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November 25

What is this physical process called?

I’ve noticed that when I add my couscous to rapidly boiling water and then rapidly cool it by removing it from the heat, the parts where the boiling was most prominent becomes little cavities once it cools. What is this process called and what is happening to produce the holes? Viriditas (talk) 00:46, 25 November 2021 (UTC)[reply]

Nucleate boiling likely applies here. Vapor bubbles are being generated from nucleation sites (probably small surface cavities that increase the fluid's contact area with the heating surface, causing water at those sites to vaporize faster) forming dense bubble columns. The fluid/couscous not in the path of these columns also should cool slightly faster, causing it to solidify around the nucleation jets. JoelleJay (talk) 06:20, 25 November 2021 (UTC)[reply]

Additionally, a larger rising steam bubble forms a channel that subsequently formed steam tends to follow as the path of least resistance, thereby maintaining it, much like flowing water may carve out a channel, or a conductive channel formed in lightning is not only used by the initial discharge but reused by quickly following strokes. So, next to possible preferred nucleation sites of the cookware, spontaneous symmetry breaking may also play a role. --Lambiam 07:54, 25 November 2021 (UTC)[reply]

Thank you. I’ve read the linked article and it says that the process is complex, they have trouble developing models, there are contradictory results, and more research is needed. Is there a reason this is poorly understood? Also, what accounts for the chaotic pattern of the bubble cavities? If I did this over and over again, and performed the same procedures, such as using the same amount of couscous, the same pot, stirring it the same number of times, etc, would the bubble holes always appear in different spots? Is there an approximate number that should arise each time? Or is this more a function of my stove and cookware? Viriditas (talk) 21:59, 25 November 2021 (UTC)[reply]

A cursory visual inspection suggests that the hole-y pattern does not emerge from a Poisson point process: there are too few holes very close to each other, and also very few close to the border. This is your opportunity to conduct a scientific study in your kitchen lab. The number and distribution of the holes may be influenced by the granularity, composition and other aspects of the couscous, the amounts of couscous and water, the heat level and the cookware. So, for a first series of tests, try to introduce as little variation between runs as possible. Intuitively, I expect the number to show little variation and to follow a Poisson distribution. If the pattern repeats – more precisely, if the similarity between successive patterns is larger than expected for a non-repeating pattern – nucleation is the most likely dominant factor. If the positions of the holes in successive patterns are not correlated, symmetry breaking is a good explanation. --Lambiam 23:23, 25 November 2021 (UTC)[reply]

Lambiam, I posted the photo of the second batch up above. Any comments? Viriditas (talk) 22:11, 28 November 2021 (UTC)[reply]

I didn't count them, but their numbers can't be very different. Just offhand, based merely on a visual impression, it appears to me that knowing the positions of the holes in one instance of the process gives very little information for where they'll appear in the next instance – apart from the fact that they are again sparse in a border zone. As before, they also seem neatly separated. It occurred to me that if bubbles, as they rise up (along a z coordinate) from a nucleation site, at the same time perform a random walk in the x and y directions, it may be very hard to distinguish this from the explanation as a symmetry-breaking process based on the lack of similarity. The main giveaway that I see is that, again, the pattern does not look like an instance of a Poisson point process. I'd expect nucleation sites to form a Poisson point field, and a random walk then only increases the Poisson-pointness, so that you expect to see more clustering, arising purely by chance, than seen in these images. --Lambiam 23:57, 28 November 2021 (UTC)[reply]

So, does this mean we are living in a simulation? Just kidding. Thanks for your help with this, Lambiam. Viriditas (talk) 22:01, 29 November 2021 (UTC)[reply]

Where do microrobots go after finish its job?

I learn that microrobots can also be used to target/ treat cancer cells from research site. Rizosome (talk) 03:33, 25 November 2021 (UTC)[reply]

From the linked article (emphasis mine): "These micromotors can penetrate the mucus of the digestive tract and stay there for a long time. This improves medicine delivery," Gao says. "But because they're made of magnesium, they're biocompatible and biodegradable." And from the original article: The micromotors have been eventually cleared by the digestive system via excrement, without any adverse effects. JoelleJay (talk) 06:28, 25 November 2021 (UTC)[reply]

This line answered my question: "cleared by the digestive system via excrement". Rizosome (talk) 01:10, 26 November 2021 (UTC)[reply]

Resolved

Mr Ruthven

Is mentioned in HMS Waterwitch (1866) as having had the idea of driving ships with waterpumps, as mentioned. Lacking a link, I seem to be unable to find which Mr Ruthven that is, or, barring the wikipedia does not have an article about said individual, I would like to inquire as to who this Mr Ruthven is - anyone an idea? Regards, --G-41614 (talk) 14:47, 25 November 2021 (UTC)[reply]

Here's something. They don't seem to be certain about the first name, though... --Wrongfilter (talk) 15:31, 25 November 2021 (UTC)[reply]

This seems interesting. According to this, John is the son of the inventor, Morris West Ruthven, see also here. I leave it to you to actually read these references and figure out who was involved with the good ship Waterwitch. --Wrongfilter (talk) 15:38, 25 November 2021 (UTC)[reply]

Thank you very much, this does help and, after only a first glance, appears to answer my question. Regards, --G-41614 (talk) 16:36, 25 November 2021 (UTC)[reply]

Day length of a retrograde planet

If a planet orbits a star every 15hours but the planet itself is rotates around itself every 30hours in retrograde direction, what is the average "day"/"night" length on the planet? Jo-Jo Eumerus (talk) 17:12, 25 November 2021 (UTC)[reply]

~~Its day should be 30 hours, while its year should be 15 hours. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:11, 25 November 2021 (UTC)~~[reply]

You may find Mercury (planet) interesting. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:23, 25 November 2021 (UTC)[reply]

That doesn't answer my question; I am not asking about the rotation period but about the day/night duration. Jo-Jo Eumerus (talk) 20:12, 25 November 2021 (UTC)[reply]

The general formula for the synodic period

T

is:

T={\frac {T_{o}T_{sd}}{T_{o}-T_{sd}}}

,

where

T_{o}

is the orbital period and

T_{sd}

is the sidereal period. The sigh of

T_{sd}

corresponds to the prograde/retrograde case. So, in this case, when

T_{sd}=-30

hours and

T_{o}=15

hours, we have

T=-15*30/(15+30)=-10

hours. Ruslik_Zero 20:52, 25 November 2021 (UTC)[reply]

Note that the formula can be rewritten in the form

\left|{\frac {T}{T_{\operatorname {o} }}}-{\frac {T}{T_{\operatorname {sd} }}}\right|=1,

which has an obvious and easy to remember interpretation: the number of turns during a period

T

in the synodic view differs by exactly

1

from that number in the sidereal view, because the respective reference frames then rotate exactly one turn with respect to each other. The tricky part is to figure out the sign of the difference. --Lambiam 22:45, 25 November 2021 (UTC)[reply]

If you're standing at a fixed point of the equator of this fictitious planet, holding a clock which allows for a 30-hour day, and forgetting about twilight and such, shouldn't a single day be exactly 30 hours? 15 hours of sunlight and 15 hours of night? ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:28, 25 November 2021 (UTC)[reply]

No because it is orbitting the star twice in the time, in the opposite direction, so the star appears to move faster in th sky. Graeme Bartlett (talk) 04:18, 26 November 2021 (UTC)[reply]

So if its sun is overhead at high noon on one day of that 30-hour clock, it won't be overhead at high noon the next day? ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:23, 26 November 2021 (UTC)[reply]

Yes if 30 hours is mean solar time not sidereal/(real) but only by definition. Retrograde and years close to 1 day (and orbit ellipticity) can really screw things up. There are even places on Mercury where the sun decelerates and goes back — setting the same side it rose (east or west, I forgot which). The Sun can only reverse direction once a year but that's like 0.5 average days between noons (net noons, not gross). Or 1.5 actual rotations. Sagittarian Milky Way (talk) 05:40, 26 November 2021 (UTC)[reply]

Would the Reference desk/Mathematics be a better place for this? -- Q Chris (talk) 08:13, 26 November 2021 (UTC)[reply]

Maybe. I'm curious how close to the sun that planet would have to be, to circle it once every 15 hours. ←Baseball Bugs ^{What's up, Doc?} carrots→ 14:11, 26 November 2021 (UTC)[reply]

Kepler's third law gives the answer: 0.014 AU, i.e. 2.1 million kilometers. We know many exoplanets with orbital periods on the order of days (e.g. Hot Jupiters). --Wrongfilter (talk) 14:30, 26 November 2021 (UTC)[reply]

So if the hypothetical object were that close to the sun, it would almost be "in" it. ←Baseball Bugs ^{What's up, Doc?} carrots→ 14:40, 26 November 2021 (UTC)[reply]

If it were our Sun (with a diameter of nearly 1.4 million kilometers), yes, but the OP specified a hypothetical planet orbiting a star (its sun), and many exoplanets may orbit stars much smaller than the Sun, including Red dwarfs and White dwarfs, and even Neutron stars which are very small indeed by comparison. {The poster formerly known as 87.81.230.195} 90.205.225.31 (talk) 15:48, 26 November 2021 (UTC)[reply]

The formula can just be rewritten as