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January 30

Why is the following value, relativistically invariant?

Here is a simple thought experiment: A photon and a massive particle, both being in a moving train, begin a race, starting from the rear of the train to its front, i.e. in the direction of the train's movement. When the photon arrives at the end of the race (i.e. at the train's front), the photon comes across a mirror, so the photon is reflected back, i.e. it bounces backward, on the same route but in the opposite direction, until this photon meets the massive particle - still on its way to the train's front.

The point at which the photon and the massive particle meet, specifies a part of the train's length. This part is a actually a proper fraction, i.e. its value is between 0 to 1. Intuitively and apparently, this fraction is relativistically invariant, i.e. it does not depend on the reference frame measuring that fraction.

David Mermin (in his book "It's about Time: Understanding Einstein's Relativity", p.37, in the first new paragraph of the page.), wants to assume (rather than to prove) that the fraction is indeed invariant, because he wants to use this assumption for - proving the relativistic velocity-addition formula without using the Lorentz transformations. Apparently Mermin regards this assumption, either as an axiom, or (more likely) as a conclusion of the principle of relativity, because this principle is mentioned (p. 29) as one of his two assumptions, the second one being the constancy of the speed of light.

But I'm looking for a more rigorous justification for his assumption (about the invariance of that fraction), again without relying on the Lorentz transformations nor on the relativistic velocity-addition formula. Unless someone explains to me how Mermin's assumption rigorously follows from the principle of relativity (if it really does). HOTmag (talk) 13:14, 30 January 2024 (UTC)

I'm not sure what you mean by "relativistically invariant" in this case. "X is invariant with respect to Y" means that the value of X doesn't change when Y does. What's Y in this case? Are you saying that the fraction is the same regardless of the speed of the train? Of the initial speed of the massive particle? Both? Something else? PianoDan (talk) 18:22, 30 January 2024 (UTC)

I mean exactly what you mean when you say that the speed of light is relativistically invariant, i.e. it does not depend on the reference frame measuring the speed of light. The same is true for the fraction: it does not depend on the reference frame measuring the fraction. HOTmag (talk) 18:52, 30 January 2024 (UTC)

Ah, OK. What you're looking for here, then is the concept of "Spacetime Interval", given for one dimension as: ${\textstyle (\Delta s)^{2}=(\Delta ct)^{2}-(\Delta x)^{2}}$. For any reference frame the computed value of delta s will be the same for two events, even though the values of delta x and delta t may vary from frame to frame.

Say you're in the reference frame of the train, and you observe the photon to collide with the particle at point P (relative to the train, so P is just the fraction in the problem times the length of the train). The two events here are "photon reaches point P" and "particle reaches point P"

In the reference frame of the train, delta x between those two events is zero, and delta t is ALSO zero. As such, the invariant spacetime interval between those events is also zero, which means it is zero in all frames.

To put it even more simply - events which are separated in space can be perceived to be simultaneous or not, depending on the frame of the observer. But events which are NOT separated in space have a well-defined meaning of "simultaneous," and are simultaneous in all reference frames.

This is also necessary for causality. If shoot a bullet at a pumpkin, different observers can disagree over precisely how long it takes between the shot and the pumpkin exploding. But the fact that the bullet hits the pumpkin AT ALL can't change depending on the motion of the observer, or you get nonsense.

PianoDan (talk) 23:00, 30 January 2024 (UTC)

Thanks. Let me quote you:

"The two events here are 'photon reaches point P' and 'particle reaches point P'...the invariant spacetime interval between those events is also zero, which means it is zero in all frames...events which are NOT separated in space have a well-defined meaning of 'simultaneous', and are simultaneous in all reference frames".

I still want to find out the logical way your argument quoted above can be used for proving that the fraction does not depend on the reference frame, although your argument can't be used for proving the same for the length, so first let me put your argument in my own words:

A given length involves its two endpoints, being two different "events" separated in space by a non-zero interval. However the fraction (I'm asking about) is only specified by a single point (e.g. the one indicating the part of the train's length), while this point is not separated in space by any pair of different points/events.

That said, your argument must still rely on the principle stating that: If all reference frames agree that a given event occurs at a single point in spacetime (e.g. a meeting point), then all of them agree about the exact value of this point, hence about the exact fraction specified by that point. But, what's the exact source/origin of this principle? Does it directly follow from the principle of relativity, if we don't want to rely on the geometrical formula you've indicated at the beginning of your response? HOTmag (talk) 09:17, 31 January 2024 (UTC)

The fraction is determined by three events, not one. It is the ratio of two lengths, each determined by two points, with one point (the collision event) in common. For the fraction to be frame-dependent those two lengths would have to be transformed differently, i.e. length contraction would have to depend weirdly on direction. You may want to look at the derivation of the Lorentz transform from the two axioms and see if there is a shortcut without going via the full transforms. The invariance of the zero-interval is the second axiom (invariance of speed of light). --Wrongfilter (talk) 09:54, 31 January 2024 (UTC)

Thanks. How does the invariance of the zero-interval, follow from the second axiom (invariance of speed of light)? HOTmag (talk) 10:20, 31 January 2024 (UTC)

For light, ${\frac {dx}{dt}}=c$, or $c^{2}\,dt^{2}-dx^{2}=0$. If the speed of light is the same in all reference frames then light follows zero-interval trajectories in all reference frames. --Wrongfilter (talk) 11:01, 31 January 2024 (UTC)

Thank you @Wrongfilter: Regardless of your "principle of existence" (which I find helpful here), do you think the invariance of the zero interval can be used - and is sufficient - for concluding that the fraction is invariant? HOTmag (talk) 11:40, 31 January 2024 (UTC)

"If all frames agree that an given event occurs at a single point, then all of them agree about the exact value of the point" is a tautology. It doesn't need to be derived further, because the two halves of that sentence say the same thing. PianoDan (talk) 15:41, 31 January 2024 (UTC)

It's not a tautology. Think about our meeting yesterday. All of the reference frames agree it was held, at a single point, and at a single moment. However, there may still be a dispute over whether this point at which we met was my home or your home. The same is true for my original question: All agree that the photon and the massive perticle met at a single point and at a single moment. However, perhaps there may still be a dispute over whether this point was at the half of the train's length, or at the quarter of the train's length. Indeed, I'm not claiming there is really a dispute, but I'm still asking if you can prove there isn't one, without your relying on the Lorentz transformations, nor on the Minkowski invariant distance in spacetime, nor on the velocity-addition formula. HOTmag (talk) 17:18, 31 January 2024 (UTC)

There are likely to be other postulates that are implicitly used. For more detail see postulates of special relativity. This thought experiment relies on spatial homogeneity. You can imagine marking the side of the train with tick marks every 1 cm. The number of tick marks between the rear of the train and the meeting point is invariant; the number of tick marks between the meeting point and the front of the train is also an invariant. The distances themselves are not invariant, because the 1 cm spacing is not invariant. Then the question is: in any given frame, are all the tick marks equally spaced? Or is there a frame where some spacings are wider and some are narrower? Spatial homogeneity says that in any frame, the tick spacing must be uniform across the length of the train. And that guarantees the invariance of the fraction you asked about. --Amble (talk) 20:59, 30 January 2024 (UTC)

There needs to be an assumption that the train is not accelerating.  --Lambiam 23:46, 30 January 2024 (UTC)

Thanks. Let me quote you: "the number of tick marks...is...invariant". What basic principle does this assumption follow from? HOTmag (talk) 08:47, 31 January 2024 (UTC)

Not sure how to phrase it — invariance of existence? --Wrongfilter (talk) 09:54, 31 January 2024 (UTC)

Thanks, but it's not only the invariance of existence alone, but rather the invariance of the number of tick marks. My question is, why should we regard this number as invariant? Maybe because this number is dimensionless? HOTmag (talk) 10:20, 31 January 2024 (UTC)

The number can only change if one or more tick marks vanishes or appears. Therefore "invariance of existence" is equivalent to "invariance of number". --Wrongfilter (talk) 10:47, 31 January 2024 (UTC)

Thanks. HOTmag (talk) 11:40, 31 January 2024 (UTC)

x tick marks has the dimension of tick marks. Unlike the velocity-dependent magnetic field, which vanishes leaving only the electric field whenever v=0, the marks are presumed to all be physically present in all reference frames. Modocc (talk) 10:53, 31 January 2024 (UTC)

Thanks. HOTmag (talk) 11:40, 31 January 2024 (UTC)

OP's summary: To sum up, regardless of the Lorentz transformations, and of the Minkowski invariant distance in spacetime, and of the relativistic velocity-addition formula, what's the simplest intuitive principle one had better rely on, for concluding that the fraction I've been asking about is relativistically invarinat, in your opinion?

Actually, what still bothers me is the following fact: a ratio between two velocities - does depend on the reference frame, so how can it be intuitively justifiable to assume that lengths are different from velocities, i.e. that a ratio between two lengths - does not depend on the reference frame? HOTmag (talk) 11:40, 31 January 2024 (UTC)

Because everything that matters for that is linear. In any reference frame an object or photon goes twice as far in twice the time. NadVolum (talk) 13:11, 31 January 2024 (UTC)

Your response has involved time, while my question has not (regardless of the velocity of the reference frame). Let's think about my hand's length and your hand's length. No time is involved (regardless of etc.). So why should we think that the ratio between those lengths (involving no time) does not depend on the reference frame, although a given length per se does depend, and although a ratio between velocities does depend? Can you explain how your argument ("everything that matters for that is linear") explains the difference between the invariant ratio between lengths, versus the variant length per se, or versus the variant ratio between velocities? HOTmag (talk) 14:19, 31 January 2024 (UTC)

So if A, B, C are three points along a lines along a line you're wondering why AB/AC should be the same in one reference system as another? Linearity in the transformation between one reference system and another is what is required for this. Time and distance are transformed linearly in that doubling in one system doubles in the other. It does not matter that velocity changes. AB/AC will be the same in any system and the time for anythng going at a constant speed from A to C to get to B will be AB/AC of the time it takes to get to C. Assuming a physical system leaving A and meeting up at B are the same event in every system and B is the same point. How velocities change given this and the invariance of the speed of light is what one would next come to but the basic asssumption is the linearity of distance and time after transformation. NadVolum (talk) 15:05, 31 January 2024 (UTC)

I take your three words "linearity of distance" as stating that any ratio between two given lengths does not depend on the reference frame. Now I'm asking, if this fact about the ratio is a third postulate of special relativity, or one can derive this fact from any of Einstein's two declared postulates (Principle of relativity and the constancy of speed of light), or one must rely on the spatial homogeneity for deriving this fact, or on...what? Of course without relying on the Lorentz transformations, nor on the Minkowski invariant distance in spacetime, nor on the velocity-addition formula. HOTmag (talk) 17:28, 31 January 2024 (UTC)

Any two distances along the same line. Yes this would have been assumed, the special theory of relativity was a minimal change needed to Newton's laws to accomodate that the speed of light is a constant. Everything changes when one get to general relativity. NadVolum (talk) 21:14, 31 January 2024 (UTC)

Your first sentence is an important correction. Thanx. HOTmag (talk) 22:36, 31 January 2024 (UTC)

What do smartphone strips do?

Thin strips attached to the phone under the cover possibly always glued to the phone. Not the ones that look like it serves no purpose besides an internals protector or loudspeaker membrane. Sagittarian Milky Way (talk) 15:45, 30 January 2024 (UTC)

I don't see how anyone can answer this without more information. Which strips and which brand of phone? Can you supply a pic? The strips around the outside are the antennae. Shantavira|^{feed me} 16:24, 30 January 2024 (UTC)

It seems like you're describing Antenna Strips or NFC Antenna or Wireless Charging Coil or Vibration Motor or some Biometric Sensor Components , can you use a pic? Harvici (talk) 13:45, 31 January 2024 (UTC)

Not wireless charging or biometric. They're actually small tabs of a matte black sheet. One tab covers what looks like a very small port (what the hell is that?). The glue side of the tabs is NOT matte black.Sagittarian Milky Way (talk) 22:00, 31 January 2024 (UTC)

Evolution and most efficient solutions

Many people seem to believe that evolution leads to the most efficient solution to a problem.
Is there any scientific evidence that confirms this? 2A02:8071:60A0:92E0:0:0:0:B32D (talk) 20:51, 30 January 2024 (UTC)

Sounds a bit pointy. Define evolution in context and define efficient. Darwin talked about fitness, not efficiency. Not surprisingly Evolution probably answers your questions. Greglocock (talk) 21:55, 30 January 2024 (UTC)

For any selection criterion, an evolutionary process consisting of many steps of small changes, retaining only changes that correspond to improvements, is like a combination of a random walk in the design space (also called a drunkard's walk) with the optimization technique of hill climbing. If it converges, it does so on a local maximum. There may be a much higher maximum that can only be reached by traversing a valley; local search methods such as hill climbing will not find it. Therefore, the belief appears to be unwarranted.  --Lambiam 23:37, 30 January 2024 (UTC)

Evolution is a pretty good way of finding solutions, but for instance it is pretty unlikey any vertibrate will ever develop eyes without a blind spot like other creatures for example cephalopods have. In computing the 'evolution' can be run past some points where in effect a real creature would quickly die so it can be a very effective ways of finding good solutions - but we have no guarantees of most efficient except in very simple circumstances. NadVolum (talk) 13:22, 31 January 2024 (UTC)

I like your example of the mammalian eye as a local optimum. Greglocock (talk) 06:09, 1 February 2024 (UTC)

The Recurrent laryngeal nerve seems to be a good example of a non-optimal solution (though each step in its evolution may have been). AndrewWTaylor (talk) 13:55, 31 January 2024 (UTC)

I wonder how many is many in "Many people seem to believe...". Most of biology seems to be non-optimal working solutions at all scales, work in progress, but not necessarily to optimize efficiency. Sean.hoyland (talk) 14:28, 31 January 2024 (UTC)

You might look at our article on genetic algorithms#Limitations. But amongst all the reasons why natural selection is not expected perfectly to optimise, let's not lose sight of the fact that many biological structures seem to be extremely well "designed" for the job they perform. I suppose that there are some nice examples from biomechanics but the example I am more familiar with is the excellent quantitative fit between predicted and observed sex ratios.JMCHutchinson (talk) 18:41, 31 January 2024 (UTC)

There are important distinctions to make between biological evolution and genetic algorithms used in engineering. In biology, there is no teleology, it is very chaotic, a mutation that does not select negatively enough but can be considered useless may be preserved, elements of selection are unpredictable, etc. In engineering, the designer is everywhere: the constraints, processes and efficiency evaluations for selection are designed with a purposeful goal. This also means that when trying to find a solution, many iterations can be run and reset. The most efficient one can then be used for the intended application. These methods are only used for problems where it is useful. —PaleoNeonate – 23:56, 31 January 2024 (UTC)

I believe evouloution is fake and NG does to I believe in a different scientific anwser. 23.24.81.162 (talk) 22:09, 1 February 2024 (UTC)

I'm not sure that I understand, but in case it may be useful, WP has an evidence of common descent article. —PaleoNeonate – 05:27, 5 February 2024 (UTC)

Wien's Displacement curves

Stephan-Boltzmann law gives:
$B=T^{4}\sigma$ in watt per square meter.
Expressed to all spectrum in wavelength, you have:
$B=\int _{\lambda =0}^{\infty }B_{(\lambda ,T)}d\lambda$
$B_{(\lambda ,T)}d\lambda$ is in watt per square meter, like $B$.
In frequency you have:
$B=\int _{\nu =\infty }^{0}B_{(\nu ,T)}d\nu$
$B_{(\lambda ,T)}d\lambda =B_{(\nu ,T)}d\nu$ both in watt per square meter, like $B$.
With Planck's law:
$B_{(\lambda ,T)}d\lambda ={\frac {2hc^{2}}{\lambda ^{5}}}\cdot {\frac {1}{e^{\frac {hc}{\lambda k_{b}\theta }}-1}}\cdot d\lambda$
$B_{(\nu ,T)}d\nu ={\frac {2h\nu ^{3}}{c^{2}}}\cdot {\frac {1}{e^{\frac {h\nu }{k_{b}\theta }}-1}}\cdot d\nu$
For $d\lambda =x$, ${\frac {c}{\nu ^{2}}}d\nu =x$
You have:
$B_{(\lambda ,T)}d\lambda ={\frac {2hc^{2}}{\lambda ^{5}}}\cdot {\frac {1}{e^{\frac {hc}{\lambda k_{b}\theta }}-1}}\cdot x$
$B_{(\nu ,T)}d\nu ={\frac {2h\nu ^{5}}{c^{3}}}\cdot {\frac {1}{e^{\frac {h\nu }{k_{b}\theta }}-1}}\cdot x$
So, you see that for a fixed value for x and T, with $\lambda$ from 0 to $\infty$ and $\nu ={\frac {c}{\lambda }}$ you get the same curves for:
$B_{(\lambda ,T)}d\lambda$ and $B_{(\nu ,T)}d\nu$
(when $\lambda =0$, $\nu =\infty$ on the same abscissa, and vice versa),
and, of course, that $\lambda _{peak}$ and $\nu _{peak}$ have the same coordinates.
I specify that in the black body experiment, the power is measured by a bolometer, $\lambda$ and $d\lambda$ are obtained by a crystal filter, therefore of a finite value. One is unable to measure the frequency, it is deduced from the wavelength.

Can you tel me what is wrong in there ?

Malypaet (talk) 23:44, 30 January 2024 (UTC)

If $\lambda$ and $\nu$ are related by the invariant relation $\lambda \nu =c,$ and it is given that, universally,

$\int _{\lambda _{0}}^{\lambda _{1}}B_{\lambda }(\lambda ,T)d\lambda =\int _{\nu _{1}}^{\nu _{0}}B_{\nu }(\nu ,T)d\nu ,$

it follows that the functions $B_{\lambda }$ and $B_{\nu }$ are related by

$\lambda B_{\lambda }(\lambda ,T)=\nu B_{\nu }(\nu ,T).$

This is just mathematics, independent of any physical interpretation. Is this what you mean by "getting the same curves"? (For the rest I didn't understand what you are trying to say; perhaps other editors are more successful in interpreting it.)  --Lambiam 11:51, 31 January 2024 (UTC)

How can you write that, knowing that $\lambda$ and $\nu$ are linked by $\lambda ={\frac {c}{\nu }}$, and $d\lambda$ and $d\nu$ are linked by $d\lambda ={\frac {c}{\nu ^{2}}}d\nu$?

From Planck 1901 article:

Therefore, designate the spatial density of the energy of the radiation belonging to the spectral region $\nu$ to $\nu$ + d$\nu$ by u∙d$\nu$. One must write u∙d$\nu$ instead of E∙d$\lambda$, c/$\nu$ instead of $\lambda$, and c∙d$\nu$/$\nu ^{2}$ instead of dλ.

What I want to write is if you use or not d$\lambda$ and d$\nu$, you get the same curves or different curves.

But which one correspond to experiment?

So, to the nature or reality based on experiment? Malypaet (talk) 13:26, 31 January 2024 (UTC)

Plancks goes from:

$\int _{\lambda _{0}}^{\lambda _{1}}B_{\lambda }(\lambda ,T)d\lambda =\int _{\nu _{1}}^{\nu _{0}}B_{\nu }(\nu ,T)d\nu ,$

to:

${\frac {4\pi }{c}}B_{\lambda }(\lambda ,T)d\lambda ={\frac {4\pi }{c}}B_{\nu }(\nu ,T)d\nu ,$

that is the same as:

$Ed\lambda =ud\nu$

If you have:

$\int _{\lambda _{0}}^{\lambda _{1}}B_{\lambda }(\lambda ,T)\lambda =\int _{\nu _{1}}^{\nu _{0}}B_{\nu }(\nu ,T)\nu ,$

What do you get out of it?
Malypaet (talk) 18:53, 31 January 2024 (UTC)

The latter formula makes no sense, mathematically.  --Lambiam 22:01, 31 January 2024 (UTC)

With me:

$B_{(\lambda ,T)}d\lambda ={\frac {2hc^{2}}{\lambda ^{5}}}\cdot {\frac {1}{e^{\frac {hc}{\lambda k_{b}\theta }}-1}}\cdot d\lambda$

With you:

$B_{(\lambda ,T)}={\frac {\nu }{\lambda }}{\frac {2h\nu ^{3}}{c^{2}}}\cdot {\frac {1}{e^{\frac {h\nu }{k_{b}\theta }}-1}}$

$={\frac {2hc^{2}}{\lambda ^{5}}}\cdot {\frac {1}{e^{\frac {hc}{\lambda k_{b}\theta }}-1}}$

The difference is related to physical dimension:
$B_{(\lambda ,T)}d\lambda$ and $B_{(\nu ,T)}d\nu$ are equal and of the same dimension as B.

When $B_{(\lambda ,T)}$ and $B_{(\nu ,T)}$ have different values and dimension , each of them and to B.

So, how can you put them on the same graph, to compare their curves?

In mathematics, you don't mix cabbages with carrots, do you?
Malypaet (talk) 22:44, 31 January 2024 (UTC)

Assuming $c$ is a known fixed quantity, you can use the same sheet for plotting two graphs, marking the abscissa with a pair of logarithmic scales for each of $\lambda$ and $\nu$ respecting the invariance of $\lambda \nu =c.$ For the ordinate you also need two scales, one for each of the two graphs. This will give two exponential curves, one ascending and one descending.

A completely different approach is to plot the locus of $(B_{\lambda },B_{\nu })$ pairs under the constraint $\lambda \nu =c.$ This will then result in a segment of a hyperbolic curve. A scale marking $\lambda$ and $\nu$ can be placed on the curve, in situ.  --Lambiam 12:27, 1 February 2024 (UTC)

Ok, I understand that.

But $B_{\lambda }$ or $B_{\nu }$ cannot exist whithout $d\lambda$ and $d\nu$ as finite value in experiment.

If $d\lambda =0$ the radiation power measured is 0.

So, for me one would also show the curve with a single ordinate in $watt/m^{2}$ for $B_{\lambda }d\lambda$ and $B_{\nu }d\nu$, and 2 abscissa respecting the relation $\lambda \nu =c$

Just put $d\lambda =1$ and the curve is easy to draw.

Many students are confused with the current presentation because they remain on the relationship $\lambda \nu =c$ in their minds. Malypaet (talk) 22:29, 1 February 2024 (UTC)

My first job was maritime electronics support in Paris. All equipment based on electromagnetic waves had a frequency band around the frequency used. The smaller it was, the more efficient the equipment. But none had a zero band around that frequency, alway with $d\lambda \neq 0$ Malypaet (talk) 13:18, 2 February 2024 (UTC)

January 31

Actual drawing of the Pentalobe screw profile

Hi. While repairing some electronics, I came across the Pentalobe screw. I noticed that Wikipedia, and many other sites[1] , use a black-and-white generalized image to represent the Pentalobe screw profile. This image is made using 5 circles, with each circle's centre located at one vertices of a regular pentagon[2].

I don't know whether it's an optical illusion or not, but this "5 circles diagram" looks vastly different than the actual profile of the Pentalobe screw. It's more apparent if you look at a zoomed in image of the screw or screwdriver[3].

If the Pentalobe screw profile is indeed different than the "5 circles diagram", is there somewhere that I can find this profile?

Anything would be extremely helpful, even non-official sources, because right now the best thing I found is just a zoomed in image of a screwdriver tip, which is less than ideal. Liberté2 (talk) 02:41, 31 January 2024 (UTC)

c:Category:Pentalobe screws has photos and diagrams of several different variations. One important variable is the size of the circle relative to the pentagon. If the circles are tangent to each other or overlap, there is a point where they meet, like File:Pentalobular.svg. If the circles are small enough that they don't touch, like File:Medida de un tornillo pentalobular.svg, what should be the profile in the space between them? A second is an apparent difference of design: starting with a circular hole, do the five circles expand it (bumped-out curves) like File:Pentalobular.svg, or contract it (bumped-in) like File:Torx plus tamper.jpg? Is there an overlap of terminology between 'pentalobe' and 'torx variant with 5 points'? DMacks (talk) 09:56, 31 January 2024 (UTC)

The shape of the indentation of (Apple-specs conforming) Pentalobe screws should be more determinative than the heads of some screwdrivers capable of turning them. As seen here for bottom case screws of a MacBook, the circles clearly overlap.  --Lambiam 10:26, 31 January 2024 (UTC)

Are free radicals electrophile?

Are free neutral radicals, like CH3 electrophile? Their are differing and opposite answers to all on internet, some saying "Free radical is neither a nucleophile nor an electrophile because it does not seek out positively or negatively charged reactants", others say "Since methyl radical contains 7 electron ( 4 from C and 3 from H) and needs 1 more electron to complete its octet, so it will accept electron from some species and will behave as Lewis acid." and others say "I don’t think that methyl free radical is a Lewis acid. According to the definition of Lewis acid, Lewis acids accept a pair of electrons". ExclusiveEditor _{Notify Me!} 17:19, 31 January 2024 (UTC)

Methyl radical can indeed accept an electron and form unstable methyl anion, which (but not the radical!) is a Lewis base. Ruslik_Zero 19:26, 31 January 2024 (UTC)

February 1

Life on Mars

Let's say that at some point in the future and after enough search scientists conclude that there is no life on Mars, at all, nor has there ever been. And then it is proposed to bring life ourselves. Of course, no animals or plants can survive in Mars as it is, but is there any bacteria that may survive and thrive there, if just released and left on its own? Cambalachero (talk) 00:49, 1 February 2024 (UTC)

There are probably no Earthly microbes that would thrive at the surface of Mars, as it is dry, often very cold, and exposed to high levels of radiation. There might be some that could endure in those conditions, though likely in a dormant or mostly dormant state. However, the considerations change dramatically if you consider the subsurface. At several kilometers of depth, the Martian lithosphere is likely warm enough for liquid water and shielded from space radiation. Conditions suggest that many of the bacteria found in the deep subsurface of Earth would also find the deep subsurface of Mars hospitable. Dragons flight (talk) 02:17, 1 February 2024 (UTC)

Not bacteria, but tardigrades could probably hack it for a while, but the lack of water and nutrients might prevent their long-term survival. See:

• Can tardigrades theoretically survive on Mars?
• Spacesuits optional for 'water bears'

Alansplodge (talk) 12:23, 1 February 2024 (UTC)

Lichen as well [4] might have a chance so it's fairly certain lots of strange single celled ones can survive no problem. NadVolum (talk) 12:45, 1 February 2024 (UTC)

There are a thing called water bears. They are real life aliens from space. 23.24.81.162 (talk) 22:07, 1 February 2024 (UTC)

Froget about the reply you did think of that. 23.24.81.162 (talk) 22:08, 1 February 2024 (UTC)

Some organisms—those that can thrive in extreme conditions on Earth—have been considered as potential candidates for life on Mars. Some extremophiles, such as certain types of bacteria and archaea, can withstand extreme temperatures, high levels of radiation, and other inhospitable conditions.

One example is Deinococcus radiodurans, a bacterium known for its ability to survive high levels of radiation. Harvici (talk) 03:07, 2 February 2024 (UTC)

As I recall, Carl Sagan talked about the possibility of terraforming Mars. It would require creating a greenhouse effect, to hold in warmth and moisture. ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:24, 2 February 2024 (UTC)

There are organizations inside of NASA tasked with Planetary Protection - (not to be confused with the similarly-named, totally-different task of Planetary Defense).

"Planetary Protection is the practice of protecting solar system bodies from contamination by Earth life..."

That NASA website has a lot of information about the objectives, methods, and accomplishments that are used to make sure that life does not accidentally spread.

From one of the resources linked, here is a 2021 COSPAR official statement of policy COSPAR POLICY ON PLANETARY PROTECTION. The goal is "to protect “special regions” on Mars, which were defined as “a region within which terrestrial organisms are likely to propagate, or a region which is interpreted to have a high potential for the existence of extant Martian life forms.”" In other words, in direct answer to the original question - scientists are actively studying specific conditions in specific areas on Mars to fully understand (and control for) any possibility of Earth life-forms that might spread there.

Nimur (talk) 18:52, 2 February 2024 (UTC)

I wanna throw in my two cents. The other respondents basically covered the topics of tardigrades and the deinococcus radiodurans, i actually don't believe lichen would make it for long... the thing is, Mars is as barren as it gets. There's literally nothing there. Said tardigrades or other minor organisms could probably survive there for years, but there's nothing to go on, any scarce supply of nutrients would be exhausted fairly quickly. It's basically sending them over there so they could barely exist. No kind of thriving would be possible, there's just nothing there. --Ouro (blah blah) 20:33, 2 February 2024 (UTC)

Dryness, a relatively high radiation level and relative coldness are not only reasons why the terrestrial life cannot survive on Mars. What is more important is that the Mars environment probably lacks any energy sources necessary for life to survive. In addition this environment is chemically very hostile to life: there is an abundance of chlorine compounds (e.g. perchlorates, chlorates and elemental chlorine) that are normally used as disinfectants on Earth. Ruslik_Zero 20:46, 2 February 2024 (UTC)

That's very narrow view. Seeing what evolution did on Earth, it is almost certain that if life started at Mars warm period, they adjusted to the conditions wherever the temp rises over 4 c in the summer. Zarnivop (talk) 03:26, 3 February 2024 (UTC)

The OP's question is not about native Martian life, but about how native Earth life would fare if released on Mars. I think Ruslik is correct: the answer to the OP's question is "not well". Double sharp (talk) 14:36, 3 February 2024 (UTC)

Well, it's clear that you need some kind of thermal energy source in order for life (as we know it) to start in order to keep the water liquid. But you need nutrients (which are energy sources that have to be digested) in any case. --Ouro (blah blah) 05:48, 3 February 2024 (UTC)

True, and both exist on Mars. On the equator, the temperature is 0 c in average. This means that during the martian day, temperatures rise above 0, and likely above 4 c - that's means water ice -> water. This is with direct dun - which provides energy. As for nutrients, primary life forms only require minerals, co2, light and water. Some primary organisms on Earth developed alternate energy production means (We actually found lithobacteria 2 km deep in rocks on Earth). Zarnivop (talk) 12:53, 3 February 2024 (UTC)

The problem is that under Martian ambient conditions the boiling point of pure water is lower that the melting point. So, the ice quickly sublimates before any water can appear. The water can exist only in form of a heavy brine saturated with various (often toxic) salts. Ruslik_Zero 13:13, 4 February 2024 (UTC)

On earth some lichen can continue photosynthesizing and growing within ice at down to minus 20 degrees centigrade. And there's some life growing in the salt crystals in the various salt flats around earth and the brine of the Dead Sea. NadVolum (talk) 17:41, 4 February 2024 (UTC)

Dead Sea's water indeed contains a small number of bacteria and archaea but usually in place where salinity is lower. But you should take into account that Dead Sea is not nearly as toxic as any possible Martian brine. So, on Mars no terrestrial microorganism can possibly survive. Ruslik_Zero 20:30, 4 February 2024 (UTC)

Hellas Planitia would barely support liquid water. Double sharp (talk) 15:17, 5 February 2024 (UTC)

There are some microbes that can grow via perchlorate reduction. Double sharp (talk) 15:21, 5 February 2024 (UTC)

The problem is not perchlorates themselves but their radiolysis products. Ruslik_Zero 20:02, 5 February 2024 (UTC)

Thanks for the paper! In that case, I agree: ClO₂ nails the door shut. Double sharp (talk) 08:22, 6 February 2024 (UTC)

Extensive exposed layers of clean ice may have been found in the mid-latitudes [5] and at 11.9 degrees from its equator there is evidence of a geothermal mantle plume [6] at Cerberus Fossae. Deeper, warmer water with fewer salts than its desert-like soils will allow would mean... Hell, it is even possible that Mars's life hitched a ride along time ago and colonized our hydrothermal vents. Modocc (talk) 16:56, 6 February 2024 (UTC)

February 3

Contradictory views

Are there any studies on the effects on people when they hold inconsistent views?
I suspect it creates a certain amount of psychological distress. 2A02:8071:60A0:92E0:0:0:0:B32D (talk) 13:56, 3 February 2024 (UTC)

IP editor. The article on cognitive dissonance has many references. Mike Turnbull (talk) 14:25, 3 February 2024 (UTC)

See also wishy-washy. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:23, 3 February 2024 (UTC)

It only creates distress when people are aware of the inconsistency. I've heard speakers demand that critics be silenced, arguing that the criticism is an attack on their right to free speech.  --Lambiam 22:13, 3 February 2024 (UTC)

Ha ha, that's really quite a good one. :-} NadVolum (talk) 00:47, 4 February 2024 (UTC)

You might find [7] interesting. NadVolum (talk) 00:58, 4 February 2024 (UTC)

Tribal thinking AKA groupthink is polarizing. However, this study "No association between numerical ability and politically motivated reasoning in a large US probability sample" and most others (see its introductory section) failed to replicate the original study that appeared to "...show that deliberative analytical... reasoning is hijacked by identity...". Instead, it affirmed that political alignment affects everyone's accuracy, but those with better analytical reasoning had better scores regardless. Modocc (talk) 02:23, 4 February 2024 (UTC)

Well that's a bit reassuring - or perhaps the statisticians results were affected by their tribal identities ;-) NadVolum (talk) 12:34, 4 February 2024 (UTC)

February 4

Planck'law and stefan-Boltzmann law relation

In a forum I found that the relation between the Stefan-Boltzmann law and the Planck law in the black body cavity hole is:
$B=T^{4}\sigma =\int _{\lambda =0}^{\infty }\pi B_{\lambda (\lambda ,T)}d\lambda$
Is that true, or is there a different equation ?
Malypaet (talk) 13:42, 4 February 2024 (UTC)

See Stefan–Boltzmann_law#Derivation_from_Planck's_law, where the integral is written in terms of frequency rather than wavelength. --Wrongfilter (talk) 13:52, 4 February 2024 (UTC)

Thanks for the link, I understand better. Malypaet (talk) 09:21, 5 February 2024 (UTC)

Organic Nomenclature

I've been unable to find the name of the CH2= group, I suspect it might be methenyl. So what would be the preferred IUPAC name for 2-ethylbut-1-ene. Taabibtaza (talk) 14:34, 4 February 2024 (UTC)

Hey, I'm not a chemist, but AIUI IUPAC rule P2 gives methylidene for CH₂=, so I'd suggest 3-methylidene pentane for CH₃.CH₂.C(:CH₂).CH₂.CH₃ which is what I think you're after for CH₂:C(C₂H₅).CH₂.CH₃ HTH, Martin of Sheffield (talk) 15:12, 4 February 2024 (UTC)

@Taabibtaza IUPAC naming conventions are a minefield. For this sort of question, I'd use Chemspider and look at the various possibilities. See this entry for 2-ethyl-1-butene which they say is alternatively called 3-methylenepentane. IUPAC wouldn't use methylidene in the case of simple alkenes where the double bond is the only functional group. Mike Turnbull (talk) 17:13, 4 February 2024 (UTC)

February 5

I have requested change

Recently, I have requested change in Talk:Potassium but no one replied. Hope someone jump in and let me know! 2405:4802:64C7:BF70:B50C:773B:1A40:16BA (talk) 02:56, 5 February 2024 (UTC)

Regardless of what the time was in Vietnam, you made your request on a Sunday morning in Europe, and in the early hours of Sunday in the USA, where the bulk of English-language Wikipedia editors live. Please remember that Wikipedia does not have a 24-hour duty roster of paid editors – we are all unpaid volunteers, who do everything in the spare time available from our everyday jobs and lives.

On the Help desks, you might reasonably expect a response in a few hours (less if you're lucky); on the Reference desks perhaps longer; on an article Talk page, you should allow at least several days. I'm sure an editor interested (and competent) in this type of correction (so not me) will evaluate your request within a week, and fulfil it or not according to what they think appropriate. {The poster formerly known as 87.81.230.195} 90.199.208.215 (talk) 03:09, 5 February 2024 (UTC)

I'm sorry, as far as I'm concerned, the Help desk is suitable only for questions involving editing and using Wikipedia, not demanded for questions about general knowledge and questions that need broad agreement. I'm afraid this doesn't fall into those categories, thus not count as a proper enquiry and may need consensus before performing the edit by my request. As a result, I seek for help in this page, reference page. 2405:4802:64C7:BF70:20BF:895B:4915:B58A (talk) 04:39, 5 February 2024 (UTC)

I wasn't advising you to ask on another Desk. I was advising you to be more patient, and wait longer for a response to your Talk page request. {The poster formerly known as 87.81.230.195} 90.199.208.215 (talk) 10:08, 5 February 2024 (UTC)

Is 9.8 meters per second squared really accurate?

Is the figure 9.8 m/s² really accurate for the gravitational acceleration on Earth? It seems overly simplistic to me. I figured that the actual value would be more variable based on an object’s mass and distance from the center of the Earth (as both are factors in calculating the force of gravity objects exert on each other). Primal Groudon (talk) 04:49, 5 February 2024 (UTC)

The article Gravity of Earth discusses the variation of g depending on various parameters. --Wrongfilter (talk) 04:55, 5 February 2024 (UTC)

Mass doesn't matter, but distance from center of the earth does. EvergreenFir (talk) 05:05, 5 February 2024 (UTC)

Yeah, now that I see it, the object’s mass cancels itself out, leaving us with just the UGC times the mass of the Earth divided by the square of the distance from the center of the Earth. Primal Groudon (talk) 06:56, 5 February 2024 (UTC)

It feels like we should include a brief review of accuracy and precision; "9.8" is accurate (with respect to the standardized gravity of Earth) to a few parts per thousand, and it's precise to two significant figures. Beyond those special adjectives, we might also say that one is engaging in a category mistake when they conflate "inaccuracy" of the standard parameter with well-understood variation in magnitude of the measured acceleration due gravity caused by location and realities of the shape and mass-distribution of Earth. This is also - yet again - categorically distinct from any theoretical question about whether such measurements ought to be sensitive to the object's mass (...it ought not be, based on pretty well-established physical relationships, but the effort to re-establish this certainty would be an interesting exploration of experimental and theoretical physics).

All of these ideas raise real and valid concerns; but these concerns are best described using words other than "accurate" or "inaccurate." Nimur (talk) 13:50, 5 February 2024 (UTC)

The section Gravity of Earth § Comparative values worldwide gives the reader a good idea of the variation one can expect while remaining earthbound. It may be assumed that the figures in the table are accurate within the precision with which they are presented.  --Lambiam 14:37, 5 February 2024 (UTC)

Exactly, and therein comes both the issue of accuracy vs precision and significant figures for the "9.8" value. According to that article, surface varies from 9.7639 m/s² to 9.8337 m/s². Of course, one shouldn't just assume that the average surface gravity is halfway between those two numbers, but we can say with a high degree of accuracy that the surface gravity is 9.8 m/s², which for most applications is going to be more than good enough. With far more measurements taken around the globe (which might already have been done), we could give that average with greater precision, but I imagine any application that would benefit from that higher precision would need to know the local surface gravity to a high precision and not the global average surface gravity. I would imagine that once you are out at a distance where a global, rather than local average of high precision is of use, you are also at a distance where you shouldn't be calculating gravity based on surface values anyways.

TLDR the needed precision depends upon the needs of the application, and the highly accurate 9.8 m/s² is precise enough for most everyday applications and, when it isn't precise enough, a highly precise global average would probably have a too low a local accuracy to be useful to said application. --OuroborosCobra (talk) 14:51, 5 February 2024 (UTC)

Naming of electromagnetism and optics standards

In Wikipedia and elsewhere we find different names like $I_{\nu (\nu ,T)},B_{\nu (\nu ,T)},B_{\nu T},M^{0}$,etc... for things that seem to be the same.
Is there a naming standard referenced somewhere?
Malypaet (talk) 09:30, 5 February 2024 (UTC)

I don't think there is, at least not among physicists (but there may be engineering regulations that I'm not aware of). Having said that, I believe that I've seen the convention that B refers specifically to a Planck spectrum, whereas I is the physical quantity "intensity" regardless of whether the spectrum is Planck or something else. --Wrongfilter (talk) 09:37, 5 February 2024 (UTC)

In the last link on wikipedia you gived me, one find:

$M^{\circ }=\sigma T^{4}$

I find also:

$P=\sigma T^{4}$

${\frac {P}{A}}=\sigma T^{4}$

$B_{\nu }(T)$ or $B_{\nu (\nu ,T)}$,etc,...

So, in fact there is no rule, it is like "Happy hour"...

I have just to give the naming convention at the beginning of my text!
Malypaet (talk) 13:39, 5 February 2024 (UTC)

Where do you see $B_{\nu (\nu ,T)}$? To me this notation looks confused. I think this ought to be $B_{\nu }(\nu ,T),$ in which $B_{\nu }$ is the name of a function and $(\nu ,T)$ is its argument. So the subscript $\nu$ is not a variable; the second $\nu$ is.  --Lambiam 14:25, 5 February 2024 (UTC)

What branch of physics discusses a given stationary system containing moving bodies?

Maybe Thermostatics? Maybe Hydrostatics? Maybe Aerostatics? I guess almost all branches of physics may discuss that, but I'm asking about the main branch. 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 09:31, 5 February 2024 (UTC)

Easy - physicists study stationary processes! Nimur (talk) 13:34, 5 February 2024 (UTC)

The adjective stationary means that something remains invariant as time progresses. What kind of system do you have in mind and which are the unchanging aspects that lead to its being characterized as "stationary"? Ergodicity with respect to its phase space? This is something studied in dynamical systems theory – not by itself a branch of physics, but applied in many different branches.  --Lambiam 14:15, 5 February 2024 (UTC)

If anything is moving inside the system, then "-statics" is not quite appropriate. Galaxies are stationary systems (approximately) even though all the stars in a galaxy are swarming around like crazy. In a way even the Solar System is stationary. Stationary systems appear in all many branches of physics; which branch is appropriate for describing any particular system depends on the characteristics of the system, for instance how many particles make up the system, which forces act between the particles, etc. --Wrongfilter (talk) 14:23, 5 February 2024 (UTC)

By "stationary" I mean "at rest". Actually, my aim is to solve a contraction I'm coping with, regrading systems at rest, so I'd like to know where the whole issue may be dealt with. See below my new thread. 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 20:35, 7 February 2024 (UTC)

February 6

Should backs of solar panels sometimes be black?

I noticed that the backs of solar panels tend to be silver or white but I was thinking, in the UK, where the climate is cold most of the time, it would be useful if solar panels were radiating as much heat as possible into the building they are attached to. Is a black coating better for this? Are passive daytime radiative cooling coatings relevant given that reflection isn't needed? It would be useful to passively cool the panels and heat the building. Doing that actively is not viable in the UK. 92.7.46.126 (talk) 00:32, 6 February 2024 (UTC)

Says here that the optimal temp for solar panels is 25°C, and hotter is bad. Abductive (reasoning) 05:20, 6 February 2024 (UTC)

Heat can only passively flow hot to cold, so you can only radiate net heat from the solar panels into the building if the solar panels are hotter than the interior of the building. That won't happen when you want to heat the building. Cooling the panels is useful, but radiative cooling isn't very important. Most cooling happens by conduction into the air, which then takes the heat away with convection. Liquid (evaporative or run-off) cooling can be effective too. Solar panels tend to perform best in unstable spring weather here in Europe: not too hot, bright sunshine, alternating with short showers to cool the panels. I don't think a black back coating will have any significant effect.

There are solar panels that are both PV and heat collectors. As most mixed devices, they're not the best at either of them: they have to be designed to run hotter than ordinary PV panels. PiusImpavidus (talk) 11:58, 6 February 2024 (UTC)

"idiopathic?" kidney damage and date rape drugs

What is known about the number of criminal convictions for using date rape drugs that have caused kidney damage to the victims?Rich (talk) 02:28, 6 February 2024 (UTC)

If kidney damage is known to have been caused by a drug, it is (by definition) not idiopathic. While the use of MDMA can lead to acute kidney failure (PMC 4220747), I suspect that the incidence of such convictions (where I assume the defendant stands accused of administering the drug surreptitiously to a victim) is too low to warrant keeping statistics.  --Lambiam 11:35, 6 February 2024 (UTC)

Nitrogen microwave heating

Nitrogen (N2) molecules are non-polar and have no molecular dipole moment. Microwave radiation induces rotational motion in molecules with a dipole moment, thereby increasing their kinetic energy and temperature. It seems that nitrogen would not be susceptible to microwave heating, yet nitrogen is heated with microwaves in many practical applications: [8], [9], [10]. How is this possible? 2600:1700:6D30:49A0:89F5:4D2B:8BCD:3F6C (talk) 11:20, 6 February 2024 (UTC)

All molecules (and atoms) have a non-zero "instantaneous" dipole moment. Although the electron cloud is symmetrically distributed, it undergoes spontaneous quantum fluctuations that break the perfect symmetry. See also London dispersion force.  --Lambiam 11:48, 6 February 2024 (UTC)

Thanks for your response. So are molecules with no permanent dipole moment heatable with microwaves in an efficient fashion in general? Or is nitrogen specifically susceptible? 2600:1700:6D30:49A0:89F5:4D2B:8BCD:3F6C (talk) 15:06, 6 February 2024 (UTC)

I don't think "efficient" (measured in power consumption) is the right term, but it is an effective method that is very easy to control.  --Lambiam 20:19, 6 February 2024 (UTC)

February 7

Planck's law 1901 article entropy and integration

In his 1901 article, Planck go to the following equation:
$\vartheta =\nu f({\frac {U}{\nu }})$
and
${\frac {1}{\vartheta }}={\frac {dS}{dU}}$
to
${\frac {1}{\vartheta }}={\frac {1}{\nu }}f({\frac {U}{\nu }})$
And integrated (that is what he writes):
$S=f({\frac {U}{\nu }})$

For me, it's too short, not clear.
How can it go from ${\frac {1}{\nu f({\frac {U}{\nu }})}}$ to ${\frac {1}{\nu }}f({\frac {U}{\nu }})$?
And what operation describes the integration that follows?

Malypaet (talk) 13:50, 7 February 2024 (UTC)

I think he uses f as a general symbol to denote unknown functions. The form of the function may differ for the various quantities, but the important thing is that the argument is $U/\nu$ or $\vartheta /\nu$ in this combination. Not particularly good style, but he's Planck and we're nobody. --Wrongfilter (talk) 18:16, 7 February 2024 (UTC)

Same as Wrongfilter, but I went to the effort of typing this out, so I'm darn well gonna post it.

I don't think you can make the assumption that the two appearances of ${\textstyle f({\frac {U}{\nu }})}$ are the same function.

It's clearer if you keep the derivative on the left:

${\frac {dS}{dU}}={\frac {1}{\vartheta }}$

with

$\vartheta =\nu f({\frac {U}{\nu }})$

clearly leads to

${\frac {dS}{dU}}={\frac {1}{\nu }}f({\frac {U}{\nu }})$

as long as you don't require the two "f"s to be the same.

The same holds for the integration - after you integrate dS with respect to dU, you have another ${\textstyle f({\frac {U}{\nu }})}$, with no reason to believe they are the SAME function. PianoDan (talk) 18:29, 7 February 2024 (UTC)

It's easy when you know the result in advance to use magic functions to build a demonstration. I did notice that he used $f$ for different functions (at least five), but until those there was a logic that I understood and that's what I'm looking for here.

And please, don't say that we are nobody in front of someone who found an equation before its demonstration and who never wanted to tell how, by accident?

So far, I have managed to rewrite his article while remaining in the context of power and replacing the resonator with an element of surface $\lambda ^{2}$, I just stumble on these last two magic functions.

For example, if you rewrite this article from the beginning staying in power context, you get:

$\vartheta =\nu f_{7}({\frac {cU}{\nu }})$ where ${\frac {c}{\nu }}\equiv {\frac {1}{h\nu }}$ in time dimension. Just by logic.

Until 2019: $h={\frac {1}{c}}...$

I'll still try to find it myself.

I don't want to learn physics at Hogwarts School. Malypaet (talk) 23:19, 7 February 2024 (UTC)

It's not a question of "magic functions". Planck doesn't give a full expression for f here, because he is trying to establish a general property of ALL possible "f"s.

The point of that section of the paper was simply to demonstrate that the entropy is ONLY a function of ${\textstyle {\frac {U}{\nu }}}$, and NOT a function of the value of c. The derivation successfully demonstrates that fact, and that is its only purpose. PianoDan (talk) 00:17, 8 February 2024 (UTC)

@Wrongfilter My name is nobody, it's for the giant Cyclops Planck!

It's easier to understand when you know the result:

$U=f_{1}(\vartheta$)	$U=ln(\vartheta$)	$f1()=ln()$
$\vartheta =f_{2}(U)$	$\vartheta =e^{(U)}$	$f_{2}()=e^{()}$
${\frac {1}{\vartheta }}={\frac {1}{f_{2}(U)}}$	${\frac {1}{\vartheta }}={\frac {1}{e^{(U)}}}$
${\frac {1}{\vartheta }}=f_{3}{(U)}$	${\frac {1}{\vartheta }}={\frac {1}{e^{(U)}}}$	$f_{3}()={\frac {1}{e^{()}}}$

"c" is a constant and that's not the problem here. If we consider that the radiation circulates in the vacuum between the atoms of the cavity wall, at the microscopic level its value remains unchanged, while at the macroscopic level the speed of the radiation is given by $c_{m}={\frac {c}{n}}$ (n for refractive index), voilà.

Thank you all, our discussions allowed me to find the answer quickly.
Malypaet (talk) 13:59, 8 February 2024 (UTC)

Why isn't rain consistent around the world?

The laws of evaporation physics should be consistent around the world, right? So how can some place rain a lot more than other places. And let's make this about large cities near lakes/oceans. In the U.S., southern California had to open a saltwater treatment plant in 2015 as they are sucking the Colorado River try, while being net to an ocean, which is true of Washington state next to an ocean but rains heavily. What is the missing variable?

Also, what is the lowest altitude for rainfall? Thanks. 170.76.231.162 (talk) 17:36, 7 February 2024 (UTC).

The surface of the Earth is not consistent, so why would we assume that the application of the "laws of evaporation physics" will result in a single, uniform result globally? Different latitudes have different temperature profiles (e.g. it's a lot warmer near the equator and a lot colder near the north and south poles). Geography itself has effects, such as rain shadow, where warm, moist air follows the breeze up a mountain range and cools as it goes up, resulting it it raining on the mountains, and dry air coming down the other side. --OuroborosCobra (talk) 19:16, 7 February 2024 (UTC)

Then there's oceanic currents, and the Earth's tilted axis, and its non-circular orbit around its star, and... Bazza (talk) 19:58, 7 February 2024 (UTC)

You are too close to the horse latitudes and the west side of continents exaggeration of the horse latitudes. This is why Afrique Nord to west India is dry and the rest of the Eurafrasian horse is wet. With a transition zone in between. The horse latitudes exist cause Earth has three pairs of convection cells (the average wind circuits of the atmosphere). The air sinks on average in the horse latitudes reducing rain. The southbound California current also reduces rain on that side by causing subsidence. While the northbound West Atlantic express boosts East Coast rain if anything. All mountains to the west of you reduce rain. So Mojave is dry, Arizona is dry. Also fossil fuels is making the driest latitude which is in Baja (not the end or Tijuana, somewhere in between) move closer to you and many other effects. Sagittarian Milky Way (talk) 23:09, 7 February 2024 (UTC)

laser cutting glass

This video[11] shows glass being cut by a laser. There appears to be some sort of "white smoke" during the process. I'm trying to understand what's going.

The melting point of Silicon dioxide is approximately 1713 °C, the boiling point is around 2,950 °C.

I'm guessing that it's a combination of the following factors:

1. The glass in the cut area is melted, and remains attached to the the two pieces. After the cut, the melted parts solidifies as a part of the two resultant pieces. The total mass of the two resultant pieces is the same as the original piece.

2. The glass in the cut area is melted and then vaporized, the glass vapour is removed by the fume extraction system. The total mass of the two resultant pieces is less than the original piece.

3. The glass in the cut area is melted, and micro particles of it (in either liquid or solid state) are carried away as a "white smoke", and micro particles of it are also splattered around the machining area. The total mass of the two resultant pieces is less than the original piece.

Which one of these factors apply in this case?

If it's a combination of two or more factors, which one is the dominant one? OptoFidelty (talk) 20:04, 7 February 2024 (UTC)

It's likely closest to #2 in your list. See laser ablation. It's not necessarily melting and then evaporating, however, it could be sublimating (direct from solid to gas). Yes, some amount of mass is being lost (or at least no longer useful to these pieces, as I suppose you could recover and reuse the smoke material for a nice atom economy with the fume extractor), although whether you say that the larger or smaller resulting pieces of glass is losing that mass (or some ratio between them) is basically up to where you designate the division between the pieces. In this case, they seem to want the smaller piece of glass that they cut out, so they would program the machine to cut it out such that the tolerances of the dimensions of the final piece are within required specifications. Rather than cutting "directly on the line," if you were, it would intentionally cut slightly outside of that line knowing the diameter of the "drill." The same would be done with a mechanical process. --OuroborosCobra (talk) 20:21, 7 February 2024 (UTC)

Thank you!

Regarding, Sublimation (phase transition), what determines whether "solid->liquid->gas" happens or "solid->gas" happens? It it related to the speed of the heating?

For example:

1. Iodine is heated at room temperature. Only "solid->gas" occurs.

2. Silicon dioxide is heated slowly above its boiling point. Only "solid->liquid->gas" occurs.

But in the situation in the video, it's not clear whether it's "solid->liquid->gas" or "solid->gas", or a mix of both. Is there a way to calculate the behaviour if all the processing conditions are known? OptoFidelty (talk) 20:45, 7 February 2024 (UTC)

It depends upon both temperature and pressure. See phase diagram. Even for pure substances (and glass is usually intentionally not pure), calculating those values is non-trivial as you end up having to factor in a lot of things, such as calculating intermolecular forces. With a lot of effort, you might be able to do that with computational chemistry software packages, but one is better off figuring out the phase diagrams for given substances experimentally, such as through various forms of calorimetry. That said, phase diagrams for common substances have largely already been figured out experimentally, so you can usually look it up. I don't know how reliable it is, but here is one for silicon dioxide. However, that gets into another topic, which is "what is glass?" Look over at glass, and you will see that it isn't any single thing that can be just called "silicon dioxide;" that's just one type (and probably not the type in the video). What specifically that glass is made of, what crystal phase it is in, whether there are additives (such as in borosilicate glass, these are all going to have an impact on the conditions for melting, boiling, and sublimation. --OuroborosCobra (talk) 21:28, 7 February 2024 (UTC)

(edit conflict) For a one-component three-phase substance it depends on the pressure $P_{\text{tri}}$ of the triple point. A three-phase substance can only be in the liquid phase at pressures greater than $P_{\text{tri}}$. If it is heated at a lower pressure, it will go directly from solid to gas.  --Lambiam 21:35, 7 February 2024 (UTC)

True, but unless you already know the triple point pressure, you still can't just calculate it easily. --OuroborosCobra (talk) 23:12, 7 February 2024 (UTC)

Ordinary glass doesn't really have a crystalline phase, nor a melting point. At least, not one that can be practically demonstrated. It does have a glass transition. And most ordinary glasses are pretty liquid at 1000°C; no need to go all the way up to the melting point of silica. Which is good, otherwise glassmaking would've been impossible in antiquity.

The glass particles in the white smoke (which could also contain contaminants) must be fairly solid. If they had been fairly liquid, they would be glowing yellow.

Glass is also brittle. Quick localised heating can break off tiny particles without making them significantly liquid or heating them to incandescence. You should be able to tell when viewing the smoke particles under a microscope. PiusImpavidus (talk) 10:31, 8 February 2024 (UTC)

Couldn't the "smoke" also contain vapourized SiO₂ (boiling point 2,950 °C)?  --Lambiam 11:27, 9 February 2024 (UTC)

Classical Mechanics: I wonder where my mistake in the following calculation is:

0. (This setion was added later). Here, I'm only referring to theoretical Newtonian situations involving no potential energy (i.e. no chemical and nuclear particle bonds, nor physical force fields, nor deformation of solids).

1. Hence, a given system's kinetic energy is equal to the sum of the kinetic energies of the system's parts, regardless of whether the system consists of two bodies or two photons or whatever.

2. Let two bodies, having the same mass $m$ each, move by the same speed $v$ in opposite directions.

3. Hence, each body carries kinetic energy of ${\frac {1}{2}}mv^{2}.$

4. Hence, per #1, the whole system, consisting of both bodies, carries kinetic energy of ${\frac {1}{2}}mv^{2}+{\frac {1}{2}}mv^{2}=mv^{2}.$

5. On the other hand, per #2, the whole system is at rest, so it's kinetic energy must be zero.

I wonder how I can settle the contradiction between #4 and #5. I guess the mistake is in #1 or in #5, but I can't find it yet.

Since, besides the first sections 0-1, each of the other sections 2-5 is supposed to derive from a previous section, so can you precisely point at the section (of the above six) which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 20:31, 7 February 2024 (UTC)

What is your definition of a system being "at rest"? The concept applied to a collection of objects is usually understood to mean that each object in the collection is at rest. For a fluid it means that the particles of which it exists are all at rest. (See Hydrostatics.) A spinning disc such as a gyroscope is not at rest, even though for each infinitesimal part having some velocity v there is a corresponding infinitesimal equi-massive part having velocity −v.  --Lambiam 21:01, 7 February 2024 (UTC)

Don't you agree, that a given collection of two photons, that move in opposite directions, is a collection at rest? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 21:42, 7 February 2024 (UTC)

You're confusing energy and momentum. Momentum is a vector quantity, so your two particles cancel each other out. The system has zero net momentum.

Kinetic energy is a scalar quantity, and so direction doesn't matter when you add it up. PianoDan (talk) 22:24, 7 February 2024 (UTC)

Since, besides the first sections 0-1, each of the other sections 2-5 is supposed to derive from a previous section, so can you precisely point at the section (of the above six) which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 07:44, 8 February 2024 (UTC)

1-4 only apply to the system's parts because the KE of any massive object is entirely dependent on its speed. To clarify, consider another example: A) A newspaper is at rest on someone's doorstep: v=0, m₀v=0, KE=0 and it has rest mass m₀. OK, and... B) Its molecules vibrate, thus: v_i>0, m_iv_i>0 and KE_i>0 for every ith molecule. Descriptions A and B differ in granularity, and because mass and energy are equivalent the sum of its internal energies E_i divided by the speed of light squared equals its rest mass m₀ even though it is still at rest on the doorstep (v=0, m₀v=0, KE=0), thus 5 is correct. Also the classical formula is an approximation and does not hold for photons. Clarifying 1 to say "...a given system's [internal] kinetic energy..." would distinguish it from the kinetic energy of 5. Modocc (talk) 00:54, 8 February 2024 (UTC)

you have discovered temperature. Greglocock (talk) 02:40, 8 February 2024 (UTC)

Since, besides the first sections 0-1, each of the other sections 2-5 is supposed to derive from a previous section, so can you precisely point at the section (of the above six) which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 07:57, 8 February 2024 (UTC)

More precisely, they've discovered internal energy (there's a connection to temperature, of course). Energy is a book-keeping device. Depending on the level of description of a system, energy can be booked to different accounts. The kinetic energy of the individual particles contributes to the internal energy (and total energy) of the two-particle system; in the rest frame of the two-particle system, it does not contribute to the kinetic energy. In a many-particle system one can distinguish between the mean motion of the particles (which make up a gas flow, for instance) and the random motion (which determines the temperature of the flowing gas). --Wrongfilter (talk) 07:01, 8 February 2024 (UTC)

Since, besides the first sections 0-1, each of the other sections 2-5 is supposed to derive from a previous section, so can you precisely point at the section (of the above six) which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 07:44, 8 February 2024 (UTC)

Number 4, which should be formulated as "the whole system ... has a total (or internal) energy of ..." (assuming there's no interaction between the particles). --Wrongfilter (talk) 07:53, 8 February 2024 (UTC)

Thank you. Are you sure, the section - which is the first one you don't agree with, is #4 - rather than #1? Note #4 directly derives from #1, so if you agree with #1 you apparently have to agree with #4. 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 07:57, 8 February 2024 (UTC)

Yes, I missed that. So in #1 it should be "The systems total energy is the sum of the kinetic energies of its constituent particles". This is true if there are no interactions between the particles, and if there are no further internal degrees of freedom apart from the translational motion of the particles (translate as appropriate for photons, where the term "kinetic energy" is also not quite right). --Wrongfilter (talk) 08:14, 8 February 2024 (UTC)

Thank you. I've just added section #0 above, which excludes potential energies. Could you point now at the section which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 08:43, 8 February 2024 (UTC)

I've already said all I have to say. --Wrongfilter (talk) 08:46, 8 February 2024 (UTC)

New question: Do you agree that #1 directly derives from #0? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 08:50, 8 February 2024 (UTC)

I've told you what I think is wrong with #1, this is independent of your #0. Please don't make these threads endless. --Wrongfilter (talk) 08:55, 8 February 2024 (UTC)

I only wanted to be sure I understood well (which sometimes needs further clarifications). So, as I understand now (due to your last clarification in your last response), internal energy is still different from kinetic energy, despite #0. Thank you. 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 09:02, 8 February 2024 (UTC)

February 8

How good could rayguns be at demolition if the possibly intractable problems were trivial?

Like if you pick the particle type, flux, beamwidth (at least .010 or .001 caliber) and energy per particle right then could a (probably impossibly low waste-heat) man-portable or even pistol-like photon beam or particle accelerator (that could convert much of its mass to beam) demolish as much stuff per second as a dude throwing grenades? Is there a beam that won't dump too much energy into the air, can cause tactical nuke-level damage a few kilometers away in 3 seconds and yet not kill the user before he can finish? Preferably you'd not want to harm the user more than current common infantry weapons (maybe trading less naked ear hearing loss for more radiation bouncing back?) or have too much recoil to aim well, what would be possible with that restriction? Would neutrons be better or worse than photons or charged particles? (I don't know if neutron accelerators are even possible) Sagittarian Milky Way (talk) 00:16, 8 February 2024 (UTC)

Particle accelerators as conventionally defined cannot accelerate neutrons, as they work by applying an electric field to a charged particle. You can't accelerate something with no charge using an electric field.

The way neutrons are generated for research is fascinating - check our article on Neutron sources.

As far as the rest of it goes, given the amount of heavy power infrastructure attached to even small medical accelerators, the idea of something man-portable seems unlikely. PianoDan (talk) 00:23, 8 February 2024 (UTC)

I know, I was just wondering how powerful a theoretical particle (including photons) black box could be before you'd be too close to target and/or beam and what kind of particle is that. Obviously there could be orders of magnitude destructiveness difference between "OSHA-compliant" and "not possible even as a suicide attack (unless you can make it hold down the trigger after you're dead or something like that)". Sagittarian Milky Way (talk) 00:36, 8 February 2024 (UTC)

If there was such a thing, then why is not in use by armies? Probably a small rocket propelled tiny tactical nuke could do what you say. But that is a big particle. Or fire antimatter bullets or pellets at the target. Otherwise generating beams from a portable device would have to be extremely efficient, not to overheat the user, and impart all energy to the target. So it is not feasible. Graeme Bartlett (talk) 12:36, 8 February 2024 (UTC)

Yes but many cool science-fiction things can already be done with a different current tech. Like the Independence Day city ray. I'm not sure if antimatter bullets would not cause user damage either as air is matter. Even if you had a force field to delay matter contact till leaving the barrel. I am not sure what kind of particle beam (including different kinds of photons) has the highest ratio of impart energy to target to impart energy to air between raygun and target and would like to know. Also some kinds of impart energy to air are worse then others, i.e. X amount of energy as ionizing radiation or boom sound is worse than the same amount as heat. Sagittarian Milky Way (talk) 18:00, 9 February 2024 (UTC)

You want something like this [12] but beefed up and stuck in a gun? There's far better weapons already for everything you say, a rocket is needed for the three seconds and nuke business but really you don't want that sort of energy being produced in something you hold! NadVolum (talk) 13:16, 8 February 2024 (UTC)

There's a reason that the atomic hand grenade has never caught on. {The poster formerly known as 87.81.230.195} 90.199.107.217 (talk) 06:08, 9 February 2024 (UTC)

Also it wouldn't fit in a hand grenade. Lots of suicide terrorists would be willing to use them though if they could get them. There is an atomic grenade launcher at the limit of feasible though. Would not meet OSHA regulations on blast nearness and recoil. Sagittarian Milky Way (talk) 18:05, 9 February 2024 (UTC)

Possibly you missed the fact that I was making a joke [clue: small type], the point of which was that, assuming one could make an atomic device that small, one would of course not be able to throw it far enough to avoid being killed by its detonation.

Grenades are battlefield weapons: a suicide mission would not require a device small enough to be throwable, one would merely carry a suitcase bomb to the target.

Fortunately, construction of such devices is probably beyond the capabilities of most terrorist organisations, but there remain the possibilities of a rogue state supplying them with one, one being stolen from a state's arsenal, or the alternative of a small dirty bomb which would be less immediately destructive but possibly even more effective in the longer term. {The poster formerly known as 87.81.230.195} 90.199.107.217 (talk) 03:37, 10 February 2024 (UTC)

There was a Davy Crockett (nuclear device) really inconvenient for one person but much less so than any chemical energy way of carrying 20 tons of TNT equivalent. 1.25 mile range. Sagittarian Milky Way (talk) 04:20, 11 February 2024 (UTC)

February 9

Neurotransmitters, part 3

Does prolactin counteract the effects of dopamine (either through suppressing its release, through blocking its signalling and/or through exerting an opposite effect on the brain)? Does it counteract the effects of oxytocin? 2601:646:8080:FC40:B8B0:B42B:BB0C:97BA (talk) 22:18, 9 February 2024 (UTC)

Oxytocin stimulates prolactin secretion, while prolactin, once a threshold is reached, stimulates dopamine release, which has an inhibitory effect on prolactin secretion (PMID 22129099, PMID 18477617).  --Lambiam 14:15, 10 February 2024 (UTC)

Thanks! And a related question: does prolactin then inhibit D4 receptors? (Yes, this has to do with my personal research about introversion -- I'm trying to see whether or not increased prolactin levels would be pleasurable for an introvert from the deep end of the scale!) Also, does prolactin either stimulate or inhibit oxytocin? 2601:646:8080:FC40:9D4E:C237:28BF:6A78 (talk) 03:59, 11 February 2024 (UTC)

Autism vs. introversion

Two questions in one: (1) how can they tell for sure between autism and just an extreme case of introversion (given that 3 of the 4 diagnostic symptoms, of which only 2 are needed to diagnose autism -- in fact, all of them except stimming -- can also be associated with introversion)? And (2) how does autism specifically affect oxytocin signalling in the brain? 2601:646:8080:FC40:B8B0:B42B:BB0C:97BA (talk) 23:14, 9 February 2024 (UTC)

As to (2), here is a review article, "Oxytocin and Autism Spectrum Disorders", PMID 28766270. Several studies have reported a tendency of lower oxytocin levels in children on the autism spectrum, but the causal connection is not clear. In summary, autism is not well understood, and more research is needed.  --Lambiam 14:25, 10 February 2024 (UTC)

Not sure why the two should be confused at all, they're very different. Being an introvert is something they would diagnose for? Do they diagnose being an extrovert as well? What is the opposite of autism if they think that way? NadVolum (talk) 20:41, 11 February 2024 (UTC)

Do you not think a psychiatrist would never mistake an extreme case of introversion (a normal condition) for autism and misdiagnose someone when in fact the person doesn't actually have anything to be diagnosed with??? After all, as I pointed out, 3 out of the 4 diagnostic symptoms for autism can also be observed in introverts -- and now that I think of it, even stimming might not always be a reliable symptom of autism, in some cases (although uncommonly) what is thought of as "stimming" might actually be just a reflexive pain response (because, in introverts, over-stimulation -- such as being at a big loud party -- can in some cases cause physical pain, such as a headache!) Not to mention that there's also the possibility of malicious diagnosis (although that's probably another topic for another time)! 2601:646:8080:FC40:8443:5BCE:DED0:5463 (talk) 03:27, 12 February 2024 (UTC)

You appear to be assuming that autism is an all-or-nothing condition; but Biology is complicated and messy.

It's often thought (rightly or wrongly) that everyone is on a "spectrum", with most (the "neurotypical") at various low values on it and some at higher values which correlate with greater interactional (with the material world and/or other people) difficulties.

If this is so, inevitably the threshold value for "having autism" (or whatever) will in part be a matter of subjective judgement that may vary between medical practioners, and will evolve as medical science advances. Moreover, some people (including myself) may consistently exhibit some cognition and behaviour which, if more pronounced, would qualify as autism. Lastly, I doubt that anyone scores exactly the same "spectral value" all the time. 90.199.107.217 (talk) 07:29, 12 February 2024 (UTC)

Right, and that brings us back to the same question: where exactly is the line between just strongly pronounced introversion on one hand and high-functioning autism on the other? 2601:646:8080:FC40:99FE:A33D:618C:8AB5 (talk) 11:51, 12 February 2024 (UTC)

Planck's law 1901 article conundrum

In my personal popularization of Planck's 1901 paper on Planc's law, I have a sticking point. I am ok up to the equation (9):
(9)     ${\frac {1}{\vartheta }}={\frac {DS}{DU}}$
But between:
$S=k${$(1+{\frac {U}{h\nu }})log(1+{\frac {U}{h\nu }})-{\frac {U}{h\nu }}log{\frac {U}{h\nu }}$}
And
Substitution in (9) gives:
${\frac {1}{\vartheta }}={\frac {k}{h\nu }}log(1+{\frac {h\nu }{U}})$
Between these 2 equation I don't find?
(After to the end it is ok)
I arrive to this point:
$S=klog(1+{\frac {U}{h\nu }})+{\frac {kU}{h\nu }}log(1+{\frac {h\nu }{U}})$
Apparently, that suppose:
$S-klog(1+{\frac {U}{h\nu }})={\frac {U}{\vartheta }}$
But I'm stuck there and it is not clearer in Planck's following manuscripts.
Any idea ?
Malypaet (talk) 23:18, 9 February 2024 (UTC)

I think you've made a math error in there somewhere. Taking the derivative of S with respect to U gives:

${\frac {1}{\vartheta }}=k[{\frac {1}{h\nu }}ln(1+{\frac {U}{h\nu }})-{\frac {1}{h\nu }}ln({\frac {U}{h\nu }})]$

Which only takes a little poking with log identities to yield the desired result. PianoDan (talk) 02:27, 10 February 2024 (UTC)

An error, sure.

But how do you go from Planck:

$S=k${$(1+{\frac {U}{h\nu }})$$log(1+{\frac {U}{h\nu }})-{\frac {U}{h\nu }}log{\frac {U}{h\nu }}$}

to your formula "taking the derivative of S with respect to U"?

${\frac {1}{\vartheta }}=k[$${\frac {1}{h\nu }}$$ln(1+{\frac {U}{h\nu }})-{\frac {1}{h\nu }}ln({\frac {U}{h\nu }})]$

You cannot simply replace $S$ with ${\frac {U}{\vartheta }}$ to get your result, isn't it?
Malypaet (talk) 09:29, 10 February 2024 (UTC)

I think I've asked before: Do you know what the derivative of a function is and how to compute it? --Wrongfilter (talk) 09:36, 10 February 2024 (UTC)

I only ask to relearn, Planck integrates where pianodan takes the derivative? Malypaet (talk) 11:17, 10 February 2024 (UTC)

Planck isn't integrating here either. PianoDan (talk) 17:32, 10 February 2024 (UTC)

${\frac {1}{\vartheta }}={\frac {{\text{d}}S}{{\text{d}}U}}$

$~~~~={\frac {\text{d}}{{\text{d}}U}}~k\!\left((1+{\frac {U}{h\nu }})\log(1+{\frac {U}{h\nu }})-{\frac {U}{h\nu }}\log {\frac {U}{h\nu }}\right)$

$~~~~=~...$

 --Lambiam 14:39, 10 February 2024 (UTC)

$=k[(0+{\frac {1}{h\nu }})log(1+{\frac {U}{h\nu }})-{\frac {1}{h\nu }}log({\frac {U}{h\nu }})]$

$=\cdots$

Many thanks Lambiam, that’s all I was missing.
Now with Planck's 1901 paper, for me everything is clear from start to finish and in detail.

I will be able to simplify it and make it more realistic, without resonators.

(°—′)
Malypaet (talk) 18:51, 10 February 2024 (UTC)

February 10

Rotating a magnetic field

I've been thinking about a superconducting electric coil with some current, creating a stable magnetic field. Now assume an axis through the middle of that coil. My intuition tells me that rotating the coil around this axis should be inconsequential. But assume an axis perpendicular to the first one. If I rotate the coil around this axis (and, I assume, the magnetic field with it), a distant observer would see a regularly fluctuating magnetic field - and, I again assume, because auf the unity of electric and magnetic fields, this would appear as an electromagnetic wave. Is this correct? If so, where is the energy carried by the wave coming from? Is the magnetic field dissipating? Or do I need to put energy into the system to keep rotating the coil? Thanks! --Stephan Schulz (talk) 13:53, 10 February 2024 (UTC)

Assuming that the setup indeed creates a changing magnetic field, by Lenz's law this induces an opposing current which (as in an induction brake) should counteract the rotation.  --Lambiam 15:00, 10 February 2024 (UTC)

An ordinary permanent magnet is a cheaper and more convenient alternative to a super-conducting coil. Spinning the magnet about its transverse axis will induce eddy currents in any nearby metal objects, which will oppose the rotation of the magnet and remove energy by resistive heating of the metal - this is induction braking. In the absence of any nearby metal objects, then there will indeed be braking associated with the energy and angular momentum carried away by the radiated field. However it is necessary to spin the magnet very fast or be very far from metal for the radiation braking to be significant compared with the induction braking, requiring something of the order of ω = c / d where ω is the angular frequency of the rotation and d is the distance from metal. catslash (talk) 18:09, 10 February 2024 (UTC)

Thank you - and of course. I should have thought of that! --Stephan Schulz (talk) 05:07, 11 February 2024 (UTC)

An example of such a strong, rapidly spinning permanent magnet with no metal anywhere near would be a pulsar. There is, however, some interesting plasma physics happening in the pretty good vacuum of the near field, so although there are no metals, there are electric currents. And pulsars do spin down. PiusImpavidus (talk) 11:57, 11 February 2024 (UTC)

February 11

Collision analysis

A bus of mass 25 tons starts to travel along a straight road from A to B at 50 mph. At the same time a bee of mass 2.5 grams starts to fly from B to A along the same road at 5 mph. Eventually the bee and the bus collide.

Assume the bee impacts a surface of the bus which is flat and at right angles to its direction of travel, such as the windscreen, and the bee sticks to this surface.

Then the bus will be slowed by a tiny amount reflecting the relative masses / momenta of the bee and bus. In addition, relative to A, the velocity of the bee changes from -5 mph to +50 (approximately) mph during the collision. So at some stage during the collision the bee must be stationary. This can only be when the bee is in contact with the bus.

The question is : Why is the bus not stationary at the same time?

Does it make any difference to the situation if the bee is replaced by either a perfectly elastic object, or a perfectly rigid object, of the same mass?

And does it make any difference if the collision is considered at the level of the individual atoms involved? Ionlywanttoknow (talk) 18:40, 11 February 2024 (UTC)

If you define the direction from A to B as positive, the bee is changing from a speed of -5 mph to +50 mph. Regardless of the exact nature of that change, in order to get from -5 to +50, at some point the speed must pass 0.

The bus's speed is changing from +50 mph to +49.999999... mph. That change does not pass through zero, so the bus is never stationary.

You could get into more details of the interaction, but that's the basic point. PianoDan (talk) 20:11, 11 February 2024 (UTC)

There's two explanations. If you're thinking of absolutely rigid bodies then the bee is never stationary, it immediately switches between going in one direction and the other. In the physical world there is always an bit of elasticity so the bus atoms can keep going forwards whilst the bees ones slow down and reverse as the bee is squashed against the windscreen. NadVolum (talk) 20:30, 11 February 2024 (UTC)

This follows from the Intermediate value theorem. Ruslik_Zero 20:35, 11 February 2024 (UTC)

There is no intermediate value of speed in the case of rigid bodies. NadVolum (talk) 21:06, 11 February 2024 (UTC)

Here is the argument: The intermediate value theorem tells you that the speed of the bee is zero at some points in time. By assumption, the bee and the bus move at the same speed when they are in contact. So the bus must move at speed zero at that point in time. So far, so good. But we are talking about points in time. Speed, being distance per time, is not well-defined for an individual point. You need an "expanse of time" to have speed. --Stephan Schulz (talk) 10:20, 12 February 2024 (UTC)

Why is the bus not stationary? Because it is moving at 50 mph! For the bus to decelerate from 50 mph to stationary in the brief duration of this collision would take a very, very large force. So large it could not possibly occur in a collision with a small object. Dolphin (t) 06:19, 12 February 2024 (UTC)

February 12

Hospital Gangrene

Is there is Wikipedia on Gangraena nosocomialis or hospital gangrene, predominant up until about 1870-1880. I can't seem to locate anything although it might on under some different name. scope_creep^Talk 09:29, 12 February 2024 (UTC)

Here ? Heihaheihaha (talk) 10:29, 12 February 2024 (UTC)

Strange "mitotic phase" of oocytes in rabbit's primary follicle

Two months ago, I posted a question on StackExchange but no one answered me there. It's about a primary follicle (Inferred from the number of layers of granular layer cells), whose oocyte seems to be undergoing division, which is supposed not to happen at this stage by my textbook. I've asked my histology teacher and got no answer. My initial idea was whether it is possible to have multiple oocytes simultaneously in the follicles of rabbits. I've tried to google and searche on PubMed, I didn't find relevant evidence there. The full size picture is provided here, which may provide more information.

The question is: how to explain this phenomenon? I did not observe this phenomenon in other many slices. Heihaheihaha (talk) 10:26, 12 February 2024 (UTC)