dmckee --- ex-moderator kitten

Some uranium ores have a powdery consistency, which poses a significant danger of inhalation after even gentle handling. That would be bad as you then have non-trivial alpha activity in the lungs.

For those who might be wondering, it only take about a week for the twitching need to answer questions on the site to fade.

@user8718165 I'm not back in any substantive sense—just leaving a trail of breadcrumbs—but hello to you as well.

After all, it would be gauche to just provide a link to the one of the various "replace stack exchange" efforts that I'm currently looking at...

In an effort to provide some kind of continuity for those who asked, I've started a blog just so there is a place to get me on the internet.

I did go home and put their whole discography on shuffle.

Yeah.

Peart was always shy of either media or fan attention.

@JMac I’ve seen them in concert four times, and have all the studio albums. I don’t actually have the Rio concert album.

@JohanLiebert A bulk non-polar substance works because evey part of it is non-polar. Any perturbation can be expected to spread.

Showing them a lunar eclipse and the transit of Venus might have something to with it. Or perhaps it was the blue glowing jack-o-lantern.

@JohnRennie at a previous residence the neighborhood kids had dubbed me “Mr. Scientist”.

And my expertise is pretty limited. I'm reasonable conversant with the kind of experiments you might do in an undergraduate lab and I know two sub-fields of experimental nuclear and particle physics pretty well (but they're both niche areas).

@NikolaPetrovic It really depends. I have a regular work schedule and a toddler at home, so my time to chat is limited.

@PM2Ring Simple and effective. I'm a little aghast that I've never seen that before, but pleased as punch to have learned about it. Thanks.

On the matter of "How AI think?", I suggesting reading about Alpha Go Zero/AlphaZero (successor programs to Alpha Go that trained without an human data set). Their approach to playing the game baffles and intrigues human go masters.

For me it means times no later than the classical Greek period in Europe and equivalent times elsewhere, but is that a common usage?

What should "ancient" denote.

Though the division of the past into named periods is a learned convention.

@ACuriousMind ::chuckles::

I don't see much use for [wave-optics] separate of [optics], but including it as a alias might be reasonable.

Right now we have 800+ posts tagged [geometric-optics] (which has been around since 2011). But we seem to have gotten along without [wave-optics] all that time. Presumbly those questions have historically simply been tagged [optics] as Kyle suggests.

At that level those tags make sense. Of course, anyone who sticks with the topic rapidly comes to a point where they can see them as expressions of the same underlying behavior. at that level there are a bit redundant.

On the issue of [wave-optics] and [geometric-optics] (which is the master that [ray-optics] aliases), in the intro level course we teach they as two distict ways to treat light (typically in separate chapters and the connection between them is rarely treated in detail).

@user8718165 Indeed, and to you, too.

Personally I have been way down on chat time since I was able to bring my family out to the location of my new job this summer.

Many regulars will have interruptions from travel and unaccustomed social events.

This week isn't a great time for chatting.

Yep. The path of the object's motion bends in the direction of the applied force, and then the force is providing a change of speed as well as change of direction.

Programmers call that "rubber duck debugging". The notion being that you should explain the problem to the rubber duck (or other totem) that you keep around for the purpose.

YEah. Sometimes just framing the question carefully to ask someone else is enough to work out the solution for yourself.

You should compare this to the case of uniform circular motion where the force is always perpendicular to the motion and the KE is constant.

OK, if you applied the force in the y-direction for an infinitesimal time it would generate no work. But if you apply it for a finite time, then for part of that time the velocity is no long purely in the x-direction, so the angle between the force and the motion is no longer 90 degrees, so the sine term is no longer zero and the work is also non-zero, and the KE rises.

Ah.

And the change coms int he form of they-component changing.

Both the magnitude and direction of velocity will change. What will be unchanged is the x-component of the velocity.

@user8718165 Don't ask about asking. Just ask.

Hello.

Both systems give the same results, but I would argue that it is easier to confuse yourself with the old system. (In fact I have argued this is presentations at minor conferences...)

This is basically a shift in terminology to follow a shift in focus away from "that which is weird and unexpected if you know Newtonian physics" to "that which is clear to explain and easy to compute with on it's own terms".

The big advantage of that approach is that "Invariant" means "the same in every frame", and it is also easy to compute in every frame: just find the square of the four-momentum

Instead we use "mass" to mean only the invariant mass (what would be called "rest mass" in t he old nomenclature).

@Knight The thing that increase is the quantity $\gamma m$. This is sometimes called the "relativistic mass". But in the modern framing of the subject we simply don't assign a name to that quantity and don't frame explanations in terms of it.

You're facing the usual problem of audience and what the audience wants to do with your words. A description which is accessible and engaging to a pop-sci audience will inevitably be incomplete; worse if those readers try to do too much reasoning from such descriptions they will get led off track because "incomplete" becomes "incorrect" for those applications. And on Physics your words get read and judged by peole with a more technical background as well. Isn't that fun?

'Atomic orbitals are only "correct" wave functions in one-electron systems such as the hydrogen atom'" While this is defensible at some level, it'd not very precises and risks being misleading. Multi-electron atoms also have stationary states; they live is a higher-dimensional space, don't allow one to say "this electron is in this sub-space and the other one in that sub-space", and aren't simply products of the of single-electron orbitals; but they still "exist" in the same sense that single-electron orbitals do.

Similarly the "do not move" comment is defensible, but could get the naive reader in trouble if taken too strongly. Orbitals are stationary states meaning they don't change in time, which is why the statement has some value. But the electrons have a well defined energy which has to be interpreted as including a kinetic component and a momentum distribution which may include zero but also includes non-zero value with non-trivial probability density.

Too bad it told me I wasn't a very good teacher in most contexts.

Yeah for data!

Mind you, I never achieved much improvement over my lectures results (right around 0.25) by using "active" methods except in the small upper-division seminars I ran flipped (mid 30s, though the statistics were not good and the instrumentation rathr incomplete).