Wikipedia:Reference desk/Archives/Science/2016 September 17

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September 17

A photon, alone between mirrors

If I put a photon between two mirrors so as it bounces from one side to the other and back. What happens to the photon? Would it bounce for ever?

And what if I approximate the mirrors until they touch. Would it just become tangled between the two mirrors?Llaanngg (talk) 00:26, 17 September 2016 (UTC)

Here's a 2006 article from Scientific American that nicely explains how mirrors reflect photons, providing a wave-theory and a quantized-theory description.

Your photon will probably bounce back and forth many times, but eventually (it's a statistical certainty) the photon will "totally disappear" during one of the subsequent interactions with the mirror-surface. In its place, the same quantity of energy will be converted into heat and/or kinetic energy of an electron in the mirror's atomic lattice.

Nimur (talk) 00:44, 17 September 2016 (UTC)

Between two mirrors you will get a standing wave. If the mirrors are moving together the photon will get a blue shift each time it bounces off a mirror. The photon energy will increase with each bounce, and as mirrors approach each other the bounces will get more and more frequent. Will this increase the energy of the photon to infinity when the mirrors meet? It seems likely, but I am not doing the calculation! It means that the force needed to push the perfect mirrors together is higher and higher, and the energy needed goes into the photon. To carry this to absurdity, a black hole will form and collapse the mirrors and Earth and the rest of the cosmos into it. So you can be pretty sure that this cannot be done! Graeme Bartlett (talk) 01:53, 17 September 2016 (UTC)

See Particle in a box for a thorough explanation of such scenarios. --Jayron32 02:24, 17 September 2016 (UTC)

I presume the black hole would give rise to a white hole Big Bang elsewhere in the multiverse. Sandbh (talk) 04:34, 17 September 2016 (UTC)

I’m sorry for interrupt, but black holes never being in this situation because all mirrors always did had an optical focus as all optics.--Alex Sazonov (talk) 09:48, 17 September 2016 (UTC)

It is always necessary to know how much optical focuses did had an optical environment such as mirrors, may did focuses in focus as specters in specter.--Alex Sazonov (talk) 10:22, 17 September 2016 (UTC)

Alex, the black hole is a bit of a joke, but a large fraction of science desk questions have to be about black holes. In reality a perfect mirror at all photon energies does not exist. And a real mirror is not going to reflect a photon right on its surface, so real mirrors cannot reduce the reflecting space between them to nothing. So a photon is not going to be compressed to anywhere enough energy. See if you can work out the force the photon exerts on a mirror, and how much work is done on the photon as the mirrors are moved together. Graeme Bartlett (talk) 13:25, 17 September 2016 (UTC)

Thanks. But I don't understand, what always doing work in optics a photons or optical focuses and specters?--Alex Sazonov (talk) 14:00, 17 September 2016 (UTC)

Graeme Bartlett, it seems me that only focuses are always been a optical force in all optics, and may been specters are too.--Alex Sazonov (talk) 14:29, 17 September 2016 (UTC)

In a quick search I found an apparent answer ... understanding it is a whole different ball of wax. [1] According to this, the Kramers-Kronig relations demand that a material that does not absorb light must have no optical activity, which I take to mean, that there is no absolutely perfect mirror hence no absolutely perfect reflection. This surprises me because mirrors seem to have been made that are really, really good. But no matter how good they are, they don't meet up to the original specification given here unless they are really without absorption at all. Wnt (talk) 11:37, 18 September 2016 (UTC)

So that are ideal focuses and ideal specters (spectrums) being? — Preceding unsigned comment added by Alex Sazonov (talk • contribs) 12:08, 18 September 2016 (UTC)

Particle in a ring

The particle in a ring article begins "In quantum mechanics, the case of a particle in a one-dimensional ring ...". Would someone please explain what a one-dimensional ring is? The article goes on to apply the model to derive Huckel's Rule but it seems to me that this refers to a ring restricted to a plane. What am I missing? Thanks. EdChem (talk) 02:46, 17 September 2016 (UTC)

I think the point is that the ring itself is one-dimensional, meaning you can only move forwards or backwards within it, not side-to-side. To embed a ring into Euclidean space, you need at least two dimensions, but those are not important to the problem. --Trovatore (talk) 03:29, 17 September 2016 (UTC)

In an ideal, infinitely thin ring, you only need one coordinate to fully describe a particle's position in the ring (an angle). Thus it is 1D. Certainly it is embedded in a 2D space, but while this 2D space is useful for actually visualising the ring, it is completely irrelevant to the problem: everything would work just as well if it didn't exist. Double sharp (talk) 03:49, 17 September 2016 (UTC)

The relevant article is n-sphere; the ring as described in the titular article is S¹, i.e. a circle, i.e. a "one-dimensional" sphere embedded in two-dimensional space. The position in a one-dimensional ring is defined by a single angle theta, whereas in spherical coordinates a S² sphere (i.e. a "sphere" in the common sense) needs both a theta and a phi angle to specify the latitude and longitude of a point. Wnt (talk) 11:44, 18 September 2016 (UTC)

The "number of coordinates" formulation is — a good place to start, but it has some problems. On the Earth, for example, you need only one coordinate to identify the poles (the North Pole has a latitude of 90 degrees north, and its longitude is undefined).

There are several different notions of topological dimension, which are distinct mostly in cases that non-mathematicians rarely need to worry about. For the sort of case raised by the question, the most relevant one is manifold dimension, which can be expressed in terms of "number of coordinates", but local coordinates rather than global ones. That is, the key fact is not how many coordinates you need to locate an arbitrary point, but rather how many you need to describe how to move around near a given point. --Trovatore (talk) 15:58, 19 September 2016 (UTC)

Electricity without mechanical work of magnetism and electromagnetism

Why only mechanical oscillations of magnetism and electromagnetism creating an electricity?--Alex Sazonov (talk) 11:50, 17 September 2016 (UTC)

What about thermoelectric generators and photovoltaics? -- ToE 12:41, 17 September 2016 (UTC)

I think that elementary particulars or magnetism and electromagnetism of them did mechanical working to creating an electricity, is I'm right? — Preceding unsigned comment added by Alex Sazonov (talk • contribs) 13:48, 17 September 2016 (UTC)

Electromagnetism, in its most general sense, is certainly involved in any form of electrical generation given that electricity is a manifestation of electromagnetism, but the photovoltaic effect or photodiode#Principle of operation is not considered a mechanical process simply because of the motion of photons and electrons. Likewise the Seebeck effect, even though heat does involve a disordered motion of the microscopic elements of a system. If any motion of electrons is sufficient to satisfy your definition, then it is true by tautology. -- ToE 14:15, 17 September 2016 (UTC)

I'm talking about that all oscillations are always considering as mechanical oscillations independently from what they are always being, the oscillations of elementary particulars or oscillations of magnetism or electromagnetism.--Alex Sazonov (talk) 14:52, 17 September 2016 (UTC)

What will be if photon oscillations being the same as electron oscillations?--Alex Sazonov (talk) 17:25, 17 September 2016 (UTC)

Try reading our article on quantum electrodynamics. I'm not sure exactly what you're asking, but it seems to be a quantum electrodynamics-related question. I hope that helps. loupgarous (talk) 07:34, 18 September 2016 (UTC)

Thank you. It can put that vectors directions of oscillating elementary particles always have a value, but not their electric charge and their mass.--Alex Sazonov (talk) 08:34, 18 September 2016 (UTC)

Mechanical models of atom such as photon and electron being importance.--Alex Sazonov (talk) 09:48, 18 September 2016 (UTC)

There are not being paradoxes in mechanical physics, what effect did it having on nuclear physics and quantum physics which are having paradoxes, if looking on them through mechanical physics?--Alex Sazonov (talk) 18:26, 18 September 2016 (UTC)

Our article Atomic theory addresses the relative merits of classical and quantum mechanics in describing atomic and particle physics interactions. You may also wish to read our article Introduction to Quantum Mechanics to gain insights on how photons transfer energy to subatomic particles such as electrons. The whole field of quantum mechanics came about because classical mechanics was not sufficient to describe interactions at the subatomic scale while describing everything observed to happen. loupgarous (talk) 16:15, 20 September 2016 (UTC)

Are there any real gas in a container systems that would probably measurably lose energy within the age of the universe?

In the manner of fake gas made of literal billiard balls in vacuum? How long could you make this fake gas of real billiard balls take to stop? The wall is made of billiard ball material of ideal thickness and shape and completely surrounds the balls (no escaping into outer space and moving forever). Diamonds are very rigid, would a "gas" of diamond billiard balls in a diamond container stay gassy for longer? Sagittarian Milky Way (talk) 14:43, 17 September 2016 (UTC)

"Very rigid" is a relative thing. Crystals, like diamonds, are rigid compared, to, say, rubber, but that in no way means they are absolutely rigid. If you strike them, they vibrate, and this dissipates energy. If you had diamond spheres bouncing around in a diamond box, floating in a perfect vacuum, I suspect that within a few hours the balls would have lost most of their momentum, relative to the box, but everything would be vibrating and somewhat hotter (the box and the balls), instead. StuRat (talk) 16:49, 17 September 2016 (UTC)

Even for diamond balls in a box, once everything settles into thermal equilibrium, the average kinetic energy of the balls should stay constant. This would be the same as that of a thermal photon in the box. If the temperature is low enough that the balls do not suffer thermal damage, the time it takes for a ball to travel a visible distance would be lo-o-ong.

Take this picture in the opposite direction: start with helium (or any other gas) in a box, and allow the contained thermal radiation to escape at one point, and seal it. During the time it takes for the photon field to re-establish equilibrium with the gas, the average kinetic energy of the gas particles will be losing energy measurably. Further, there may be other fields (e.g. the neutrino–antineutrino field) that may not be in thermal equilibrium to start, assuming a perfectly sealed box. Since the coupling to this field is very weak (but nonzero), a similar effect would occur here, but over considerably longer time spans. The conditions inside a neutron star would make this time very short, though. —Quondum 17:28, 17 September 2016 (UTC)

Ah, I forgot, container and contents is at thermal equilibrium with its surroundings (though it might be hard to find places that have constant temperature for 14 billion years (AND are the standard temperature of STP so that all substances classified as gas can exist as gas)) Sagittarian Milky Way (talk) 18:11, 17 September 2016 (UTC)

Some gases will undergo changes, eg radon will decay, ethylene or boron monofluoride will polymerise. Graeme Bartlett (talk) 00:28, 18 September 2016 (UTC)

Feynman Lectures. Lecture 13

Link: http://www.feynmanlectures.caltech.edu/I_13.html

Quote
As we know, the force is −GM/r2 times the mass m, where m is the mass that moves.
This one-dimensional case is easy to treat because we know that the change in the kinetic energy is equal to the integral, from one end of the motion to the other, of −GMm/r2 times the displacement dr

Why Feynman writes that force = −GMm/r2?
We have the formulae $\int _{1}^{2}\mathbf {F} \cdot d\mathbf {s}$ and $\mathbf {F} =-{\tfrac {GMm\mathbf {r} }{r^{3}}}$ . What should we do to write next integral expressions? Should we use some axis and a projection of the force? But if an axis will be arbitrary, the force projection value can be both positive and negative. If on Fig. 13–2 we direct an axis to the right, then we will have the projection value with minus sign, as Feynman writes, but if an axis will be directed to the left, result will have plus sign. Username160611000000 (talk) 17:44, 17 September 2016 (UTC)

${\tfrac {\mathbf {r} }{r}}$ is a unit vector indicating the direction of the force. In the one-dimensional case Feynman is talking about, there aren't that many directions, so Feynman omits this unit vector. Directing the axis to the right (away from the big, "central" mass) is the standard convention. Note that if the direction of the axis were changed, you'd need to change the signs of both the force and of dr and would end up with the same integral for the change in kinetic energy. Huon (talk) 18:17, 17 September 2016 (UTC)

Note that if the direction of the axis were changed, you'd need to change the signs of both the force and of dr - Can you prove it? $dr$ is a projection of $d\mathbf {s}$. But sign is hidden inside, like with this quote:

Quote
No matter how the object moves in those circumstances, falling in a parabola for example, F⋅s, which can be written as Fxdx+Fydy+Fzdz, has nothing left of it but Fzdz=−mgdz, because the other components of force are zero.

If axis is directed to the right : $\int _{1}^{2}\mathbf {F} \cdot d\mathbf {s} =\int _{r_{1}}^{r_{2}}+F_{r}\cdot dr=\int _{r_{1}}^{r_{2}}-{\tfrac {GMm}{r^{2}}}dr=+GMm({\tfrac {1}{r2}}-{\tfrac {1}{r1}})$; $(r_{2}<r_{1})$; ${\tfrac {1}{r2}}-{\tfrac {1}{r1}}>0$. Result is positive.

If axis is directed to the left: $\int _{1}^{2}\mathbf {F} \cdot d\mathbf {s} =\int _{r_{1}}^{r_{2}}+F_{r}\cdot dr=\int _{r_{1}}^{r_{2}}+{\tfrac {GMm}{r^{2}}}dr=-GMm({\tfrac {1}{r2}}-{\tfrac {1}{r1}})$; $(r_{2}>r_{1})$; ${\tfrac {1}{r2}}-{\tfrac {1}{r1}}<0$. Result is positive.

If we will in addition change the sign of$dr$, we will get different results.

I doubt the thought process here (as intended for the student, at least) is really that sophisticated. The universal gravitational law may provide a vector force, but it has a scalar magnitude; a person can simply look at this simple system and see which way the gravity should pull along the axis the way it is drawn. Wnt (talk) 19:47, 17 September 2016 (UTC)

First, we are considering work, but not force. Second, mathematics must give correct answers without any look on any drawings or real objects. Just write down how we should expand integral formula $\int _{1}^{2}\mathbf {F} \cdot d\mathbf {s}$, what are the rules and where can I find them. And where can I find the rule that if axis direction is opposite then minus appears before $dx$.Username160611000000 (talk) 08:34, 18 September 2016 (UTC)

First, the sign is chosen in the force; what you calculate from the force after that is irrelevant. Second, this is a philosophical issue I'm not so convinced about. I know there is a deep-rooted Principia Mathematica school of thought that everything can be logically defined from the ground up with axioms. But at the same time, it seems rare for a physics lecture to pass without someone discarding a "non-physical solution". I remain agnostic on that issue in theory, but in practice, well, the brutish way of thinking remains common and intuitive. Wnt (talk) 11:52, 18 September 2016 (UTC)

For physicist mathematics is the good servant. Physicist must be a good mathematician. So according dot product we have:

$\mathbf {F} \cdot d\mathbf {s} =F_{x}dx+F_{y}dy+F_{z}dz.$

All these vector components are positive. Only when we will substitute values for components we will add sign.

$\int _{1}^{2}\mathbf {F} \cdot d\mathbf {s} =\int _{1}^{2}(F_{x}dx+F_{y}dy+F_{z}dz)=\int _{x_{1}}^{x_{2}}F_{x}dx+\int _{y_{1}}^{y_{2}}F_{y}dy+\int _{z_{1}}^{z_{2}}F_{z}dz,$ where point 1 has coordinates $(x_{1};y_{1};z_{1})$ and point 2 has coordinates $(x_{2};y_{2};z_{2})$. Then if $F_{x}=0$ and $F_{y}=0$, only $\int _{z_{1}}^{z_{2}}F_{z}dz$ remains.

We can replace $F_{z}$ by its value with sign, but $dz$ remains as the vector component without sign.

Is it correct? — Preceding unsigned comment added by Username160611000000 (talk • contribs) 15:22, 18 September 2016 (UTC)

gravitation - oscillations

What about a gravitation - oscillations vectors of elementary particulars in magnetic or electromagnetic waves, did a new vectors directions of oscillating elementary particulars giving them new opportunities?--Alex Sazonov (talk) 20:25, 17 September 2016 (UTC)

If it did, what about a new opportunities of magnetic and electromagnetic waves?--Alex Sazonov (talk) 20:35, 17 September 2016 (UTC)

Long-term transmission of non-STDs in semen

There has been much news of transmission of Ebola virus and Zika virus in semen. What I'm not very clear about is: does this represent a coincidental uniqueness of their biology, or people taking a better look at viruses recently studied? The theory is simply that the testes are subject to immunological privilege, which seems like it should be a very broad phenomenon. I see very little on PubMed about influenza, norovirus, coronavirus, enterovirus etc. in semen, and what little there is being in relation to animal studies with little direct epidemiological applicability. I'm wondering if common viruses could be using semen as a way to hide away for months or even years so that they can jump from season to season, outbreak to outbreak, and people just haven't looked at them? Wnt (talk) 19:41, 17 September 2016 (UTC)

Non-STDs can not be transmitted in semen – if something is transmitted in semen, this is by definition a STD. Ruslik_Zero 20:49, 17 September 2016 (UTC)

I think such a categorical approach might obscure the question. Sexually transmitted infection says: "Sexually transmitted infections (STI), also referred to as sexually transmitted diseases (STD) and venereal diseases (VD), are infections that are commonly spread by sex." So if the predominant vector is not via sex, it would not normally be labelled as an STD. Because of the distinction, Wnt's question is sensible: are there other infections that exploit this route of infection? —Quondum 21:19, 17 September 2016 (UTC)

Also, it potentially might not be a true sexual route; the particles might have to get from semen to the digestive or respiratory tract to be infectious. I'm really wondering about very rare events here because they can be instrumental in the spread of a virus. For example, smallpox couldn't spread between continents by the usual person-to-person route; it had to hide somewhere - people assume in some sort of clothing, container, whatever, but in theory, why not in immunologically privileged sites inside the body of someone apparently long recovered? And polio, somehow, had to get from one summer to the next - when it was epidemic, it could do that by statistics or in the tropics, but what about during the much longer period when it was not so common? I've just gotten suspicious ... Ebola and Zika don't have much in common aside from being RNA, so hearing these results for the two viruses, well, it's like you dug a little hole on one side of a house and hit a buried screw, and dug a hole on the other side and hit the same kind of screw. You'd start wondering if the whole building site was covered in buried screws. Wnt (talk) 11:26, 18 September 2016 (UTC)

It's not especially odd that ebola and zika don't have much in common - nor do, say, HIV and hepatitis C virus. This doesn't necessarily mean that all viruses are semen-borne, only that it's a very beneficial mutation for a disease that wants to spread (it may explain why other ebolaviruses haven't had the same sudden spread). The two forms of Herpes simplex virus are extremely similar, and yet they seem to prefer different parts of the body, and only one is primarily an STI (neither is found in semen - rather, they seem to be transmitted by mucous membranes). To give an another example - hepatitis viruses are well studied, and we know that Hepatitis C virus can be transmitted in semen, but Hepatitis A virus (which, to be fair, is not closely related to Hep C) is only rarely found in bodily fluids except faeces. I couldn't find much about polio reservoirs, but it certainly seems to be the case that transmission is by the fecal-oral route - this paper is pretty good if you want to see how scientists try to work out where poliovirus hides between seasons. Smurrayinchester 11:21, 19 September 2016 (UTC)

Yeah, I was musing out loud in that last response without really thinking through much. It is certainly true that viral tropism can evolve rapidly, but some sources have implied that with semen there is something inherent in the immune privilege that makes it a ready reservoir, rather than specific adaptations for these two viruses to spread that way. Wnt (talk) 11:53, 20 September 2016 (UTC)

Marburg virus was observed to spread laterally during sex in semen and to cause hemorrhagic injury to the testes during the first known outbreak in Marburg, Germany in 1967. That technically made it an STD, but it's not considered as such because its main mode of transmission is by other means. loupgarous (talk) 16:57, 20 September 2016 (UTC)

Effect of low air pressure on musical instruments

Would/does lower air pressure than is common at ground level have an effect on the acoustics - either produced or perceived of musical instruments? My initial thought was related to how instruments would sound on a pressurised commercial airliner, but I'm also interested more generally as well. Thryduulf (talk) 23:35, 17 September 2016 (UTC)

There wouldn't be much difference in frequency, which is more likely to be affected by temperature. They would tend to be quieter for a given level of vibration in the case of things like guitars. Greglocock (talk) 00:10, 18 September 2016 (UTC)

(edit conflict)Strictly in terms of sound propagation, air pressure has little effect (at least at pressures where humans can survive). There are some potentially interesting effects in terms of sound production and reproduction. For example, the acoustics of speaker enclosures depend on the mass of air within the enclosure compared to the mass of the driver (usually a speaker cone); stringed instruments also will tend to ring longer because there is reduced damping by the air; and so on. These effects are strictly due to air density rather than pressure but at the same temperature air pressure and density are proportional. If you're comfortable with math this article is very interesting. Shock Brigade Harvester Boris (talk) 00:12, 18 September 2016 (UTC)

Um, no, the equation for acoustic radiation from a radiating panel contains rho, hence my claim about amplitude. Greglocock (talk) 05:52, 18 September 2016 (UTC)

Not sure I see the contradiction. Can you clarify? Shock Brigade Harvester Boris (talk) 15:28, 18 September 2016 (UTC)

I was reacting to your first sentence, a reduction in sound pressure level is a pretty big effect to ride roughshod over. Greglocock (talk) 23:03, 18 September 2016 (UTC)

Oh yes, of course. I had in mind speed of propagation, and sound propagation being non-dispersive. I should have been clearer. Shock Brigade Harvester Boris (talk) 00:55, 19 September 2016 (UTC)

The primary effect would be in how long it would take me to boil the water, prior to throwing it on anyone playing violins in what I call "the nails on the chalkboard style". :-) StuRat (talk) 01:18, 18 September 2016 (UTC)