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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]
Article: abscission. I believe this is not any "disorder", but normal physiological responses by the plants to seasonal changes in light levels. See Abscission § Mechanisms. Living plant leaves are living tissue, despite them not performing acrobatics to impress the watching hairless monkeys, and are constantly turning over molecules and responding to the environment. Did you note whether the patterns of green leaves corresponded to things like sun angle, which, remember, also changes seasonally? As abscission says, the varying leaf colors in different plants are due to the presence/absence and concentrations of various pigments made by the plants, such as polyphenols and carotenoids. (If you noticed the word's similarity, yes, carotenoids are what give carrots their colors.) --47.147.118.55 (talk) 04:57, 17 October 2022 (UTC)[reply]
Addendum: often these pigments are always present in the leaves, but their color is overwhelmed by the green color of the much greater amount of chlorophyll. But when deciduous plants start shutting down chlorophyll production in their leaves, the other pigments are then unmasked. The other pigments have various purposes, such as serving as photosynthetic accessory pigments to capture wavelengths of light that chlorophyll doesn't for photosynthesis, or photoprotection of the leaf against damage by short-wavelength light. --47.147.118.55 (talk) 12:19, 17 October 2022 (UTC)[reply]
When you say due to changes in seasonal light levels. What if I took such a tree to an Equatorial region, then, will that tree have leaves stay green year-round? Without dropping? Or, it will stay green, but still drop. Or will it still turn red, but not drop, it will just turn green again in the spring? ;o 67.165.185.178 (talk) 23:53, 17 October 2022 (UTC).[reply]
Sorry for a little delay... Good question! It varies among plants. See deciduous plant, semi-deciduous, evergreen. Some plants also drop their leaves in response to water stress, so in the topics they shed them during the dry season. Some plants that are seasonal in temperate climates indeed shift in the tropics to shedding various leaves on an ongoing basis rather than all at once. --47.147.118.55 (talk) 03:37, 21 October 2022 (UTC)[reply]
I'm hearing now that leaf changing colors guarantees dropping, but dropping does not guarantee color change. So you can't really have trees where the leaves turn red, don't drop, and then in the spring, revert back to green? Would be dope if we could genetically modify a random pattern for colors of leaves heh. 67.165.185.178 (talk) 03:52, 21 October 2022 (UTC).[reply]
Hairless monkeys suffer from a similar disorder during the summer but luckily keep their body parts and recover during the winter. 😁 The pigment protects the innards from ultraviolet and for leaves allows the plant to extract any goodness from the leaves in an orderly manner before discarding them while they are unnecessary and might be a drain. The extracted resources can then help with growing new leaves for instance in the spring. NadVolum (talk) 12:25, 17 October 2022 (UTC)[reply]
When was stereoscopic photography first practiced?
From the book Instruments and the Imagination: "But in 1849, the Scottish natural philosopher and steadfast opponent of the wave theory of light Sir David Brewster came up with a convenient and inexpensive lenticular stereoscope (see fig. 7.3).21 George Lowden of Dundee constructed several models based on this design for Brewster. After a disagreement with Lowden—a common event in many of Brewster’s professional relationships—he unsuccessfully searched for another British manufacturer. During the spring of 1850, Brewster took one of Lowden’s models with him to Paris, where he showed the device to the opticians François Soleil and Jules Duboscq. Within a short time, they began producing Brewster’s stereoscope and accompanying stereoscopic daguerreotypes.22"[2] The text goes on to describe that the invention caused a sensation at London's Great Exhibition of 1851, whereupon a craze ensued, as evidenced by over-the-top rave reviews ("a divine gift"). --Lambiam06:07, 17 October 2022 (UTC)[reply]
@Sandbh: I assume you mean "not" coloured for N and O? The answer is that the colour results from the electron transition between the HOMO and LUMO orbitals, which for Cl2 and F2 is an energy/absorption in the visible range, with the trend continuing for bromine and iodine. But I suspect you already knew that, given your membership of WP:ELEM. So are you tempting us down the why rabbit-hole and asking why the HOMO and LUMO gap for O and N is larger? Mike Turnbull (talk)
Oxygen gas is slightly blue from photons hitting oxygens while they're extremely close, nitrogen I don't know and both should look red when backlit through enough hundreds of miles and blue from any direction when lit by the Sun at high angle. Due to the large ratio of naked eye light wave to gas molecule size presumably all gases would do this unless they're colored something else from non-Rayleigh scattering reasons and strong enough to notice. 14:59, 17 October 2022 (UTC) Sagittarian Milky Way
Thank you @Michael D. Turnbull: Yes, I was effectively wondering why the HOMO-LUMO gap is larger for N and O. Could it have something to do with much higher bond strengths in N and O? The triple bond in N = 945 kj/mol; double bond in O is 498; single bonds in H is a mighty 436; F = 157; Cl = 242; Br = 194. Sandbh (talk) 06:05, 18 October 2022 (UTC)[reply]
That's puzzling. It's not clear to me if these lines apply to the atoms or the molecules. Thus, this article says:
"In the case of the oxygen molecule all the low-lying states have been observed or their positions have been quite accurately fixed by theoretical calculations. The six lowest energy states of oxygen are shown in Figure 1. The first excited state (labeled 'A,) lies 0.98 ev above the ground state (labeled %2,-). A transition between these states gives rise to a weak absorption at 12,690 Å which is in the infrared region."
Liquid oxygen is visibly light blue. I think this Chemistry Stack Exchange answer explaining why liquid O2 shows colour but gaseous O2 doesn't might interest you, even though it doesn't point to a reference. Apparently the problem is that the process creating the colour involves two oxygen molecules and one photon, which is understandably harder to achieve outside a condensed phase. I'm speculating here, but I wonder if this could also explain why liquid fluorine is bright yellow while gaseous fluorine is only very pale yellow. Double sharp (talk) 22:17, 18 October 2022 (UTC)[reply]
The same JChemEd Sandbh cited, doi:10.1021/ed042p647, explicitly identifies the blue color of the liquid as a reaction of 1 photon with 2 O2 molecules. "The pair of molecules taking part in this process are not bound to each other but are simply a colliding pair." DMacks (talk) 02:40, 19 October 2022 (UTC)[reply]
Let's try to answer a single question: why is O2, at the concentrations found in the atmosphere, colourless despite having absorption lines in the visible region? (And, yes, those lines are for the molecule, not the atom.) The key paper seems to be this 1972 review and on p 430 it says "O2 is a weak light emitter because, for most of its excited states, transitions to the ground state are strongly forbidden", which may or may not be relevant. The rest of that article is well beyond my level of comprehension but maybe @Sandbh can make something of it! Mike Turnbull (talk) 12:18, 19 October 2022 (UTC)[reply]
Well, nice to see DMacks, Michael Turnbull, and Double sharp in the same thread. It's been a while for that? All we're missing is WyzAnt or so. And I'm looking at the StackExchange link, it seems the person is asking why is oxygen colorless when liquid oxygen and solid oxygen has some colors? Am I missing something here, like is that supposed to be an anamoly? Are most colorless things supposed to be colorless in all 3 states? 67.165.185.178 (talk) 00:45, 20 October 2022 (UTC).[reply]
Rich (1965, pp. 115–116), in his book Periodic correlations, has the following to say about the colours of nonmetals:
"…the heaviest elements are the ones that absorb light at the lowest wavelengths and they are also the ones that, although least easily reduced, are most easily oxidized. This is reasonable because the absorption of light in these cases (for example, Br) partially removes an electron from a bonding to an antibonding molecular orbital farther from the nuclei. The antibonding feature is related to molecular dissociation, as in the photo-induced reactions of chlorine or bromine with hydrogen or hydrocarbons.
A factor causing the greater intensity of color in the heavier elements is the larger size of their molecules, This effect is seen in all kinds of wave motion. Both sound waves and water waves bypass objects much smaller than their wavelengths, but are stopped by larger ones. Even with a given element, we find that the allotrope having larger molecules is more intensely colored, except in the case of diamond and graphite…Organic dyes also illustrate the connection between molecular size and color."
Our article on chlorine says:
"Specifically, the colour of a halogen, such as chlorine, results from the electron transition between the highest occupied antibonding πg molecular orbital and the lowest vacant antibonding σu molecular orbital (Greenwood & Earnshaw, pp. 804–809)."
G&E (p. 806) say:
"...the violet colour of I2 vapour can be seen to arise as a result of the excitation of an electron from the highest occupied MO (the antibonding πg level) into the lowest unoccupied MO (the antibonding σu level)."
At p. 606 they refer to oxygen:
"The blue colour of oxygen in the liquid and solid phases is due to electronic transitions by which molecules in the triplet ground state excited to the singlet states. These transitions are normally forbidden [why?] in pure gaseous oxygen and in any case, they occur in the infrared region of the spectrum at 7918 cm-1…and 13 195 cm…However, in the condensed phases a single photon can elevate 2 colliding molecules simultaneously to excited states, thereby requiring absorption of energy in the visible (red-vellow-green) regions of the spectrum."
Judging from Rich (1965) it seems that my guess that the absence of colour had something to do with bond dissocation energy was in the right ball park, although I'm not sure if this is all there is to oxygen being colourless, nor H2 and N2). — Sandbh (talk) 12:35, 20 October 2022 (UTC)[reply]
1 more day left, but, I recall in 11th grade chemistry class, pulling a plastic syringe until the pressure inside got so high, something broke in the syringe. By that logic, the color of the air inside should be more visible. Was I supposed to have noticed a color change right before the syringe broke? 67.165.185.178 (talk) 18:51, 23 October 2022 (UTC).[reply]
With the help of the above responses to my opening question, and further research, I've added this to the nonmetal article:
"Nonmetallic elements are either shiny, colored, or colorless.
For B, graphitic C, Si, black P, Ge, As, Se, Sb, Te and I, their structures feature varying degrees of delocalised electrons that scatter incoming visible light, resulting in a shiny appearance (Wiberg 2001, p. 416).
The colored nonmetals (S, F, Cl, Br) absorb some colours (wavelengths) and transmit the complementary colours. For Cl, its "familiar yellow-green colour...is due to a broad region of absorption in the violet and blue regions of the spectrum" (Elliot 1929, p. 629). The absorbed light may be converted to heat or re-emitted in all directions so that the emission spectrum is thousands of times weaker than the incident light radiation (Fox 2010, p. 31).
For the colorless nonmetals (H, N, O, and the noble gases) their electrons are held sufficiently strongly such that no absorption occurs in the visible part of the spectrum, and all visible light is transmitted (Wibaut 1951, p. 33)."
I hope this is a reasonably accurate summary of the situation?
According to orders of magnitude (luminance), the atmosphere being only 8 kilometers thick if it was compressed by a lid to the average density of sea level, and Mauna Kea having very clean air with record low zenith extinction around 0.1 astronomical magnitudes, more than a few thousand kilometers or miles of nitrogen or oxygen would be COMPLETELY OPAQUE. As distance increases the view of the Sun through a 1 bar nitrogen tube orbiting the Sun would go from noon-like to sunset-like to dimmer than sunset and deep blood red then by a few thousand miles of air no amount of dark-adaption would allow a human eye to see it. It'd look pitch black. Compare with intergalactic space you can see quasars through billions of light years of it in a $999 telescope on a dark ranch with the eyeball. Sagittarian Milky Way (talk) 02:46, 24 October 2022 (UTC)[reply]
@Sandbh I think you are indeed "reasonably accurate" but I'm still bugged by the visible spectrum of oxygen, which clearly refutes that "no absorption" statement. Also, Beer–Lambert law#Application for the atmosphere does imply that oxygen needs to be taken into account, although, again, is perhaps a small contributor.
My best guess, given the sources provided, is that O2 should be coloured, but very faintly. The reasoning is that the reaction of one photon with two O2 molecules can just as well happen in the gas phase: it is just much rarer. This is why I asked if anyone knew the colour of oxygen under pressure. What SMW says does rather tend to confirm this hypothesis. One of those Stack Exchange answers claims it is blue too, but without a reference. Double sharp (talk) 16:08, 24 October 2022 (UTC)[reply]
"Note that the light absorption requires 2 molecules of O2 and a photon, a 3-body process. If we now consider the probability of this 3-body process occurring in the gas phase, we can see that, since a gas is much more dilute than a liquid, the probability for of these 3 items coming together at the same time will be much smaller. Hence, the probability of photon absorption in the gas phase is much reduced. To the human eye oxygen gas will appear colorless (at normal and reduced pressures). However, if you place the gas in a cell and record its visible spectrum you will still be able to detect this absorption." Sandbh (talk) 07:34, 25 October 2022 (UTC)[reply]
Fox, M (2010). Optical Properties of Solids (2 ed.). New York: Oxford University Press.
Wibaut, JP (1951). Organic Chemistry. New York: Elsevier Publishing Company. p. 33. "Many substances...are colourless and therefore show no selective absorption in the visible part of the spectrum."
Wiberg N 2001, Inorganic Chemistry, Academic Press, San Diego; Wiberg is here referring to iodine.
I read in Wikipedia "Planck's law" publication that the Planck's radiation law given in the SI units of Bν are W·sr−1·m−2·Hz−1.
But for me there is a time problem there:
watts are joules emitted during 1s, but when you divide by Hz you remove this time of 1s, keeping only the value of the number of periods during 1s.
So, effectively if this value (number of periods) is dimension less in this equation, however it is related to the time interval of 1s, no time but time dependent !
You have the same problem in Planck's Einstein relation.
How should we understand, interpret or justify this conflict? — Preceding unsigned comment added by Malypaet (talk • contribs) 22:48, 17 October 2022 (UTC)[reply]
The point of this law is to show have much power is emitted at different frequencies, so that is why the Hz−1 is there. You can work out the power in a frequency band by integrating from the low to the high frequency of the frequency band. If you look at the Planck's law article you can see that there is also an expression of this by wavelength. If you want to eliminate "Hz−1" and the solid angle, you can use the Stefan–Boltzmann law instead which gives power per square meter. Perhaps other ways you could reduce the time dimension is to look at the energy per unit volume in the blackbody radiation field. Or you could look at the average energy per photon. Or you could look at the energy in one cycle of the emission. But that last one will still depend on the frequency and bandwidth. Graeme Bartlett (talk) 23:40, 17 October 2022 (UTC)[reply]
OP: I think you're confusing time dependence with time as a part of dimensional analysis. Time dependence means that a system evolves over time; which is to say at different points in time, it exists in different states. For example, light with a constant frequency we say is time independent, because the frequency doesn't vary at different points in time. This is true even though dimensional analysis tells us that the units of frequency, s-1, contains a unit of time. --Jayron3213:44, 18 October 2022 (UTC)[reply]
Yes for your example, but in a radar pulse each cycles are moving with the pulse, so you can write that at different point of time the cycles are in different locations,the same for photons (I worked on radars by capturing echoes on a memory oscilloscope...) Malypaet (talk) 15:48, 18 October 2022 (UTC)[reply]
Again, you're making the same mistake you did earlier: the 'cycles' of the pulse (which is to say the periodic emission of a burst of EM radiation) are unrelated to the properties of the photons and/or light waves involved in the pulse. Lets simplify it: Imagine you have a light bulb that emits pure yellow light of 580 nm. That light has a frequency of 5.16 x 1014 Hz. Now, imagine you take your light switch and turn it on and off once per second. You're creating pulses of light with a frequency of 1 Hz. There is no relationship between the frequency of the light wave, and how often you turn it on and off. Those are not the same thing. --Jayron3220:36, 18 October 2022 (UTC)[reply]
Yes, there is a relation: in your pulses of yellow light at 1Hz: you have half of 5.16 x 1014 cycles of this yellow light in each of your pulses. He who does not want to see, cannot see. Malypaet (talk) 04:31, 20 October 2022 (UTC)[reply]
