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August 17

Transmutation

I'd like to see a video of an element transmuting into another element. I quote Richard Dawkins and Yan Wong in their wonderful book 'The Ancestor's Tale', "The half-life of carbon 15 is 2.4 seconds. After 2.4 seconds you'll be left with half of your original sample. After another 2.4 seconds you'll have only a quarter of your original sample..." I realize that carbon-15 is not something you are going to readily find in a laboratory. I'm guessing it would appear to be sublimating into a gas. Ifyoucrydon'tlose (talk) 11:21, 17 August 2024 (UTC)[reply]

Does he say what it transmutes into? ←Baseball Bugs ^{What's up, Doc?} carrots→ 11:24, 17 August 2024 (UTC)[reply]

No he doesn't. In the Wikipedia article 'Isotopes of Nitrogen', nitrogen-15 is said to come from carbon-15. Ifyoucrydon'tlose (talk) 11:33, 17 August 2024 (UTC)[reply]

Even more interesting would be a video of a solid transmuting into another solid. But when I search for videos on the subject, all I find are chalk talks. Ifyoucrydon'tlose (talk) 11:43, 17 August 2024 (UTC)[reply]

Our article Isotopes of carbon agrees with the half-life. It transforms into nitrogen-15. But it would not be possible to make a visible solid piece of carbon-15 as it decays so fast, and in decaying releases so much energy. I could imagine a solid lump would just explode into a hot carbon plasma fireball in a fraction of a second. An example of a video could be of a supernova, where nickel-56 decays into iron-56 over about a month. Graeme Bartlett (talk) 11:38, 17 August 2024 (UTC)[reply]

You're right. These isotopes are unstable and are involved in long chains of reactions. Ifyoucrydon'tlose (talk) 11:46, 17 August 2024 (UTC)[reply]

I'd like to see a video though of the transmutation of solid lead into solid gold. --Lambiam 21:27, 17 August 2024 (UTC)[reply]

It could be done with some isotope with a half-life of a few decades, then make a time-lapse video. But you have to put the highly radioactive sample in front of a camera (or automatically position is in front of the camera for a few seconds every week), instead of safely locked away in some nuclear waste storage facility. A risky move for a time-lapse video that will only be seen by the next generation. PiusImpavidus (talk) 08:35, 18 August 2024 (UTC)[reply]

Any time you look at something that is radioactive, or at a nuclear explosion, you're seeing it. I think the interesting examples would be when the substance changes appearance. Watching one metal (or metal-oxide) become another metal...are you actually seeing a change? Something like a solid to a gas, or evidence of releasing a gas from a solid (alpha particles) would be nice. But it is indeed an annoying trade-off between how long you want to spend making a time-lapse video vs how willing and able to are to handle extremely hot isotopes. Sevearl isotopes of radium have a nice balance...easily handled by those who do that sort of thing, a half-life of a few days, and radium metal becomes radon and helium gasses. "All you need" is a large enough sample to see it and not mind wasting it. Iodine has an available isotope with a half-life of hours and another with a half-life of days, which both decay into tellurium, so "purple gas or condensed phase becomes metal", but I would assume it would be microscopic particles rather than a noticeable solid aggregate. DMacks (talk) 09:22, 18 August 2024 (UTC)[reply]

What happens when kids don't get enough sugar?

What happens when kids don't get enough sugar? 2A00:23C5:1C00:C001:8DB5:7D3A:7C98:F54A (talk) 11:24, 17 August 2024 (UTC)[reply]

see Protein poisoning if there is no carbohydrate in the diet. Carbohydrate does not have to be in the form of "sugar" but could be starch. Graeme Bartlett (talk) 11:41, 17 August 2024 (UTC)[reply]

The name of the article is misleading. This is not poisoning due to proteins. Also, reading the article, it seems it's more the lacking of fats rather than carbohydrates. 2A0D:6FC0:982:9000:411B:FF20:C5F4:B172 (talk) 18:21, 17 August 2024 (UTC)[reply]

The health problems, metabolic disturbances such as hyperammonemia and hyperinsulinemia, are caused by the excess digestion of proteins due to a lack of metabolically less problematic energy sources. So it might be more accurate to speak of poisoning by protein metabolites, but for many poisons it is not the compound as ingested that impacts a patient's health, but metabolites of the ingested substance. --Lambiam 21:18, 17 August 2024 (UTC)[reply]

Basically, nothing. Especially if sugar means simple sugars or disaccharides, where the result would likely be better health and teeth. But unlike the other two macronutrients, proteins and fats, carbohydrates are not essential to human diet, and communities have lived for millennia, and millions of people today with close to zero carbohydrate consumption. This refers to those beyond infancy, Breast milk of course has substantial sugar/carbohydrate. Removing sugars from an infant's diet might lower insulin levels and inhibit growth, just as modern high sugar consumption (together with ample fat and protein) is thought to promote growth, sometimes deleteriously.John Z (talk) 00:44, 19 August 2024 (UTC)[reply]

Some amino acids are constructed directly from sugar. So most enzymes (which needs these amino acids) can not be constructed when low on sugar. If the protein is rich in all these amino acids there is no problem. But some amino acids are not abundant in protein, but often needed in enzymes (Tyr). That will lead to health problems in the long run if not specifically countered by using specific protein for nourishment. 176.0.148.248 (talk) 14:58, 30 August 2024 (UTC)[reply]

Gluconeogenesis normally goes on at all times and provides sufficient glucose for normal adult needs, including what is needed for amino acids and enzymes the body can construct. The universal modern problem is excessive sugar consumption, leading to dangerously high or erratic blood sugar levels, not insufficient levels. Lower blood sugar levels have repeatedly been clearly associated with health and longevity as in this study. As I said, many human communities have subsisted for millennia consuming very few carbohydrates- but maintaining adequate blood sugar levels with no known negative consequences. Many more, perhaps a majority, have subsisted that way for months at a time every year. So without a reference, skepticism is warranted here. Aside from growth the only time carbohydrates have a real usage for which fat is not a near-perfect substitute is after "bonking" or "hitting the wall". That is, when exercise has depleted glycogen stores faster than they can be resupplied.John Z (talk) 23:40, 1 September 2024 (UTC)[reply]

I'm a bit confused about the formal definition of the work done by a constant force: Is that work defind, as a product, or as an integral?

Regarding the work W done by a constant force ${\vec {F}}$ along a displacement ${\vec {s}}$ in a straight line in the direction of the force, our article work (physics) claims, that if the force is constant then the work done by that force is $W={\vec {F}}\cdot {\vec {s}},$ whereas if the force is variable then the work done by that force is $W=\int {\vec {F}}\cdot d{\vec {s}}.$

For simplifying my question, let's assume the force is constant, so that it exerts a constant acceleration $a$ on a given mass $m$ being constant (in classical mechanics).

1. Using the first formula $W={\vec {F}}\cdot {\vec {s}},$ we receive the potential energy only: $W={\vec {F}}\cdot {\vec {s}}=(m{\vec {a}})\cdot {\vec {s}}.$

2. Using the second formula $W=\int {\vec {F}}\cdot d{\vec {s}},$ we receive the change in the kinetic energy only: $W=\int {\vec {F}}\cdot d{\vec {s}}=\int ma\cdot d{\vec {s}}=\int m{\frac {d{\vec {v}}}{dt}}\cdot d{\vec {s}}=\int m{\frac {d{\vec {s}}}{dt}}\cdot d{\vec {v}}=\int m{\vec {v}}\cdot d{\vec {v}}=\Delta ({\frac {1}{2}}mv^{2}).$

3. To sum up: It seems there are apparently two kinds of work (assuming that the force doing the work is constant): The work equivalent to the potential energy should only be defined according to the first formula $W={\vec {F}}\cdot {\vec {s}},$ whereas the work equivalent to the change in the kinetic energy should only be defined according to the second formula $W=\int {\vec {F}}\cdot d{\vec {s}}.$

I wonder where my mistake is. HOTmag (talk) 20:43, 17 August 2024 (UTC)[reply]

One issue is that you are using the notation of indefinite integrals, although the integrals here should definitely be definite, giving the change in energy. By the conservation of energy, for an object in free fall,

|\Delta E_{\text{pot}}|=|\Delta E_{\text{kin}}|.

Apart from the sign, they are the same. If the object is not in free fall, you may not replace

g

by

{\frac {d{\vec {v}}}{dt}}.

--Lambiam 21:03, 17 August 2024 (UTC)[reply]

As for the notation, you are right, but I followed the notation in our article work (physics), and I thought it was clear what this notation meant.

As for the "change" in the kinetic energy: Thanks to your comment, I've just added the word "change" (See above).

As for your last remark about objects in free fall: Please notice I've only replaced the acceleration

a

by

{\frac {d{\vec {v}}}{dt}}

in #2, that discussed the kinetic energy (in free fall), so what was wrong there?

Let me ask you that more explicitly: What is the formal definition of the work

W

done by a constant force

{\vec {F}}

along a displacement

{\vec {s}}

in a straight line in the direction of the force? Is it

W={\vec {F}}\cdot {\vec {s}},

or

W=\int {\vec {F}}\cdot d{\vec {s}}?

HOTmag (talk) 21:15, 17 August 2024 (UTC)[reply]

If

{\vec {F}}

is constant, you can take it out of the integral:

W=\int _{start}^{end}{\vec {F}}\cdot d{\vec {s}}={\vec {F}}\cdot \int _{start}^{end}d{\vec {s}}={\vec {F}}\cdot {\vec {s}}

so both definitions are equivalent. If

{\vec {F}}

is not constant, only the version with the integral works. So the version with the integral is the proper way to do it, the version without is a simplification possible with a constant force. PiusImpavidus (talk) 23:13, 17 August 2024 (UTC)[reply]

On the other hand, if the force is not constant, then from the proper definition (in that case):

W=\int {\vec {F}}\cdot d{\vec {s}},

we receive the change in the kinetic energy only:

W=\int {\vec {F}}\cdot d{\vec {s}}=\int ma\cdot d{\vec {s}}=\int m{\frac {d{\vec {v}}}{dt}}\cdot d{\vec {s}}=\int m{\frac {d{\vec {s}}}{dt}}\cdot d{\vec {v}}=\int m{\vec {v}}\cdot d{\vec {v}}=\Delta ({\frac {1}{2}}mv^{2}),

yet we cannot receive the change in the potential energy - from that definition, unless we assume the conservation of the sum of both energies, am I right? If I am, then the kinetic energy has apparently a "technical advantage" over the potential energy, mathematically speaking, right? HOTmag (talk) 08:04, 18 August 2024 (UTC)[reply]