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October 10

Why not travel to Mars in a straight line?

My understanding is that when we send a space craft to Mars, the ship goes into an orbit around the Sun like this and the ship's orbit eventually intersects with Mars' orbit. Why not go in a straight line from Earth to Mars? That would certainly be the shortest distance. I'm guessing that takes more energy? Pealarther (talk) 21:57, 10 October 2022 (UTC)[reply]

Going in a proper straight line would take very much energy indeed. And you would need to aim very carefully not to miss. But more generally, for any plausible spacecraft, such an approach does not work. Remember that you don't only have to match position, you also need to match speed. In other words, both when departing Earth and when arriving at Mars you must necessarily have the same position and the same speed as the corresponding planet. Usually, the most efficient way to do that is a Hohman transfer orbit. --Stephan Schulz (talk) 00:19, 11 October 2022 (UTC)[reply]
Yes, the path shown uses the least energy. Bubba73 You talkin' to me? 05:44, 11 October 2022 (UTC)[reply]
Hello, Pealarther. Start by reading Orbital mechanics several times until you understand it. Then read the references in that article until you understand them. Then you will understand the answer to your question. Cullen328 (talk) 05:49, 11 October 2022 (UTC)[reply]
See also Interplanetary Transport Network.  --Lambiam 06:09, 11 October 2022 (UTC)[reply]
  • The ideal path to Mars is the one that follows a geodesic, which is (in some sense) a "straight" path through curved spacetime, which has been curved through gravity (see General relativity for how this works). The analogy of a geodesic in terms we can visualize is a great circle path on a globe. The most efficient path between two points on earth is along a great circle. This path is not straight and curves when viewed in three dimensions, but in a sense it is straight along the 2D curved manifold that forms the surface of a sphere. In the same way, the most efficient path between two points in 4D spacetime is a geodesic. If you want to think about it in non-GR terms (i.e. not thinking about curved spacetime), then the most efficient path to Mars (the one that uses the least fuel) is the one where you spend the least amount of energy fighting against the gravity. Since the earth, the sun, and other planets all exert gravitational forces on your spacecraft, fuel spent fighting against those forces unnecessarily is wasted fuel. The greatest effect is going to be that of the sun; since the spacecraft leaving the Earth already has the earth's momentum, it's going to want to continue to orbit the sun in an ellipse under the sun's gravity. The most efficient path to Mars is one that doesn't fight against that tendency, and so will be a curved path. --Jayron32 12:29, 11 October 2022 (UTC)[reply]
Hohmann transfer orbit example
Pealarther, here's my best take for explaining why... If you just go on a straight line, you will need a ton of energy to overcome the Sun's gravitational influence. We already need a lot propellant just to get to Earth's orbit, and we want to save the remaining propellant as much as humanly possible. So instead, it would be more clever if we just boost the spacecraft's orbit and make use of the Sun's gravity, just enough so that close to the furthest point in the orbit (apogee) the spacecraft would reach Mars. This is a transfer orbit that requires the least amount of energ and it is called the Hohmann transfer orbit.
In practice, unless you are really, really short on fuel, the spacecraft will be boosted more than what the Hohmann transfer requires to reach Mars faster. And yes, if you are in a hurry and want to move around space like the Expanse, see Space travel under constant acceleration, where the first half of the journey is used to accelerate the spacecraft, and the second half is used to decelerate. CactiStaccingCrane (talk) 15:43, 16 October 2022 (UTC)[reply]
To chime in and hopefully add some help in inuiting this: the problem is that you, Earth, and Mars are all constantly moving—as observed from a Sun-centered reference frame—which makes the notion of "going in a straight line" somewhat incoherent. Your goal is to start at Earth and "hit" Mars. To do this you have to aim at where Mars will be when you intersect its orbit. This takes some calculus to work out, and since you don't control the positions of Mars and Earth you are forced to accommodate their trajectories. The thing you do control is when you leave, and people sending probes to Mars often wait for favorable times when Mars is close to Earth in their orbits to reduce fuel and travel time. Our intuition, "designed" for life as African plains apes, misfires here, because to us it looks like the heavens are sitting there nice and stately, but in reality we're whizzing at fantastic speeds around a nuclear fireball. Try to stand on a moving platform, throw a ball, and hit something on a different platform moving at a different speed. ‐‐47.147.118.55 (talk) 03:58, 17 October 2022 (UTC)[reply]

Photon duration

Today we know how to generate photons individually or a few units. So a photon is emitted at a time t1 and from a time t2 it is considered to have left (like a car with a front and a rear), if so, does its length correspond to its wave length ?

In this Nature article [1] they used "pulse of light", so a pulse have a Start time t1 and end time t2, right ? Malypaet (talk) 22:06, 10 October 2022 (UTC)[reply]

References

  1. ^ https://www.nature.com/articles/nnano.2017.218
A photon exhibits wave–particle duality. Viewed as a wave, it has no definite extension in space, and viewed as a particle, it has no definite wavelength. The wavelength of a photon is directly proportional to the inverse of the magnitude of its momentum. Position and momentum are complementary variables. So, by the uncertainty principle, the more precisely we determine the wavelength, the less we know about its position.  --Lambiam 05:54, 11 October 2022 (UTC)[reply]
The photon either has left, or it has not. There's no time when it's in the process of leaving. PiusImpavidus (talk) 09:11, 11 October 2022 (UTC)[reply]
This is like saying, either Schrödinger's cat has died, or it has not. Welcome to the world of quantum weirdness.  --Lambiam 15:04, 11 October 2022 (UTC)[reply]
When you take a measurement, the photon has either left or not. Otherwise, it can be in a superposition of having left and not having left (in which case the photon hasn't been created), but the photon is never in the state of leaving, which is the state OP implied. Schrödingers cat however can be dying, in addition to being dead and alive at the same time. PiusImpavidus (talk) 15:13, 12 October 2022 (UTC)[reply]
Photons, insofar as they are particles, are point particles. As such, they have no dimensions, and take no time leaving. They don't have sides or edges as such. --Jayron32 12:21, 11 October 2022 (UTC)[reply]
Viewed as wave packets, however, photons do have a spatial extension.  --Lambiam 14:56, 11 October 2022 (UTC)[reply]
(edit conflict)Kinda-sorta-in-a-way-but-probably-not-really-depending-on-how-you-think-of-it. The wave in this sense is a probability distribution of some property of the particle in question (its wave function), and while "location" (Δx in one dimension) is one of those properties, the "wave packet" in this case is a complex function, and while it is common to treat the "real part" of this function as something like "location", eh? It's really that the wave packet represents the uncertainty in the location of the photon itself; when you do any measurement of the photon, either at emission or absorption, it acts as a point particle. It doesn't have any dimensions when you measure it, so it doesn't take any "time" to absorb or emit; it doesn't behave as though it has a front or back edge. --Jayron32 16:25, 11 October 2022 (UTC)[reply]
(post EC comment) Or what Nemur said below, which is much better than my comment. --Jayron32 16:25, 11 October 2022 (UTC)[reply]
Or, to resolve the condundrum, (if we're being really rigorous), we have to stomp on the notion and shout loudly in to the abyss: "photons don't work like that!" And then we can rest easy knowing that we don't really know...
Really, the issue comes down to a semantic problem; an issue of trying to use natural English language to describe something for which natural English language is a poor representation.
The original post used an interesting word-choice - it asked about when a photon had left. What exactly does that mean? What exactly does it mean for a photon to "leave"?
If we can use some physics-ese to define precisely what "leaving" means for a photon, we can probably answer with scientific rigor. Has a photon "departed" or "left" from some well-defined point in space only when the expectation value for the electric field falls below some threshold? If this is what "departure" means for a photon, we can answer precisely - albeit, couched in a statistical, predictive sort of manner - by solving for some quantum-mechanical wave equation, or something. The details of this boring equation, I leave to the physicists, who already know how to solve it (it's trivial!)
In my experience, most of the conundrums of quantum mechanics stem from trying to force a vague, plain-language description onto the system. In large, every-day macroscopic systems, the vagueness is not quite so apparent - but when we try to translate plain language descriptions onto the microscopic scales that we care about for most of the text-book problems of quantum mechanics, those ambiguous, plain-language words make for ill posed problems. This is why many physicists quip: "shut up and calculate."
Now, to first order, that sounds like we're being dismissive about a sophisticated question; we're saying "stop asking that!" And that's not a very nice or educational thing to do. But... there's nuance, here. So let's rephrase "shut-up-and-calculate" into something that is both more polite, and more aligned to what we're trying to say: "The words you're using, in the English language, are not a good way to describe photons. We'd rather teach you to use a different language - mathematics - and then we will politely ask you to phrase your question using this language - because this language will be better at expressing questions and answers relating to the way photons really behave."
And that's why if we pull out any of our books-about-photons - I mean, take your pick! I've got a shelf full of them, if you'd like to read them! - as soon as they get into the nitty-gritty details about photon behavior, they jump out of English and codeswitch into physics-ese mathematical language. We don't talk about "when photon has left." Rather, we reference Equation 26.3 in Section 12, and in order to understand it, you need to have learned how to read it in its native language.
Fundamentally, this is not so different from any other academic pursuit. We can teach an introductory class on Roman history in English, but an advanced study of the topic invariably requires learning at least a little Latin. To English, certain things, well, do not translate, well.
Nimur (talk) 16:14, 11 October 2022 (UTC)[reply]
I did mention some books, for further reading -
The Griffiths' series -
Jackson's famous book -
...
Of course, As I've said many times before, in many years of responding on the reference desk - don't underestimate this stuff.
I quote myself, here:
Normally, people read this book after they have completed a full four or five years of prior full-time preparation as full-time undergraduate physics students, because many very smart people find this mathematical content to be just at the limit of their mental capacity for comprehension after four or five years of full-time preparation. If you are not presently writing and solving wave equations for electrodynamics in conventional cases, you probably are not going to have great success writing and solving wave equations for nontrivial relativistic cases; and if you plan to understand general relativity without solving any equations, you're not going to get very far.
So - and I really don't want to be dismissive - the answer is, it ain't easy to answer questions about quantum mechanical behaviors of photons - not correctly and concisely, at least. Very smart people typically spend years trying to understand the complexity.
Nimur (talk) 16:30, 11 October 2022 (UTC)[reply]
A radar which emits a pulse of an electromagnetic beam of a certain energy for one microsecond between a time t1 and t2 is very real and trivial. Either we have n waves (E=hv) at // and there we know the beginning and the temporal end of each wave (t1 and t2), or we have a flux of n photons (E=hv) at // and there they must each be framed within an interval < t2 - t1? Sorry, but a point is in the realm of math, not physics. This impulse, when it encounters an obstacle, comes back to us at a time t3, it reacts like classical mechanics, simple ballistics. There's a very old cartoon from the beginning of computing where engineers disassemble a huge computer looking for a fault, next to them is a cleaning lady leaning on her broom and showing them the plug unplugged: "Is that what you're looking for?" Malypaet (talk) 20:41, 11 October 2022 (UTC)[reply]
But you asked about a single photon, not a pulse. A photon has no size and cannot be known precisely in location and momentum simultaneously. Also it might not even exist if you don't look at it, according to this year's Nobel prize winners (or something like that) Rmhermen (talk) 04:59, 12 October 2022 (UTC)[reply]
Exactly! The reformulated question posed an "either/or" scenario - and Lambian already described earlier (by bringing a cat into the mix) that this kind of "binary" yes/no certainty is at least inapt.
Even if we restrict to macroscopic systems, simply determining whether we did receive a signal devolves into a game of statistics and probability and confidence.
Every real signal has noise. Every signal! This is a stronger statement than it might initially appear. Noise isn't some mere annoyance that we can will into oblivion. It's not even some "practical detail" that we can usefully ignore in theoretical physics. We cannot construct a perfect (frictionless, while we're at it!) machine in our lab that observes signals with zero noise - not even in a purely theoretical construction. Some noise is an annoyance of unrefined engineering; we can hire better designers or purchase better parts to take this noise out of our machine (real or theoretical). But... other noise is inherent to the physics, and we may not ignore it!
At the most fundamental level - the regime of quantum mechanics - we have quantization noise - the inherent uncertainty involved in correctly counting a small integer number of physical events. If the number of events is "either zero or one," the best we can say is that we "either did or did not observe one single event, with some level of confidence."
How can we boost our confidence? We can re-take the measurement - but you see, that would literally entail observing a second time, for which the event either did or did not occur! If the first event had uncertainty and never even arrived, the second experiment might be the the first observance of the single event; or it might be a second repeated observance of two identical events, or maybe we failed to observe either, or maybe we saw the event only on one of two tries (first, or second attempt?...) and so ... this is noise. It cannot be avoided. As I said, even if we define "photon present" as equal to some measured level of electric field ... how do you know that electric field came from your photon and not from my other thermal photon that I (a notoriously troublesome nearby emitter of white noise) fired toward your detector at exactly the instant to interfere with your finely-crafted imported artisinal photon detector?
And this bizarreness - this "quantum weirdness" - as Lambian said, and as Rmhermen said, is really important stuff. Great minds have spent a lot of cycles stewing on it. As Rmhermen also said - this has something to do with some prestigious development in modern physics.
What we can do here is point our readers toward resources, and maybe inspire them to be enthusiastic about the miraculous complexity of even the most simple small particle of our universe.
We cannot, however, completely, correctly, concisely explain "the photon." It is too mysterious an animal to describe in a few words.
Nimur (talk) 15:37, 12 October 2022 (UTC)[reply]
The radar pulse is a group of photons and is a very concrete example contradicting the answers given to me. And for the photon that doesn't exist if you don't look at it, that's science fiction. So for a blind person, the moon does not exist? And yet it exists, regardless of whether we can see it or not. It is the same for a photon in a radar pulse, we are sure that it travels in this pulse. Here we know its speed, its wavelength and its position at least somewhere in the pulse. Malypaet (talk) 21:29, 12 October 2022 (UTC)[reply]
What Rmhermen said. If you change your question midpost than the answer will of course change too. One photon is a point particle and has no dimensions, and takes no time to interact with what is absorbing it. A group of photons has a spatial component, and does take an amount of time to interact with what is absorbing them. --Jayron32 10:50, 12 October 2022 (UTC)[reply]
The easiest way to simplify the description of one complicated thing is to add lots more of them!
Yeah, we treat individual photons differently than we treat groups of photons, ... but I'm already feeling that the original question has become less coherent as it progresses forward along its light cone...
Nimur (talk) 15:50, 12 October 2022 (UTC)[reply]
Less coherent? Do you frequently make such apt puns? --Jayron32 11:34, 13 October 2022 (UTC)[reply]

October 11

Unidentified fruit

Mystery fruit...
Cross-section if that helps.

Can anybody help identify this fruit on a shrub found at Abbey Gardens in Bury St Edmunds (East of England) today? Fruits are about 2 cm across and slghtly downy, a bit like peach fuzz. A taste test was not unpleasant but inconclusive and I'm still alive at the time of writing. Alansplodge (talk) 16:35, 11 October 2022 (UTC)[reply]

Quince? Bazza (talk) 16:39, 11 October 2022 (UTC)[reply]
Ah yes, you may be right. I was thinking that it was far too small to be a quince, but I see that some flowering cultivars - like this one - have smaller fruits. Dissecting the sample I brought home matches the illustration in our article and it is indeed "fragant". Alansplodge (talk) 16:53, 11 October 2022 (UTC)[reply]
You might also consider one of the Chaenomeles. DuncanHill (talk) 17:00, 11 October 2022 (UTC)[reply]
Oh, I see you have linked to one of them. The Flowering Quinces are not cultivars of the Quince, they are a different genus. DuncanHill (talk) 17:03, 11 October 2022 (UTC)[reply]
We have Quince#Cultivars which are of the same tribe, but Pseudocydonia and Chaenomeles are not, despite all are being known as "quince" in English. I can't see an exact match so far, but won't be losing any sleep if we can't pin down an exact species. Alansplodge (talk) 17:18, 11 October 2022 (UTC)[reply]
The stem and leaves are very citrus-like. It is definitely Citrus trifoliata. Thriley (talk) 18:51, 11 October 2022 (UTC)[reply]
Sounds plausible. I have added a cross-section - it does look a bit citrus-like inside and has the "fuzzy texture" mentioned in our article. Alansplodge (talk) 21:38, 11 October 2022 (UTC)[reply]
Thank you for the photo. You can see the Juice vesicles found in citrus fruit. Thriley (talk) 21:52, 11 October 2022 (UTC)[reply]
I am persuaded. Many thanks. Alansplodge (talk) 12:38, 12 October 2022 (UTC)[reply]
Resolved

Energy delivered by a machine gun

I read this online: "kinetic energy of a bullet = k kinetic energy of n bullets = nk According to the law of conservation of energy, the kinetic energy of the bullets must be equal to the work done by a machine gun per second".

So can I write that the energy "E" delivered by a machine gun with a firing frequency "f", "F" = "f" without time, is : "E = kF" ? — Preceding unsigned comment added by Malypaet (talk • contribs) 21:07, 11 October 2022 (UTC)[reply]

No, you can't. The addition "per second" makes no sense in the "online" statement. Conservation of energy means "kinetic energy = work". When time is brought into this, you talk about power. --Wrongfilter (talk) 08:07, 12 October 2022 (UTC)[reply]
So if I reformulate, I can only consider the kinetic energy of the group of bullets over one second "E=kF", (even if they were emitted in sequence at different times). But for the machine gun I have to consider its power "P=kf"? On the other hand, for a target (a regid ball with a mass a thousand times greater for example) which receives this energy over one second, it is indeed energy, isn't it? Malypaet (talk) 08:47, 12 October 2022 (UTC)[reply]
What does "F" stands for? Is it a dimensionless quantity? You ask several questions of the nature of "which is better/more correct?". The answer depends on what you are seeking to achieve, which is not entirely clear. In either case, depending on what is given and what is wanted or needed, one can consider either energy or power, or both.  --Lambiam 09:32, 12 October 2022 (UTC)[reply]
Compared to earlier questions, this time our OP has asked a surprisingly easy question: "can I write...?"
Yes! You can write this! It's free country, (I think! I'm not sure what country you're in, but it sounds like it has an alarming amount of machine-gunnery, which causes me to wonder about the relative merits of "freedom" in so-called free countries)... but let's assume (with, perhaps, some loss of generality) that it is a free country. You can write this mathematical expression!
Does it help you to write this expression?
Do you find it therapeutic? Does it help you to understand photons, or machine guns? Does it help you to communicate your insight with others who study photons (or machine guns)? Does it help you design a better photon, or a better machine (gun? ... Please don't - we've got plenty of fine-enough of those machines already!)
Really, though, I read a lot of philosophy into the question. For what purpose do you express these ideas? For a practical one? If so ... the science reference desk is not a place to participate in practical design and engineering of machines (or guns).
Is it for a thought experiment of some kind? If so ... we can point you to better thought experiments, constructed by smart, well-respected authors. (I did link several introductory texts, above!)
Is it some form of artistic expression? If so, I commend it... I think! I applaud artistic expression, even if I don't particularly subscribe to its aesthetic. In this case, I interpret it as some kind of reactionary postmodernism extolling the absurdity of technological progress and the self-destructive nature of the human. But, ... the science reference desk might not be the place for it. Besides, as they say, viewers impose their own meaning on to art, so... let's not fall victim to pro-institutionalist interpretations. This is a free country encyclopedia where anyone may contribute. "Can I write...?" Yes! But Why?
In seriousness, our OP can write any equation they want - some equations are useful, some writings serve to encapsulate useful content, and the rest we call art, I think. Art, in all its forms, including open-form prose on public internet websites. Art, an abstraction that can provide a medium to reflect complex ideas and stimulate the minds of the viewer... easier to make and yet harder to define than science...
Nimur (talk) 16:16, 12 October 2022 (UTC)[reply]

October 12

Metal object on the top of styrofoam balls

A metal object on the top of a box of styrofoam balls (the kind used for filling a package) would not necessarily sink to the bottom (unless some shaking is applied). I imagine that that's because the friction between styrofoam balls, that locks them in place. Is this intuition right? And could you put some object on the top of a liquid with lower density in a similar way (without it sinking)? Bumptump (talk) 20:08, 12 October 2022 (UTC)[reply]

Water certainly has a high surface tension that allows you to float heavier objects carefully on it. Rmhermen (talk) 00:45, 13 October 2022 (UTC)[reply]
But could we say that, for example, a needle floating in water it's the same physical phenomenon as in the styrofoam balls case above? Bumptump (talk) 01:00, 13 October 2022 (UTC)[reply]
No, it is a different phenomenon; it has some superficial similarities, but (perhaps obviously!) solid materials differ from liquid materials (and perhaps non-obviously!) this manifests in profound ways when we study the complicated dynamics of such systems.
The behavior of a group of solid spheres in profoundly different from the behavior of liquid molecules - and despite the appeal of using analogy to model liquid molecules as "very very small" solid spheres, that's just not quite right.
We have an article on granular materials, which are an actively studied area of applied condensed matter physics.
Nimur (talk) 03:10, 13 October 2022 (UTC)[reply]
Part of the reason why the metal object stays on top of the styrofoam balls is friction. The other part is that the metal object can only move down when the styrofoam balls move up. The geometry may be such that the styrofoam balls must initially rise much faster than the metal object sinks, so we get a local minimum in the potential with the metal object still on top.
As you correctly observe, with a bit of shaking, the metal object will sink. In water, there's always shaking: Brownian motion. If we stop Brownian motion by freezing the water, the metal object can stay on top again. PiusImpavidus (talk) 09:42, 13 October 2022 (UTC)[reply]
A B52
Surface tension is sufficient if the object is small enough; things like paper clips and small needles and the like can be floated on water if one is careful to not submerge them under the surface. --Jayron32 11:02, 13 October 2022 (UTC)[reply]
This discussion reminds me of many interesting non-scientific evenings. MinorProphet (talk) 21:20, 15 October 2022 (UTC) [reply]
The trick (relevant to the original question here) would be to get the layers in the reverse order. A fun demo is to freeze a solid bottom layer of something that is darkly colored and less-dense (when it's liquid, but slightly more dense as a solid), then put a liquid that is pale, transparent on top. If the glass temperature, layer temperatures, and densities are right, one can watch wisps of the bottom layer rise high in the top layer. DMacks (talk)

Ships with no atmosphere

If Earth's atmosphere disappeared suddenly, would steel ships continue floating in water? (The role of air isn't addressed in Buoyancy.) On one hand, I think "no", because the vacuum would have essentially no mass, and I think the air filling the ship helps to make the whole thing lighter in relation to the water, and the ship's slight buoyancy in air makes it weigh less than it would in a vacuum. On the other hand, I think "yes", because the water would weigh much more than the vacuum, and the weight of ship + vacuum would still be less than the weight of the water that fills the same volume. 175.39.61.121 (talk) 21:00, 12 October 2022 (UTC)[reply]

The whole thing would be moot, water can not stay in liquid state in the vacuum. It will either freeze or become vapor. Cambalachero (talk) 21:24, 12 October 2022 (UTC)[reply]
While it's technically true that you can't have liquid water in equilibrium at zero pressure, the triple point pressure of fresh water is only about 612 Pa, less than a hundredth of an atmosphere, and it goes down as you add salt. I conclude that, if all the air suddenly vanished, the oceans would start to boil, but not very fast. There would be plenty of time to observe a ship floating. --Trovatore (talk) 06:44, 13 October 2022 (UTC)[reply]
I think any potential observer would have more important things to worry about than watching ships float.... --User:Khajidha (talk) (contributions) 17:25, 13 October 2022 (UTC)[reply]
"air filling the ship helps to make the whole thing lighter": Why would that be? I believe if you created a vacuum in a dome above water and put an empty ship inside (completely empty, without air) its walls would crumble with the water pressure outside. It was not designed for that. Empirical data would be needed to determine if this is right. Bumptump (talk) 22:05, 12 October 2022 (UTC)[reply]
OK, but the answer is that yes they would float. It isn't the air that makes them float it is the absence of heavy stuff. Greglocock (talk) 22:17, 12 October 2022 (UTC)[reply]
A vacuum tube floats in water. So, an object with as much of a vacuum as we can create on earth still floats. This is not about having a vacuum inside the ship. 97.82.165.112 (talk) 00:51, 13 October 2022 (UTC)[reply]
Actually, the air inside a ship has weight, so it makes the ship heavier. When the air is pumped out of an airtight-sealed floating container, it rises ever so slightly.  --Lambiam 05:27, 13 October 2022 (UTC)[reply]
The answer may be more complicated than it looks at the first glance. Air actually adds to buoyancy – the above-water parts of a ship experience buoyancy force (upwards) from air in which they are immersed. On the other hand, air also exerts pressure downwards on parts, whose lower surface is under water (mostly the bottom of the hull). So, if the whole ship is made of materials heavier than air, then disappearing of air will make the ship's draft to grow, the hull will sit deeper in water. However, if you mount a helium or hydrogen tanks (baloons) in it, they will be adding net buoyancy in air and net weight in vacuum. I can't estimate in my head what size the balloon should be compared to the hull itself to reduce the ship's draft on atmosphere removal. Let's imagine a normal floating toy balloon with a small weight attached to it. You can choose the weight small so that it floats just by touching the water surface. If you remove air, the balloon would fall and the weight would immerse in water. However, you cant make such 'ship' to sink this way: if the balloon's buoyancy force in air is (almost) enough to fly with the weight, it obviously will prevent the weight from sinking in water. --CiaPan (talk) 06:58, 13 October 2022 (UTC)[reply]
It can go either way. The structure of the ship above the waterline (excluding air-filled spaces) experiences buoyancy from the air, which will disappear in a vacuum. In the air-filled spaces below the waterline, air provides less bouyancy than vacuum. If the volume of the ship above the waterline is less than the volume of air below the waterline, the ship will rise when put in a vacuum. In a fully metal ship this is always the case, or it wouldn't float. If however the ship is loaded with something less dense than water (like oil), the ship may sink a little.
If the vacuum is high enough, the top few decimetres of water will be boiling, until it freezes. The vapour bubbles lower the density of the water, which can cause the ship to sink a bit, but not much. PiusImpavidus (talk) 10:27, 13 October 2022 (UTC)[reply]
  • If we want to spherical cow this and ignore all of the complicating factors of the oceans boiling away and all the rest, of course the ships float. They float better, because air is matter and matter has mass. A ship filled with air is heavier than a ship filled with nothing. The ship filled with nothing will thus displace a smaller volume of water than a ship filled with air will. It will thus float higher in the water. --Jayron32 11:00, 13 October 2022 (UTC)[reply]
    Just to run this into the ground, that argument applies only to the part of the air in the ship that's below the waterline. The weight of the air above the waterline is cancelled by the buoyancy due to the air outside the ship. --Trovatore (talk) 16:41, 13 October 2022 (UTC)[reply]
    That's true, but because the ship is less buoyant in air than it is in water, there will always be (in air) a volume inside the ship which has air below the waterline. If we replace that space with vacuum, it weighs less. That means the volume of the steel+void now has a lower mass than a similar volume of water than did the volume of steel+air, and as such, will displace less water. Thus, the ship floats higher in the water, QED. --Jayron32 12:21, 14 October 2022 (UTC)[reply]

October 13

Planck relation and time

The kinetic energy of an asteroid is independent of time and energy in general, otherwise we speak of power in Watts. In Planck's relation the radiant energy depends on a frequency which is temporal. How is it possible ? — Preceding unsigned comment added by Malypaet (talk • contribs) 08:40, 13 October 2022 (UTC)[reply]

Sorry, what? Radiant energy is a function of temperature. See blackbody radiation. It is unrelated to kinetic energy. The first is about the motion of individual atoms and molecules, the second is about the motion of the entire object. Also, it's vague and unclear what you mean by "depends on". Dimensional analysis will also tell you that all measurements of energy have time embedded in their units; a joule is a kilogram meter squared per second squared (kg m2 s-2), or if you want to think in terms of unitless measure, energy is mass times distance squared divided by time squared. So your first statement is nonsensical; time is embedded in the measurement of energy. A watt, by the way, is a kilogram meter squared per second cubed, or power is mass times distance squared divided by time cubed. --Jayron32 10:54, 13 October 2022 (UTC)[reply]
Sorry, but 1 W is equal to 1 kg m2 s−3.  --Lambiam 12:42, 13 October 2022 (UTC)[reply]
So corrected. --Jayron32 12:08, 14 October 2022 (UTC)[reply]
Effectively radiation energie (not radiant) and when you have "E=hv" you don't agree that "E" depends on "v" ? Malypaet (talk) 21:46, 13 October 2022 (UTC)[reply]
That's the energy of one photon. Your question is still unclear, which is why it is hard for people to respond meaningfully. Can you please explain again, using more details and in different words, what you are trying to understand? --Jayron32 12:17, 14 October 2022 (UTC)[reply]
Perhaps you meant asteroids are not losing or gaining energy with time. That is true for KE + PE (potential energy), but not KE alone due to differences in orbital speeds. In fact, since Compton's experiments 99 years ago scientists have known that radiated photons are particles that carry energy that gain or lose energy during collisions with matter. Photons also have an EM quantum wave nature that interacts with matter so as to impart or acquire their characteristic frequencies such as with Doppler effects. Modocc (talk) 21:32, 13 October 2022 (UTC)[reply]
If we compare with a bullets flow from a machine gun with firing frequency "f" (I'm peaceful, it's just a thought experiment), so that we can write the kinetic energy equation for a group of bullets "Ec=k 1/2mv2 f" this group must have been emitted over 1s with the constant "k=1s". In this case we can also write "h=k 1/2mv2" and "v=f", we then find "Ec=hv". There it is coherent for me as it would also be if we replaced the bullets by the peaks of an elementary wave, whereas for the Planck relation I cannot manage to find coherence with the frequency. Malypaet (talk) 22:25, 13 October 2022 (UTC)[reply]
If a pulse of light (possibly a single photon) has a certain energy Ep, then n such pulses have, together, the energy nEp. If these pulses are fired with a frequency f, then there are fs pulses per second, giving a power of fEp. This holds equally if we fire bullets instead of light pulses. The only difference is in the energy Ep. The photon energy of a single photon equals hν, whereas the kinetic energy of a single bullet equals ½mv2. There is no relation between the frequency f of firing, and the frequency ν of the photon.  --Lambiam 06:16, 14 October 2022 (UTC)[reply]
Per Lambiam, and to clarify: The vibration of a light wave is unrelated to how often a photon hits a target. While both could be described as a "frequency", they are completely different and unrelated things. --Jayron32 12:13, 14 October 2022 (UTC)[reply]

October 14

Human surrogacy of animals

Basic idea: if you implanted a fertilised animal egg into a human womb, could it grow and be birthed (whether naturally or via cesarean). I think something similar was done with dogs and wolves, which are obviously closely related. My guess would be that you couldn't do it with say a human and a cat because of differing hormones, blood chemistry, etc. -mattbuck (Talk) 15:18, 14 October 2022 (UTC)[reply]

Dogs and wolves are not merely "closely related", they are different varieties of the same species. Dogs were domesticated from wolves, and can readily interbreed with them; the distinction between dogs and wolves is largely linguistic, with dogs being "wolves who we bred and trained to not eat our babies". All that being said, WHAAOE, see Interspecific pregnancy. That should help direct you to your answers. --Jayron32 15:46, 14 October 2022 (UTC)[reply]
That is the best description I have ever read about the domestication of the dog. I may have to steal it. --User:Khajidha (talk) (contributions) 18:06, 14 October 2022 (UTC)[reply]
Thanks. The wit there was inspired by CGP Grey's video on animal domestication. See Why Zebras are Terrible Horses. If my quote gave you a chuckle, CGP Grey's video will have you rofling your waffles. --Jayron32 18:18, 14 October 2022 (UTC)[reply]
Article: interspecific pregnancy. Unless the species are closely related, it doesn't work because the purpose of the immune system is to identify and attack things that are "not-self", and a fetus trying to xenograft itself to the host qualifies. Successful mammalian pregnancies involve a complex dance between the fetus and parent that establishes parental immune tolerance, and this is very species-specific. For instance, different mammal lineages have different types of placenta: see placentation. Even within the same species, it frequently fails, leading to things like miscarriage, premature birth, and preeclampsia. There is the theoretical possibility it could work if the prospective parent's immune system were suppressed enough, but obviously this would be hideously unethical. --47.147.118.55 (talk) 04:21, 17 October 2022 (UTC)[reply]

mushy cough drops

I bought a bag of cough drops earlier this year. They are basically hard candies with menthol or something like that added. Since then there has been a heat wave here in CA (temperature reached maybe 33C on some days) though I don't know if that relates to the current situation: the cough drops are now somewhat mushy, like hardened chewing gun, underneath the outer layer but above the center. They are still usable but it is a bit annoying.

Any idea what has happened? Can they be fixed, such as by putting the bag in the freezer for a while? Current indoor temperature is reasonably cool. The cough drops are individually wrapped and idk if moisture could have gotten into them. The outer bag is not sealed. Maybe I should have put them into a sealed container. Thanks. 2602:24A:DE47:B8E0:1B43:29FD:A863:33CA (talk) 19:57, 14 October 2022 (UTC)[reply]

ObPersonal: in my experience (over three decades, in the UK) this always happens to such cough drops if left long enough, even in a fairly stable environment. 'Long enough' might be a few months on an open shelf or similar in a hot summer, or a few years inside a more enclosed cupboard. I don't think the softening (and eventual leakage) materially effects the cough drops' effectiveness, but it does make them considerably more difficult to use. Freezing them might help with the latter (if you don't mind spitting out bits of wrapping paper).
In this recent Covid-enforced regime, with limited social contact and (until recently) mask-wearing in the proximity of others, I haven't had a cold or cough for nearly 3 years, so all my stored cough drops have deteriorated in this way. {The poster formerly known as 87.81.230.195} 90.195.172.49 (talk) 20:50, 14 October 2022 (UTC)[reply]
Hard candy mentions that it's harder if amorphous and softer if crystalline. That sounds similar to what can happen to liquid honey over time. You might try heating them up, rather than freezing them, and then letting the result set again, although that will give you a single bar of something like cough toffee, unless you get fancy and create a lot of little molds somehow.  Card Zero  (talk) 00:26, 15 October 2022 (UTC)[reply]
Thanks both. Hmm hard=amorphous soft=crystalline is the opposite of what I would have guessed. I will try heating them still in the wrapper. I hope the wax in the wrapper doesn't melt all over. I'll try with one or two of them first. Might use hot water, or a heat gun, or microwave. I wonder why they crystallized to start with. I see the bag they came in is the resealable type, but I haven't been sealing it because I figure that's to carry the bag around without spilling them, and this bag is sitting in one place. But, I may try sealing it, or making a point of keeping the next bag sealed. This is unfortunately a pretty large bag that will last a while, but it is ok. 2602:24A:DE47:B8E0:1B43:29FD:A863:33CA (talk) 05:53, 15 October 2022 (UTC)[reply]

Decline in insect populations

Could it be caused by microplastics? Rich (talk) 22:13, 14 October 2022 (UTC)[reply]

Microplastic Polystyrene Ingestion Promotes the Susceptibility of Honeybee to Viral Infection, Deng et al (2021) thinks it might be, at least in honeybees. It's the first paper listed here (NB: it's a pdf that will want to download, which I have, but I haven't read it in full yet). There are other relevant papers in the list (not surprisingly) – have a browse!
You will have seen that the Wikipedia article Decline in insect populations doesn't mention microplastics (yet). {The poster formerly known as 87.81.230.195} 90.195.172.49 (talk) 03:16, 15 October 2022 (UTC)[reply]
Most insects don't eat plastic or small particles of plastic as many will use smell to decide if it is food. There are exceptions though, like wax worms eating polyethylene. Graeme Bartlett (talk) 21:39, 15 October 2022 (UTC)[reply]
But microplastic particles can be so small that they permeate some foods that insects do eat, and therefore are ingested anyway, just as if you or I eat any fish, we are certainly ingesting plastic microparticles within it, without noticing. {The poster formerly known as 87.81.230.195} 90.195.172.49 (talk) 08:21, 16 October 2022 (UTC)[reply]

October 15

Seeing colors and animals questions.

1. Most birds typically see colors from the range orange to UV. So, they cannot see red. This is why red flowers don't typically appear in certain latitudes as fewer species of birds pollinate red flowers, but what does red look look like to birds under light, same as what infrared looks like to humans in light. So my question is, if there was a bucket of paint of the color "infrared" what color does it look like to humans under light? As well as paint of "ultraviolet" color?

Now the following are all animal questions so I doubt anyone here knows aside from Google searching, but I'll throw them in here just in case.

2. What animals can see IR - blue/indigo, so would not be able to see violet? Although I did a Google search for "what animals can see IR" it doesn't say if they can also see violet. Though it appears goldfish are the only animals that can see from IR to UV range. Certain animals can partially see some of the IR range.

3. I did Google searches for "can rabbits see color?" "can alligators see color?" "can frogs see colors?" and a whole bunch of those animals, they are missing the red rod. They have blue and green rod, but are missing red. To me, that is not red-green colorblindness, but simply red-colorblindness, because they can see blue and green the same way we see blue and green, but what does red look to them, seems to be a matter of debate. For rabbits, the ability to see blue seems to evolve from being attacked by predators from the blue sky. And maybe green for the ability to eat plants, and not seeing red, for fear of blood? But I'm wondering what animals have red and blue, or red and green rods instead of blue and green. It appears turtles have all 3. 67.165.185.178 (talk) 13:40, 15 October 2022 (UTC).[reply]

For 1 - if a paint absorbed visible light and reflected infrared you would see it as black. However near infrared can be detected by eyes if it is bright enopugh. So for example an 850nm remote control led can appear red. Graeme Bartlett (talk) 21:35, 15 October 2022 (UTC)[reply]
A remote control looked dim pale violet to me (2 photons ganging up on 1 photoreceptor molecule dozens of times a second looking like 425nm light?). I don't know if this can cause permanent eye damage. Sagittarian Milky Way (talk) 22:22, 15 October 2022 (UTC)[reply]
The remote control is quite safe to look at. I also have some IR camera filters. If I look through the 720 nm filter at a sunlit scene it is dim red. Through a 760 nm filter it is very dark red. And for 850, 950 and 1000 nm to me it looks black. Don't look at the sun through these though! Graeme Bartlett (talk) 01:28, 16 October 2022 (UTC)[reply]
If the 720 or 760 is between the eye and sunlit green plants what does it look like? Sagittarian Milky Way (talk) 02:25, 16 October 2022 (UTC)[reply]
IR photo comes out quite similar to what a human eye sees. The leaves on the tree are bright.
Trying out the 720nm with the eye, most plants are very bright; English box, rose, clover, elm, and moss are all about as bright as cream paint or clouds. However cypress is dimmer, and grass is quite dark. The sky is very dark. Graeme Bartlett (talk) 03:23, 16 October 2022 (UTC)[reply]
If a material (like on a wall) is of IR or UV material or a paint of that color, does that mean it will reflect say, UV light when light is shine on it, causing you to get a sunburn? Since people don't seem to get sunburns from a wall of a UV-color, I wonder if there is something in physics that prevents it to exist past certain wavelengths? Like a microwave color? 67.165.185.178 (talk) 03:01, 16 October 2022 (UTC).[reply]
You are talking about a reflectance spectrum here We probably have articles on this. But it depends on the material. Metals tend to reflect electromagnetic radiation from radiowaves through to ultraviolet light. But coloured metals like copper or gold don't reflect the short wavelength. (eg Webb space telescope cannot see blue or ultraviolet reflected from gold). Small particles with high refractive index, like titanium white reflect a lot at their surface, but if the wavelength is big compared to the particles, the radiation will penetrate. Most materials have some absorption cutoff where shorter wavelength radiation than some amount cannot penetrate, because it move an electron across a band gap. Some ridiculous materials like solid helium can let ultraviolet though deep into the vacuum ultraviolet. But air will stop your vacuum ultraviolet. Molecules in solids, or covalently bonded substaces also have absorption caused by vibrating atom to atom bonds, either bending or stretching. These kinds of materials may let some radiation in and reflect some back after going through the substance, so it will have the infrared bands subtracted from what it reflects. So this would include most common non-metallic solid things you see like rock, paper, skin, dirt, paint binder. Graeme Bartlett (talk) 03:23, 16 October 2022 (UTC)[reply]
If a wall is painted in UV, it will reflect UV whenever UV falls on it. It won't cause sunburn, unless daylight already causes sunburn. PiusImpavidus (talk) 11:42, 16 October 2022 (UTC)[reply]
For 3, you are talking about subjective experience. See Qualia for an article on this. To get the experience yourself, see if you can get a cyan coloured filter that blocks red light and look through it. After a couple of days of this you brain will get used to it and it won't look tinted, and the experience could be similar to that of red-blind animals. Graeme Bartlett (talk) 01:28, 16 October 2022 (UTC)[reply]
Just don't get cyan-eyed. ←Baseball Bugs What's up, Doc? carrots→ 03:56, 16 October 2022 (UTC)[reply]
Ba-dum tish! {The poster formerly known as 87.81.230.195} 90.195.172.49 (talk) 08:25, 16 October 2022 (UTC)[reply]
1: Bird colour vision is somewhat different from human colour vision. Insect colour vision is way more different, as it's shifted quite a bit towards the UV. For that reason, blue flowers are typically pollinated by insects and red flowers by birds (except red flowers with a yellow heart, like roses, as this gives a very strong contrast in insect eyes). Pollinating birds would have great difficulty surviving winters, so these red flowers are a bit rare in places like Europe. A nice example of a plant with red flowers to attract pollinating hummingbirds is the schlumbergera from Brazil.
If we were to paint something in infrared, reflective only on longer wavelengths than the human eye can see, it would appear black to us. The same for ultraviolet.
2: Our article on colour vision states that "It is a myth that the common goldfish is the only animal that can see both infrared and ultraviolet light; their color vision extends into the ultraviolet but not the infrared." It would be quite useless anyway. Water is pretty much opaque to IR, so the IR part of sunlight is absorbed very close to the surface, so there isn't any IR to see. In air, the IR cut-off of sunlight isn't very sharp. Brightness of solar IR rapidly decreases with increasing wavelength and thermal noise increases, so it rapidly gets less useful. On the UV side, the hard limit is about 310 nm. Below that, the ozone layer absorbs all, so there's nothing to see. Apparently humans with normal vision can see the 365.4 nm mercury line, although sensitivity is quite low there. So different species and different individuals of a single species have different ranges of vision with fuzzy boundaries, but the differences aren't huge. And what exactly do you define as IR and UV?
3: It seems you missed the difference between being able to see a wavelength and being able to distinguish it from other wavelengths. All mammals can see blue, green and red, but most cannot distinguish green from red. A mouse cannot distinguish a red fox from green grass, making the fox well camouflaged (even though the fox is very obvious to humans). If the mouse hadn't been able to see red, the fox would have appeared black and not camouflaged. This is because the red and green cones are both sensitive over the entire visible spectrum, but the exact sensitivity and sensitivity ratio varies. So the human eye detects the signal ratio between the red and green cones – or rather, the difference between the logarithms of the signals. Many animals have better colour vision than mammals, not because they have larger range, but because they can distinguish more colours with more types of cones. The price they pay for this is less sensitivity in the dark, less temporal resolution, less spacial resolution or bigger eyes. PiusImpavidus (talk) 11:42, 16 October 2022 (UTC)[reply]
1. I would say blue flowers are pollinated by insects and birds, whereas for red flowers, just birds. Bumblebees cannot see red for example, however it is possible they cannot see orange either. But good post. 67.165.185.178 (talk) 14:20, 16 October 2022 (UTC).[reply]
3. Also I meant most animals completely lack the red cones, so green still looks absolutely green to them. So if red looks black to them, then their ability to distinguish red from green is the same as red from blue. So I think what you're talking about includes cats, which might have all 3 in various ratios. 67.165.185.178 (talk) 14:26, 16 October 2022 (UTC).[reply]

By the way I am curious, where does brown and pink fall in the EM-spectrum? As well as, a color that is half-red and half-purple, such as maroon. 67.165.185.178 (talk) 01:15, 17 October 2022 (UTC).[reply]

Pink and brown are not spectral colors, so do not appear in the spectrum. See the linked article which specifically describes both of those colors as examples of extra-spectral colors: Any color obtained by mixing a gray-scale color and another color (either spectral or not), such as pink (a mixture of a reddish color and white), or brown (a mixture of orange and black or gray). CodeTalker (talk) 01:21, 17 October 2022 (UTC)[reply]

October 16

Why is salt water more "fizzy" than regular water?

When you shake or pour salt water, it seems to create more bubble and has a longer fizz sound than regular water. I think it has to do something with surface tension, but I'm not sure that's the only reason. If surface tension is the sole factor, then why shaking mercury do not produce an even longer fizzing sound or more bubbles? CactiStaccingCrane (talk) 15:26, 16 October 2022 (UTC)[reply]

Evidently, it's complicated. A summary:
  • Ball, Philip (4 April 2016). "Why salty water foams". Chemistry World. --136.56.52.157 (talk) 21:44, 16 October 2022 (UTC)[reply]
I thought it was all the dead mermaids. Iapetus (talk) 23:33, 16 October 2022 (UTC)[reply]

October 17

Tree leaf coloring disorders for autumns.

I've seen some trees where on 1 half side is red leaves, other half side is predominantly green colored leaves still. Is there a term for this kind of disorder? Then, I've seen trees where the top 1/3rd leaves turn red, bottom 2/3rd still green. 2 different disorders. Also, are there terms for trees where the leaves turn red, then fall out, and where they turn yellow, then fall out? And some may turn dark reddish-purplish then fall out, is there a scientific reasoning for the differences? Thanks. 67.165.185.178 (talk) 00:16, 17 October 2022 (UTC).[reply]

When was stereoscopic photography first practiced?

Not covered in Stereoscopy. I tried asking on the talk page there about 3 weeks ago, but no dice. - Jmabel | Talk 00:17, 17 October 2022 (UTC)[reply]

Probably 19th century. Sagittarian Milky Way (talk) 01:27, 17 October 2022 (UTC)[reply]
The first stereoscopic device, invented by Charles Wheatstone in 1838, used drawings; the type developed in 1849 by David Brewster could use paired daguerreotypes, and was popular until the 1930s.[1] The earliest compact stereoscopic camera that I found: Vérascope, was introduced in 1893 by Jules Richard. --136.56.52.157 (talk) 02:10, 17 October 2022 (UTC)[reply]
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