Science desk
< August 2	<< Jul \| August \| Sep >>	August 4 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

August 3

Wording in article Gravity

Why the sentence in Newton's theory of gravitation of article Gravity is "most modern non-relativistic gravitational calculations are still made using Newton's theory because it is simpler to work with and it gives sufficiently accurate results for most applications involving sufficiently small masses, speeds and energies." instead of "most modern non-relativistic gravitational calculations are still made using Newton's theory because it is simpler to work with and it gives sufficiently accurate results for most applications involving sufficiently small masses, speeds or energies."? Does it mean that at least the three prerequisites must be met simultaneously then Newton's theory can be applied? - Justin545 (talk) 05:49, 3 August 2020 (UTC)[reply]

It just means that Newtonian gravity is perfectly adequate except in some highly specialized applications. Unless you're measuring the defect in the orbital precession of Mercury or making fine adjustments to atomic clocks on satellites (and most people aren't) then Newtonian mechanics works fine. --Jayron 32 06:24, 3 August 2020 (UTC)[reply]

Mercury moves at low speed (compared with the speed of light) but has a large mass and large kinetic energy (if I understand correctly). According to the sentence, Newtonian gravity is not adequate to accurately predict Mercury's orbit since the mass and energy criteria are not met, isn't it? Because of the "and" in the sentence, 3 criteria (mass, speed, energy) must be ALL met (all of them must be small) or Newtonian gravity cannot give sufficiently precession, right? In other words, Newtonian gravity is inadequate for precisely calculating anything about Mercury because the mass of Mercury is not small in all cases (even if it has small speed and has small energy), right? - Justin545 (talk) 08:36, 3 August 2020 (UTC)[reply]

See Tests of general relativity#Perihelion precession of Mercury. It's the distance to the Sun that leads to a measurable deviation; essentially the gravitational attraction is close enough for deviations to just barely be noticeable with Mercury. It's very small, but real, on the order of 40 arcseconds per hundred years; given that an arcsecond is 1/1296000 of a revolution, we're talking a deviation of a few dozen ppm per 100 years. What's more impressive is not that GR is needed to explain the deviation, but that the deviation was measured in the 19th century. You're focusing too much on the individual words here. Newtonian gravity works in cases except where the values become large enough for the deviations between GR and Newton to show up in significant figures of whatever the tolerances of your measuring devices you need to use are. For you, dropping a ball on earth, and making measurements with a chronometer and a ruler, this is way out of the tolerances of your measurement, so use Newton. Even for the Earth-Moon system, Newton works fine in most cases. Look at the equation of gravity according to Newton: F = G * (m₁*m₁/r²). The only values that matter are m and r. G is a constant that just makes the units work out, so we can ignore it. The only times where General Relativity needs to be used is when 1) The values of m and r are extreme (large values of m and small values of r) and 2) Where relativistic effects significantly alter the values of m and r (in things like time dilation or length contraction, or mass-energy equivalence) and those values only show up at high speeds (which is the same thing as high kinetic energy) or high potential energy (high forces). Since gravitational force (potential energy) increases with high masses and short distances, that just the same as what I said at #1. If you're wondering how GR deals with these forces, it doesn't. It doesn't treat gravity as a force, it treats it as a warping of spacetime itself, and treats objects with a large mass at a close distance as though they are moving in straight lines at a constant speed through a distorted spacetime. There is no force of gravity in GR. This kind of warping of geometry is fantastically mathematically complex; it took Einstein something like 10 years to formalize General Relativity because he had to wait for people like Minkowski and Hilbert and Lorentz and the like to literally invent the mathematics that he needed to do the calculations right. So, if you want to do gravity calculations, you can either do this insanely complicated tensor calculations, or you can do simple algebra. If the difference between the two methods doesn't show up on your measuring devices, and you literally could not tell the difference between the two results in a real physical sense, then use Newton's equation. The various relativity equations only shows up at speeds close to the speed of light, and near very large masses. The reason why satellites need to correct for this is because they are at a different distance from earth than you are, so their clocks run at a slightly different rate than yours; on the order of a few ppm, which seems small but can amount to several meters in a day, and several kilometers over the course of a year. When you're using those satellites for doing things like guiding your car on a road, or guiding a bomb to its target, you don't want it to drift that much. --Jayron 32 18:02, 3 August 2020 (UTC)[reply]

Okay ... basically, I though every sentence in the article is supposed to be correct, so readers are not misled by the contents. Because I am not familiar with relativity and the conjunction looked odd to me, so I was trying to confirm it here. I would directly modify the article and correct the conjunction if it were confirmed to be wrong. Although the condition is that they need to be small, I didn't see an explicit statement is about how small should they be. Which may be the source of ambiguities where formulas and equations should kick in. - Justin545 (talk) 07:40, 4 August 2020 (UTC)[reply]

English is not as restrictive as boolean operators in terms of the meanings of conjunctions, and I don't find the use of "and" confusing here. I don't generally read "and" to mean "only and" unless the additional qualifiers are there. --Jayron 32 15:43, 4 August 2020 (UTC)[reply]

If your concern is in regard to grammar, then either wording is fine. One can be calculating speed and energy; or, one can be calculating either speed or energy. In either case Newtonian (non-relativistic) physics equations provide "sufficiently accurate results for most applications...". 2606:A000:1126:28D:E00A:A68D:43D6:44E1 (talk) 07:23, 3 August 2020 (UTC)[reply]

Yes, my concern is the conjunction. I want to know that it should be "and", "or" or "and/or"... - Justin545 (talk) 08:50, 3 August 2020 (UTC)[reply]

All three must be met. If the mass of an involved body is huge, the warping of spacetime may become non-negligible. Large differences in potential energy show up as gravitational time dilation not accounted for by Newton's Laws. And high speeds give rise to velocity time dilation. The last two are large enough that they have to be taken into account to make the GPS system work. So the appropriate conjunction is indeed “and”. --Lambiam 08:34, 3 August 2020 (UTC)[reply]

Thanks for the answer. - Justin545 (talk) 08:50, 3 August 2020 (UTC)[reply]

Newton is good enough that no one thought anything was off till about 2 centuries of good telescopes. Degenerate stars need relativity for short term naked eye stuff. Sagittarian Milky Way (talk) 14:22, 3 August 2020 (UTC)[reply]

Excuse me Justin, but if you are not familiar with relativity, how can the conjunction seem odd to you? You must assume that the sentence is right and conclude that an inclusive 'and' is meant (which as a matter of fact is true: if speed is large OR mass is large OR energy is large then newtonian equations will not deliver accurate results, so mass AND speed AND energy have to be small). But here I see a bigger problem: every statement in WP is supposed to be based on some source, so if any source was given in this article, you would have checked there whether 'and' or 'or' is correct. But if you had to ask here, there is probably no source given and the sentence is an original research of the original editor, which is forbidden in WP. Of course many editors do find that their own O.R. is not O.R. but a direct conclusion from general knowledge. I, more an user of WP than an editor, I do prefer that every claim in WP to be validated by some source. 2003:F5:6F0C:E600:487C:DA02:59F5:AE62 (talk) 20:08, 8 August 2020 (UTC) Marco PB[reply]

...Nobody dared to raise a concern about the Oxford comma? Nimur (talk) 17:28, 3 August 2020 (UTC)[reply]

Earlobes and figure

Recently, I have noticed that nearly any person with attached earlobes I have seen either in real life or in the media was slender, as a matter of fact. Now, as I was unable to find any pertaining scientific evidence on this issue, is there anybody here who can explain if there possibly could exist a corresponding link between earlobe shape and bodily figure (apart from the mere fact that attached earlobes are a recessive genetic trait, of course)?--Hildeoc (talk) 09:30, 3 August 2020 (UTC)[reply]

I've never heard of such an association, and the first person who came to my mind with attached earlobes (an old acquaintance of mine) is chubby bordering on obese. 2601:648:8202:96B0:0:0:0:5B74 (talk) 09:53, 3 August 2020 (UTC)[reply]

Earlobes are either attached to the side of the head, or detached and only connected to the ear itself. Yes, free earlobes are the dominant trait while attached earlobes are the recessive trait. Researchers have looked at 34 genetic markers that influence earlobe type. This gives references. 84.209.119.241 (talk) 13:05, 3 August 2020 (UTC)[reply]

All of them are "attached". I've never seen one yet that was floating in space. ←Baseball Bugs ^{What's up, Doc?} carrots→ 15:44, 3 August 2020 (UTC)[reply]

^{See V.vGogh. 84.209.119.241 (talk) 20:36, 3 August 2020 (UTC)}[reply]

Unfortunately, Van Gogh didn't have Jesus around. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:27, 3 August 2020 (UTC)[reply]

More like van Gagh 93.136.213.247 (talk) 21:46, 4 August 2020 (UTC)[reply]

You've never seen an astronaut's earlobes, then? --47.146.63.87 (talk) 21:48, 3 August 2020 (UTC)[reply]

Codeine, promethazine, and CYP2D6

Working through some older edit requests, but I don't know what to make of this one at Talk:Lean (drug)#Semi-protected edit request on 21 July 2020. The article states that promethazine induces CYP2D6, which doesn't seem to be true ([1]). As I understand it, CYP2D6 turns codeine into morphine. So does promethazine inhibit CYP2D6, meaning that less CYP2D6 can metabolize codeine, so there's more of that in the bloodstream, making people feel more high? But I thought morphine makes people feel high? I would appreciate it if someone can also take care of the edit request, but I've been thinking about this one for a while that I need to know the answer too. This is not what I studied in college! ◢ Ganbaruby!  (Say hi!) 13:49, 3 August 2020 (UTC)[reply]

Codeine is itself inactive; it's a prodrug. It's converted by the liver (by the CYP2D6 enzymes in liver cells) into morphine, which is of course active. This conversion happens slowly, which makes the effects of codeine different from those of directly administering morphine. And co-administration of drugs (in this case promethazine) often means they interact with each other, which can lead to different effects from when the drugs are administered by themselves. If promethazine inhibits the activation of codeine, that generally means the blood concentration of the active drug will be lower. It also often means the duration of effect will be longer, since the body is metabolizing the drug ("clearing it out") more slowly. --47.146.63.87 (talk) 21:46, 3 August 2020 (UTC)[reply]