# Wikipedia:Reference desk/Archives/Science/2011 November 10

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# November 10

## Infrared-to-optical wavelength coating

Hi. In incandescent lightbulbs, most energy is wasted as heat. Would it be thermodynamically correct and/or feasible to apply an optical coating that absorbs the heat and re-emits it as visible light? Or is there a reason why coatings cannot increase the wavelength, or would heat build up within the lightbulb? Thanks. ~AH1 (discuss!) 00:34, 10 November 2011 (UTC)

A possible method could be through a photoluminescent coating which absorbs infrared and emits in the visible range through the process of double absorption and single emission. Plasmic Physics (talk) 01:40, 10 November 2011 (UTC)
The higher the intensity of infrared radiation, the higher the intensity of the visible emited radiation. The exiton only exists for a short amount of time before colapsing, in that short amount of time, it must absorb another photon before it can emit in at a higheer energy. Plasmic Physics (talk) 01:48, 10 November 2011 (UTC)
To be more acurate, I should rather say an increased visible to infrared ratio, than intensity of visible radiation. Plasmic Physics (talk) 02:27, 10 November 2011 (UTC)
• The waste heat from an incandescent light bulb is of a long wavelength longer than about 700 nanometers, so your coating would need to absorb long wavelength photons and emit short wavelength (visible) photons. This would be a decrease in wavelength, to between 400 and 700 nanometers, and a corresponding increase in frequency. I don't know the insurmountable obstacle, but I suspect that if that coating existed, they would have started using it over 120 years ago. Find it and patent it and you have made your fortune. Edison (talk) 05:15, 10 November 2011 (UTC)
It is difficult to go from longer wavelengths to shorter ones. To do that, your coating has to absorb multiple photons and merge them together to emit single photons with shorter wavelength (higher frequency/more energy). There are such materials, but they tend not to be very efficient unless the light intensity is very high. Most forms of photoluminescence go the other way: they "split" single photons into multiple photons with less energy (longer wavelength).--Srleffler (talk) 05:56, 10 November 2011 (UTC)
Certain types of nonlinear optics - that is, special types of crystal lattice materials, exhibit frequency doubling at optical wavelengths. This is the operating principle behind certain types of lasers. Thermodynamics dictates, however, that there must be an additional energy input (the same law dictates that heat can't flow from cold objects to hot objects, and amplifiers can't amplify without power, and perpetual motion is infeasible). Here's an article from the Encyclopedia of Laser Physics, frequency doubling, also available in textbook form at a considerably greater price. I am not aware of any "coating" or other material that can passively double frequency, and I sort of have a hunch that it is physically impossible. Second-harmonic generation discusses some more applications. Nimur (talk) 03:06, 13 November 2011 (UTC)

## Sodium electrolysis

Is is possible to electrolyse an anhydrous solution of a sodium compound so that reduced sodium forms? Plasmic Physics (talk) 01:25, 10 November 2011 (UTC)

Sure. Zap molten sodium chloride with a bunch of electricity, vent off the chlorine gas, and you should get sodium metal. This is almost exactly what happens in the process that extracts metallic aluminum from bauxite. See Hall–Héroult process. This method should work for any ionic compound, it's pretty expensive and brutal so it only needs to be used in cases like extraction of aluminum, but it should work for extracting sodium from sodium chloride. --Jayron32 04:52, 10 November 2011 (UTC)
And more research leads to more details. I thought I had this knocking around in my brain. See Downs cell and Castner process for two electrolytic processes used to produce metallic sodium. --Jayron32 05:03, 10 November 2011 (UTC)

Note, I'm inquiring about solutions, not melts. Plasmic Physics (talk) 05:35, 10 November 2011 (UTC)

As in, a room temperature solution? Probably not. Any room temperature liquid sufficient to dissolve any ionic sodium compound would also be reactive enough to sodium metal to prevent its formation, akin to water. There are some solutions which utilize very high temperatures, the Hall–Héroult process cited above uses molten cryolite flux; but the flux acts as a purifying agent more than a solution; molten (or "fused" as is common in this context) ionic compounds. If you are trying to get solid sodium metal, you're going to need to do it using molten sodium compounds of some sort. See Ionic liquid for some background; the reason why molten salts are often used in eletrolytic smelting is that they are electrolytes; that is they are composed of ions much as if they were in the solution phase, but without any pesky solvent molecules around to screw stuff up. --Jayron32 06:03, 10 November 2011 (UTC)

OK. I'm trying to refine my technique for isolating sodium. My method is based on electrolysing molten sodanol hydrate in a ceramic crucible with two graphite electrodes and a variable power supply. I thought that perhaps, I can stop the liquid reduced sodium from being rapidly oxidised by burrying the electrolyte with fine silica. Once the reaction is complete, how do I seperate the sodium and cover it in parafin oil in one fowl swoop? Will covering it in sand even work? Plasmic Physics (talk) 07:11, 10 November 2011 (UTC)

Sand will let oxygen and nitrogen through which would react. Since no gas is made in this reaction it would suck in the air through the porous sand. You had better have the liquid sodium under an argon (or inert) atmosphere. Are you seriously attempting this? Or just doing a paper study? Graeme Bartlett (talk) 11:55, 10 November 2011 (UTC)

I did before, and was successful, but most of the sodium oxidised and was lost. What if I use parafin wax, the heat should melt it to completely exclude oxygen. An have another parafin oil denser than sodium but lighter than than the wax, to seperate the sodium from the reation. Plasmic Physics (talk) 20:07, 10 November 2011 (UTC)

Sodium hydroxide can also be used:

"In the laboratory, with careful control of conditions, sodium metal can be isolated from the electrolysis of the molten monohydrate in a low temperature version of the Castner process, according to the following reaction:

4 NaOH·H2O(l) → 4 Na(l) + O2(g) + 6 H2O(g)

The monohydrate does not need to be heated in order to melt, as the process produces enough heat due to ohmic heating. However, it must be initiated with a small quantity of liquid water to create an electrically conductive electrolyte. As the system's temperature increases, the monohydrate will start to melt at about 65 °C as stated above. Only when the temperature reaches about 100 °C can sodium be isolated. Below this temperature, the water produced will react with the sodium, above this point, any water formed will be driven off in the vapour phase, creating an essentially anhydrous reaction. While this process has some advantages over other electrolytic processes, it is not preferred by most chemists for several reasons: a marginal quantity of sodium produced boils at the electrode interface, the vapour thus given off consists primarily of fumed sodium oxide, which tends to settle on any surface in close proximity with corrosive consequences."

Count Iblis (talk) 20:26, 10 November 2011 (UTC)

Yup, that's the one. Maybe, should swap out my graphite anode for an iron one. Sodium interclation is disintegrating it at a rate of a centimetre an hour, and those things don't come cheap. Plasmic Physics (talk) 20:43, 10 November 2011 (UTC)

In a past attempt, one of my anodes violently expanded with force for some reason. Plasmic Physics (talk) 23:08, 10 November 2011 (UTC)

So, does anyone have an idea about the parafin? Plasmic Physics (talk) 08:25, 11 November 2011 (UTC)

The paraffin should melt and float on top of the sodium, which is also molten at 100°. But you do have a serious risk of setting it on fire. Graeme Bartlett (talk) 11:25, 11 November 2011 (UTC)

OK, what do recommend then? Plasmic Physics (talk) 11:55, 11 November 2011 (UTC)

I don't have argon at my disposal. Plasmic Physics (talk) 12:05, 11 November 2011 (UTC)

Then what about nitrogen? 67.169.177.176 (talk) 22:44, 11 November 2011 (UTC)

No, no gas. In any case nitrogen reacts as well to produce sodium nitride. Plasmic Physics (talk) 03:53, 12 November 2011 (UTC)

You can buy an argon gas cylinder for TIG welding, perhaps from a hardware shop, a car parts shop, and definately from a gas supplier. Helium will do too, but will cost you more. Graeme Bartlett (talk) 21:32, 13 November 2011 (UTC)
According to Sodium nitride, Sodium nitride is extremely unstable.RJFJR (talk) 23:18, 13 November 2011 (UTC)

The thing is, I don't want the cost of anything that I'm using outweigh the value of the sodium that I'm producing. Plasmic Physics (talk) 00:29, 14 November 2011 (UTC)

## Proof of Conservation of Momentum

Is there any way to prove the Law of Conservation of Momentum without using Lagrangian Mechanics (such as in Noether's Theorem)? Preferably using something more accesable such as Newtonian Mechanics, or perhaps an intuitive argument based on the geometry of space? — Trevor K. — 05:06, 10 November 2011 (UTC)

Momentum#Conservation_of_linear_momentum discusses the connection between conservation of momentum and Newtonian physics. --Jayron32 05:17, 10 November 2011 (UTC)
Conservation of momentum is simply an extension of the conservation of energy. This doesn't realy answer your question, but it's useful to know. Plasmic Physics (talk) 07:23, 10 November 2011 (UTC)
I'm not sure that deriving from conservation of energy is helpful. In Newtonian mechanics, conservation of momentum is a much more fundamental principle, being derived, at elementary level, directly from Newton's laws of motion. Dbfirs 07:31, 10 November 2011 (UTC)
I've seen that notion before, however same question still stands for Newton's laws. Is it possible to prove ${\displaystyle {\frac {d{\vec {p}}_{\text{a}}}{dt}}=-{\frac {d{\vec {p}}_{\text{r}}}{dt}}}$ by a geometrical basis (a=action, r=reaction). Also, I would be interested in learning more about how conservation of momentum is derived from Energy -- that one I haven't been introduced to before. — Trevor K. — 15:21, 10 November 2011 (UTC)
Newton's second law is F = dp/dt and the third law is Fa = −Fr. In Newtonian mechanics, there's no deeper geometrical meaning to momentum conservation; it's true by fiat. In modern physics, momentum conservation follows from the least action principle and translational symmetry of space. I don't know how to turn that into an intuitive geometrical argument, but I should know, because it's certainly a geometrical result. Energy conservation is proved the same way (from translational symmetry of time). If you somehow only know about energy conservation, you can prove momentum conservation using Lorentz invariance ("the principle of relativity"), because energy and momentum are almost the same thing, just associated with different directions in spacetime. -- BenRG (talk) 20:31, 10 November 2011 (UTC)
No, Newton's 3rd law cannot be derived from geometry. That's why it must be stated separately as an additional law.
Even so, energy existed before forces did. Plasmic Physics (talk) 08:00, 10 November 2011 (UTC)
I don't think that is a scientifically defensible claim, nor even one that makes sense. I may be wrong (QM is all weird), but if so, please explain how. --Stephan Schulz (talk) 09:23, 10 November 2011 (UTC)
Go back to the Big Bang, to a time when forces were not differenciated. Maybe, I've got it wrong - possibly there was always force, things just were diffr=erent back then. Plasmic Physics (talk) 09:47, 10 November 2011 (UTC)
Well, the fact that forces were unified does not mean that no forces existed. --Stephan Schulz (talk) 10:11, 10 November 2011 (UTC)

The derivation of conservation of momentum from conservation of energy proceeds in the smae spirit as Noether's theorem, you use that physics s invariant under translations. Suppose that you observe a collision between particles where kinetic energy is conserved (if it were not, you didn't consider all the particles involved in the collision). So, in your frame, the following equation holds:

${\displaystyle \sum _{j}{\frac {1}{2}}m_{j}v_{j}^{2}}$ = ${\displaystyle \sum _{j}{\frac {1}{2}}m_{j}u_{j}^{2}}$ (1)

where the ${\displaystyle m_{j}}$ are the masses of the particles, ${\displaystyle v_{j}}$ are the velocities before the collision and the ${\displaystyle u_{j}}$ are the velocities after the collision. Then consider another observer that is moving w.r.t. at velocity U. In the frame of that observer, the velocities before and after the collision are

{\displaystyle {\begin{aligned}v_{j}'&=v_{j}-U\\u_{j}'&=u_{j}-U\end{aligned}}} (2)

Then assuming that conservation of energy holds in all frames, we must have:

${\displaystyle \sum _{j}{\frac {1}{2}}m_{j}v_{j}'^{2}}$ = ${\displaystyle \sum _{j}{\frac {1}{2}}m_{j}u_{j}'^{2}}$ (3)

Now, we can also use Eq. (2) to write the conservation of energy Eq. (1) in your frame in terms of the velocities in the frame of the other observer. You can just look at what happens to some arbitrary term in the summation:

${\displaystyle {\frac {1}{2}}mv^{2}={\frac {1}{2}}m\left(v'+U\right)^{2}={\frac {1}{2}}mv'^{2}+mU\cdot v+{\frac {1}{2}}mU^{2}}$

Eq. (1) can thus be written as:

${\displaystyle \sum _{j}\left[{\frac {1}{2}}m_{j}v_{j}'^{2}+m_{j}U\cdot v_{j}'+{\frac {1}{2}}m_{j}U^{2}\right]=\sum _{j}\left[{\frac {1}{2}}m_{j}u_{j}'^{2}+m_{j}U\cdot u_{j}'+{\frac {1}{2}}m_{j}U^{2}\right]}$

Using conservation of energy in the moving frame Eq.(3) and conservation of mass, you then get:

${\displaystyle \sum _{j}m_{j}U\cdot v_{j}'=\sum _{j}m_{j}U\cdot u_{j}'}$

If we choose the original frame differently but keep the moving frame fixed, then that amounts to changing U. So, the above equation must be valid for arbitrary U. This means that momentum is conserved:

${\displaystyle \sum _{j}m_{j}v_{j}'=\sum _{j}m_{j}u_{j}'}$

Count Iblis (talk) 16:50, 10 November 2011 (UTC)

Even switching to a new, $48.85 router (the Linksys E1200) did NOT solve the intermittent kick-off issue. I've made numerous calls to Cox Communications and Dell; even got Cox to send technicians to my apartment a few times, and whatever they tell me to do to fix it, or the techs do to fix it themselves, only takes care of the problem for maybe a day, at the most, and this pattern repeats once again. My modem is a Scientific Atlanta. The wireless signal, "Sanders Exclusive," is passworded, and the newest router is 802.11n. This constant, intermittent kick-off of the network started shortly after I moved a metal-frame futon into my apartment. I don't suppose that the last location of this futon was James Bond's MI6-owned apartment or office, hence has an intermittent, embedded comm jammer, does it? I am suspicious that something about the futon interfered with the wireless signal but only because the woes coincided with my moving the futon in. But futon or not, how do I solve this issue once and for all? --70.179.178.145 (talk) 12:06, 10 November 2011 (UTC) I had a similar problem, and my hubby who is a techno geek managed to track the problem down to a rogue router in the neighbourhood, which was broadcasting a signal strong enough to jam my reception. Maybe one of your neighbours has recently installed a new router and doesn't know how to manage it properly? --TammyMoet (talk) 12:16, 10 November 2011 (UTC) Do you have somethinbg like Inssider so you can at least have a look at the networks around you and see what channels they're using etc ? I have similar issues sometimes. I switched channels. I've never quite figured out what is going on. Sean.hoyland - talk 12:52, 10 November 2011 (UTC) If it's a wireless connection, try connecting the router to a computer by cable and see if this fixes it. That should at least narrow down the location of the fault. I had the same problem for six months and my ISP refused to take responsibility. In the end they suggested I ask the phone company to check my line, and to tell them it was crackly even though it wasn't. The problem was fixed within a couple of hours.--Shantavira|feed me 16:32, 10 November 2011 (UTC) I second this.. it seems quite likely it's just the nature of wireless that's causing the problem. The futon could possibly have something to do with it, if it's near the router, where you commonly use your wireless devices, or in between. So could fish tanks, large picture frames, microwaves, and older cordless phones. Nevard (talk) 21:45, 10 November 2011 (UTC) Why don't you move the futon out and see if that fixes the problem? I highly doubt it will, but there's no harm in trying. --140.180.3.244 (talk) 07:08, 11 November 2011 (UTC) ## How do I mimic phone static and hang up under the cover thereof? One time, I was told of this offer that I had "won" - a cruise to the Bahamas, that would cost$598 and would need to be redeemed in 18 months. (Seemed part of a Readers Digest sweepstakes or something like that; it's been a long time.)

I told them that I had nowhere near that amount in my bank account, but then they asked for how much my overdraft protects me. I told them $300, but my gut told me that they needn't know this anyway. They then lowered the fee for me, to$298. (However, they said I had to pay for my own transport down to Ft. Lauderdale, and a separate $59 fee to leave port for an international destination. My gut told me then, that "wouldn't a REAL win include those expenses paid as well? A real win IS "all expenses paid," isn't it?") When I made clear my hesitations, they tried harder, expounding the benefits that I'd get from being on this trip ($50 "free gambling credit," etc.) and I knew that despite how powerful they were with their negotiation skills, I had to get off the phone anyhow, and wasn't exactly in the mood to be rude to anyone that day. Hence, I wanted to fake static noise, and hang up once I've made the noise too loud.

I dragged the phone across the carpet.

However, they didn't sound fazed; when I talked while dragging said phone, they still reacted like there was no static at all.

Then I decided to try a "Plan B," which was disconnecting at mid-sentence somehow without hitting the end call button. (They'd likely hear an "end call beep" just before the dial tone if I did, which would automatically sound rude to them too.)

Which is why I took off the battery case and removed the battery. (That was with a vintage 2007 Palm Treo 700wx. I now have the Sony Ericsson Xperia Play, whose battery takes a slightly more convoluted process to remove.)

So this comes to the questions:

1. Does removing the battery make the call termination sound like an innocent disconnection? Or would any average caller be able to tell that this was deliberate?

2. What are all the best ways you know of, to make fake, albeit convincing static sounds that hint at an imminent disconnection? (Keep in mind that it must be equipment that is easily obtainable and kept in typical homes and apartments, so no high-ticket equipment normally found in tech labs, for example.)

Thanks, --70.179.178.145 (talk) 13:06, 10 November 2011 (UTC)

Could I gently suggest that if you do not wish to continue a phone conversation, you should merely hang up, optionally announcing that you're done with the conversation before hitting the end-call switch. If you really feel you need to imitate the sound of static, then I'd do so with your vocal tract ... I'm sure it's possible to make some sort of staticky type noise if you try hard. But really, especially when you're dealing with scam marketing calls, the most expedite way is to say "sorry, not interested, bye". It's your life, and you are under no obligation to humour phone drones shilling for abominable businesses. --Tagishsimon (talk) 13:15, 10 November 2011 (UTC)
I can't feel confident that using my vocal tract will sound convincing. When I tried this just now, it sounded like I had attempted to relieve a food allergy itch. --70.179.178.145 (talk) 13:28, 10 November 2011 (UTC)
But the central question is, why do you feel the need to invoke the static excuse? And if you do _really_ need an excuse, try "you're breaking up ... sorry, I'm only hearing every third word ... no, sorry, I'm going to have to hang up." --Tagishsimon (talk) 13:32, 10 November 2011 (UTC)
No need to be so polite, or even make any excuse. Just say "I done talking to you. Don't call me back." then hang up without waiting for a reply. Astronaut (talk) 13:34, 10 November 2011 (UTC)
I also don't get why you don't want to be impolite with people who want to scam you. Telephone scammers are well-known for using the inability of the people to say NO to screw them. And, on the top of that, the scam above seems worth reporting to the police. 88.14.195.138 (talk) 13:35, 10 November 2011 (UTC)
We get regular calls, presumably from overseas, in an oriental accent, asking to speak to Mr "Whoever", (usually using a credible English surname) When we say he doesn't live here or whatever they ask "Ah, who am I speaking to?" at which point we say "Sorry I can't continue this call" and hang up. Never do they call back - except in about 3 months when the whole routine starts again! If the person is trying to persuade you to do something you do not wish to be involved with you have every right to politely warn them and terminate the call. If you don't know the person and likely they don't know you then hang up. I agree with the above poster that you should consider reporting it to the police.
@ Tagishmon "phone drones shilling for abominable business" Beautiful! Richard Avery (talk) 14:48, 10 November 2011 (UTC)
basically, a phone scammer wants something from you, which you are not interested in. The fact that these people call you on your phone trying to persuade you to buy something you do not want/need or probably never get (despite having paid for it) is definitely one of the more extreme forms of rudeness. In my opinion this calls for rudeness in return. So why fake connection problems? Besides, if you do fake an interrupted connection they might believe that and call you again tomorrow/next week, which is something you surely dont want, or do you??? So just tell them where to stick it, so they know they are not only wasting your time, but theirs as well. Phebus333 (talk) 15:32, 10 November 2011 (UTC)
If you're looking for a nice passive aggressive way to hang up, just say, "oh, sorry, I've just sat down to eat, call me back later?" and they'll be happy to. Then just don't pick up from that number ever again. Obviously this only works on a phone where you can see the incoming number (like a cell phone), but I've found they're super excited when you actually sound like you want them to call you back. Eventually they give up. Just my experience, as someone who is polite to a fault on the phone. --Mr.98 (talk) 16:44, 10 November 2011 (UTC)
If you want passive-aggressive, ask them to hold on for a moment while you get the home owner. Then put the phone down (without hanging up) and go out for dinner. They get paid per sale, so wasting their time is an effective means of payback. Matt Deres (talk) 18:07, 10 November 2011 (UTC)
This is just me, but I enjoy trying to get them to hang up on me. How you ask? Refuse to discuss the details they want to discuss, instead ask lots of irrelevant questions. The stupider your questions are, the better. Don't let them steer the conversation. After a while, they will either realize they are being screwed with or will just give up. As has been mentioned, they gat paid based on how many people they scam per hour, so scam 'em back. Beeblebrox (talk) 21:42, 10 November 2011 (UTC)
While I agree with the above posters that the scammers deserve neither tact nor mercy, the technical problem would be worth figuring out. But the problem I have with this is, what kind of phone connection gives out with static? To me it seems like they end just as if you had hung up, or the sound stops and starts intermittently ... never static. I don't know what you're mimicking. Wnt (talk) 04:42, 11 November 2011 (UTC)
As noted by others here, it's your phone, you pay the phone bill. You don't owe anything to a telemarketer. In fact, if you say, "Not interested", it will actually help them, as it will allow them to get to their next call all the more quickly. If you want to "mess with them", though, there is no end of possibilities. A more convincing trick than static might be to somehow invoke the sound of a dying battery, i.e. a repetitive "beep". For that matter, the OP could claim he has a call on his call-waiting, and ask the caller to hold while he clicks over to the non-existent call. However, it's hard to beat this cruel and hilarious trick played by a radio guy named Tom Mabe on an unsuspecting telemarketer:[1]Baseball Bugs What's up, Doc? carrots→ 06:37, 11 November 2011 (UTC)
If you blow into the phone, not really hard, but notreally softly, it sounds a lot like static. Heck froze over (talk) 13:53, 11 November 2011 (UTC)
These kinds of hare brained tactics just mean they'll call again. Say the words "put me on your do not call list" and (assuming you live somewhere with such a thing) the problem will be solved. --Sean 16:58, 11 November 2011 (UTC)

Another strategy that works pretty well in my experience (at least here in Germany and if you get called on your mobile phone) is to ask them how the heck they got your number, because most of the times they obtain these information by illegal means (as long as you dont post your contact details on every website you can find). They will try to evade directly answering that question and if you keep riding it they will just hang up. Phebus333 (talk) 15:39, 12 November 2011 (UTC)

## planck distribution function

I wanted to draw planck distribution function in Graph (software) and animate it with respect to T as the varying constant, but even though I put everything correctly, it doesn't draw it. Why?--Irrational number (talk) 13:44, 10 November 2011 (UTC)

Hello?--Irrational number (talk) 18:00, 10 November 2011 (UTC)

Hard to answer your question since nobody really knows what you did. Dauto (talk) 18:06, 10 November 2011 (UTC)
OK, how do I draw it in graph, is that more clear?--Irrational number (talk) 18:23, 10 November 2011 (UTC)
You may get a better answer on the Computing reference desk. --TammyMoet (talk) 18:33, 10 November 2011 (UTC)
thanks, I just re-posted it there--Irrational number (talk) 18:49, 10 November 2011 (UTC)

is the Fukushima nuclear plant still producing radioactive iodine — Preceding unsigned comment added by 84.244.183.116 (talk) 14:16, 10 November 2011 (UTC)

I'm not sure it's clear, but here's why. Iodine-131 is a short-lived (8 day half life) fission product. It would only be at Fukushima if there was still uranium fission happening. Is there still fission occurring? Nobody seems to know for sure. There are conflicting technical reports, some of which say there are still criticalities occurring, and others saying there aren't. In any case, any iodine-131 present from before or during the early parts of the accident has since decayed into something else. I don't find it very likely that there is much being produced, simply because I don't find it likely that there is much fissioning going on (if there is some, it's not a lot), but we're still in the "we're not entirely sure what's going on stage of things." (If that seems like a long time, be aware it took over a year before they really understood what was going on inside Three Mile Island's damaged reactor core, and that was a much less significant accident.) --Mr.98 (talk) 16:41, 10 November 2011 (UTC)

## Experimental drug slims obese monkeys

In reference to this,[2] where do the calories go? Defecation? A Quest For Knowledge (talk)

Mostly breath and urine. Defecation only accounts for eliminating solid waste which is otherwise undigetible; in other words it is the stuff you ate which never got absorbed into the body. Its also a one-way deal; other than digestive enzymes and the like, nothing goes into the digestive tract from the rest of the blood; it only goes into the blood from the digestive tract. Any metabolic wastes pretty much exist in one of two forms: carbon waste, which is exhaled as carbon dioxide and nitrogen waste, which is excreted in the urine. So, if you lose weight because you metabolise your own fat, it's pretty much all getting breathed out. --Jayron32 17:46, 10 November 2011 (UTC)
I concur — Preceding unsigned comment added by 203.112.82.2 (talk) 18:16, 10 November 2011 (UTC)
I dissent. The question is not what happens to the products of metabolism of the fat. The question is what happens to the calories. That is a bit hard to answer without reading the original article. From looking at some other representations of the story in the media, it seems to me that killing fat cells with the drug releases fat into the bloodstream, which has the effect of lowering your appetite. So, you lose weight because you eat less. Duh. --Itinerant1 (talk) 21:12, 10 November 2011 (UTC)
The drug is adipotide, which fuses a prohibitin binding peptide to target the vasculature feeding white fat, with an apoptosis inducing sequence. [3] The sequence is CKGGRAKDC-GG-D(KLAKLAK)2; the CKGGRAKDC targets the prohibitin and the D-amino acids (KLAKLAK)2 cause apoptosis - compare [4]. So the whole effect is to reduce blood flow going into white fat, which oddly enough seems to cause weight loss while actually reducing diabetic effects. (Makes me wonder if those ridiculous belt-shaking devices from the 1930s-ish films might actually have damaged blood vessels and had a similar effect?) Anyway, the point of the first reference above is that the reduced circulation to the fat somehow reduces food intake independently of leptin or POMC in mice fed a high fat diet. Wnt (talk) 04:38, 11 November 2011 (UTC)