Wikipedia:Reference desk/Archives/Science/2017 July 11

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July 11

How hot can CCFLs get before breaking?

How hot can a CCFL get before sustaining damage? (note that CCFL is different from a CFL). I'd like to use about 50 of them (from broken laptop monitors) in a project as close together as possible and expect ~200 W waste heat. --145.255.245.88 (talk) 07:48, 11 July 2017 (UTC)

Same as a CFL - are you talking about the tubes, or their drive electronics? The electronics generate most of the heat (especially for CFLs, less so for CCFLs) but the tubes are prety much insensitive to this. Although they're a "cold cathode", that only means that they're not a directly heated cathode, it doesn't mean that theyir cathodes have to be kept cold. Andy Dingley (talk) 10:25, 11 July 2017 (UTC)

If this project is intended to be viewed by the public, then simply use an electric fan or two so as to avoid roasting the public. If the public feel comfortable so will the monitors. Also, it would help if you explained the spacial arrangement of said monitors. 75 degrees gives max efficiency. Ie. the most light for the energy converted into light. Aspro (talk) 10:39, 11 July 2017 (UTC)

In general, the CCFLs I've seen have operating temperature ranges that go up to 50 or 60 C. If you know what parts you are using, you can often find a specification online. That will tell you want the manufacturer recommends. As to how hot your installation will get, that will depend on the airflow, the mounting materials (a metal backing would remove heat more effectively than a wood one, for example), and other aspects. If 200 W is the total dissipation for 50 monitors, that is actually not very much energy, and unless they are completely enclosed and lacking in airflow, you are probably fine. Dragons flight (talk)

Thanks for everyone's responses. I'm talking about taking the CCFLs out of the monitors and lining them all up next to each other for maximum density. I expect to be able to fit all 50 CCFLs in a space about 25 cm x 35 cm. I was hoping to passively cool them by attaching to a thin (~4 mm) aluminium plate but the way each tube is terminated precludes physical contact between the tube and the metal plate. I don't want to use a fan because I hope to use this light for (non-professional obviously!) video recording. 145.255.245.88 (talk) 17:03, 11 July 2017 (UTC)

Hotel body soap, hand soap, and shampoo

I looked through the archives and didn't find what I was looking for.

In hotels, they have all three. I'm guessing they're all pretty much the same, but difference smells and colours. Is this about right? Anna Frodesiak (talk) 07:48, 11 July 2017 (UTC)

Soap and shampoo have a couple of differences: manufacture and use.

Their use is that "soap" ought to feel soapy and should lather well. A "good" soap is perceived as being the one that feels "luxuriant" on this basis. A difference with body soap (as a rarely made distinction) is that hands are washed quickly, bodies more leisurely. So a hand soap isn't quite so richly lathering, as you have to get it on and off without waiting around for it. Shampoo is different - you want the minimum amount of lather, as that has to be rinsed clean in turn. It takes longer to rinse hair than to wash it clean. Also, if conditioning the hair afterwards, any remaining shampoo will make the conditioning ineffective, so there has to be a four stage wash-rinse-condition-rinse cycle.

Their manufacture is that they can be either soaps or detergents. Most are increasingly detergents rather than soaps. Local product labelling laws may require non-soap detergents to be labelled as "handwash" rather than soap. It's rare to see a bath "soap" that's a detergent, as they just don't feel like luxury. There are though plenty of in-shower bodywashes based on detergents. These are mostly pitched as "invigorating" rather than "luxurious" and they sell on the basis of fragrances and spicy tingles (don't miss the 'flaps on fire' post (non-FB link) - I have this stuff, they're not joking). Liquid hand "soaps" are almost all detergents, because they're quicker to rinse. Shampoo is all detergent based, not soap, because although soap is usable for hair washing (and used to be, as there was little else) it takes a huge amount of rinsing.

Incidentally, the "two in one" shampoo and conditioner formulas work by mixing an easily rinsed shampoo with a lot of conditioner, and making this conditioner persistent against washing and rinsing. By the time you've rinsed the shampoo out and the conditioner can get to work, there's still enough left to do it. Andy Dingley (talk) 10:36, 11 July 2017 (UTC)

Andy, I don't know how you know all this, but I am well impressed. I had no idea. I cannot view 'flaps on fire' because facebook is blocked here in China. Thank you so much for the very educational answer! I am grateful. Anna Frodesiak (talk) 11:40, 11 July 2017 (UTC)

Note that soap is made from fats, via saponification, whereas detergents are not. StuRat (talk)

Back in the old days, chemists and pharmacists weren't as protective of their trade secrets as one might think. Sure, they'd have their own tonic recipe that they kept secret, but for the hundred mundane things they had to make and sell anyway, there were standard industry formularies like The Chemical Formulary and Spons' Workshop Receipts. These listed a great many standard perfumes, soaps and shampoos. Often bulk soap would be made in large factories (a smelly process, not popular in towns) and large blocks at different grades (from faces to nits to floors) would be sold to a retail chemist who would blend them and package them.

These days I sell at craft markets, which often involves talking to high-end soap makers. To sell artisan soap today, you have to find a way to make it special.

Really I don't know much about this stuff at all, but I keep a house chemist who does. There's little other work for an industrial chemist. Andy Dingley (talk) 12:20, 11 July 2017 (UTC)

and "flaps on fire" is a cautionary tale, unaccountably omitted from Struwwelpeter, about bathing in Mint and Tea Tree Shower Gel. The warning is addressed to women but, from the replies, is also applicable to men. Hopefully it isn't on sale in China. Thincat (talk) 12:35, 11 July 2017 (UTC)

Ah, I found news items about the facebook page. I will avoid such products.

Thank you all again. And I guess all soaps in liquid form are detergent here, but am not sure. Anna Frodesiak (talk) 15:40, 11 July 2017 (UTC)

Almost all liquid "soaps" that are transparent will be detergents. It is possible to make liquid soaps (that are soaps), but they're rare these days, because detergents are cheaper and also liquid soaps store badly, having a tendency to set in the nozzle or pump. Detergents are more efficient (less mass needed for a wash) and so are much cheaper to manufacture in the same apparent quantity. Most people use much more detergent than they really need.

Transparent soap has an interesting history of its own, but I think it's solid, not liquid (I don't know of any transparent, liquid, true soaps). They're made from glycerine (although not all glycerines are transparent). Pears soap was the first such soap, the first branded soap and one of the world's first major retail brands. At a time when product adulteration was rife and some badly-made soap was still caustic, the transparency, brand identity and general high quality of Pears' were factors in making it an easily recognised premium brand. Andy Dingley (talk) 16:04, 11 July 2017 (UTC)

Yes, when you say "...much cheaper to manufacture...", it convinces me that hotel body soap, hand soap, and shampoo are all just detergent. Anna Frodesiak (talk) 18:14, 11 July 2017 (UTC)

For liquid "soap", yes, but I'm not even sure if it's possible to make a "detergent bar" that resembles a soap bar. StuRat (talk) 21:45, 12 July 2017 (UTC)

(Caution: Do not try this at home. Lye and potassium hydroxide are dangerous).American backwoods folk of a bygone era made lye from dripping water through wood ashes, then used that with rendered animal fat to make soap. The lye tended to be weaker than commercially made lye, and if so the resulting soap was likely to be liquid rather than solid. I think this latter advice was in one of the Foxfire books. The weak lye liquid soap was still said to clean. Another site says potassium hydroxide is used in place of sodium hydroxide to make liquid soap. The example is certainly not clear. Soap does not work well in hard water for cleaning clothes or hair, in that it does not lather well without a softening agent or the use of water caught in a rainbarrel. Edison (talk) 03:45, 14 July 2017 (UTC)

Rotating an object with circularly polarized light

Suppose you have a beam of circularly polarized light, and it strikes an object. If absorbed, it should tend to rotate the object, though only to a very small degree as the Planck constant of momentum per photon is small: 6.6E-34 J*s (= N*m*s). Even a meter-long microwave photon carries relativistic mass-energy of 1.2E-6 eV = 2E-25 J, so I think to apply the equivalent of a middling 1 N*m torque screwdriver for one second (i.e. 1 N*m * s) you need 3E+9 J of energy, which our joule article describes as the energy of a one-ton vehicle moving at 100 mph. Times three, plus another factor of two because photons have *half* a Planck unit. Does that sound right to you?

However... photons don't have to be absorbed. They could reflect, and mirrors can be up to 99.999% reflective. That means that in concept you could recycle a photon 100,000 times, getting double the angular momentum transfer each time ... provided, that is, you can come up with a really good mirror, perhaps of some metamaterial, that works at the right (long) wavelength and which reverses the polarization of the photon striking it. Though I think this is actually the norm - not reversing it would actually be unusual! That would knock down the required photon energy to 1.5E+4 J, more like the solar radiation striking the Earth each minute.

I also get curious about the linear momentum (or impulse (physics) ... hmmm, is there a meaningful difference?) transferred in this system: photon says h/lambda (i.e. wavelength); for those one-meter microwaves that is 6.6E-34 N*s per photon. In the mirror system we needed 1.5E+4 J/2E-25 J = 0.8E+29 photons, so that would be still something like 5E-5 N*s of momentum, not much... wait, no... the mirror reduces the amount of energy needed, but not the force of each impact -- so actually it's the full 3E+9 J/2E-25 J = 1.5E+34 photons (that should be twice the inverse of the Planck number) that hit, carrying 2 N*s of kick, certainly not negligible. It could be made smaller with radio wave photons...

Anyway, the questions:

• a) can you point to any good lab demonstrations of turning an object with circularly polarized light?
• b) how good do low-frequency mirrors (microwaves, etc.) get?
• c) with perfect lens technology or the like, could you confine the microwaves to a cavity without completely enclosing it? How far could you get? In other words, could you use them to transmit torque between two objects at a distance from each other?
• d) the linear momentum issue is also interesting -- with that, the energy for "kick" is constant, so I think that any sufficiently "perfect" set of mirrors (any frequency, and not much better than those presently available!) should be able to physically levitate objects... at least, until a mote of dust gets in. Is anyone trying to do this?

Wnt (talk) 13:16, 11 July 2017 (UTC)

• Couple should suggest why this is a tiny effect and hard to measure - but it is real. I think Poynting was the first to recognise this, around 1909, a couple of years after Einstein's annus mirabilis. It took until the 1930s before it was successfully measured though. Andy Dingley (talk) 13:31, 11 July 2017 (UTC)

Easier to measure the effects at microwave wavelengths, BTW. Andy Dingley (talk) 13:35, 11 July 2017 (UTC)

You are basically describing some sort of movement-out-of-nowhere mechanism: a single photon strike a mirror A, makes it rotates, and restart the other way with reverse polarization, strike another mirror B just in front of A, makes it rotates, too, than restart again toward A with just the same rotating power than the first time, so it increase the rotation, etc.

congratulation for your new perpetual motion device, first time i read about this one.

Gem fr (talk) 13:45, 11 July 2017 (UTC)

@Gem fr: The "perpetual motion" issue is something I thought about, but from the other end: it is clear that the energy for the rotation has to come out of the light (which as I explain above, contains considerable energy) but how? For one thing, bear in mind that the energy gained by the target object is not going to be simply proportional to the angular momentum - they're two different quantities. I was still thinking about that part, perhaps for a follow-up (I have to work out how energy and moment of inertia relate, etc.), but first I was hoping to hear more about the state of the art for photon energy transfer. Note that bouncing photons between mirrors is often addressed in the linear sense - there it is clear that bouncing off a receding mirror redshifts the photon, deducting the energy fee. Real ordinary mirrors do reverse circular polarization ... hmmm. I just ran into an issue thinking about that.

If I visualize the photon as a rotating disk striking a wall, the direction that the disk rotates relative to its forward motion does reverse when it hits the wall ... but the angular momentum doesn't change. So can a photon change angular momentum in order not to give up angular momentum? Would a mirror somehow rigged (metamaterials and handwaving...) to keep the same circular polarization actually be what gets angular momentum transferred to it? I realize I am quite confused about this. But I know that some angular momentum sure as hell has to go somewhere if you bounce the photon at a 90-degree angle, because it can't keep the same plane it came in with. So the question is valid, I just don't know what I'm asking. ;) Wnt (talk) 16:06, 11 July 2017 (UTC)

I think you need to check redshift. The energy of a photon is tied with the frequency of the photon (energy = h*frequency). So, since Planck's constant h cannot change, when energy is lost, frequency is reduced. Visible light becomes shifted towards the red end of the spectrum. The photon will keep shifting to a lower and lower frequency, losing energy. Since you can't have a mirror that reflects every possible frequency of electromagnetic radiation, the photon will eventually be absorbed (or pass right through the material). 209.149.113.5 (talk) 16:55, 11 July 2017 (UTC)

I didn't intend to disagree with any of that. There are, nonetheless, some things I'm presently confused about. Bear in mind that linearly, kinetic energy is obtained by integrating momentum, by which I suppose I mean impulse as I described before. It makes intuitive sense that the faster something is moving away, the more redshifted the photon is, because in the plane of the mirror the interaction needs to be symmetrical. A bouncing photon should be redshifted by the same amount - which depends only on that difference in velocity - so it will lose just as much energy to any mirror receding at a velocity v, regardless of its mass; it will also give just as much momentum to the receding object as any other mirror moving at v hit by a similar photon. And yet ... the object's energy is 1/2 mv^2, and it doesn't seem like that increases the same for large and small m given an equal change in momentum from the photon. Now rotational energy is obtained by integrating angular momentum in a very similar manner; the velocity along the radius of gyration changes in the same manner, and I have the same confusion. Wnt (talk) 20:02, 11 July 2017 (UTC)

It isn't the case that a photon will lose the same amount of energy to any mirror that is receding at velocity v - that is only an approximation for large mirror masses. In the general case, the velocity of the mirror changes as well, by a different amount depending on the mirror mass; and that change in velocity occurs at the same time that the frequency of the photon changes.

Basically it's just a totally elastic collision of a photon and a mirror (what is conserved are the relativistic versions of energy and momentum). Icek~enwiki (talk) 20:24, 11 July 2017 (UTC)

@Icek~enwiki: That's true - the math is gone into here. But what troubles me is that if you bounce a photon off a 1-ton mirror, the delta v will be twice what it is if you bounce it off a two-ton mirror, which means that the "delta 1/2 mv^2" will be twice what it is for the two ton mirror. But the term in the paper I link for the change in photon frequency, as you'd expect, is a small inverse term for mass that is negligible compared to the velocity term. So I'm not convinced this actually is relevant to the issue. Wnt (talk) 21:31, 11 July 2017 (UTC)

For the change in kinetic energy of the mirror, we have

$\Delta {\frac {1}{2}}Mv^{2}={\frac {1}{2}}M(v+\Delta v)^{2}-{\frac {1}{2}}Mv^{2}=Mv\Delta v+{\frac {1}{2}}M(\Delta v)^{2}\approx Mv\Delta v$

neglecting the small quadratic term.

This gain in energy, linear in $\Delta v$, is balanced by the loss in energy described by the middle term of the right hand side of equation (6) in the article that you linked to.

But your question pertains to this neglected quadratic term, I guess. It is cancelled by the rightmost term of the same equation, just that in the next step the author neglects that term as well.

Icek~enwiki (talk) 22:17, 11 July 2017 (UTC)

@Icek~enwiki: I shudder to get into math markup, but that equation 6 is:

Change in photon energy = $\hbar \omega -\hbar \omega '={\frac {\hbar v}{c}}(\omega cos\alpha +\omega 'cos\beta )+{\frac {\hbar ^{2}}{2Mc^{2}}}(\omega cos\alpha +\omega 'cos\beta )^{2}$ ... for our purposes, $\alpha$ and $\beta$ can be 90 degrees; they are the angles light moves relative to the mirror taking into account that the incidence = reflection only in the mirror's rest frame. Now, there's a problem with this equation that it's obviously not solved for $\omega -\omega '$. Still, we know that its limit as M tends to infinity is $\omega '=\omega {\frac {1-\nu cos\alpha /c}{1+\nu cos\beta /c}}$. Now your expression $Mv\Delta v+{\frac {1}{2}}M(\Delta v)^{2}$ can, with a slow-moving heavy mirror, be written under the assumption that the momentum is conserved - without angles this is ${\frac {\hbar (\omega +\omega ')}{c}}=M\Delta v$ ... Well, I can't say phooey, because now it seems like it works out, and I'm not sure why it didn't before. I am not getting top marks in math today. Wnt (talk) 23:45, 11 July 2017 (UTC)

I scanned to see if anyone had made this point; apologies if I missed it.

User:Wnt, assuming you're right that photons flip their circular polarization when they reflect, that actually means they don't transfer angular momentum.

That's because the circular polarization of light is defined by whether the angular momentum is in the same direction as the velocity (applying the right-hand rule), or the opposite direction.

So if you shoot, let's call it a "right-handed photon" headed at your mirror, that means that it's "spinning clockwise" (of course there's nothing that really spins AFAIK) as you look at it in its direction of motion.

Now it reflects, and its direction of motion is reversed, but it's still spinning in the same direction as you look at it. But now it's coming back at you rather than going away from you, so now that same spin is counter-clockwise for someone looking towards it in its direction of motion, and hit has become a left-handed photon.

And yet no angular momentum has been transferred. --Trovatore (talk) 20:19, 11 July 2017 (UTC)

@Trovatore: Well, as I said above, you can surely... maybe... make them transfer some angular momentum by bouncing them at a 90-degree angle. Because then the angular momentum out isn't what came in, but in a whole different plane. True, that seems to be transferring partial Planck units, which is a no-no, but what I guess it to mean is that you must end up racemizing the photons to a random mix, with the momentum necessarily going into the mirror. Wnt (talk) 20:25, 11 July 2017 (UTC)

@Wnt: Angular momentum can be a little painful to deal with at the quantum level. The problem is that the operators corresponding to the components of angular momentum in different directions do not commute with one another. So in general, a particle won't have well-defined x-, y-, and z-components all at the same time. I think maybe for a photon they can, if z happens to be the direction of motion, because the other two are zero so it doesn't matter that they don't commute. (For particles with half-integral spin, like the electron, they can basically never have well-defined angular momentum components in all directions simultaneously.)

See Clebsch–Gordan coefficients for more information. Not exactly a light read, and I don't have time right now, though I would like to get this straight in my head. --Trovatore (talk) 23:39, 11 July 2017 (UTC)

@Trovatore: As I recall, photons actually have angular momentum of sqrt(2) Planck constants. Like in a lot of systems, a smaller amount (1 Planck unit) can be measured in a given direction. Intuitively I assume that if you know a photon has a spin along its axis, it must have a "topspin" also in some unknown direction, sort of like an electron in a p orbital that really has an angular momentum of sqrt(l(l+1)) = sqrt(2) in that case. You know it is going round and round the nucleus in the complex solution, but it's also doing something else you don't know (there's a picture in magnetic quantum number). Wnt (talk) 23:53, 11 July 2017 (UTC)

I think Wnt has made this point himself already. As both of you say, an ordinary mirror does not change the direction of rotation of the circularly polarized wave, i.e. it does change the polarization (the relation between the rotation and the direction of propagation).

But if you have an interface between two materials of which at least one is birefringent, you can have a reflection without a change in polarization: The circularly polarized photon can be represented as superposition of 2 linearly polarized states, say a vertically polarized state and a horizontally polarized state, with a phase shift of π/2 between them.

When light would pass from a material A to a material B and is instead reflected, the phase of the light changes by π if the refractive index of B is larger than that of A.

What we need is that for the vertically polarized light, the refractive index of B is larger than that of A, and the other way around for the horizontally polarized light. Then the direction of rotation of the reflected light is inverted, and thus the polarization stays the same.

Because we want to make it easy to measure the angular momentum transfer, material A should be a vacuum; so we need a material that has a refractive index larger than 1 for one linear polarization and smaller than 1 for the other.

Icek~enwiki (talk) 20:24, 11 July 2017 (UTC)

bottom line: as stated above by @Trovatore:, NO momentum is transfered in the reflection process. The photon just reverse course, but keeps the very same angular momentum. Same if reflected with an angle.

there is no real energy / momentum issue, anyway: a mirror is a macroscopic object with close to infinite energy(thermal, etc.) relative to a photon. The reflected photon is just a minute contribution to the regular emission of this macroscopic object, that is, it makes no process difference in energy /momentum for the macroscopic object, whether it reflects incoming light (white body), or absorbs it then emits it as per Planck's law ([[black body}}), or anything in between (grey body): in any case, Radiation pressure applies, but the special photon considered is just one in the ocean of photons that the mirror floats in/contributes to.

Basically, you cannot cope with your question with the particle avatar of the Wave–particle duality of incoming light.

Furthermore, entropy prevents the small photon to rotate the whole macroscopic mirror

Gem fr (talk) 21:45, 11 July 2017 (UTC)

@Gem fr: I don't think physics lets you get away with "small" accounting errors. ;) Wnt (talk) 00:04, 12 July 2017 (UTC)

statistical physics. But actually the "small" accounting error would be to consider the photon, as if it mattered. it doesn't. Mirrors do not exist at the particle level, it is a wave/macroscopic conceptGem fr (talk) 07:30, 12 July 2017 (UTC)

There is some truth to what you say, in the sense that AFAIK a line of particles in the mirror bounce photons off them in random directions, and constructive interference from a large number of particles is needed to convert this from being more like a diffraction grating to being something that sends all the light out in one single direction. That said, a single photon with a poorly known position might reflect off many particles and interfere with itself in this same way, much like a double-slit experiment. Alternatively, I assume that a QM analysis of the mirror with many photons whose localization is well known would find that the interference between them ends up working out that the individual photon has a high chance of leaving on the route it is supposed to leave from a mirror, with its angular momentum still being meaningful. Wnt (talk) 12:25, 12 July 2017 (UTC)

OK, I'm convinced the math works out for linear momentum of a photon bounce. The way it works out is that the mirror yields more when there is low mass, causing more of a change in the photon energy. But ... how does the mirror yield more because it has a low moment of inertia, when a quantum of angular momentum is transferred? And there are still a bunch of leftover questions above... Wnt (talk)

Suppose we have a mirror that reverses the z component (along the direction of propagation) of the angular momentum of the photon. It has to absorb z angular momentum amounting to $2\hbar$.

If the mirror wasn't rotating before the reflection, it is going to begin to rotate. Because the photon is circularly polarized, that changes the frequency of the photon and in this way reduces the energy of the photon.

If we have another mirror, just like the one before except with a lower moment of inertia, the z angular momentum transferred will be the same, but the resulting rotation rate of the mirror will be higher, and thus the frequency shift of the photon will be higher.

Neglecting the transfer of linear momentum, and assuming that the inertial tensor is diagonal in our coordinate system (the z direction is the direction of propagation of the photon), the math looks like this:

Angular momentum of incoming photon:

$L_{z\gamma i}=\hbar$

Angular momentum of outgoing (reflected) photon:

$L_{z\gamma o}=-\hbar$

Angular momentum of the mirror after the reflection:

$L_{zM}=2\hbar$

Angular velocity of the mirror after the reflection:

$\Omega _{zM}={\frac {L_{zM}}{I}}={\frac {2\hbar }{I}}$

The relevant energy (here I use $\omega$ meaning $2\pi$ times the frequency of the photon; $\omega '$ is the frequency after the reflection) before and after reflection:

$E=\hbar \omega =\hbar \omega '+{\frac {L_{zM}^{2}}{2I}}$

From this, we get

$\omega '=\omega -{\frac {2\hbar }{I}}$

Icek~enwiki (talk) 13:46, 12 July 2017 (UTC)

@Icek~enwiki: Your math is impeccable, and yet ... it is saying that the photon, in the act of bouncing "instantaneously" off a mirror, is able to measure the size of the mirror, which could be meters, before recoiling from the surface. Wnt (talk) 22:33, 12 July 2017 (UTC)

@Wnt: You are right; consider the macroscopic analog of a plate that reflects a ball in an elastic collision. There is a spring on the backside of the plate connecting it to a heavy mass. If the natural oscillation time of the system (the spring, the plate, and the heavy mass) is much longer than the interaction time of the ball bouncing, then the speed of the reflected ball is to a good approximation the same as after the reflection from a free plate. The "mirror" then gains some vibrational energy.

In the case of a photon being reflected by a mirror, the angle between the path of the reflected photon and the surface normal being the same as between the incoming photon and the surface normal, it is always a coherent interaction between the photon and many atoms/electrons of the mirror, but of course not necessarily all of them. I guess that the effective mass from which the photon bounces off is in general the higher the larger the coherence length of the photon is. And the formula resulting from the simple elastic collision model is a good approximation if the whole mirror interacts significantly with the photon, like in the case of the reflection from an uncoated glass plate. Icek~enwiki (talk) 09:49, 13 July 2017 (UTC)

That makes sense, though I doubt the photon can couple with all or most of the mirror. If it turns a smaller part, the energy it gives up is greater, and as you say, that energy ends up in little vibrational/rotational whirlwinds that move about in the atomic lattice of the mirror. But some of the rest of the mirror can make itself known to some degree in the rigidity of the bonds where the torque is actually applied. I suppose the process should be quite inefficient for light with a wavelength much smaller than that of the entire mirror. Wnt (talk) 15:34, 13 July 2017 (UTC)

@Icek~enwiki: your math says nothing about this being real or not. Let's take a macroscopic equivalent: if you launch against some rubber wall a cylinder, its axe being parallele to its motion, and rotating clockwise from your point of view: it will bounce and come back to you, but still rotating clockwise relative to you. Write all the math you want stating that it comes back rotating counterclockwise, it won't make it happen. Gem fr (talk) 02:10, 13 July 2017 (UTC)

It is straightforward to construct a mirror that reflects circular polarized microwave radiation with the same handedness (opposite absolute rotation) as that incident. A planar screen consisting of fine parallel wires, each several wavelengths long and spaced about a tenth of a wavelength apart will reflect microwaves plane-polarized parallel to the wires while being transparent to microwaves plane-polarized perpendicular to the wires. The mirror consists of such a screen placed one quarter of a wavelength infront of, either a similar screen rotated 90 degrees with respect to the first, or else in front of a flat metal sheet. A circularly polarized wave striking the front screen is split into two plane-polarized components. The component polarized parallel to the wires is reflected by the front screen, the other component is reflected by the rear screen. The component reflected by the rear screen has a round-trip which is half a wavelength longer than the other, so it suffers a relative 180 degree phase-shift - or equivalently (an additional) reversal of polarity. The superposition of the two reflected waves constitute a circularly polarized wave of the desired handedness.

Because the mirror depends on the λ/4 spacing of the screens, it will only work at one frequency, but it is not beyond the wit of microwave engineers to construct a relatively broadband mirror.

Classically the mirror feels no torque within the incident beam, but would feel a transverse force at the periphery of the beam (because there is some longitudinal magnetic field that there, which results in a Lorentz force on the current-carrying wires). However, the total torque depends on both the length of the periphery of and on the lever-arm (the radius of the beam), and so is proportional to beam the cross-sectional area. --catslash (talk) 23:01, 12 July 2017 (UTC)

nice. i won't call it "straightforward", but nice

however, neither the first nor the second screens would be applied any torque, nor any lateral force, just a push rearward as per radiation pressure, and so would the pair of them as a whole. So i don't see where the torque is supposed to come from is this clever arrangement.

Gem fr (talk) 02:10, 13 July 2017 (UTC)

A torque comes from a force on the charges that accumulate at the ends of the wires due to being pushed there by the component of parallel polarization. For a given half-oscillation, positive charge accumulates at one end and negative charge at the other. Now the electric field of the component of perpendicular polarization acts on those charges.

I thought about that, but for the other half-oscilation, it works just the opposite (negative where positive were, and vice versa), for a null average effect, doesn't it?Gem fr (talk) 10:33, 13 July 2017 (UTC)

For the other half-oscillation, the locations of negative and positive charges are swapped, but the electric field of the component of perpendicular polarization now points the other way, so the torque is still the same. Icek~enwiki (talk) 12:01, 15 July 2017 (UTC)

By the way, conserved quantities like energy and angular momentum etc are (at least as far as all successful theories go) conserved exactly, not just in an approximate or macroscopic way as you seemed to suggest earlier (it reminds of an anecdote from about 1930 when the energy spectra of beta decay seemed puzzling and some famous physicist (Niels Bohr, in a different version Ernest Rutherford) suggested that the conservation laws aren't exact, but Wolfgang Pauli objected and suggested an invisible particle (later called neutrino) instead). Icek~enwiki (talk) 09:49, 13 July 2017 (UTC)

Of course conserved quantities are conserved quantities, i didn't suggested otherwise. My point was exactly of the same nature of your anecdote: there has to be something forgot, as the suggested device (a pair of mirror and a photon bouncing in between) would generate rotational movement out of ... where?Gem fr (talk) 10:33, 13 July 2017 (UTC)

@Gem fr: That's the easy part - the energy comes out of the photons, which is why they get redder with every bounce. This is not conceptually much different than running a motor on the Space Station and watching the one end start spinning one way while the back end spins the other, except here the photons are our battery and drive shaft. Wnt (talk) 17:01, 13 July 2017 (UTC)

@Wnt:indeed. But the energy is the easy part, too (it is still an issue, as this means a "rotating" photon would redshift more than a "non rotating" similar one that just bounce --radiation pressure--. WTH?). The main issue is the "rotation" of the photon, which basically switch side in the scenario: why, how? I surely can imagine a scenario where the whole of a particle's energy is turned into rotationnal energy of a macroscopic object, but it seems to me that entropy doesn't agree (since you require a hell of negentropy to coordinates the mirror molecules into a nice rotation).Gem fr (talk) 06:57, 14 July 2017 (UTC)

It's interesting, isn't it? I'm not sure if the angular momentum of photons seems neglected out of disinterest or if the experiments are just too hard -- I'm still hoping someone can tell me more about state of the art experiments actually being done. First, my assumption is that all photons are "rotating", and a plane polarized photon is merely one in a quantum superposition of states. So there's no distinction between them except which way they're rotating. We have discussed what happens one way -- the other is interesting though, since if a photon decreases the object's rotational energy, it should actually be blueshifted by this to some extent. (Bear in mind though that as described above there is potentially a lot of inefficiency by getting small portions of a mirror to rotate independently as a vibrational energy) Now bouncing a beam of racemic photons at a rotating object and getting back two different frequencies of reflection with opposite circular polarization would be a very persuasive experiment, and one that seems like it might be doable given how awesome the spectroscopy studies of exoplanets etc. seem to be. I even wonder ... can a rotating object spontaneously create a pair of circularly polarized, very long wavelength photons de novo to carry away its energy and momentum of rotation? It seems like in the sort of ridiculously far future scenarios people look at that something like that could spin down the surviving supermassive black holes and the like. Wnt (talk) 13:30, 14 July 2017 (UTC)

I cannot point you to relevant publications (at least not at the moment), but an experiment to detect the frequency change of circularly polarized light reflected from a rotating mirror should be easy to do even with relatively small rotation rates; there are laser spectroscopy systems achieving a frequency resolution of something like 1 Hz or better.

By the way, if you bounce linearly polarized light off a rotating mirror, you get what you describe with racemic photons, just that the phase shift between the 2 circular polarizations is known and constant at the mirror. The reflected light can be thought of as a clockwise polarized beam with a frequency ω+Ω and a counterclockwise polarized beam with a frequency ω-Ω. Assuming that the mirror's angular frequency Ω is small compared to ω, it is still approximately linearly polarized light of circular frequency ω; but because you have a beat between the 2 circular polarization modes, the plane of polarization slowly changes as you move away from the mirror.

If you can make your mirror rotate 10⁷ times per second, the plane of polarization should make a full 360 degree turn per 30 meters (speed of light divided by frequency). I.e. the polarization would be perpendicular to the original one after 7.5 meters. If the mirror is only a few micrometers in diameter, the mechanical stress is low enough; there should be a way of magnetically levitating it and also spinning it up magnetically.

Icek~enwiki (talk) 12:01, 15 July 2017 (UTC)

@Icek~enwiki: I really like this "macroscopic circular polarization" where the apparent plane polarization of two oppositely circularly polarized light beams with different frequencies slowly changes. It reminds me of Rydberg atoms in a vague way. It seems like both very small differences between the light, accomplished by rotating a mirror on the way out, and much larger differences, ?perhaps? accomplished by splitting a pulse laser and bouncing the halves off a mirror that is synchronized to recede or approach the pulse, and circularly polarizing them before recombining them?, might be very useful.

Picture a spacecraft orbiting, oh, Eris. It wants to know exactly where it is now, but bouncing a round trip signal to Earth would take a while. So it has a pair of very sensitive detectors for a small frequency band, one behind a filter linearly polarized one way and one behind a filter polarized the other. On Earth, you have a big laser very precisely pointed, and send out a distinctive timing pulse, then a bunch of other pulses with this macroscopic circular polarization, same base frequency (or more than one if the satellite has more than one detector) but with different beat frequencies for each pulse in the series. One filter or the other on the spacecraft will light up most depending on how many meters it is away, and the succession of pulses each can provide different beat wavelengths, so even if it has interrupted viewing. By using these sort of like digits (more like a Mayan calendar, maybe) the satellite can quickly run through them all and determine precisely how far it is to the base station - possibly down to the meter, provided the Earth-based beat production can be made accurate enough. Set up three base stations (each with its own allotted places in the series after the timing pulse; they'd have to time their broadcasts to match up when they meet at the spacecraft's approximate position) and now it can triangulate. I'm daydreaming the thing could be so accurate that it could measure the stretching of space by the gravity wells of passing asteroids. You might need to program a good GR model of the Solar System just to get maximal accuracy out of its position finding. (probably not, but I can dream!) Does anyone do anything like this? Wnt (talk) 17:08, 15 July 2017 (UTC)

That's a nice positioning system, though I don't think you could do it with pulsed lasers unless you do something fancy with mode-locking. That's because a short pulse means a broad frequency spectrum. Then you have different frequencies rotating the plane of polarization. E.g. if you have a frequency separation between the 2 circular polarizations of 10 MHz, and a spectral width of 1 Hz, then the different frequency components of each of the 2 circular polarization will mess up the helical pattern after 10 million rotations or earlier - and as one rotation is 30 meters, that would be 300,000 km, less than the distance to the Moon. For Sedna you need the a pulse last a day or so and be very stable in its frequency. And a separate frequency for each "digit".

The number of such digits needed depends of course on how accurate you can measure the angle of polarization. There is a limit to the accuracy in terms of photon number (for a single photon, you can only see if it does or doesn't pass through a polarization filter and then conclude that its polarization was likelier to have been closer to the horizontal or vertical direction of the filter).

If there are N photons per digit (given a certain distance from the transmitters, a certain collecting area, and a certain time limit for determining the position), the uncertainty of the polarization angle should be proportional (correct me if I am wrong here) to $N^{-{\frac {1}{2}}}$, at least if N isn't too small. So, if you have a constant factor of polarization rotation lengths between adjacent digits, and you consider leaving out every other digits by increasing the power of the transmitter in order to increase the number of photons, then you'd now need N² photons instead of N photons. That makes it energetically favorable to rather have many digits with lower power than few digits with high power. Icek~enwiki (talk) 19:43, 15 July 2017 (UTC)

Phooey! I knew there'd be something ... should have remembered that a precise frequency implies something happening over and over enough to have one. And making a day-long pulse is no solution because the probe keeps moving, Eris keeps moving for that matter - the hope was to get an "instantaneous" distance reading. Unless there's some fancy math thing to do to figure out the polarization of interfering copies of a pulse despite the lack of a clear wavelength, I think I'm sunk here. Wnt (talk) 20:17, 15 July 2017 (UTC)

(sorry for editing the archives...) If the photon density is high enough, and the frequency is defined well enough, there should be no problem in measuring the polarization within a far shorter timespan than the length of the laser pulse - basically you just still just measure the orientation of an oscillating electromagnetic field. Icek~enwiki (talk) 14:45, 16 July 2017 (UTC)

Energy consumption in different contexts

You walk for an hour on a certain piece of land in temperatures that require little thermoregulation by your body. Some days later, when you're in the same physical condition, you perform an identical walk: same route, same duration, etc., but it's significantly warmer, and you're sweating a good deal. Are you likely to consume the same amount of energy, or does the extra thermoregulation require additional energy?

Related question: you walk for a much longer period of time on the cooler day, doing that same route ten times without significant interruption. Of course, by the end, you're more tired and haven't stopped for more than bathroom breaks and drinks. Since you're more exhausted, does it require a greater expenditure of energy to move your body the same distance, especially if you push yourself to walk the same speed? It feels much more difficult, but difficulty is fatigue-related; it doesn't have a 1:1 correlation with energy expenditure. Nyttend backup (talk) 13:54, 11 July 2017 (UTC)

human power is about ~100 W ;

osmotic power, the one used in drinking and sweating, is about ~1 kWh/m³, that is, ~1 W if you sweat 1 liter in a hour.

so thermoregulation does require some energy, but not an amount significant enough to be felt.

What you will feel, however, is muscle fatigue, lack of water if you run out of water, overheat if thermoregulation is overrun, and other health risks

Gem fr (talk) 14:44, 11 July 2017 (UTC)

However, generating heat burns lots of energy, with 10 minutes of shivering burning as much as an hour's exercise: [1]. Also note that there's the myth that high temps burn lots of calories, based on the immediate weight loss due to dehydration. But that's absolutely the wrong way to lose weight. StuRat (talk) 15:40, 11 July 2017 (UTC)

Feynman Lectures. Exercises PDF. Exercise 8-4 JPG. Lecture 8 & 9

. .

Here is my solution png. But I do not fully understand the resolution into components. Feynman speaks about this after eq. 8.11. At a moment of Lecture 8 we do not know about vectors. According to Feynman we can get directly only the components for $ds$. I have shown in solution that the velocities obey an analogous resolution with cosines of the same angle. But where is the guarantee that accelerations do obey the resolution? We should prove the possibility of acceleration resolution and the formula $dv_{x}=a_{x}dt$. How to do that? Username160611000000 (talk) 15:10, 11 July 2017 (UTC)

hell, you make it so complicated... it is not, since you have no horizontal acceleration ($dv_{x}=a_{x}dt=0$ ; BTW this is just definition of acceleration, no need to "prove" it), and constant (g) vertical acceleration.

Your answer is obviously not stupid since you find 0 distance traveled for both horizontal and vertical shot, and distance traveled proportional to V² and inverse of g, as dimensional analysis suggest.

Indeed it is correct (but you knew it)

Gem fr (talk) 15:57, 11 July 2017 (UTC)

In the ex. 8-4 we need only components for velocity (so there is no problem yet, but the problem will appear in Lect. 9 with components for acceleration). In the ex. 8-4 the angle Θ is given. Between what is this angle measured ? I think between path (i.e. ds) and X-axis, so we should prove that it is same angle between v and its components.

I have a general question , not directly connected with ex. 8.4: If we have acceleration direction and magnitude and XYZ rectangular coordinate system, then we can project magnitude on axes. Do these 3 projections coincide with the acceleration 3 components received from table of x, y positions? Username160611000000 (talk) 16:48, 11 July 2017 (UTC)

$\definecolor {myorange}{rgb}{0,0.52,0.37}\color {myorange}dv_{x}=a_{x}dt=0$ ; BTW this is just definition of acceleration, no need to "prove" it -- a = dv/dt is the definition, but Feynman have said nothing about components. E.g. we have a table with columns t, x, y and infinite number of rows for each particle position. Then from this table we know dx and dy, therefore we can find v_x, v_y (organize 2 more columns). Then we can find dv_x, dv_y and therefore a_x, a_y. We know only $ds={\sqrt {dx^{2}+dy^{2}}}$ from assumption that trajectory can't be very curved over small $ds$ and $v={\sqrt {v_{x}^{2}+v_{y}^{2}}}$ as proved, but we know nothing about dv, dv_x, dv_y. How can we calculate resultant $a$ and angle $\angle (a,a_{x})$ and check these values with directly measured ones by the accelerometer? If such formula as ${\text{magnitude}}={\sqrt {{\text{x-component}}^{2}+{\text{y-component}}^{2}}}$ is axiom, then why does Feynman prove it for $v$?Username160611000000 (talk) 20:34, 11 July 2017 (UTC)

Again, you turn very simple things into horrible. dv = dv_x + dv_y, that's a definition, too. Not a definition of v, but of a coordinate system. This answers your "general question" above.

there is no such thing as an assumption that "trajectory can't be very curved over small $ds$"; physicists are simple-looking, pragmatic, persons, that first try simple things and are happy with them as long as they works. Here, the idea is just to try to find a solution in the realm of the simplest, that is, "not very curved". If it works (and it does, it eventually appears), why look for complicated solution in the realm of erratic, brownian, fractal bang-around kind of motion, where the word "trajectory" lose relevance? If, however, it appeared that so such solution could be found, or if it didn't match the observations, well, some other, more complucated solution would be looked for. happily this is not needed here.

An accelerometer is not a mean to measure angle. It will just give you g, not the current angle of the trajectory relative to the flat ground. This you'll have to calculate with usual trigonometry.

Gem fr (talk) 22:35, 11 July 2017 (UTC)

dv = dv_x + dv_y, that's a definition, too. -- it is a vector form, which we officially do not know. If v = (v_x² + v_y²)^0.5 is definition, then why does Feynman prove it by formulae 8.11 - 8.15?Username160611000000 (talk) 03:56, 12 July 2017 (UTC)

facepalm. Feynman is teaching, for god sake. Teaching implies saying things so that even the dumbest get it. That's why. Gem fr (talk) 07:09, 12 July 2017 (UTC)

First, please use dry scientific voice. Second, such answer does not agree with Feynman's approach. Feynman gives some mathematics subjects but not in a style of writing down redundant data.

The Eq. (9.4) and the same in Lect.8 are also redundant information for poor students, are not they? Username160611000000 (talk) 09:36, 12 July 2017 (UTC)

IMHO, saying that v_x is the horizontal component of v is a definition. Coming up with a Pythagorean Theorem solution for the relationship of its magnitude with v is a calculation. Wnt (talk) 12:44, 12 July 2017 (UTC)

It seems in Ch. 8-5 we do not know at first that v²=v_x²+v_y², because all these values are calculated independently (it could be that v²≠v_x²+v_y² as it is for Δs, Δx and Δy). Username160611000000 (talk) 15:14, 12 July 2017 (UTC)

In the limit, as Δs, Δx and Δy → 0, then Δs² → Δx² + Δy² and so (Δs/Δt)² → (Δx/Δt)² + (Δy/Δt)². But the limit of Δs/Δt is v, the limit of Δx/Δt is v_x and the limit of Δy/Δt is v_y (by definition). So v² = v_x² + v_y². Gandalf61 (talk) 15:34, 12 July 2017 (UTC)

@Gandalf61: I think we are finally on the right track. So Feynman proves v²=v_x²+v_y² not because he is teaching, but because it is necessary step. Then how can we prove the analogous for acceleration? For trajectory we have used the graph and ds² = dx² + dy². Is it possible to plot v_x - v_y graph? E.g. if X-Y graph is like this then the v_x - v_y graph will be png and we can use the same approach, can't we?Username160611000000 (talk) 16:24, 12 July 2017 (UTC)

copyright

Point of order: the link above is actually to "The Swiss Bay", which apparently is a The Pirate Bay knockoff of some sort (I recognize the ship). I don't think there's any need to resort to this when the text is readily available at Caltech, and those links might be more stable. In any case, there is no earthly reason not to simply reprint Question 8-4 here in its entirety for the reader's convenience: "8-4. A projectile is fired over level terrain at initial speed V, at an angle $\theta$ with the horizontal. (Neglect air resistance) Find the maximum height attained and the range." Wnt (talk) 12:35, 12 July 2017 (UTC) Thank you. "The Pirate Bay" was a torrent-files hash base, but not a file - server. If The Exercises are available on Caltech site, please write the link. I'm still in suspense about copyright, Medeis said that typing the exercise was bad. I think if I give a link to a public resource, it is not copyright violation. Username160611000000 (talk) 14:03, 12 July 2017 (UTC) I'm not going to argue Wikipolicy here; it's off topic, though I'll note some people are fanatic about it. But from a practical standpoint, you're generating a database of calculations that could lose the associated questions if the local shamans put out a fatwa against Swiss Bay. The individual exercises are short, straightforward questions and you can use Fair Use to simply write them here as I did, and it will make your questions much more usable long term. Wnt (talk) 14:10, 12 July 2017 (UTC)

I think we might solve the height by determining the kinetic energy at the beginning, subtracting the kinetic energy of the horizontally moving round at its highest point, then determining where that amount of potential energy puts it. Or we might divide components entirely, then solve them separately. You seem to be pushing for a parametric approach, which is possible but more complicated. So if we take that all the way:

theta(t) is the angle of the round; it starts at a known positive value at t=0, has the same negative value at t=T when the round lands, and has a value of 0 at t=T/2.

v(t) is the speed (scalar) of the round; it starts at a known positive value at t=0 and has the same value at t=T.

We expect dv/dt = -g sin theta(t), i.e. gravity pulls back to the extent the round is going up.

Dimensional considerations make me think dtheta/dt is something like -g cos theta(t) v(t)^-2 (we want inverse seconds). Unfortunately I actually have to think about this one. ;) If we can get through this, and the diffy-q's that follow, then we can have a true parametric solution. Wnt (talk) 14:07, 12 July 2017 (UTC)

... OK... after a short delta t with perpendicular gravity g cos theta(t), the object will be moving at its original velocity, but also downward by g delta t cos theta(t); this divided by v is the sine of delta theta(t), which for small values is delta theta(t). So delta theta(t)/delta t = g cos theta(t)/ v(t), which has the appropriate dimensions at least (i.e. none).

Now I have to dust off what I know about solving these, which is in a sorry state of decay. I'm thinking the first equation is a nonlinear separable differential equation that can be rewritten dv = -g sin theta(t) dt and integrated... erm, not so easily as I thought. Hrm. Well, if I put the sines and cosines together I know that (dv/dt)^2 + (v dtheta/dt)^2 = g^2. Which means ... well, what it means is I ought to go relearn differential equations. Meanwhile, this exercise demonstrates that the only way I can see to do this "without vectors" is to 'intuitively' make so many vector assumptions (like perpendicular velocity and splitting gravity into components) that you're just fooling yourself. The interesting part in all this is I keep feeling that there is some angular momentum like property about the moving shell, despite its mass and orientation not being an issue, as evidenced in the second term in that last equation above. That equation is sort of like a KE + PE calculation, with the mass cancelled out, but instead of PE we're looking at how much the trajectory turns. I'm wondering if there's some physics concept I've overlooked here. Wnt (talk) 12:18, 15 July 2017 (UTC)

Energy loss, planes and ships

How much energy planes and ships lose by generating a vortex behind them? — Preceding unsigned comment added by Tyief (talk • contribs) 17:24, 11 July 2017 (UTC)

See lift-induced drag. It isn't a simple "how much" question. There are a lot of variables. 209.149.113.5 (talk) 17:49, 11 July 2017 (UTC)

see also

Drag coefficient

finesse: if it is 12, for instance, the plane travelling a distance L lose the same energy as it gains by falling 1/12 L, that is, mgL/12 (where m is its mass and g the acceleration of gravity). The higher the finesse, the lower energy loss.

Reynolds number

Gem fr (talk) 23:06, 11 July 2017 (UTC)

Boats, ships and aircraft all leave wakes consisting of vortices and turbulence which represent kinetic energy of the water and air. That kinetic energy is all eventually transformed to heat as the kinetic energy subsides to zero. Dolphin (t) 21:29, 12 July 2017 (UTC)

But if another vehicle is riding in the wake, or draft, it can reclaim some of that energy. StuRat (talk) 21:42, 12 July 2017 (UTC)

Yes, and not only other vehicles. If a ship's wake washes sand or debris higher up a beach it represents kinetic energy transformed to potential energy. Vortices and turbulent wakes can also transform kinetic energy to sound energy. Dolphin (t) 07:59, 13 July 2017 (UTC)

Relative calorie burn in air and water

Hello. How many calories would a person standing in water burn per minute, compared with the same person standing on dry land at the same air temperature? Twice as many? Thrice as many? Thanks.--90.69.12.160 (talk) 18:00, 11 July 2017 (UTC)

Assuming that the air and water are the exact same temperature and there is absolutely no air or water motion (impossible), then the only difference would be weight change due to buoyancy. You weigh less in water (mass is the same, weight changes) because you float a little. Therefore, your muscles have to hold up less weight. You should burn less calories, but not much less. How much less is dependent on the person. Some people are more buoyant than others and body mass is highly variable. 209.149.113.5 (talk) 18:07, 11 July 2017 (UTC)

Assuming that the temperature of the environment (water or air) is less than body temperature, the body loses heat. The loss of heat is larger in water, as the heat transfer coefficient from skin to water is larger than from skin to air, and the thermal conductivity of water is larger. The heat has to be produced by the body, so more energy is converted to heat in the case of water (the picture is made more complex by many factors, e.g. the constriction of blood vessels close to the skin as the skin is colder in the case of water). Icek~enwiki (talk) 20:32, 11 July 2017 (UTC)

you need to consider not just temperature, but also speed of the air/water, and Thermoregulation in humans.

I remember (no ref at hand, sorry) that confort temperature (that is, when thermoregulation is mostly in idle state, the loss being more or less equal to basal production, around 100W) is around 20-22°C in the air, but 26-28°C in water

also, you can easily stand -40°C in the air (with no wind) for as long as enough food is available , while a mere 2°C in water will kill you in few minutes (hypothermia).

Gem fr (talk) 23:23, 11 July 2017 (UTC)

I think that being correctly dressed would also assist in surviving at -40C. I don't think -40 in a t-shirt and jeans is going to do it. I would like to see a source for the idea that you could survive at -40 provided you had food and were correctly dressed. CambridgeBayWeather, Uqaqtuq (talk), Sunasuttuq 00:04, 12 July 2017 (UTC)

I seldom if ever try to argue from experience/authority. But User:CambridgeBayWeather I think hails from Cambridge Bay, and so I respect their thoughts on human survival in cold weather and climates ;) SemanticMantis (talk) 00:38, 12 July 2017 (UTC)

Turns out this is hard to search for. I suspect because nobody has tried it. I just realised that without water you aren't going to live too long. Melting snow in your mouth will just cause hypothermia to set in quicker. This also means that you are putting frozen food in your mouth and that's going to lower your body temperature. By the way it is possible to survive at least 18 minutes in −1.7 °C (28.9 °F) water, see Lewis Pugh#North Pole. I uploaded a couple of pictures of people swimming in the Arctic Ocean, File:Swimming in the Arctic Ocean 01.JPG and File:Swimming in the Arctic Ocean 02.JPG. CambridgeBayWeather, Uqaqtuq (talk), Sunasuttuq 04:22, 12 July 2017 (UTC)

I remember a documentary about these things, picturing a man swimming in cold water (without swimsuit). But the man was trained, and the usual "don't do this, you may kill yourself" message was clear, as the body of the man had adapted in very interresting ways (hence the point of the documentary). He could swim and hence generate heat in conditions the normal person just dies. Wasn't this one [2]; may be this one [3] (i didn't check)

Henri Guillaumet was not trained, but he nonetheless survived 1 week in harsh climate above 3000m

Gem fr (talk) 10:42, 12 July 2017 (UTC)