Wikipedia:Reference desk/Archives/Science/2012 September 1

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September 1

Alkane isomers

What's the general formula for the number of isomers of the alkane with n carbons? --168.7.232.202 (talk) 01:04, 1 September 2012 (UTC)

There is no known mathematical relationship between the number of carbons in an alkane and the number of possible isomers of that alkane, but it rises as a rate that resembles an exponential function (see here for a list of alkanes 4-15,20]. The problem is further complicated by the fact that some isomers that can be represented by a line diagram are not actually stable molecules (you start running into these at C16). People have written algorithms to calculate the number of isomers, and you can find one here that can be told to remove unstable isomers. Someguy1221 (talk) 01:12, 1 September 2012 (UTC)

See our article entitled "Alkane", which has a section "Isomerism" that discusses the theoretical calculation issue. Might be a good place to add a mention that the calculated result is a theoretical maximum, whereas stable reality is more limited. DMacks (talk) 04:41, 1 September 2012 (UTC)

Delta formation

I am not able to understand how is delta formed. Why is delta formed near the mouth of the river, not in its middle course? I want to know the reason for this according to geographical as well as scientific view. Sunny Singh (DAV) (talk) 02:16, 1 September 2012 (UTC)

Who says that Deltas don't form midcourse? Have you read the article River delta? It covers inland deltas and how they form. --Jayron32 02:18, 1 September 2012 (UTC)

I know but most of the delta is formed near mouth.Sunny Singh (DAV) (talk) 03:15, 1 September 2012 (UTC)

Delta formation requires the flow speed of the river to slow enough that it can no longer keep sediments suspended. The land formation that permits such a thing to happen is simply unlikely to exist near a river. In my own original researchy hypothesis, I would assume this is because rivers tend to erode themselves a riverbed through which to travel, preventing a sufficient loss of speedy. Someguy1221 (talk) 02:40, 1 September 2012 (UTC)

I read the following statement in a Chemistry book -

"Delta is formed when river water comes in contact with sea water for a long period.
River water is mostly muddy. These mud particles are charged colloidal particles. When river comes in contact with sea, the dissolved salts present in sea water provide ions with charge opposite to the charge on mud particles. These particles get uncharged and combine with each other to form bigger particles. They settle as solid mass. Over the years, deltas appear at these places."

This is a short explanation of delta formation. I want to know more about this. Sunny Singh (DAV) (talk) 03:15, 1 September 2012 (UTC)

This is the first time I've heard that theory. The far simpler explanation above seems more likely, that river water slows when it hits the ocean, and all the sediments it carried then settle to the bottom. StuRat (talk) 03:35, 1 September 2012 (UTC)

The theory sounds silly on the surface, anyway. If the river is filled with charged colloidal particles, it must also be filled with oppositely charged particles, colloidal or otherwise. The same goes for the ocean. Either Sunny misparaphrased, or whoever came up with the theory doesn't understand even the most fundamental of chemistry or physics. Someguy1221 (talk) 03:53, 1 September 2012 (UTC)

I have written the same thing written in the book. According to you, book's statement is wrong. This means the answer you gave earlier is the real cause of delta formation. Sunny Singh (DAV) (talk) 04:16, 1 September 2012 (UTC)

For the muddy particles in suspension in river water, flocculation does occur when it encounters seawater, and is one of the reasons that mud is deposited in deltas [1]. Mikenorton (talk) 08:07, 1 September 2012 (UTC)

For coarser particles (not muds), then it is the reduction in flow velocity that causes the sediment to be deposited, as already stated. Mikenorton (talk) 09:47, 1 September 2012 (UTC)

You might be interested in this paper, "Modeling river delta formation". Also, the River delta article isn't bad. Sean.hoyland - talk 10:28, 1 September 2012 (UTC)

What Sunny Singh said is exactly right (Someguy1221 jumped the gun). The colloidal are water soluble in fresh water but become insoluble when the charges present at their surfaces is cancelled out by charges present in the salty water. Dauto (talk) 23:50, 1 September 2012 (UTC)

I still hate that explanation. It's not the presence of any oppositely charged particles that renders the suspension insoluble, but specifically ones that can stably bind to the suspended particles. Someguy1221 (talk) 00:56, 2 September 2012 (UTC)

Water - oil mixture

Why does water not get uniformly mixed with oil? One reason I read is that water is heavier than oil. Then, why does water get mixed with alcohol and doesn't make separate layer as in case of oil. Sunny Singh (DAV) (talk) 02:24, 1 September 2012 (UTC)

We have an article on this phenomenon, at Hydrophobe. You may also be interested in Chemical polarity. The general way I have seen this presented to beginning chemistry students is that polar molecules, such as water, bind extremely well to themselves and other polar molecules, very much unlike non-polar molecules such as oils. The immiscibility of water and oil is a consequence of the water molecules basically binding to themselves to the exclusion of the oil molecules. Someguy1221 (talk) 02:36, 1 September 2012 (UTC)

And no, differences in density are not the reason, in the case of water and oil, although two fluids of different densities might take a long time to mix, just because one would be at the bottom and the other on top, with a limited area of interface, which would slow the mixing. This can happen, for example, with fresh water and salt water. Even though they would mix completely if stirred up, under the proper conditions they can stay in layers for a long time, due to density differences. StuRat (talk) 03:31, 1 September 2012 (UTC)

Given that they are not miscible, they form layers even if stirred up because of different density: they "separate" rather than dissolve because of immiscibility, and the separation is top-vs-bottom because denser stuff sinks and less dense stuff floats. Two separate (sorry:) issues that have various combined effects. If you pick two immiscible materials that happen to have the similar density and mix them well, they can form an even suspension or colloid that does not visibly re-separate. Milk (especially whole homogenized milk) and Mountain Dew (thanks to [[brominated vegetable oil) are two common examples. The closer the density and/or the better you mix them, the longer time or stronger gravity it will take for them to separate. DMacks (talk) 04:51, 1 September 2012 (UTC)

Even with mixtures where the densities are somewhat different, you can create metastable suspensions that last a very long time with the help of extreme agitation or emulsifiers. But regarding the water-oil problem, let me confirm StuRat has already said: the bonding energy you would gain back by forming water-oil bonds is not enough to overcome the energy needed to seperate water molecules from each other, or to get fancy, the energy of solvation is too low between oil and water. Part of the situation with the immiscibility of water and oil is that oil is composed of essentially nonpolar molecules, and the polarized water needs to have some sort of electric charge to "hold on to". There are no localized charges in oil molecules, so water is far happier holding tightly to other molecules. --Jayron32 05:00, 1 September 2012 (UTC)

It's a shame there's no such thing as butter or mayonnaise. μηδείς (talk) 23:30, 1 September 2012 (UTC)

Exactly. Butter is a metastable mixture of fat and water, and it forms via extreme agitation. Mayonaise contains eggs, which contain lecithin, which is an emulsifier. It seems odd that you would confirm exactly the point I made about metastable mixtures of fat and water, and then take a tone as though you would be disagreeing with me. That is quite weird. --Jayron32 01:38, 2 September 2012 (UTC)

It's a shame you're not American, or you would recognize kidding sarcasm when you come acrost it. But yer not. Some sorta weird island of literal zombies called Carolina, apparently. μηδείς (talk) 01:55, 2 September 2012 (UTC)

The individual molecules in a liquid are electrically polarised - they have an unequal distrabution of electronic charge across the molecule. This means that different portions of a molecule is more negatively charged than the rest. I assume that you know that oppositely charged objects attract; molecules attract each other due to their polarity.

Water molecules are more strongly polarised than oil molecules; the attraction force between charged objects increases as the charge on the object increases. Meaning, water tends to associate more strongly with itself than with oil. Relative to water, oil doesn't associate strongly with itself at all, there is just no other alternative.

The differen density between water and oil has nothing to do with it, density just determines the order of the fractions as it sepperates due to gravity or centrifugation. Alcohol mixes with water, as even though it has a weaker polarisation, it is comparable to that of water.

Note that nothing is truely insoluble, just diminishingly soluble. Plasmic Physics (talk) 01:33, 2 September 2012 (UTC)

A lot of beginner chemistry students make the mistake of thinking that "polar" and "non-polar" are opposite terms, more accurate terms should be "weakly polar" and "strongly polar". Plasmic Physics (talk) 01:45, 2 September 2012 (UTC)

Tumor Lysis Syndrome seen in antibody dependent cell mediated cytotoxicity?

Tumor Lysis Syndrome (TLS) is of course a major adverse event in anti-cancer treatments such as Rituxan. As the monoclonal antibody kills B-cells, their contents are spilled into the bloodstream.

What I'm trying to determine is whether TLS would be a possible consequence of anti-cancer treatment that is based in antibody-dependent cell-mediated cytotoxicity, such as the kind you might find when using a therapeutic cancer vaccine.

If not, what is the difference between what activated T-cells are doing to the targeted cancer cells and their contents vs what an immunotherapeutic like Rituxan is doing?

Or God forbid am I even making sense? I'm a health/medical writer but cred-wise I'm just a lay person.

--Wolfgangus (talk) 06:25, 1 September 2012 (UTC)

Yes, TLS is listed as a possible consequence of Rituxan, see http://wiki.riteme.site/wiki/Rituxan#Adverse_events. You should also read the manufactuer's Patient Information sheets, and note the advice given. Wickwack58.164.230.22 (talk) 06:46, 1 September 2012 (UTC)

A few mechanisms of action (including ADCC) of rituximab (anti-CD20) have been proposed but it's not clear which one dominates, as described in this recent review PMID 22621628. It's not clear that the mechanism of action will be the same for different antibodies and cell types. Much may be learned from situations in which these treatments stop working, e.g. PMID 22368276 -- Scray (talk) 20:16, 1 September 2012 (UTC)

I may not have been entirely clear in my question. TLS is generally a consequence of the manner in which an anti-cancer drug kills a cancer cell--the cell's contents are released into the bloodstream. My question is whether TLS might be expected when the treatment involves enlisting the body's own T-cells for an immune response, as is the case in the emerging area of therapeutic anti-cancer vaccines. If the body's T-cells are responsible for killing a cancer cell, what does it do with the cell? If chemotherapy induces cell death during the cell cycle and the contents spill into the blood, what does an activated T-cell do with a cancer cell that it has killed? Does it just carry it back to the lymph nodes like any other foreign pathogen? If so, TLS doesn't seem possible. Wolfgangus (talk) 22:18, 2 September 2012 (UTC)

See the WP article I linked in my 1st post. TLS is unlikely with the usual run of chemotherapy agents, which interfered with a cell's divison/reproduction process. In standard chemo, if a cell is not dividing, it does not die, it just carries on. If it would otherwise divide, chemo either prevents it from dividing, or makes the divided (daughter) cells unviable. All cells are affected, not just cancer cells - fortunately almost all cells in the body only divide during growth. Cells that do divide, such as hair cells, finger nail beds, etc tend stop dividing during chemo, and because the brain cannot make new neurons, learning a complex subject may be difficult. Rituxan/Rituximab is quite different. It is not a chemotherapy. It is an antibody targeted to a specific cell type, whether cancerous or not, and leads to all cells reached getting destroyed, subject to imune system capability. Thus TLS is quite unlikely with chemo, but can occur with Rituxan. Wickwack58.170.164.118 (talk) 23:01, 2 September 2012 (UTC)

Cytotoxic T cells usually kill their targets by induction of apoptosis (programmed cell death), which may result in tumor lysis syndrome. This paper reports a case of modified T cell treatment for chronic lymphoid leukemia where delayed tumor lysis syndrome occurred. --202.28.181.200 (talk) 08:48, 4 September 2012 (UTC)

Insulation value of aluminium foil

Hi, I'm currently building an ice-box in my boat, and I intend to insulate it on all sides with 100mm of rigid urethane foam. Friends have suggested I include a layer of reflective foil around the outside of the foam. If I did, the foil would be tightly sandwiched between urethane foam on the inside, and plywood on the outside, and I can't understand how this would reduce the heat flow into the ice-box. Can someone explain this to me please?124.191.178.49 (talk) 07:02, 1 September 2012 (UTC)

I think the idea is that the aluminum foil will reflect radiated heat. If so, it should be placed on the foam, with an air gap between it and the plywood (a vacuum would be better, but I assume that's not possible here) . However, I agree that jamming it in so it fills the air gap isn't good, as that will allow the aluminum foil to conduct heat directly from the plywood to the foam. So, if there isn't room for it, leave it out (or make the plywood box a bit bigger).

Another concern is that you could get condensation on the foil, which may then soak the wood, allowing it to rot. If the air inside was completely dry and sealed from the outside, this wouldn't happen, however, an airtight seal would be difficult, so an air gap and drainage holes at the bottom are a good alternative, to allow the water to drip down and out. Put a pan under the drainage holes to collect the water. StuRat (talk) 08:48, 1 September 2012 (UTC)

Agree with Stu, foil won't do much in this situation. Careful construction of the insulation with seams minimized (and staggered if using several layers) should give best results. Ssscienccce (talk) 14:19, 1 September 2012 (UTC)

Perhaps your friends were confusing aluminum foil with mylar foil, which IS used effevtively as a thermal insulator. Though mylar is often shiny and metallic looking, it isn't interchangable with the stuff you wrap your leftovers in. Mylar may be useful for this, but I can't imagine Reynolds Wrap as being a good idea. --Jayron32 14:50, 1 September 2012 (UTC)

Aluminium foil works just as well as insulator, it reflects about 98% of radiant heat and light (source). However, aluminized Mylar is a better choice because it does not tear as easily. - Lindert (talk) 15:30, 1 September 2012 (UTC)

Yes, but foil's value as a radiative reflector is more than compensated by its value as a contact conductor of heat. That is, while it may reflect radiated heat, it is also an excellent conductor, and as such, if it is in contact with what you are trying to keep cold/warm, its reflective ability doesn't really come into play. --Jayron32 15:49, 1 September 2012 (UTC)

Yes, if it not separated by air or empty space, it is basically worthless. In this case the space should not be between the cooled object and the foil, but between the object/foil and its environment. The reverse is true if keeping something warm. The same is true if you use mylar foil though (which btw is just plastic foil with an aluminium layer). - Lindert (talk) 15:52, 1 September 2012 (UTC)

No, no, no. The aluminum foil will be useful (By reflecting radiation) even if touches the plywood layer. It would conduct heat, off course but it would still reflect radiation back to the plywood. You don't need to separate the foil from the environment with vacuum or air. You can separate with urethane foam and plywood which are good insulators. Dauto (talk) 23:40, 1 September 2012 (UTC)

It will reflect radiation, but it will be of no significant use because the amount of heat transfer due to conduction far exceeds the amount due to radiation. - Lindert (talk) 12:32, 2 September 2012 (UTC)

Many answers on Ref Desk are very good answers. However, this question is a good example of well-meaning folk jumping in without thinking, and giving opinions on things they know little about. The reflective properties of aluminium foil are of no help here, as wood is not transparent to heat radiation. Rigid urethane foam also has negligible transparency to heat radition. Therefore, only heat conduction is important, and, of course aluminium is an excellent heat conductor. However, Dauto may have intened the aluminium foil to go on the inside of the foam, not between the foam and the plywood. In that case, the low emissivity of the aluminium may help, provided none of the contents of the ice box touch it, as Jayron said. However, the OP's friend may have been confused, not by aluminsised mylar, whose properties in this regard are not much different to alauminium foil, but by the use of aluminium foil as thermal insulation in houses and shops. Here the aluminium is istalled uder the rafters, so there is a 125 mm air space between the aluminium and the roof outside material, and below it is the air in the ceiling space. Here, the low emissivity of aluminium does work with the air above and below it to lower heat transmission. It is not anywehre as good in this respect as 100 mm of foam or fibreglass batts on top of the ceiling. I can't see condensation affecting the outer wood being an issue in the way StuRat said. Rather, the aluminium will act as a moisture barrier (which is a common use of it) and protect the wood. However the OP will need to prevent condensation from inside of teh ice box from getting into the fowm, ruining its insulative properties. Ratbone121.221.217.156 (talk) 03:33, 2 September 2012 (UTC)

Re: "negligible transparency to heat radiation", yes, this is true, but irrelevant. The heat passes through the plywood (slowly) by conduction, not radiation. Once on the inside edge of the plywood, if there is an air gap, it can not efficiently transfer to the foam by conduction (and a vacuum would stop convection/conduction completely). However, radiation of the heat from the plywood to the foam can still occur. This is where the foil comes in. Yes, a non-conductive material which is reflective would be better, but aluminum foil would still be better than nothing, provided there is an air gap between it and the plywood to prevent conduction. StuRat (talk) 03:57, 2 September 2012 (UTC)

It's very relavent - its the crux of what diffrent posters are on about. The OP specifically said "tightly sandwiched" between the foam and the wood. However, you are correct in that if an air gap was intrduced, the aluminum would help.

Agreed, and I said so in my initial response. So I'm baffled at why you said this Q is an example of "well-meaning folk jumping in without thinking, and giving opinions on things they know little about". StuRat (talk) 04:52, 2 September 2012 (UTC)

You had it right in your 1st post, but others did not:-

Jayron32 said that mylar film would be good, implying it would be good using the way the OP said, sandwiched between the foam and the wood. It would not be good. While the thermal resistivity of mylar is very much greater than that of aluminium, in film form it is too thin to matter.

Lindert thought that the reflectance of the film was important. It is not, because the foam and the wood will block radiant heat. Only thermal conductance is important, unless there is an air gap, or vacuum gap. But after prompting from Jyaron, he got it right, only for....

Dauto said, no, film without an airgap will work. No, it won't. And he went on to imply that an airgap between the foam and the wood will improve it because they are insulators. Well, an airgap will improve things, but NOT because the foam and wood are insulators. An airgap will always improve things, regardless of materials, because it is itself a thermal insulator. An airgap will be less important if teh materials are insulators.

Ratbone58.170.164.118 (talk) 11:20, 2 September 2012 (UTC)

As for condensation, this happens when moist, hot air seeps into the air gap, hits the cool aluminum foil, and the water vapor condenses on the foil. If it's in contact with the plywood, the wood would then get wet. If not, then drops will form and fall to the bottom of the aluminum foil, and, hopefully, down to a drainage tray below. If the foam is thick enough to keep the foil from getting too cool, or the box is sealed well enough to keep out hot, moist air, then this won't be a problem. Also, if this was in a desert, then moist air would be rare, but, in a boat, it should be common. StuRat (talk) 04:01, 2 September 2012 (UTC)

Actually, although wood is not a good conductor of heat, a likely thickness of wood will conduct vastly better than the OP's 100mm of foam. Therefore nearly all the temperature drop will be across the foam, not much across the wood. Only if the foil temperature is below the dew point will condensation on the wood side occur. It should be easy to seal the wood with varnish of whatever, as you would anyway on a boat, to reduce the diffusion of moisture to a level that does not matter. Wood ALWAYS has a significant moisture content anyway, as any cabinet maker will tell you. Wickwack124.182.13.142 (talk) 04:45, 2 September 2012 (UTC)

Sealing it doesn't seem so practical, to me. First there's the door to consider, where you would need to seal between the foam and plywood. Then there's thermal expansion to consider, as well as the wood swelling, shrinking, and warping with humidity changes (say dry dock versus out on the water). Loading up the freezer will also tend to cause it to sag under it's own weight. All of these factors will make maintaining an airtight seal rather difficult. If an airtight seal is the goal, then plywood is the wrong material. StuRat (talk) 04:56, 2 September 2012 (UTC)

As I explained, you don't need to seal it to air tight / gas tight degree. You only need to seal it to bring down the moisture diffusion to a tolerable level. Plywood does not have significant swelling/shrinking in the legth and width, only in thickness. Plywood does not warp, as the plys balance each other out. Wickwack124.182.13.142 (talk) 06:09, 2 September 2012 (UTC)

I disagree that plywood does not warp: [2]. To balance out, all of the plies would need to be exposed to identical temperatures and humidity, which is definitely not the case here. StuRat (talk) 06:13, 2 September 2012 (UTC)

Thank you all very much for your responses, now I can clearly understand the differance between radiant heat transfer and conduction, so I won't be wasting effort on foil. (OP)124.191.178.49 (talk) 05:05, 2 September 2012 (UTC)

You're welcome, but don't overlook the suggestion to add an air gap. Any good thermos will have a gap, although they are able to maintain a vacuum inside it. StuRat (talk) 05:25, 2 September 2012 (UTC)

Weight of atom

What will be the weight of a atom if its size expanded to a cricket ball?--Sunny Singh (DAV) (talk) 10:46, 1 September 2012 (UTC)

It would depend on the atom, since they each have different masses and sizes. The mass varies by element and isotope, and to a much lesser extent by charge. The size of an atom is a bit harder to define, as the outside edge is an electron probability envelope.

But, if we assume the atom in question is a typical atom from which a cricket ball is made, then the mass would be about the same. This is because in a solid, the atoms are packed essentially right next to each other. There is still lots of empty space, between the nucleus of each atom an the electron cloud, but that exists in both the individual atom and the collection of atoms in the cricket ball, in the same proportion. I'd say the single atom would have a slightly greater mass, since packing is never perfect with a collection of atoms. The atomic packing factor runs around 68-74% in a crystal, and might be a bit worse with an amorphous solid, let's say 50%. In that case, the single atom would have twice the mass, and so would weigh twice as much, too. StuRat (talk) 11:17, 1 September 2012 (UTC)

This is a none question because expanding atoms is not possible. It can only be answered hypothetically and depends what one assumes for the properties of the hypothetical atom. In other words the answer can be anything you like. For instance, if one postulates an atom that composed of normal protons and electrons, but has somehow contrived to place the electron shell at the diameter of a cricket ball, then the mass would be that of a single "ordinary" atom - almost nothing. On the other hand, if one postulates an atom with a nucleus the size of a cricket ball but with the same density, which is about the density of a neutron star, the weight would be something like that of the Alps. SpinningSpark 20:01, 1 September 2012 (UTC)

You seem to be assuming an atomic nucleus the size of a cricket ball--which would weigh 100,000^3 the mass of a cricket ball--but that is not what the OP said. μηδείς (talk) 20:09, 1 September 2012 (UTC)

I think we can assume that the question was asked hypothetically. 86.146.107.246 (talk) 21:03, 1 September 2012 (UTC)

An atom the size of a cricket ball (let's assume 10 cm radius for convenience, and a Helium atom, since that is the example at our atom article--the proportions of carbon twelve are the same) would have a "solid" nucleus 1/1000th of a millimeter across, containing 99.94% of the atom's mass (pretty extremely dense, huh?) and the remainder would be made of a cloud of electrons. It would in no sense resemble a cricket ball with a uniform density and a solid surface. μηδείς (talk) 20:06, 1 September 2012 (UTC)

This is an easy one, just simplify it to finding the atomic density for a given atom using its atomic mass and Van der Waal volume, and multiplying it by the standard volume of a cricket ball. Plasmic Physics (talk) 22:11, 1 September 2012 (UTC)

The Van der Waal volume being based on the size of the electron cloud, yes, that's what I said. μηδείς (talk) 23:27, 1 September 2012 (UTC)

Just a small aside, his name is Johannes Diderik van der Waals. With the 's'. Otherwise, yeah, that's how i would solve it as well. --Jayron32 00:30, 2 September 2012 (UTC)

• The atomic radius is n atomic mass units and something on the order of 100 picometers. A men's cricket ball is about 160 grams and 226 mm circumference = 36 mm radius. Taking 100 pm as our benchmark, expanding it to 36 mm is a 0.36 x 10⁹-fold increase in diameter; if we imagine that mass increases per the increase in volume, that's 0.047 x 10²⁷-fold increase in mass multiplied by 1.66 x 10^-27 kg per amu. This gives us 78 grams per amu mass of the atom - to be corrected by dividing by the cube of the atomic radius over 100 for the element chosen. So phosphorus has 100 pm diameter, and would weigh 78g * 31 = 2.4 kg. Hydrogen has only 25 pm radius, so we take its 78 gram mass but multiply by a factor of 4³ (because we have to blow it up 4x larger in diameter) getting us 78g * 64 = 5.0 kg. Likewise radium is 78g * 216 / 2.15³ = 1.7 kg. Now what's odd about this is that they're all coming up at least 10 times heavier than an actual cricket ball, and StuRat's logic seems convincing - true, the article says cricket balls are made from cork, which is not a solid material, but even so, seems a little off... there may be some aspect of the definition that I'm not properly considering. Wnt (talk) 06:15, 2 September 2012 (UTC)
  • Here's how I tried to figure it out. Imagine if a cricket ball were made of solid lead. Using your numbers, I get the volume of a cricket ball to be about 40.6944 cubic centimeters. If it were made of lead, that would weigh about 461 grams. Lead is face centered cubic crustal structure, so that's a packing efficiency of 0.74. So, a single lead atom, scaled up to the size of a cricket ball sized chunk of lead, would weigh 461/0.74 = 662 grams. I did this because a single atom would have a packing efficiency of 1.00: there would be no gaps between atoms to consider if the cricket ball were a single atom, but the individual atoms would have the same mass and density as in the real example. If we choose a different atom, like carbon, then carbon has a hexagonal crystal system with a packing efficiency of 0.74 also. Using the density numbers for graphite, I get a ball of graphite to weigh 92 grams, which makes our mega-atom of carbon would weigh 92/0.74 = 124 grams. So it really depends on which atom you're using. This makes some sense, given that atomic weight and van der Waals radius doesn't scale, Antimony is about 10x the mass of carbon, but the van der Waals radius of Carbon atom is 170 pm , and that of an Antimony atom is 206 pm. So heavier atoms should be a LOT more dense than lighter atoms, as this calculation bears out. Someone should check my math, but this feels intuitively right. --Jayron32 06:38, 2 September 2012 (UTC)

Hmmm! the way I did it, the mass of lead is 207 amu and its radius is 180, so taking 78g*207/1.8³ gets me 2.77 kg, which is high by a factor of 4.18 (or a radius off by 1.6 fold). Trying to get at the discrepancy, I find [3] which I take to indicate that 495.08 pm * sqrt(2) / 4 is the closest distance between lead atoms in it (along a diagonal on a face of the cube). That's 175.0 pm .... which doesn't account for any difference! And as you say the fcc packing is efficient at 0.74-fold difference in mass. Dang it, I just went over the math a third time and I still haven't found the bug. Wnt (talk) 15:49, 2 September 2012 (UTC)

Balanced wheel

Could I check a couple of things?

Suppose a wheel is oriented horizontally (i.e. the rim of the wheel lies in a horizontal plane), and it balances perfectly under gravity when it is resting stationary on a point (say on the tip of a spike). Does that also mean it's perfectly balanced when it rotates on that axis at any speed in any orientation?

Is it always possible to balance any wheel by adding one point mass to the rim?

86.146.107.246 (talk) 19:27, 1 September 2012 (UTC)

A wheel that was balanced on a point when horizontal will only still be balanced when tilted if it has zero width. A real wheel with thickness will have its cg shifted slightly to the side which is tilted down causing it to tilt further and then fall. The same applies to a wheel rotating slowly. If the wheel is rotating fast enough, gyroscopic action can overcome the unbalance and hold it in place. SpinningSpark 19:49, 1 September 2012 (UTC)

The guy talked about rotation along the plane of the wheel (I think) - to be precise, along the horizontal plane that is perpendicular to down - not about tilting, which wouldn't make sense. So, he says: if you can get the wheel to balance when it's still, can you rotate it 'along the horizontal plane' and it it still balanced? For me, the answer is obviously "yeah, you can still rotate it" as this is no different from "rotating the point" that it is on, since a point can rotate and it is the same. (Alternatively, you can simply walk around the wheel, rotating your view, and the effect is the same as if you had rotated the wheel along the horizontal. So, you can always rotate it and it will remain balanced.)

As to the other question I should think the answer is yes but I'm not sure. --80.99.254.208 (talk) 20:14, 1 September 2012 (UTC)

To clarify, by "when it rotates on that axis at any speed in any orientation" I am talking about putting the wheel on an axle in the same position and orientation as the spike on which it was resting. So, it's two different setups: the first resting horizontally on a spike which is where the axle will be positioned, the second with the wheel in a normal configuration on an axle, with the axle oriented in any direction. Sorry that was not very clear. 86.146.107.246 (talk) 21:01, 1 September 2012 (UTC)

A wheel running on a perfectly rigid axle fixed in space does not need any balance weights, as any twisting torque on the axle will be absorbed, by definistion. But a real wheel of finite width on a real axle can be a very different story. To see why, visualise two coupled wheels, near each other, on a common vertical axle. Let's say the lower wheel is perfectly balanced, but the upper wheel is not. As the wheels rotate, the lower one will have no effect on the axle, but the upper wheel will obviously try to displace the axle horizonatally, twisting it round. Therefore, a real wheel of a certain width may require balancing with more than one weight. For this reason, while most car tyres are balanced by adding a single weight to the inside of the rim, you sometimes see that the tyre fitter has used a larger weight on one side of the rim, and a smaller weight on the other side. A wheel that is unbalanced on the center plane (the plane orthogonal to the rotation axis) causes a paralllel displacement of the axle. It is called static unbalance because it can be detected in situ without continuous rotation. A wheel having an unbalance mass to one side of the center plane causes a twisting force on the axle - this is called dynamic unbalance. Wickwack58.167.241.235 (talk) 02:43, 2 September 2012 (UTC)

There's also an effect that the force of gravity is different at different altitudes. So, theoretically it should be impossible to balance a vertical wheel without moving weights around as it rotates, since the top will always weigh a different amount than the bottom. However, even in the case of the largest diameter wheels (a Ferris wheel ?) this effect is so negligible as to be completely overwhelmed by the gyroscopic effect. StuRat (talk) 04:10, 2 September 2012 (UTC)

Not true. Its' mass that has to be balanced, not weight. If mass is balanced, then gravity can not upset that, as no matter what the rotatory position of teh wheel is, the pull on each part of it will always summ the same. Wickwack124.182.13.142 (talk) 04:34, 2 September 2012 (UTC)

I disagree. Imagine a wheel 10,000 miles in diameter, with a lump of extra mass causing a slight imbalance. Horizontally, at a uniform height of 5000 miles, it would work fine, with the gyroscopic effect overcoming the imbalance. But when rotated vertically, so the top is 10,000 miles into space, the gravity acting on the bottom would be far more, causing a magnification of the imbalance as the lump went around the bottom, and a reduction when it went over the top. This variable imbalance would be worse than a constant imbalance. StuRat (talk) 05:20, 2 September 2012 (UTC)

The question is whether a object, perfectly balanced when horizontally rotating about, is then perfectly balanced rotating in any plane on that point, and whether in order to balance an object, only one added weight is required. And the answer is 1) yes it is, and 2) but it may need more than one balance weight. You have it wrong, Stu. Wickwack124.182.13.142 (talk) 06:04, 2 September 2012 (UTC)

I think we're just talking about different things. You're talking about perfect balance, while I'm talking about the real world, where that is impossible. StuRat (talk) 06:07, 2 September 2012 (UTC)

No is the answer (if I have understood the question). If the stationary wheel will balance (in a uniform gravitational field) on a point at the centre of its shaft, then the wheel's centre of gravity lies on the centre-line of its shaft. This means the wheel has static balance. If it doesn't have static balance, then this can be corrected by a single weight. However this does not mean that spinning the wheel about its shaft will not result in shaking or vibration. For smooth running of the wheel you need dynamic balance, which means that (1) the centre of gravity is centred on the shaft and (2) one of the wheel's principle axes of inertia is aligned (parallel) with the axis of rotation. You can't in general correct dynamic balance with a single weight - in particular you can't balance the wheel with a single weight if it already has static balance, since any additional weight would destroy the static balance. --catslash (talk) 14:46, 2 September 2012 (UTC)

Almost Correct. You are entirely correct in the context of the OP's question. However (as a fine point), in parctice many wheels on axles have their axles pratically constrained against turning at right angles to their axis, while not so constrained against lateral dispalcement. In such cases only static balance is necessary, particularly because in practice static unbalance is larger than dynamic anyway. Wickwack58.170.164.118 (talk) 23:04, 2 September 2012 (UTC)

Could this "dynamic imbalance" versus "static imbalance" distinction arise if the wheel has a negligible thickness (i.e. a thin disk, approximately like a CD for example), or is it only a function of the non-uniformity of the wheel over the "thickness" dimension? 86.177.105.185 (talk) 23:49, 2 September 2012 (UTC)

A wheel with negligible thickness that somehow still has significant mass cannot by definition have any dynamic unbalance. The vibration consequences of dynamic unbalance of wheel mass is proportional to the offset of the mass unbalance from the plane orthogonal to the raotaion axis and passing thru its centre of gravity - yes, it is a function of the uniformity over the "thickness dimension". Wickwack121.215.159.205 (talk) 01:58, 3 September 2012 (UTC)

Thanks, that makes sense. Thanks everyone for the replies. 86.177.105.185 (talk) 02:01, 3 September 2012 (UTC)

can I see the observer effect in action?

Is there an experiment like the double-slit, where there is an interference pattern unobserved, but it disappears observed, and if so I would like to watch a Youtube video of someone (i.e. a piece of machinery) alternately either not observing the single photons and observing htem.

I would like to see the observer effect "in action" like toggling the interference pattern by observing or not. To be clear, I don't want to myself observe: I want to observe the *experiment (+ observer) OR (+ 0) = outcome* where the middle term is switched on and off and I can see, by the changing outcome, the observer effect. --80.99.254.208 (talk) 20:11, 1 September 2012 (UTC)

This is an interesting question. Note that if you have a photon-detecting device at one slit with, say, its readout hidden behind a door so you can't see it, opening the door is not going to make the interference pattern disappear. All that's necessary to make the interference pattern disappear is recording the passage of the photon somewhere, even if only as electrical signals in the bowels of the machine. That aside, the idea makes sense in theory, but there are practical problems:

• A milliwatt laser at a visible wavelength emits upwards of 10¹⁵ (1,000,000,000,000,000) photons per second. That's pretty dim light. To make this experiment work you need photons to pass through the slits at a rate slow enough that you can detect each one individually. So you're not actually going to be able to see the light on the screen at the far side of the slits.
• Practically speaking I don't see how you could detect a photon's passage through the slit without absorbing it, but that makes the experiment pretty boring since you're simply blocking one of the slits. The detector would need to be connected to another laser at the slit which would emit a photon whenever one was detected. Quantum mechanically it doesn't matter, but you might find the experiment less convincing this way.
• Practically speaking I think the only way to prevent measurement of the photon's passage would be to physically remove the detector (and re-emission laser) from the slit. Again, this might make the experiment less convincing, especially since single-photon detectors are not very accurate in the first place and the mere presence of one in the slit is going to make a big difference, measurement effect or no.

-- BenRG (talk) 22:33, 1 September 2012 (UTC)

Some of the difficulties pointed out by BenRG would be less problematic if you use electron instead of photons in the double slit experiment. Dauto (talk) 23:22, 1 September 2012 (UTC)

I may not know the actual way of things, but as far as I was taught in university level Special Relativity, they observer theory concerning quantum superposition of states, appears to be incomplete or faulty - everying in the universe is an observer, everything contantly communicates information through the fundamental forces. Observation should more likely to be statistically determined than definitively determined, but how? Plasmic Physics (talk) 02:12, 2 September 2012 (UTC)

Note that quite ordinary photomultipliers (http://wiki.riteme.site/wiki/Photomultiplier) can detect single photons. Measuring the precise time of photon arival is another matter. Wickwack58.167.241.235 (talk) 02:50, 2 September 2012 (UTC)

There's a logic error in your idea, since you are an observer, too, so can't watch the pattern change based on who observes it. Anytime you are watching, it is being observed. StuRat (talk) 04:13, 2 September 2012 (UTC)

Stu, which poster are you refering to? The OP? Plasmic? Me? Wickwack124.182.13.142 (talk) 04:33, 2 September 2012 (UTC)

This Q actually has proper indentation, so, since I indented once from the OP, it should be apparent that my response is to them. StuRat (talk) 05:11, 2 September 2012 (UTC)

This whole business of wave function collapse resulting from mere observation is simply folly. Superstition in the guise of science, what it is. 66.87.127.241 (talk) 05:07, 2 September 2012 (UTC)

I do believe that it is a result of mere observation, however, I also believe that observation is currently ill defined - the collapse of wavefunctions would continue without the presence of life in the universe. I sounds like flat earth logic, full of holes. Plasmic Physics (talk) 05:27, 2 September 2012 (UTC)

Yes, I never understood why we should suppose that Schrödinger's cat is both alive and dead, as opposed to definitely having one state, but that state being unknown to us. StuRat (talk) 05:40, 2 September 2012 (UTC)

It should be noted that Schrödinger thought the exact same thing. He devised the cat problem to show how silly quantum mechanical principles would be if applied to something like a cat. It was supposed to be reducing the problem to absurdity. --Jayron32 06:03, 2 September 2012 (UTC)

(edit conflict)That is not what I mean, I definitely believe in superposition. According to my understanding, the box would be observing the state of the cat. If the box was a closed experiment, then the superposition should collapse on its own.

My question is: what amount of exchange of information qualifies as an "observation"? Clearly, in the double-slit experiment, a superposition persists, despite the environment of the experiment exchanging information. Plasmic Physics (talk) 06:08, 2 September 2012 (UTC)

See double-slit experiment. Note that observing which hole the photon goes through is different from observing the pattern which is why the experiment is meaningful, and the effect of this is indeed really observable; it's not just a thought experiment. Wnt (talk) 05:46, 2 September 2012 (UTC)

Yes, but only under certain interpretations, such as the Relational interpretation, does this mean the observation actually changed the system. StuRat (talk) 05:59, 2 September 2012 (UTC)

OP here - stating my assumptions

The above responses are on a bit of a tangent. Let me state my underlying assumption. 1. The double slit experiment shows a point source of photon emisions passing through two slits like this: if the equals sign is a photon gun it goes pew-pew-pew = · · · · x | where the vertical bar at the end is a screen, and the x is either a | with two small slits in it parallel to each other or one.

2. If we repeat the procedure on a large scale with marbles, the marbles will hit the screen in two different locations: one where the ray passing from the emission point source through one small slit hits the screen, and the other where the ray passing from the emission source through the other small slit hits the screen. (i.e. the finer the slit's width / closer the screen to the slits the more narrow the bands are.) This makes sense from a particle interpretaiton

3. If we repeat the produce on a large scale with the slits in water and the point source bein ga source of ripples, the water goes like this: = ) ) ) ) ) x 3 3 3 3 | where instead of a point · we get a ripple, and the ripple passes through the double-slitted x and turns into two ripples which interfere with themselves. At the point of | you get an interference pattern.

4. Returning to the photon scale, I am told that if you slow the photon gun to emit one photon at a time (and not a stream) it will still produce a wave pattern (as if each individual photon were interfering with itself) like this: x 3 3 3 3 | showing an intereference pattern on the screen. But if you measure which slit of x it goes through it turns into · · · · x · · · | simply landing on one of the two bands that you get from connecting the point source with the slit and continuing to draw the ray. In other words, no ripple effect.

These are my assumptions.

5. Therefore, if one were to alternately check which slit the photon goes through and not check, you would alternate from the effect x · · · | to x 3 3 3 3 |

Is this right? If so I would like to see number 5 in action. I don't care if it's time-lapsed over hours. I just want to see the observer effect. --80.99.254.208 (talk) 12:55, 2 September 2012 (UTC)

That's almost correct—the only mistake is that the particles diffract at the slits whether or not they're measured there, and therefore you get one wide blob on the screen when there's no interference, not two narrow blobs. What appears or disappears is the modulating pattern of interference fringes within the blob. You could get the same visual effect by using a bright monochromatic light and putting linear polarizing filters at each slit. If they're oriented the same way, there will be an interference pattern. If you rotate one of them by 90°, the interference pattern will disappear. That's exactly what you would see in a real double-slit experiment if it were possible to see it directly. It's a classical effect (predicted by Maxwell's equations), but you can explain it in the language of the quantum double-slit experiment: using different polarizations at the two slits leaves a record of which slit the photon chose, and that's enough to destroy the interference pattern.

Do you just want a teaching video that could be shown in a classroom (in which case a simulation would be fine), or are you not quite convinced that the effect is real and want to see more direct proof (which as I said above is hard and will depend on what counts as "direct")? -- BenRG (talk) 18:22, 2 September 2012 (UTC)

Yeah, it's stupid and I don't believe it and want to see the effect -- JUST the observer efect - in action. I don't believe you could make an interference appear or disappear just by looking and not looking (even in not quite so literal terms) at the thing that produced it. let me ask this way. If the slits are far enough from each other, we could put a repeater that is activated by a photon and release a single photon "along the same" trajectory. (I don't personally care if the repeated trajectory is a bit off). Alternatively, the repeater could be moved out of the way, so that the photon is not caught. If I understand the claim correctly, then if you measure the repeater's signal (and this may not even be necessary as long as the signal is there somewhere) then if we do a time-lapse video where a die is thrown, even means the repeaters are in the way, odd means they're in the way, and a 3 is thrown, we see the pattern from 6 hours of time-lapse (no interference pattern, 3 means they repeaters measure and are in the way), then a 5 is thrown, again no int. pattern, a 1 thrown, again no int. pattern, a 4 is thrown, they're put in the way and the effect suddenly disappears, a 3 is thrown, they're moved again, etc. This would be quite convincing. Not QUITE as convincing as if the electron had been measured in transit without a repeater, but if that's the best science can do it will do. Basically I want to see the observer effect "in action" like a 'minimum difference' type setup. --80.99.254.208 (talk) 19:38, 2 September 2012 (UTC)

It's much easier to "passively" detect an electron than a photon because it's electrically charged, so this experiment might actually be possible without the "repeater". Unfortunately all I can find online is an experiment by Akira Tonomura and collaborators (web site, Youtube), but it makes no attempt to detect electrons at the slits. They don't seem to have published, either.

As far as the theory goes, it's a property of all waves (quantum or not) that you only get interference between paths that lead to the same final state. For light that means the same location and same polarization. In the example above with polarizing filters, when the polarization is different there's no interference (and there's never interference between light reaching one part of the screen and light reaching a different part). What's unique about quantum wave behavior is that the final state includes not just the position and polarization of the light but also everything else. If there's something in the world that ends up in one state when the light goes through one slit and another state when it goes through the other, there's no interference on the screen even when the final position and polarization are the same. On the other hand, if your repeater's final state is unaffected by whether it absorbed and re-emitted a photon/electron (which is theoretically possible), the interference pattern will remain. I know this is strange, but it's not that much stranger than classical waves... -- BenRG (talk) 00:00, 3 September 2012 (UTC)

of course it's much stranger than classical waves. You're asking me to believe an interference pattern will disappear if something else (a passive measurement device) is not the same if the electron goes - unintercepted - through one slit as it is through the other. You're asking me to say that if nothing is affected it goes through both. This is like saying, as long as my door is closed, every piece of furniture in my room is in every position, and only when I open the door does the waveform collapse into every piece of furniture being in its own position. This is nutso. (on its face). Since you say electroncs can be measured passively, where is the experiment that turns the passive measurement off and on and lets me see the appearing and disappearing interference pattern? Because I don't buy it. I don't have a problem iwth the double-slit experiment, but I don't believe it disappears under the observer effect, which can't rationally exist except in bizarro world. Extraordinary claims require extraordinary evidence. --80.99.254.208 (talk) 05:10, 3 September 2012 (UTC)

I haven't followed this whole thread, but your reply above caught my eye. You're missing something. QM does not require you to believe that a passive measurement device can affect whether or not an interference fringe forms. Rather, it requires you to understand that there is no such thing as a passive measurement device that can identify which path the photon followed without disturbing it. This is not some mysterious process—there is always a solid physical interaction between the photon and the observing device, usually through the electromagnetic field. Every device you can devise that will determine which path the photon took will have an effect on the photon. Without QM, one might think that a better measuring device could be devised, but the uncertainty principle proves that the problem is in fact fundamental. You can't observe the photon with a passive device because there is no such device, and conversely a measuring device that is passive will not tell you which path the photon took.

It should be possible to make a video of the experiment being performed, contrary to the objections above. I think some of the respondents misunderstood the question.--Srleffler (talk) 17:26, 4 September 2012 (UTC)

I'm not sure why you're talking about wave function collapse—I never said anything about that. You could record the photon's path by splitting it and sending one copy off into space, and the interference pattern would disappear, but no one would claim that a collapse took place. You can prove that it didn't by other measurements on the split photons. Likewise you can record the path by altering the polarization of the photons, and there's no question of collapse there either.

All of this is a necessary consequence of the basic quantum rules which are implicitly tested in every modern particle physics experiment. You're not the first person to think it's crazy, but the world does work that way. I like the idea of doing a simple double-slit setup for real and posting it on Youtube, but I'm at the mercy of what I can actually find on the net. I don't think it would convince you in any case because it would be easy to find ways to criticize the setup. The whole video could be faked. -- BenRG (talk) 20:18, 3 September 2012 (UTC)

As for not believing a video - I'm not that incredulous. I just think that if an electron is supposed to behave one way unobserved (goes through two slits and interferes with itself) and you change nothing but "we will now also record which slit it goes through" claiming that the interference pattern will desiappear ought to be backed up by a video of it happening. This is 2012, not 1712. We don't have to imagine what an experiment looks like, you can link a youtube video. --80.99.254.208 (talk) 04:43, 4 September 2012 (UTC)