Wikipedia:Reference desk/Archives/Science/2017 August 6

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August 6

How to determine electric and magnetic susceptibility

What are the methods to measure electric susceptibility (or polarizbility) and magnetic susceptibility of a given material experimentally? — Preceding unsigned comment added by Sayan19ghosh99 (talk • contribs) 05:23, 6 August 2017 (UTC)

See Electric susceptibility and this learning package. One might measure by constructing a capacitor with the test material as dielectric.

See Magnetic susceptibility which mentions experimental methods to determine susceptibility that include the Gouy balance and Evans (or Johnson-Matthey) balance. Blooteuth (talk) 14:27, 6 August 2017 (UTC)

Mounting Hole Diameter for M1.5 spring pin

I have a M1.5 spring pin and I'm trying to determine the correct hole diameter for it. So far I found this chart[1], which is for imperial sizes roll pins. I just need a similar chart but for metric sizes. Mũeller (talk) 06:06, 6 August 2017 (UTC)

Lokking at the imperial chart it looks as though the pins are sized for the quoted diameter, ie in your case 1.50mm. here's atable, it says 1.50 to 1.60 https://technifast.co.uk/recommended-hole-sizes Greglocock (talk) 06:21, 6 August 2017 (UTC)

Thanks! Mũeller (talk) 09:09, 6 August 2017 (UTC)

Resolved

Flowers' identification help

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  6

All these pictures of flowers were taken at the Fitzroy Gardens Conservatory (apart from 6, which was outside in the gardens). Can anyone here help identify them please? --SuperJew (talk) 13:08, 6 August 2017 (UTC)

For convenience I should note the conservatory is in Melbourne near the southern extreme of the Australian mainland. I haven't made many attempts to identify flowers, so I should start by pointing at plant identification, which has some resources listed at the end; this guide to 50 most common families may also be useful; searching the web for australia flower identification turns up more specifically useful resources like [2] [3] (an image database, more for confirmation I would think). The government of Australia claims to have resources [4], but ends up pointing you at a selection of copyrighted "CD keys" written for archaic operating systems... with many other hits also, it looks like beating off the ads to find the usable resources is the tough part.

I would encourage people to try to explain how to do good plant identification with limited data we see in a case like this, and to upgrade the plant identification article to match. Wnt (talk) 15:08, 6 August 2017 (UTC)

"The government of Australia claims to have resources [5]" is a duplicate of the previous link. I don't think this was intentional since none of the links there point to anything to do with CD keys. Nil Einne (talk) 15:43, 6 August 2017 (UTC)

The fact that the conservatory is in Melbourne doesn't necessarily mean the plants are local, and the fact that they are in a conservatory means that the local weather isn't a factor either. --SuperJew (talk) 16:47, 6 August 2017 (UTC)

It looks to me like 1 could be some form of Aeonium, perhaps Aeonium arboreum. Definitely it looks like some kind of succulent. (P.S. If this right, I worked it out partially based on a Google reverse image search. It looked familiar but I didn't know what it was, but it looked distinctive enough it seemed worth giving it a shot. I tried the same thing on 4 but didn't see anything that hopeful although didn't look that carefully.) Nil Einne (talk) 17:20, 6 August 2017 (UTC)

@Nil Einne: That does look very promising (and reinforces my theory that they don't have to be local flowers). A Google reverse image search is a great idea! Thanks --SuperJew (talk) 17:31, 6 August 2017 (UTC)

I agree there's a reasonable chance they're not local since it doesn't seem to be the focus of the gardens or the conservatory (unlike say the Royal Botanic Gardens, Cranbourne). That said, I wouldn't say it's useless information to know it's in Melbourne, especially since we are talking about Australia and our article suggests they're state and city owned. I have a strong feeling given those circumstances, it's unlikely they're going to intentionally display anything that's a noxious weed (or whatever terms they use for problem invasive species) in Melbourne/Victoria unless they have a very good reason. And actually a quick search finds [6] which supports my suspicion. Nil Einne (talk) 14:48, 7 August 2017 (UTC)

I'm wondering maybe 2&3 are Impatiens walleriana?

Perhaps 4 is Hydrangea macrophylla, like in this pic?

Maybe 5 is a form of Fuchsia, perhaps Fuchsia boliviana or something similar? --SuperJew (talk) 18:13, 6 August 2017 (UTC)

No. 6 could be a variety of clivia. Richard Avery (talk) 07:41, 7 August 2017 (UTC)

2 & 3 are a variety of Impatiens hawkeri ("New Guinea impatiens"), I think. William Avery (talk) 11:49, 7 August 2017 (UTC)

Weight loss depression

Is depression a common occurrence in people who lose a lot of weight very quickly? Weisqusestion (talk) 14:05, 6 August 2017 (UTC)

“ Reduced nutrition can cause weight loss. The more hyperstimulated the body becomes, the more weight we can lose. These are just a few examples of how increased stress and anxiety can cause weight loss. ” — Jim Folk, Marilyn Folk, BScN., http://www.anxietycentre.com/anxiety-symptoms/weight-loss.shtml

50.4.236.254 (talk) 16:31, 6 August 2017 (UTC)

Lack of sleep can cause more calories to burn during the waking hours. Hence weight loss. Selective serotonin reuptake inhibitors, which are used to treat depression, may cause weight loss in older adults, but that's kind of speculative. Selective serotonin reuptake inhibitors (SSRIs) may produce side effects, including insomnia, nervousness, restlessness, and diarrhea. Nervousness and restlessness may produce more physical agitations, and hence, more fuel is burned. Diarrhea may cause malabsorption of nutrients. Depression is also heavily associated with bulimia.

Source: Jae Allen, http://www.livestrong.com/article/330360-why-does-depression-cause-you-to-lose-weight-even-though-you-eat-a-lot/

50.4.236.254 (talk) 16:46, 6 August 2017 (UTC)

The above sources suggest that the symptoms of depression may induce weight loss. It is possible that intentional weight loss may induce depression, as shown here. Source: http://abcnews.go.com/Health/w_DietAndFitnessNews/diets-depression-learn-mice/story?id=12357575 50.4.236.254 (talk) 16:58, 6 August 2017 (UTC)

Understanding temperature

The definition of temperature is

${\frac {1}{T}}={\frac {\partial S}{\partial U}}$

So in order to calculate $T$, one must express $S$ in terms of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle U} .

The definition of entropy is

$S=-k\sum _{i=1}^{\Omega }p_{i}\ln p_{i}$

and the definition of internal energy is

$U=\sum _{i=1}^{\Omega }p_{i}E_{i}$.

But it should be clear that it is not necessarily possible to express $S$ in terms of $U$.

For example, suppose $\Omega =3,E_{1}=\epsilon ,E_{2}=2\epsilon ,E_{3}=3\epsilon$.

Then

$\{p_{1}={\frac {1}{3}},p_{2}={\frac {1}{3}},p_{3}={\frac {1}{3}}\}$ and $\{p_{1}={\frac {1}{4}},p_{2}={\frac {1}{2}},p_{3}={\frac {1}{4}}\}$

give the same value for $U$, but different values for $S$. So how can you define temperature? PeterPresent (talk) 14:58, 6 August 2017 (UTC)

Not an expert, but by my quick reckoning both of your examples have U = 2e. The first has evenly split energy states and entropy -k ln (1/3), while the second has -k ln (1/sqrt(8)). The second has less entropy because we have a good guess that a particle will be in the second energy state. Now of course S/U doesn't mean much by itself; we want the derivative of S/U, which means we have to look at how energy can increase. Which means that we need to look at how much S changes when energy is increased. And you're at kind of a funny point in that regard, like halfway to a negative temperature, in the sense that with one system you've defined three energy states and put the lower one as less likely than a higher state. So if we change 1/4 1/2 1/4 to say 1/4 3/8 3/8 the entropy drops from 1.03 k to 0.24 k, i.e. the entropy drops as energy is added increases to 1.08 k -- sorry, that isn't a suitable example after all -- ... at least if it is added this particular way. I'm not sure how you decide in what ratio of ways the energy is added. We'll need someone better at physics to go on from here... Wnt (talk) 15:29, 6 August 2017 (UTC)

Temperature is only well defined in thermal equilibrium, which means that you can only consider probability distributions that maximize S. Here you are free to impose constraints that prevent a global maximum from being reached, but you then have to specify these constraints (e.g. take a gas in a container of a fixed volume, the outside is a vacuum, if the volume were not fixed then the maximum entropy would be reached in the limit of an infinite volume). Now, in this case you have 3 probabilities, the sum of the probabilities is equal to 1, the internal energy provides another constraint, so there is one degree of freedom left. You should then fix this by either maximizing S or this is going to be fixed due to some other constraint the system is subject to. You then have a well defined S as a function of U, so the temperature is then well defined. Count Iblis (talk) 15:34, 6 August 2017 (UTC)

The formula for entropy expressed through $p_{i}$ pre-supposes that those probabilities $p_{i}=\exp(-E_{i}/T)/Z$ i.e. they obey Gibbs distribution ($Z$ is canonical partition function). Only in this case the (inverse) temperature can also be expressed as the first derivative from the entropy by the total energy. Ruslik_Zero 19:29, 6 August 2017 (UTC)

It makes sense that, given equal energies, entropy would reach a maximum as @Count Iblis: says. And as @Ruslik0: says, we expect to see a Gibbs distribution of energy states under normal circumstances. But are these things compatible, and can we derive the formula from the assumption? I suppose if we have a p_i, p_i+1, p_i+2 and we assume that the energy differences between them are equal, then converting x particles from p_i and p_i+2 to p_i+1 will involve no change in entropy. The total entropy from those three states only will be proportional to (p_i - x) ln (p_i - x) + (p_i+2 - x) ln (p_i+2 - x) + (p_i+1 + 2x) ln (p_i+1 + 2x). We want to minimize that negative number to maximize total entropy (-k multiplied by it). The derivative d/dx, for the limit of small x, should be - ln(p_i) - ln(p_i+2) + 2 ln (p_i+1) - p_i (1/p_i) - p_i+2 (1/p_i+2) + 2 p_i+1 (1/p_i+1) At least at a hasty consideration, I think the second terms of each derivative product cancel out in the end. Now we want an extremum of entropy, so this derivative should be 0, i.e. ln(p_i + ln(p_i+2) = 2 ln (p_i+1); taking e to the each we get p_ip_i+2 = p_i+1². This indicates an exponential relationship, at least, among the populations of energy levels, though it would take (at least) some normalization to get to the Gibbs distribution. Wnt (talk) 14:19, 7 August 2017 (UTC)

You can derive it quite easily using Lagrange multipliers. At the maximum the partial derivative of the entropy w.r.t. p_i should be a linear combination (independent of the index i) of the partial derivatives w.r.t. the constraint functions. The sum of the probabilities is one constraint function as this must equal 1 and the partial derivative if this function w.r.t. p_i equals 1. The expectation value of the energy is the sum over p_i e_i is another constraint function, the partial derivative w.r.t. p_i equals e_i. So, you then find that the logarithm of the probability is a linear function of the energy,so we're led to the Gibbs distribution. Count Iblis (talk) 19:37, 7 August 2017 (UTC)

@Count Iblis: I don't understand how Lagrange multipliers would fit in. What I derived above shows that the energy states have to follow some rule of occupancy Ae^BE. Normalizing the outer part (A) just involves summing up all the probabilities and making them equal 1; our article on the Gibbs distribution uses that summation without further processing. The part I didn't get to was the inner bit - how do you show that this constant B is actually -1/kT? Though I suppose k may be "by definition" (whatever makes the units work out), once you can show some measure of T is involved. Wnt (talk) 20:14, 10 August 2017 (UTC)

What would happen if you brought a horse to a big pressurized lunar dome?

Would it learn the way to move from A to B with least energy or most speed? Would that gait involve jumping? How fast could it go? Sagittarian Milky Way (talk) 18:22, 6 August 2017 (UTC)

What have you found on Google so far? ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:55, 6 August 2017 (UTC)

Somehow I doubt that there're any experimental results floating around (no pun intended, probably) regarding horses on the Mun... Whoop whoop pull up ^{Bitching Betty | Averted crashes} 07:37, 7 August 2017 (UTC)

[7] Sagittarian Milky Way (talk) 19:06, 6 August 2017 (UTC)

I strongly suspect that the only way we'd ever find out would be to actually take it there – from the literature and and lectures I've read/attended in the area of possible Crewed Mars missions, we're still uncertain as to how humans will best perambulate in Mars' gravity (very roughly twice that of the Moon's). One possibility might be a mentally broken, catatonic horse. It might also depend on whether the horse was taken as an adult or raised from embryo/early infancy on the Moon: Arthur C Clarke once suggested that a canary hatched in a zero-gravity space station would easily adapt to flight there, but I don't think he had any experimental data. {The poster formerly known as 87.81.230.195} 90.202.208.125 (talk) 20:51, 6 August 2017 (UTC)

Language peeve: twice that of the Moon's what? that in this phrase means 'the gravity'; and I don't think it's normal English to say "the gravity of the Moon's" in the same way as "a friend of Alice's". Or – since Moon's is parallel in form to Mars's – maybe it's twice [the something else] of the Moon's [gravity]? —Tamfang (talk) 05:37, 7 August 2017 (UTC)

Some thoughts:

1) We do know that horses are capable of adapting their gait to different situations. For example, they learn to move more slowly when wading through water, or to step higher when walking through thick mud or heavy snow.

2) So, it's reasonable to assume they would also adapt their gait to less gravity.

3) However, bouncing higher into the air with each step might be scary, causing them to use less force to take smaller steps. This would be especially true if they fall over, say by rotating when airborne, when using full force.

4) They may eventually learn to use full force again safely. Or, their muscles may atrophy from less use. Perhaps something in-between is the most likely outcome. StuRat (talk) 19:49, 7 August 2017 (UTC)

Wireless telephone in 1922

Sinclair Lewis mentions wireless telephones in his novel Babbitt, published 1922. What does he mean by that? The wireless telephone article redirects to cordless telephone, but these telephones were not invented until the 60s. Extract:

"Though he no longer spoke of mechanical engineering and though he was reticent about his opinion of his instructors, he seemed no more reconciled to college, and his chief interest was his wireless telephone set. "

--Hofhof (talk) 20:56, 6 August 2017 (UTC)

The device is closer to the mobile radio telephone of the 1950s and 1960s. These were VHF voice radios, about 150MHz and FM modulation, which could be used from within a moving car. They were also simple enough to operate, with simple crystal-controlled channel selectors, to not require "tuning" and thus a separate operator. They connected to a regional base station and operator who was able to patch them into the national telephone network. Andy Dingley (talk) 21:03, 6 August 2017 (UTC)

Wireless telephones existed by the 1910s. Here's a 1916 song about the subject.[8] ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:44, 6 August 2017 (UTC)

And here's a 1914 photo of one (not quite a "mobile" phone):[9] — 2606:A000:4C0C:E200:4890:FAC5:8AED:1CC9 (talk) 23:09, 6 August 2017 (UTC)

In that era, telephones was a word used for what we now call headphones, at least in Britain. See the testimony of Harold Bride, wireless operator of the Titanic, when he gave evidence at the American inquiry into the sinking. The relevant testimony is here and you need to scroll up about one screen page from the bottom. Akld guy (talk) 00:31, 7 August 2017 (UTC)

And again, here, again about one page up from the bottom. Akld guy (talk) 01:16, 7 August 2017 (UTC)

@Baseball_Bugs, after I listened to the song you've posted, I said: "This guy is a genius. He provided an answer, an incredible historical reference and an entertainment. I would like to get the lyrics, though. Can you do it? Thanks, - --AboutFace 22 (talk) 12:36, 7 August 2017 (UTC)

We have an article: Hello, Hawaii, How Are You?. The wireless telephone in question is (presumably) this. AndrewWTaylor (talk) 13:21, 7 August 2017 (UTC)

Resolved

With kudos and thanks to Baseball Bugs for his pertinent, well referenced response from the world of entertainment. Blooteuth (talk) 14:41, 7 August 2017 (UTC)

• Remember that at the time "wireless" was a synonym for "radio", but most radio signals were Morse code, so the term "telephone" was needed to specify that the radio was sending and receiving voice, not Morse code. The name of the firm "Cable & Wireless" dates from that era. -Arch dude (talk) 04:21, 9 August 2017 (UTC)