physics, Volume I

I have a question about the spin and the magnetic field vector. First, according to the description of the experiment the cobalt-60 nuclei were put in upward magnetic field causing the nuclei to line up and orient their spin vector upward also (rotation to the right png) . Nucleus is positive charged object, when it rotates this produces the small circle currents and magnetic field. Before the decay, nucleus spin = 5. After decay, the nucleus spin = 4. Electron spin = 1/2, neutrino spin = 1/2. Total angular momentum must be 5, so electron must rotate to the right also. If it rotates to the right it acquires magnetic field directed downward, because of negative current. What happens next (will electron repel by external field or will it perform the spin flip)? How did Feynman get electrons rotating to the left? Username160611000000 (talk) 19:47, 12 April 2018 (UTC)[reply]

Feynman is glossing over a lot of very difficult quantum mechanical physics, because the audience for this lecture had not yet studied it.

You get the spin of the electron by applying the angular momentum operator to the wave function for the system. This is not an easy task and it is not conducive to a short-form explanation.

@Nimur: Feynman writes It is easy to see from this that the electron must carry its spin angular momentum aligned along its direction of motion, the neutrino likewise. If it is easy, I think it can be explained using classical mechanics (with some warnings). Here [1] is written Particles with spin can possess a magnetic dipole moment, just like a rotating electrically charged body in classical electrodynamics. Username160611000000 (talk) 03:29, 13 April 2017 (UTC)[reply]

There is always a well-known solution to every human problem — neat, plausible, and wrong. - H. L. Mencken Tigraan^{Click here to contact me} 11:46, 13 April 2017 (UTC)[reply]

Here's a link to Griffiths' Quantum Mechanics text book. If you buy it online, pay special attention to the author's advice about retailers who sell older versions, and to his published errata.

One typically will use Griffiths' book after completing several years of advanced study of quantum mechanics at and beyond the university level.

Nimur (talk) 20:44, 12 April 2017 (UTC)[reply]

Thank you for the suggestion, but I have neither money nor time for other textbooks. Feynman Lectures contain quantum mechanics. I like Lectures and I will continue to read them.Username160611000000 (talk) 03:29, 13 April 2017 (UTC)[reply]

As for what happens next, unless you have a large apparatus or very strong field, the electron exits the external field essentially unchanged because it is moving at near-relativistic speeds. Cobalt-60 decay liberates 0.3 MeV in its most common beta mode, and a significant chunk of that goes into accelerating the electron. To the extent that it does interact with the field, the electric charge and high velocity is more important than the magnetic moment in determining the change in the motion of the electron in the magnetic field. The spin flip timescale will depend on details of the magnetic fields but anything from roughly < 1 ns to 10s of microseconds could occur depending on the nature of the apparatus (strength of field, shape of field, static/pulsed/oscillating, etc.). Dragons flight (talk) 14:02, 13 April 2017 (UTC)[reply]

Krane, Introductory Nuclear Physics, works this example problem for the cobalt beta decay and studies the apparent parity violation. Just as Feynman says, this is an easy problem - it's in an introductory textbook. The worked math is several pages long and is preceded by three hundred pages of preparatory material. Nimur (talk) 14:20, 13 April 2017 (UTC)[reply]

Somebody ripped off somebody here: the Feyman lectures go through the exact same examples - including the Wu experiment that Krane cites precisely (Phys. Rev. 105, 1413, 1957), and the Martian handshake analogy that Krane cites precisely (The Ambidextrous Universe, 1964), and attributes to Martin Gardner. So - the "Feynman lecture" appears to be the contents of Krane, Chapter 9, only presented in a sloppier and less physically-accurate fashion and without citing sources. Have I mentioned before that the "Feynman Lectures" were not actually written by Feynman, but were in fact "published essentially because an international publisher issued an ultimatum intending to republish those copyrighted works without permission"? Oh, I did mention that, and I even cited sources! Nimur (talk) 14:31, 13 April 2017 (UTC) [reply]

I was curious... how sure are people that the parity violation is part of physics itself? I mean, the Earth sits in the middle of a galaxy of fast-moving dark matter - can physicists rule out that some kind of interaction with a chiral dark matter particle could create the left-right asymmetry in these weak interactions? For example, have people been able to look at clouds of gas around rogue stars far from any galaxy and tell for sure that the radioactive decay rate (and ideally, ratio, e.g. around a rogue magnetar) is the same as on Earth? Wnt (talk) 23:53, 13 April 2017 (UTC)[reply]

Cats and dogs

Which breeds of dogs are the most cat-friendly (or maybe I should say "cat-tolerant" -- I don't think any of them are actually cat-friendly in the strict sense) and which are the most cat-unfriendly? And on this scale, where do bulldogs fit in? 2601:646:8E01:7E0B:4C74:BE2F:701E:8F7E (talk) 06:34, 13 April 2017 (UTC)[reply]

Anecdotally, my golden retriever thinks cats essentially are dogs. She is interested in them and does her normal "play with me" motions that she uses when approaching a new dog. Cats generally respond with a range of indifference, fear, or aggression, and mostly just want her to leave them alone. At least in her case, I would say she is quite friendly towards cats, though we've yet to meet a cat that is interested in being friendly with her. [And I keep worrying that she is going to approach the wrong cat some day and get her face scratched up]. So, depending on what you are looking for, it may depend not only on the type and personality of the dog but also how dog-friendly (or dog-tolerant) the cat happens to be. I doubt there are any absolutes when it comes to this, as it will depend both the personalities of the particular animals and their life experiences. A dog/cat that has had negative experiences with other cat/dog is unlikely to be friendly to a new one, so if possible it is probably better to introduce a new animal when it is young and do so gradually under controlled conditions. For example, start by having them in separate areas where they can see each other but not interact. After a while, when they are calm with that situation, gradually increase the contact in ways that aim to be as unthreatening and calm as possible. Dragons flight (talk) 13:28, 13 April 2017 (UTC)[reply]

Google "cat friendly dogs" and you'll find lots of info. Richerman (talk) 14:32, 13 April 2017 (UTC)[reply]

You might choose a small breed of dog, that way they will have a "balance of power", so the dog won't attack, partially out of fear of reprisals. Raising them together since puppies/kittens is a good way to establish a bond. Of course, they need to be the same age to do this. StuRat (talk) 15:41, 13 April 2017 (UTC)[reply]

You want to take the breed's purpose into account as well. Dogs bred for hunting or ratting (or otherwise harassing small animals), like terriers, are better avoided, all other things being equal. - Nunh-huh 05:20, 14 April 2017 (UTC)[reply]

Dog breeds that have neotenic traits such as floppy ears, curly tails, piebald coloration, shortened vertebra, large eyes, rounded forehead, large ears, and shortened muzzle, as exemplified by Cavalier King Charles Spaniel may be perceived by a cat as unaggressive. Terriers, sighthounds and herding breeds are most likely to initiate a confrontation. Click for advice on introducing dogs to cats. Cats operate with triple choice fight-or-flee-or-pointedly ignore agendas that are incrutable. Scratches to the face are the deterrance to any dog from a wary mother cat with kittens; a serious attack by a cat will be from ambush (usually unsuccesful and often aborted) to grasp and hold the dog by the neck. Blooteuth (talk) 16:50, 13 April 2017 (UTC)[reply]

Another thought is that you may want to get a dog with a similar activity level as a cat. Something like a basset hound may sleep as much as a cat. A cat may find a more active dog to be annoying. StuRat (talk) 05:55, 17 April 2017 (UTC)[reply]

Laser pointer reflected light harmful to eyes in close proximity?

At a community workshop I sometimes use is a laser cutter and it has what appears to be a red laser pointer used to indicate the location of cuts (the cutting laser is infra-red and can't be seen). I was staring intently at the red dot produced by the laser pointer and a fellow workshop member told me to avoid damaging my eyes by limiting my time spent looking closely at this red dot. In my opinion he misunderstands one of the warning labels talking about scattered radiation which I think is a reference to the 40 W IR laser and does not apply to the red indicator. Obviously a laser pointer used in a presentation presents less harm because the meters of distance mean large dissipation. I'd say that if the manufacturer used a laser pointer harmful to eyes in close proximity they designed a stupid indicator system. --129.215.47.59 (talk) 14:09, 13 April 2017 (UTC) [reply]

I'm aware that archives should not be edited. However, I am strongly concerned that some of the commentary below may be taken at face value without due diligence being done. To that end, I'm boldly hatting the discussion below with this comment. I will leave the original question unhatted. Please drop by on my talk page if you have questions. In the interests of health and safety, I would really appreciate the hatting be left as is. Blackmane (talk) 23:13, 1 November 2017 (UTC)[reply]

The following discussion has been closed. Please do not modify it.

Jeez. Don't people care about themselves. Are there written instructions for its use? Are you wearing the proper protective goggles for the machine? Dmcq (talk) 14:56, 13 April 2017 (UTC)[reply]

There are no goggles; the machine has a lifting cover which is placed down when cutting so the IR laser is not an issue. My question is about the visible laser indicator. --129.215.47.59 (talk) 16:08, 13 April 2017 (UTC)[reply]

It depends on the Laser safety#Classification of the red indicator laser. I checked a few saws with laser guides, and they were all Class IIIa (<5 mW maximum). DMacks (talk) 15:05, 13 April 2017 (UTC)[reply]

Rather than have us speculate irresponsibly about a system we have never seen, it would be better to tell us precisely what class laser is used and/or what warning labels are present. This information should be present on the device and its accompanying documentation. Dragons flight (talk) 15:28, 13 April 2017 (UTC)[reply]

I've sent an e-mail to the manufacturer of the machine to ask them about it. 129.215.47.59 (talk) 16:07, 13 April 2017 (UTC)[reply]

I would say it could cause eye damage, so why risk it ? Avoid looking directly at it for extended periods. (If you look at a white object right after you will probably see a green dot.) Does anybody make goggles with a darkened spot in the center, just for this case (it could even specifically block only red) ? StuRat (talk) 15:35, 13 April 2017 (UTC)[reply]

It's not like I'm looking at it to have a good time! It's kind of necessary when trying to line up material to be cut. I guess a pair of any old sunglasses will at least reduce the amount of radiation experienced. I'd rather the manufacturer simply distributed machines with indicator lasers set to an appropriate intensity (I just told them so in an e-mail). --129.215.47.59 (talk) 16:07, 13 April 2017 (UTC)[reply]

The manufacturer has to set the power relatively high if the red dot is to be seen on many different materials. Super black says (without reference, but from my memories of optical measurements it is about right) that common black paint will reflect about 2.5% of incident radiation, probably (?) in a close-to-lambertian manner (i.e. the apparent light intensity is the same in all directions). On the other hand, if you are cutting a material that reflects a lot more of the incident radiation and has most of the energy concentrated near the specular angle, which is the case of many polished metals, the light intensity (in the preferential direction) may be higher by three or more orders of magnitude.

There is another parameter at play related to the biology of the human eye, but I could not find it with a quick web search: the range of light intensity between "I cannot see the dot" and "It hurts my eyes". If that range is too small, then the manufacturer has no way to make a laser that is seen comfortably on all materials under most angles of incidence, but it is not clear to me that the range is indeed small. Tigraan^{Click here to contact me} 17:30, 13 April 2017 (UTC)[reply]

I have read: 100 trillion times. x: You barely see the light source with full dark adaption. 100 trillion x: You can barely read black newspaper headlines-type text on dark(ish) paper with nothing whiter in your field of view cause it hurts and if dark paper looks white and hurts imagine turning around and seeing the light source. And that's if the light source only produces visible light. Maybe someone can find the source. Sagittarian Milky Way (talk) 06:13, 14 April 2017 (UTC)[reply]

Maybe you can tape some red plastic sheets to your safety goggle to reduce the amount of red getting through, without impairing your view of the work area. Or, if you do just use regular dark glasses, be sure to increase the general light level. Also, the darker the surface it reflects off of, the less will reflect. So, you could use black marker on the area first, if it won't damage a show surface. StuRat (talk) 17:23, 13 April 2017 (UTC)[reply]

Maybe you can tape some red plastic sheets to your safety goggle to reduce the amount of red getting through - you are technically correct, but certainly misguided. If the plastic sheet is red, it is because red is the color it does not absorb well. So you are trading (made-up numbers) 10% of the laser light attenuation for 80% of the rest of the spectrum, hence impairing the view more than reducing the red dot; a blue plastic sheet would be less worse.

Of course, in any case, if there is a laser-related eye injury possibility, "tape some plastic sheet" is never, ever going to be the correct procedure. Tigraan^{Click here to contact me} 17:37, 13 April 2017 (UTC)[reply]

Compared to nothing, it sounds like a distinct improvement. If he can find an official OSHA-approved set of safety goggles that filter out red light only, then he should go for it. If not, any solution is better than what he has now. As for the color of the plastic sheets, if they reflect red only, then they should appear red. This is what I had in mind. StuRat (talk) 03:36, 14 April 2017 (UTC)[reply]

If you look for laser protection glasses (example) you will find that the protection levels indicated by "OD" (see optical density and [2]) mean the relevant laser harmonic is cut by a factor of 10^-6 to 10^-9. Big OR guess, but plastic sheets are nowhere close to these levels of transmittance (I would be surprised if they cut by more than 10^-2). When you say it is "better than nothing", it is only marginally so if anything, and such security theater may give operators a false sense of security which would increase the risk by a substantial amount.

As for the color of filters, you seem to imply that the red color of plastic sheets is linked with a high coefficient of reflexion in red wavelengths. If that was the case, a red object seen through such a sheet would appear black (since the light would be reflected on the backside of the sheet), and the reddish color would be of uniform intensity of the sheet (since the red light does not come from objects behind, but from reflection on your side of the sheet). You can try it yourself, the object actually appears comparatively brighter than similar objects of different colors, so it is because in the red, transmission is high.

STuRat, such answers are why you regularly get roasted for your RefDesk answers: (1) your "red is due to reflection, hence we should use a red sheet to filter out red" thinking was made up on the spot; in itself, a reasonable guess is not a big issue, but you need to say what part comes from a source and what part is your own guess; (2) when we are talking about something that could even remotely touch to safety, it is not only a RefDesk and Wikipedia guideline but also a moral duty to avoid "reasonable guesses" altogether. Your "tape some plastic sheet" advice is particularly problematic in that regard; a short search for "laser safety" or equivalent would have told you either that you don't understand the details of optical absorbance and stuff, or that no readily-available materials have optical properties even close to what is needed, both of which should have stopped you from typing the above. Tigraan^{Click here to contact me} 12:02, 14 April 2017 (UTC)[reply]

When talking about reducing the light level by a million or billion times, that doesn't sound like it's dealing with the small amount reflected from a sighting laser, but rather the level of protection needed against a direct hit on the retina by a serious laser. That's not what we're talking about here. This Q is about whether any protection is needed at all. Indeed, if you blocked that much of the sighting laser light, it would no longer function to show where the cut will occur, putting the operator at risk of being injured by the cutting laser. Thus, your advice is seriously dangerous. Please provide a source that states that that level of protection is recommended for the reflection from a sighting laser. StuRat (talk) 16:02, 14 April 2017 (UTC)[reply]

@StuRat: Please point to the place I have advised to use laser-class protection glasses for that particular use, or actually any positive advice to take a particular course of action. I haven't, because we shouldn't (WP:RD/G/M); except maybe to gather further information before taking some protective action, such as your suggestion of maybe you can tape some red plastic sheets to your safety goggle. That is clear advice (unless you meant it as a physical possibility, as in "maybe you could sing O Sole Mio before machining"). Do you realize the irony in asking me to provide a source?

Furthermore, you know fully well the source you request cannot be found, because we have no idea what is the power of the laser guide, its wavelength, the type of materials it is used on (reflection coefficient and spatial distribution of reflected energy, see other answers), the distance and angle at which the operator stands from the laser spot. Which is, of course, another excellent reason not to give any advice as to the course of action to follow. Tigraan^{Click here to contact me} 17:39, 16 April 2017 (UTC)[reply]

You provided sources with laser protection glasses that cut the light by "10^-6 to 10^-9". Then you said "Big OR guess, but plastic sheets are nowhere close to these levels of transmittance (I would be surprised if they cut by more than 10^-2)". It's pretty obvious to me you are saying that this is not enough and they should cut the transmittance by more than that, which is dangerous advice, given that this is sighting laser, and they don't want to lose sight of it. StuRat (talk) 04:23, 17 April 2017 (UTC)[reply]

If you're concerned, you could probably find the manual for the machine online and read the warnings and/or operating instructions yourself.

It sounds like an over-reaction, but there may be a non-obvious concern about reflections or something.

It certainly seems unlikely that the machine would be manufactured with a guide that you weren't supposed to look at! But I suppose it's possible that it originally came with some safety equipment that has since been lost or improperly replaced. ApLundell (talk) 18:03, 13 April 2017 (UTC)[reply]

There are three hazards. 1. Cutting beam to the body. Not good, avoid this by keeping the lid shut, and having safety interlocks so that there's no beam if the lid is open.

2. This is the dangerous one. Staring at the bright white light of the cut taking place. This will damage your retinas, if you persistently stare at it, just the same as staring at the sun will. If you do, do it briefly. Or look at it sideways, not by staring at it in the centre of your vision. It's worse for some materials (engraving refractory materials) rather than cutting others (cutting MDF you're lucky to see it through the smoke).

3. The red dot pointer. Here you need to appreciate the difference between a diffuse and a specular reflection. A "laser pointer" grade laser is slightly hazardous as a narrow beam, not hazardous as a diffusely reflected beam. If you have a specular (i.e. shiny and mirror like reflection), then that's the same as staring straight into it. If you bounce it off MDF or plywood though, anything more than a foot away is eye-safe for credible operation. So don't have polished metal inside the laser box.

Few other tips. Don't have safety goggles around the place. If you must (maybe you have a nasty laser that needs them for alignment) then keep them under lock and key and only the "service person" who needs them gets access to them. Safety goggles are rarely safe. They have a bunch of associated risks and they encourage sloppy working. You are far better off just keeping the laser safely shut in its working box.

Keep the laser in its box. Make sure the box can't leak laser anywhere. Make sure the box is big enough to enclose the whole tube, especially if you upgrade the tube and fit a longer one. Make sure (and regularly test) that the safety interlocks cut the laser power off. Don't allow those safety interlocks to be defeated.

Replace the "red dot" laser with one that works. Either a beam combiner, so that the red dot travels down the same optical path as the cutting beam (serious money to do this, but it's how real laser cutters do it). Or else replace the one dot with two line laser modules ($5 each) arranged so that the alignment pattern is a cross of two radial beams. This way the "spot" stays on centre even as you adjust the focus up and down. Especially when you upgrade to a 4" lens for cutting work.

I have yet to see a Hackspace with a good operations manual or safety guide. Andy Dingley (talk) 18:16, 13 April 2017 (UTC)[reply]

I'm impressed by that last, but some of the earlier comments above do not inspire me with confidence. For example, I'd expect wearing sunglasses to be only marginally effective because the eyes will dilate further to compensate for lower light, and unless you're actually fitting the laser beam through your iris with room to spare, that means an increase in injury to largely cancel the decrease in room brightness. Now if the sunglasses/goggles are so dark you're complaining you can't see (you've put your irises all the way open and it's not enough), then you might really not see as much laser, and not get damaged as much. But when I look up real laser safety goggles I see absorbance curves like this: [3] i.e. they do block certain frequencies very well, and not by guesswork with red tape!!! (But there's still some guesswork if the laser owner does his homework wrong) But how much do you want to bet on that...? Wnt (talk) 23:49, 13 April 2017 (UTC)[reply]

I'm curious about the assumptions of the answers here. Nearly all answers are based on shining a laser directly into your eye. The question claims to be looking at a little guide dot or line on a cutter. In other words, the laser is pointed at the item being cut. Is there a different type of cutter in which you shine a laser in your eye while trying to cut something? 209.149.113.5 (talk) 14:17, 14 April 2017 (UTC)[reply]

The problem is that many people don't read the Q completely before answering. They read "laser cutter" and "harmful to eye" and just assumed the Q had something to do with taking a high-powered laser directly in the eye. StuRat (talk) 16:58, 14 April 2017 (UTC)[reply]

I assumed people were worried about the possibility of a stray reflection. From trying to cut something shiny. ApLundell (talk) 18:08, 14 April 2017 (UTC)[reply]

There are not many shiny materials that are laser cutttable, or engravable. We're not talking about metal-cutting lasers here.

Also this type of laser cuts flat sheets - if anything is reflective (and mirror acrylic is the most likely), it will reflect directly upwards, not off to the side where the operator is looking. Andy Dingley (talk) 19:10, 14 April 2017 (UTC)[reply]

Well, what if a wooden substrate has a staple in it from an old No Trespassing sign, or a steel pellet from a grouse someone shot at before loggers ravaged the local gameland? I am not entirely confident about this. Wnt (talk) 19:22, 15 April 2017 (UTC)[reply]

Then you will get a flash of red light in your eye. And as you have a blink reflex to visible light, and the incident power is still pretty low, then the momentary exposure to this will be harmless. The big hazard for low powers is always invisible light, because of the lack of blink reflex or aversion.

Also this requires a specular reflection, and "old staples" will be anything but. Andy Dingley (talk) 20:24, 15 April 2017 (UTC)[reply]

The blink reflex is precisely one of the reasons some visible lasers could have a high enough power to do bio/physical damage to the eye but are at a lower class: the blink reflex limits the exposure-time so the actual amount of energy reaching any spot on the retina is low. DMacks (talk) 14:05, 17 April 2017 (UTC)[reply]

Adhara (second brightest star of Canis Major)

1. Why's it so obscure in the US and presumably Europe for something magnitude 1.50 and only -28°58'?

2. What's the millennium when it straddles the 1st magnitude/2nd magnitude border? Sagittarian Milky Way (talk) 15:13, 13 April 2017 (UTC)[reply]

-28°58' is still very low for a significant part of the Northern Hemisphere. For instance, at 55° North latitude it will be near the horizon. Ruslik_Zero 20:39, 13 April 2017 (UTC)[reply]

US star charts are often 40°N since it's close to average. I live north of that and have no problem seeing it in polluted New York City but I didn't know it's name. Sagittarian Milky Way (talk) 05:41, 14 April 2017 (UTC)[reply]

Star names are generally obscure (with the exception of the very brightest stars), I wouldn't be surprised at that. As to the second part of the question, with a current apparent magnitude of m_V=1.5, it would have to decrease in brightness by 0.5 magnitudes. This corresponds to an increase in distance by a factor of

{\sqrt {10^{0.4\times 0.5}}}=1.26

. Starting with a current distance of 130 pc (as derived from the parallax; the absolute magnitude in the table is an old value and doesn't fit with the parallax), it would have to be at a distance of 164 pc to appear with magnitude 2. With a radial speed of 27.3 km/s, this will be reached in about 1.2 million years. --Wrongfilter (talk) 10:05, 14 April 2017 (UTC)[reply]

When was it exactly 1.5000 though? That's when it stops being one of the only ~16 visible first magnitude stars (the non-visible ones being mostly deep in Centaurus or in the Southern Cross) Sagittarian Milky Way (talk) 16:48, 14 April 2017 (UTC)[reply]

That level of precision is meaningless, in particular if you're talking about visibility with the naked eye. To start with, you'd need to be able to measure its current brightness to the same precision, and that is hard to impossible even with a well-defined and well characterised filter and electronic detector at a large telescope. Take a look at this abstract: They establish a V magnitude for Vega of 0.026±0.008 — Vega is pretty much the photometric standard star, and the error on its apparent magnitude sets the bar for all other stars. In fact, the uncertainty on Vega's apparent magnitude is one of the main uncertainties in the measurement of cosmological parameters from observations of high-redshift supernovae! Incidentally, Stellarium tells me that the atmosphere right now extincts Adhara down to 2.09 mag, just past culmination in my location (48°12′N) — 1.5 is the apparent brightness outside the atmosphere. --Wrongfilter (talk) 17:10, 14 April 2017 (UTC)[reply]

I believe Vega was zero by definition at one point but like the metre being 1/40,000,000th of Earth's polar circumference that didn't work out. How'd the zero point evolve from Pogson to today anyway? I never learned. Sagittarian Milky Way (talk) 17:30, 14 April 2017 (UTC)[reply]

2. "millennium"?? —Tamfang (talk) 07:43, 16 April 2017 (UTC)[reply]