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October 16

Manhole question

why is that a person going in a manhole had met with accident and blood came out of his body —Preceding unsigned comment added by 117.197.116.74 (talk) 05:04, 16 October 2008 (UTC)[reply]

Why did they go into the manhole? To chase an alligator, perhaps? If so, that would be a clue. --Scray (talk) 05:35, 16 October 2008 (UTC) [reply]

Are you sure it's not a hole in a man, that would leak blood. Graeme Bartlett (talk) 05:39, 16 October 2008 (UTC) [reply]

Didn't you two read the question? It's not how the person met with a bloody accident, it's why. If you are a religious fundmentalist, the answer is probably because he/she pissed off God. If you are an atheist, it's because he/she was not fit for going into manholes due to genetics. :-P

Right. For the latter, the genetic non-fitness was that he was too stupid to stop traffic before working in a manhole in the middle of the highway:) DMacks (talk) 18:25, 16 October 2008 (UTC)[reply]

Who knows? Maybe the person took a printed copy of the wikipedia reference desk and was squinting to look at one of the answers and collided with a wall... Clearly a strong Darwin award candidite, if the person then falls into the sewer water, gets a deadly infection and dies.Nil Einne (talk) 09:35, 17 October 2008 (UTC)[reply]

If he was working in an electrical vault, it could have been due to the arcing of 480 volt conductors. That has been known to cause bleeding, third degree burns, and death. Edison (talk) 02:42, 18 October 2008 (UTC)[reply]

Thank you to the contributors

Hi everyone. I used to contribute to this reference desk occasionally, but that has fallen off. Nonetheless, I wanted to start a section to thank the contributors of the Science Reference Desk. Not only do you help people everyday in their pursuit of knowledge, you make for some darn good reading. I could put this on the talk page, but you all deserve a slightly more public recognition. Cheers! Eric (EWS23) 09:59, 16 October 2008 (UTC)[reply]

I too would like to thank everyone for their contributions. Some of the discussions going on here are quite interesting and often shed a lot of light on subjects that the Wikipedia articles themselves fail to do. 12.10.248.51 (talk) 13:13, 16 October 2008 (UTC)[reply]

I agree. One of the often unrecognized goods i asociate with the rd is that a lot of article improvement and expansion is generated here. --Shaggorama (talk) 10:06, 17 October 2008 (UTC)[reply]

Heat effects

You know sometimes you get sun shining through a window onto a wall, and if it is a hot day you get sort of wavy shadows in the sunlight hitting the wall? What is this called, and is it linked to the effect you get when heated air distorts what is behind it. What is this effect called as well? I'm sure I have the word somewhere in my head but I can't recall it right now. 88.211.96.3 (talk) 10:45, 16 October 2008 (UTC)[reply]

Heat wave? The shadow is simply the same thing as the shadow is just cause by light. 194.221.133.226 (talk) 11:00, 16 October 2008 (UTC)[reply]

No that doesn't look right, that seems to be an article about exceptional large area high temperatures. 88.211.96.3 (talk) 11:44, 16 October 2008 (UTC)[reply]

It's hard to guess the word you are thinking of - so let's toss in some gratuitous explanation and maybe we'll hit the right word by luck!

So the effect you are thinking of is caused by the air temperature variations causing subtle shifts in refractive index. Glass is a better conductor of heat than air - so the air nearest to the window is likely to be at a different temperature than the air in the remainder of the room. As convection causes warmer air to rise and cooler air to sink, there is likely to be a certain amount of turbulence - although laminar flow is also possible, this would not cause the effect you are seeing. A bubble of turbulent air that is at a different temperature to its surroundings (and hence has a different refractive index) acts like a lens - focussing or dispersing the sunlight that is cast onto the wall - increasing the light intensity in some areas at the cost of decreasing it in others. The areas of decreased light intensity look like shadows and the turbulence makes tham wave around. Hence "wavy shadows".

Words to describe this phenomena would be "heat haze" (a term I've never been very happy with), or perhaps "shimmer". It's possible that you are also thinking of a mirage. But that's a different effect - although it's also caused by temperature variations in the air and consequent changes in refractive index. A shimmer due to turbulance will be more obvious in the case of a mirage. You might also be thinking of the scintillation of stars (more commonly called "twinkling") - or perhaps the word you are after is "dispersion". The pattern of light on the wall might also be called a "caustic" - although that term is more commonly used for things like the pattern of light on the bottom of a swimming pool. The root cause of caustics in a swimming pool is the same as with the shimmering air though - regions of different refractive index (hot and cold air - or air and water) are moving relative to each other and focussing the light in peculiar ways. A plot of the light concentrations in an idealised caustic takes the form of a nephroid.

SteveBaker (talk) 12:31, 16 October 2008 (UTC)[reply]

Thanks! I still wonder if wikipedia has an article on the effect. I just thought of another example of what I am (well was, it is raining now, bloody weather) seeing. You know when you have a large fire, and you get distortion. I am sure this is the same effect as what causes the shadows on the wall, and what you describe, but surely it has an actual name? I'd call it heat distortion or something. 88.211.96.3 (talk) 12:46, 16 October 2008 (UTC)[reply]

It's the same thing - the heat from the fire is now the source of the temperature variation - and the convection and refractive index changes that result from that are exactly the same as in the case of the window. SteveBaker (talk) 12:49, 16 October 2008 (UTC)[reply]

Yes but does it have a name? If it does someone should maybe write a wikipedia article on it, and I may even have a pop at it. 88.211.96.3 (talk) 12:52, 16 October 2008 (UTC)[reply]

Schlieren effect. DMacks (talk) 15:01, 16 October 2008 (UTC)[reply]

Recommended book reading if you're interested in the many effects that air and moisture can have on light: "The Nature of Light and Color in the Open Air", also known by other title variations such as "Light and Color in the Outdoors", by Marcel Minnaert. --Anonymous, 21:36 UTC, October 17, 2008.

Stability of carbon

I came across the following passage: "The rarest carbon isotope is carbon-14, with eight neutrons. Unlike the other two isotopes, carbon-14 is unsta- ble". Is ¹³C stable? Why is C-14 unstable?Mr.K. (talk) 12:17, 16 October 2008 (UTC)[reply]

For smaller elements, the general trend is that elements that deviate greatly from a 1:1 proton-to-neutron ratio are generally less stable. The "stability line" generally trends more towards the neutron side as elements get larger; the actual stability line has a somewhat parabolic character. There are also some trends over the stability of "even-even" proton-neutron nuclei as well. See Isotope#Nuclear properties and stability and Stable isotope for more information on this. The general understanding is that the balance between the three forces which work to hold the nucleus together, which are the nuclear force, strong interactions, and weak interactions, require a certain balance of neutrons to protons (this relationship is not necessarily linear, however). Carbon-14 is "out of balance" with regard to this relationship, and so is unstable. The other two carbon isotopes, C-13 and C-12 are stable. --Jayron32.talk.contribs 12:29, 16 October 2008 (UTC)[reply]

The nuclear force is the same thing as the strong interaction. The relevant forces are electromagnetism and the strong nuclear force (the weak force may be involved as well, but as the name suggests, it quite weak!). --Tango (talk) 18:33, 16 October 2008 (UTC)[reply]

Actually, the nuclear force is not the same as the strong interaction. Read the articles for more on this. What we call the nuclear force used to be called the strong force, however under Quantum Chromodynamics, the strong interaction (as carried by "gluons") is restricted to cover internal forces inside the neutron and proton holding the quarks together. The nuclear force (as carried by pions) is what holds the nucleons to each other. It is generally assumed that the pions contributed by the neutrons are necessary to overcome the force of electrostatic repulsion to be felt by neighboring protons. Pions themselves are made of quarks, so the strong force holds their internal structure together as well. --Jayron32.talk.contribs 21:00, 16 October 2008 (UTC)[reply]

It looks like I'm just not familiar with the terminology (it's not a subject I've studied in depth). They are manifestations of the same fundamental force, aren't they? --Tango (talk) 22:28, 16 October 2008 (UTC)[reply]

Yes; the simple answer is that the force between nucleons is a residue of the force between the quarks that, unlike the "real thing", can act on things (nucleons) that have no color charge, much as the van der Waals force is a residue of the electromagnetic force in atoms/molecules and can act on things (atoms/molecules) that have no electric charge. --Tardis (talk) 01:06, 17 October 2008 (UTC)[reply]

HIV and hemodialysis

Could a HIV virus load be removed through some sort of hemodialysis ? Mr.K. (talk) 12:23, 16 October 2008 (UTC)[reply]

No - dialysis is able to remove certain undesirable substances from the blood by physical and chemical means - but these are typically simple chemical byproducts. The complex biochemistry of a virus makes it hard or perhaps impossible to distinguish and remove by such a simple process. Being a virus also means that HIV has insinuated itself inside the cells of the host's body. Some of those (white blood cells, for example) could perhaps be removed by some hypothetical dialysis process - but there is a limit to the number of white blood cells you can remove and still remain protected against all of the other infectious agents in the environment. But HIV invades all sorts of other cells that are not a part of your blood - so even in theory, with some amazingly high-tech science-fiction blood cleaning machine, you couldn't remove more than some percentage of the HIV load. Viruses (not just HIV - but viruses in general) invade the inside of functioning cells and actually insert their DNA into your DNA. This makes them very hard to eradicate. Viruses replicate by having your own body make new copies - and even if you remove every single independent "virus" from your body - their DNA will still be tucked away inside your DNA waiting for your own cells to produce new viruses the next time the cell replicates. SteveBaker (talk) 12:48, 16 October 2008 (UTC)[reply]

Steve is (as always) right on the money - just thought I'd add a fascinating reference to a study of plasmapheresis for HIV and Hepatitis C virus: [[1]]. As Steve said, even removing a large percentage of a virus from the blood still leaves more than enough inside and outside the cells to maintain the infection. --Scray (talk) 12:58, 16 October 2008 (UTC)[reply]

Steve's answer is a good one, but a minor clarification would be good here. While many types of virus will make use of some of their host's DNA/RNA processing equipment, most viruses don't insert their DNA into the genomic DNA of the host. Group VI reverse transcribing viruses – including HIV – are the key ones that do; in addition to AIDS, these viruses can also be responsible for an assortment of other unpleasant diseases, including some cancers.

Interestingly, some of these viruses have become permanently incorporated into our genome, and may serve a host of useful purposes. See endogenous retrovirus for more details. TenOfAllTrades(talk) 14:29, 16 October 2008 (UTC)[reply]

Note however that even if a virus doesn't incorporate into the human genome, it may still remain latent in the body via episomal latency. See virus latency for some information. For example the Varicella zoster virus, as with other herpes viruses, establishes episomal latency in neurons. After chickenpox resulting from the primary infection it can re-active sometime later in life and result in shingles (something I've personally experienced). Nil Einne (talk) 09:30, 17 October 2008 (UTC)[reply]

Fair enough, though I'm not aware of a true RNA virus that has been shown to have a latent form (not talking about clinical latency). --Scray (talk) 01:51, 18 October 2008 (UTC)[reply]

Effects of masturbation

I am too embarrassed to ask anyone this question face to face, so I thought this might be a good place find an answer. I just wanted to know what the effects of masturbation are in teenage boys. --203.81.223.178 (talk) 15:41, 16 October 2008 (UTC)[reply]

Ejaculation, usually. A feeling of guilt maybe, if you grew up in a culture that still treats masturbation as some vile evil practice that makes your hair fall out. Certainly there are no serious negative health effetcs -- Ferkelparade π 15:47, 16 October 2008 (UTC)[reply]

...unless you do it often enough to get raw ;-). Seriously, it's good exercise, may decrease the risk of getting prostate cancer, will have no obvious side-effects, and everybody does it. --Stephan Schulz (talk) 15:55, 16 October 2008 (UTC)[reply]

Indeed - the general advice "if it hurts, stop" applies to masturbation just as to anything else. If you're doing it right, it won't hurt! --Tango (talk) 16:00, 16 October 2008 (UTC)[reply]

In fact there are possibly some benefits. It won't give you hairy palms, you can rest assured there. If you want more detailed advice, however, you really do need to speak to a doctor (don't be embarrassed, it's a perfectly normal thing to be curious about and doctors will keep your question entirely confidential, you can generally ask specifically to see a male doctor if that would make you more comfortable), we can't give medical advice. I'll give you some general advice, though: People in the playground generally don't have the faintest idea what they're talking about! Ignore any advice you get from a unreliable source, it's probably worse than useless. --Tango (talk) 16:00, 16 October 2008 (UTC)[reply]

There is an addiction side-effect. Addiction is often overlooked as a health problem, but it is a problem. Just noticed that orgasm makes no reference of any kind to endorphin. I was going to point out that becoming addicted to an opioid compound is not difficult. -- kainaw ™ 16:01, 16 October 2008 (UTC)[reply]

Now, be careful. See Chuck Negron. While our article (perhaps rightfully) doesn't cover it, Mr. Negron's penis exploded due to massive overuse of said organ. [2] He reports the incident in gory detail in his own autobiography Three Dog Nightmare (the phrase "split open like a hot dog" appears in the book). He had been warned by his doctor that his level of sexual activity was damaging his L'il Chuck, but he unwisely ignored the advise of medical professionals, and continued to have sex like if he stopped for 10 minutes he might die. He reports an incident that led him to an Emergency Room in Oklahoma; the incident involved an audible tearing sound, lots of blood, and a groupie who will likely be in serious, intensive therapy for the rest of her life. Gives new meaning to the song Mama Told Me Not to Come, now don't it? Moderate levels of any sexual activity is perfectly safe (and some medical studies indicate that a regularly emptied scrotum is vital to helping prevent prostate cancer: [3]), but as with anything, you CAN overdo it. The above does not constitute medical advice in any way. --Jayron32.talk.contribs 16:03, 16 October 2008 (UTC)[reply]

Maybe Mr. Negron unwillingly pressed some Tsubo spot, like in the Hokuto Shinken martial art. Seriously, nobody hurted himself using hands. Some people are injured because they tried drugs or dangerous tools. But if you are worried, talk to a doctor. PMajer (talk) 10:12, 17 October 2008 (UTC)[reply]

ON the contrary, as mentioned by other above, the general friction, even lubricated, of overubbing can cause inflamation and discomfort or injury... Rub the same spot on your arm for hours on end, day after day, and you are likely to wear a sore into your arm. Now, apply that same test to your penis... SOund like fun? The general cautionary tale is that regular masturbation isn't harmful, but obsessive masturbation can be. --Jayron32.talk.contribs 12:13, 17 October 2008 (UTC)[reply]

Sorry but this analogy with rubbing one's hand is not convincing... The mechanics is totally different, in particular there is no friction at all on the skin in masturbation. Among all activities of a teenager, I would say this is the less dangerous (and let's add that normally it is the penis itself who decides when it is enough, and just goes to sleep). Of course too much is bad, but mainly because one misses other important things (reading, studying, socializing, making sport etc). An excessive masturbation activity is often due to a situation of stress, this is the point. Should one understand that his son is doing it too much, I would suggest to check if he is under stress, and why. Just repression would only add other stress. PMajer (talk) 16:45, 17 October 2008 (UTC)[reply]

there is no friction at all on the skin in masturbation - you are clearly basing that view on your own experience, PMajer. I can personally attest to the contrary. If the skin of the palm and penis is totally dry, there's no problem. And if it's well lubricated, there's no problem. But if there's only a small amount of moisture present, that does present a friction problem, and can lead to inflammation. -- JackofOz (talk) 22:38, 17 October 2008 (UTC)[reply]

Then I agree. In particular, it should be recommended not to handle chili peppers before. This is also based on one ancient personal experience :( --PMajer (talk) 19:25, 18 October 2008 (UTC)[reply]

Is it just me or "masturbation in teenage boys" is ambiguous? Indeed, it can be quite destructive to the personality of teenage boys. Mr.K. (talk) 16:11, 16 October 2008 (UTC)[reply]

I don't see an ambiguity and it's only likely to destroy someone's personality if they become addicted, which is unlikely. --Tango (talk) 16:38, 16 October 2008 (UTC)[reply]

I think Mr K. is playing on the word in there. As in, "The sixty year old felon who performs masturbation in teenage boys...". Personally, I think the terms masturbation and teenage boys are pretty much redundant. Matt Deres (talk) 17:33, 16 October 2008 (UTC)[reply]

Are you suggesting that adults and girls don't masturbate? I think the statistics would disagree with you there... --Tango (talk) 17:38, 16 October 2008 (UTC)[reply]

It's a superset/subset, not equivalency. He didn't say which term is redundant (and could be omitted without changing meaning) vs which term is the more limited in scope. Given "a hot fire", consider "a fire" vs "something hot" (assuming you have more experience with flames than with chicken-chokin' and...whatever the comparable female slang would be:). DMacks (talk) 18:16, 16 October 2008 (UTC)[reply]

Ok, I suppose that makes sense - since said both were redundant I assumed he meant they were mutually redundant (ie. equivalent), it makes more sense if he only meant one of them was redundant given the other. --Tango (talk) 18:27, 16 October 2008 (UTC)[reply]

Saying that teenage boys masturbate and other people masturbate too is like saying bullets are dangerous when fired from a machine gun and also when thrown freehand. Matt Deres (talk) 13:28, 17 October 2008 (UTC)[reply]

You should maybe redirect your question to the Reference desk/Masturbation in order to get an expert opinion. PMajer (talk) 13:12, 17 October 2008 (UTC)[reply]

The expert? DMacks (talk) 20:30, 17 October 2008 (UTC)[reply]

Why does pressure increase with depth

If pressure is caused by experiencing the sum total of elastic collisions of fluid molecules against an arbitrary plane (my words as I'm writing them)...

but given my current definition for studying fluid mechanics, this doesn't explain why the collisions are "stronger" at deeper depths because the density is the same and the temperature is the same.

If water molecules are crashing into the vertical wall of a dam then why should forces coming from above affect the frequency or impulse of collisions? If the impulse was greater, that would imply greater kinetic energy in the fluid molecules, which means they have more energy.

Take a 100 meter high dam, the horizontal collisions near the bottom should not depend on molecules above them, which act in a perpendicular direction, right? Water has the same density, and lets stipulate that the fluid is iso-thermic

How does defining pressure this way not explain why at greater depths, pressure is higher? I think it is because pressure exists in all directions right? So the upward force = the horizontal force. Right? So does that mean that the collision frequency is higher? or that the impulse per collision, greater? Or am I just completely wrong. Sentriclecub (talk) 17:49, 16 October 2008 (UTC)[reply]

I'll take a stab at it (though admitedly by stat-mech/thermo is a bit rusty… Pressure really does exert "in all directions" as you suggest. Molecules don't just bounce "up and down or side to side", but at all other angles as well. So vertical pressure (weight of what's above, or the effect of vertical collisions each pushing down upon the next) can exert a non-vertical force if any molecule has a non-zero horizontal component to its motion. DMacks (talk) 18:08, 16 October 2008 (UTC)[reply]

Additionally, I've spent 5 days studying the topic, and know it can be explained by Pascal's law, but I want to know why it can't be explained or deduced from the method of looking at pressure as ultimately a result of elastic collisions on an arbitrary surface. There's only two ways to increase the average force of a series of impulses... either increase the frequency, or increase the impulse per collision. How does the random translational motion increase with depth, if density and thermal energy and temperature stay the same? I've already asked on yahoo answers, and got a starred question award, and 2 wrongs answers ("because density increases") and the other answers are not helpful, and they use weasel words "well could be..." Sentriclecub (talk) 18:13, 16 October 2008 (UTC)[reply]

DMACKs, Yes I understand they exert a force downward onto other molecules, but the bottom line is that somehow there is a resultant increase in collision frequency or an increase in impulse magnitude. I fully understand Pascal's law, but my book doesn't answer my specific question, nor the related wikipedia articles. Sentriclecub (talk) 18:18, 16 October 2008 (UTC)[reply]

So far the best answer from the other website is...You asked a very good question. You can't quite compare gas and fluid but for simplicity, I will. Water is NOT incompressible. As you go deeper, its density increases with pressure. Before you go down, there is very little room between the molecules. But you go down 30 m, there is now even less room. This means that the number of collisions will go up dramatically. So the pressure increases. The water, however, keeps the same kinetic energy and temperature. BTW: The reality is a little different. The water molecules vibibrate against each other, but they also impinge on each other's electron clouds. Kinda of a spring effect, so it is not quite so straight-forward.

My next guess is through statistics and a normal distribution. Maybe if the layer of water molecules on the bottom of the fluid can only bounce between 0 and 180 degrees, then the second layer has an extremely small chance of colliding straight down because of the concentrated upward range of the the molecules directly below it. In other words, near the bottom, the statistical distribution of all possible collisions that can occur in 3d space isn't equally probable. Molecules on the bottom have zero probability of being able to return from a collision against the ground with a downward momentum. That is a majority of molecules near the very bottom can only have an upward momentum at any given time. This "fact" must be balanced by the second layer of water molecules who must pass the "fact" along, and give this problem to the layer higher up, who must eventually "resolve" the fact, by being basically free to have zero net momentum as observed in the top layer. This is just a educated guess, based on creativity, not science. Sentriclecub (talk) 18:46, 16 October 2008 (UTC)[reply]

To put in better terms, 100% of molecules on the bottom layer must have a upward momentum immediately after a collision. 50% of molecules on the bottom layer have an upward momentum before a collision. Thus the net momentum of the bottom layer is 75% upward, if a reasonable assumption is that all the bottom layer fluid molecules are returning from a collision, and fixing to collide again. A net downward force would need to act on the bottom layer to counteract the upward momentum as suggested by the 50%/100% theory. Still this is all a guess, but I have been unable to finish writing my notes because I hate the feeling of a lingering doubt on pressure. Sentriclecub (talk) 18:52, 16 October 2008 (UTC)[reply]

Just a teaser, and I plan to write a long response this afternoon when I get a chance. However, some of the confusion stems from drawing a false analogy between pressure in gases (mediated primarily by kinetic energy) and pressure in liquids/solids (mediated primarily by bond energy). Dragons flight (talk) 18:58, 16 October 2008 (UTC)[reply]

I think I'll take a wild guess. How about gravity? Looking at a free water particle outside of a gravitational field, any of the 360 degrees are equally possible. Within Earth's gravity, however, it becomes more probable that the particle has some "downward" component to its velocity vector. These add up the deeper you go, resulting in higher pressure. The interesting question would be: If you had a jug of water (maybe a large pool-sized jug) in space, would the pressure change the "deeper" (more towards the center) you go? I don't really have an answer to that. --Bennybp (talk) 19:50, 16 October 2008 (UTC)[reply]

Actually I didn't really answer the question (why it would be omni-directional, and not just downwards). It's probably a pretty flimsy theory, but - Due to the pressure from the molecules coming from the top, the molecules at the bottom would be given more side-to-side velocity, since 1.) They have the same average speed (same temperature), and 2.) There's a limit on their direction of motion ("up" is no longer available, so all others become more probable). Man, I think I'm just digging myself a nice hole. I eagerly await Dragons' answer :) --Bennybp (talk) 20:03, 16 October 2008 (UTC)[reply]

The false analogy here is that the treatment of "pressure" for the gas phase is the same as for "condensed" phases. It isn't. For gases, there is roughly a linear relationship between pressure and density; double the pressure, double the density (I say roughly; its exact under the Ideal gas law; the Van der Waals corrections alter this slightly). That is because, for gases, the intermolecular distance is roughly 1000x greater than the molecular radii, meaning that gases are compressable (you can simply "push" the gases molecules together). The entire question by the OP assumes this compressable model of pressure; it works fairly well for gases. However, liquids and solids are condensed phases; for all intents and purposes the intermolecular distance between them is essentially nil. (they can be compressed slightly; water at 100 m below the surface is slightly more dense than the water at the surface). In the case of condensed phases, pressure is determined by weight of the bulk material above you. That's it. Pressure is only force per unit area, and weight is only a force. With a gas, the force is primarily determined by collisions, since the gas molecules don't remain in contact with the surface for any meaningful length of time. With condensed phases, the surface is essentially in constant contact with the molecules (and for a liquid, the exact molecules shift places because it is a fluid, but in bulk, essentially the entire surface where the pressure is measured is totally covered with the molecules). Consider two thought experiments:

Imagine burying someone under a 100 m tall pile of sand. What happens? We say they are crushed to death because of the weight of the sand on top of them But this is merely convention. The weight is a force, and it crushes them by pressing in on the surface of their body? What is a force distributed over a surface? Pressure... Now, replace "pile of sand" with "depth of water". Its the exact same problem.
Imagine the same person, with appropriate breathing apparatus, encased in a cube of water 100 m on a side, but the person and the water are in a zero-g situation, like floating in space. What happens? Nothing. They survive fine, because the water doesn't press them because there is no gravity to force the water in any one direction. Even if they swim to one side or the other of the cube, there is no net pressure effect, because there is no gravitational force. No force equals no pressure.

Does that make sense to everyone, the OP's question is in error because it makes assumptions about pressure which are incorrect to the situation; liquids are fundementally different than gases WRT molecular organization, and that fundemental difference affects the way that the molecules exert a "force" on their surroundings. --Jayron32.talk.contribs 20:26, 16 October 2008 (UTC)[reply]

Hooray for that answer (well, for the first part - the second part is muddled ;-). People are (mostly) water, so with "appropriate breathing apparatus", we survive fine under a hundred meters of water, even under gravity. See SCUBA. We are crushed only if the pressure does not come from all sides equally, or of there are unequalized air spaces in the body. --Stephan Schulz (talk) 20:43, 16 October 2008 (UTC)[reply]

I think you are mostly correct. There's still a little bit of explaining to do as far as it acting on all sides at once, with equal pressure. Consider a perfect structurally-sound sphere. In my mind, under 100m of sand (with plenty on the sides, too), it would be crushed flat. The pressure is not equal all around. In water, in my mind, it would implode - the pressure being equal on all sides. This would be due to the fact that water cannot be treated completely as a bulk material - there still is plenty of molecular motion to account for - sand is held in place with friction. If the sand were frictionless, it would be just like macroscopic water (ignoring hydrogen-bonding, van der Waals forces, etc.)

I believe modeling water as a pseudo-gas is correct. The amount of material above you does cause the increase in pressure - for condensed phases and the gas phase as well. On Earth, our air pressure is omnidirectional - we are under a bunch of atmosphere. Same with water at the bottom of the ocean. Where does the "sideways" pressure come from? I guessed at that above. (I say authoritatively, but I might run to my p-chem book to find some equations) --Bennybp (talk) 20:57, 16 October 2008 (UTC)[reply]

In answer to Bennybp's question about water in zero-g, the water will attempt to form a sphere since that's the least energy configuration (because of surface tension, I believe - a sphere has the least surface area for a given volume). The pressure will increase as you move towards the centre because of the gravity of the water itself, however for a swimming pool sized amount of water that gravity (and therefore the pressure) would be minimal (in fact, it may not even be enough to keep the water together depending on initial conditions and outside influences - the water may break up into lots of smaller spheres). For a planet sized amount of water, the pressure at the centre will be very great. --Tango (talk) 20:40, 16 October 2008 (UTC)[reply]

Thanks Tango. I feel I knew that at one time. The more I read the answers above and below the more I realize I think I have the right answer but to the wrong question. Ah well, better luck next time :) --Bennybp (talk) 15:38, 17 October 2008 (UTC)[reply]

DF answer

Sentriclecub, your problem is mostly one of a false analogy. In gases, particles act more or less independently. In that case, pressure is primarily the result of elastic collisions and is proportional to density times temperature/kinetic energy. Hence increasing pressure implies increasing either the density or the energy per particle.

However, liquids and solids are not gases. Pressure is mediated not by kinetic energy but rather primarily by changes in the intermolecular potential energy. To see how this works, let's consider the other extreme: solids. A typical solid at fixed tempertaure can resist very large pressures with very little appreciable deformation (i.e. very little change in density. It does this because the bonds between atoms resist being compressed, and hence provide a spring-like quality to the matrix of the solid. It is that springness, essentially a result of potential energy stored in compressed bonds, that provides the force to resist compression.

Now consider what happens at the interface. For a gas, the force it applies to a wall is caused by particle bouncing off. For a solid, the force is more direct than that. It is caused by the electrostatic repulsion of the electrons in the solid interacting with atoms in the wall at the point of contact. That is a direct steady-state impact, that occurs independently of how rapidly the molecules might be moving.

Like solids, the pressure in liquids is also dominated by the effects of intermolecular "bonding", though it is harder to understand because such bonding in liquids is transitory and mutable. Molecules in a liquid can slide past each other and be rearranged, but nonetheless they try to maintain a roughly constant distance from each other due to the electrostatic interactions between molecules. It is that constant interaction with their nearest neighbors that allows liquids to resist compression and have a preferred density.

So that answers part of your question. Rather than being related to molecules bouncing off of the wall (as in a gas), the pressure the water exerts is caused by the compressed intermolecular interactions within the water and in continuous contact with the wall.

You also ask why is the pressure the same in all directions. Again I am going to start with solids and work my way over to liquids.

In a solid there is a stiff arrangement of atoms that can thought of like masses seperated by little springs. For the sake of discussion let's start with a 1-D set of mass and springs:

 O -vv- O -vv- O

If you apply a force along it's length, you compress it:

 O -w- O -w- O

As long as that is held perfectly linear, the forces balance and cancel. But the real world is not linear. We have vibrations and other imperfections, and given the opportunity, the middle mass would like to relieve the tension in the springs by skewing out of the line.

    |- O -|
 O -|     |- O

For a solid, linear compression creates a relatively small sideways force because other atoms in all the other directions are approximately fixed and restrain this sideways skew. If you kept applying force though, the solid would bulge out sideways and eventually break.

So how does that apply to a liquid. Well like a solid, if you apply a linear force, that tends to squeeze things in the middle. However, unlike with a solid, the individual atoms are free to move. So if you try to squeeze water it will squirt out sideways immediately. The only way to stop it from doing so is also apply an additional sideways force to restrain it. It may not be obvious, but the amount of lateral restraining force ends up being exactly the same as the amount of compression force being applied. Or, more simply, the amount of force in all directions must be the same in order to keep the liquid stationary. And, as if by magic, we arrive at the concept of pressure, i.e. a force acting equally in all directions. Dragons flight (talk) 22:10, 16 October 2008 (UTC)[reply]

Perfect explanation. This whole time I was thinking that pressure is just a corollary from the concept of force. This is exactly like the time I sought a relationship between impulse and work, because I just knew there was some deep intuitive relationship, and after several days and dead-ends I finally saw that they are related by average velocity. This explanation through the analysis of force, is exactly the way my brain needed to understand it. It makes sense that if the additional sideways force wasn't as large as the linear force, then a very slight disturbance to the unstable equilibrium would further propel that unequilibrium in that continuing direction.

In my own notes on Pascal's principle, I wrote: An enclosed liquid is extremely inefficient at resisting pressure. If force is applied at one point, the fluid responds everywhere as an equal & opposite reaction, as if the pressure was thought to be coming from everywhere. because this is intuitive and explains how a tall narrow straw filled with water could add several billion newtons of pressure to an enclosed liquid in a multi million gallon rigid tank, according to an ideal fluid. Your answer meshes perfectly with my notes from yesterday, and now I can expand on my notes using the same analysis method proving everything through force. What a swell day! In order to maintain the 3-d matrix, the force must exist everywhere continuously, and varies with respect to height, in order to "balance all the equations" and not violate any force laws which stipulate that an imbalance would cause an acceleration to some part of the system.

The combined answers today are definitely the best help I've had from the ref desk. I don't have a professor or t.a., just a physics book, and youtube[4], and the ref desk. My passion for learning is fueled by making connections between concepts allowing such insights into nature that I feel fortunate for being invited to see. Just as chess grandmasters have their priceless insights into chess, I cherish mine into nature(though I'm still a noob), as nature reveals herself through physics and everything else. Sentriclecub (talk) 22:38, 16 October 2008 (UTC)[reply]

Hi. Going back to the OP and the issue of a molecular explanation for pressure in liquids, the stiffened equation of state section of equation of state might be useful for understanding the compressibility of water. Water is "like air that is already under 20000 atmospheres of pressure" which explains why it is essentially incompressible under daily life (taking water from 1 to 2 atmospheres is like taking air from 20001 to 20002 atmospheres). HTH, Robinh 07:12, 17 October 2008 (UTC)

bladder TCC insitu with bone mets

a patient has been diagnosed bladder TCC insitu for many years, is recently found metastasis to bone. is it possible that a bladder TCC insitu patient has a bone mets?

Thanks,

George —Preceding unsigned comment added by Gxu (talk • contribs) 21:22, 16 October 2008 (UTC)[reply]

Sorry, no medical advice allowed here (see top of page). --Scray (talk) 22:53, 16 October 2008 (UTC)[reply]

Sounds like home work. We do not do homework either. Look up your text book.--GreenSpigot (talk) 00:29, 17 October 2008 (UTC)[reply]

Both the above comments are unnecessary. No medical advice was asked for: medical information was ("Can in situ carcinoma (as opposed to invasive carcinoma) metastasize to bone?"). And it's clearly not homework, nor would it need comment, especially one like "look up your text book", if it were. - Nunh-huh 00:53, 17 October 2008 (UTC)[reply]

Actually it could be homework. Cancers preferentially metastasize to certain other tissues - that would be a valid "homework" question, although someone at that stage of education would hopefully not be checking Wikipedia (though I know a doctor who regularly checks eMedicine for more reliable, but still background, info). And it could also be a request for "advice", as it's a pretty specific question. If we answer anything other than "maybe, ask your doctor", we are providing specific information - but we don't get to see the biopsy results, X-rays, MRI's, diagnostics from the bloodstream, nothing. So - maybe, see your doctor. Franamax (talk) 06:09, 17 October 2008 (UTC)[reply]

I've been through every course it could conceivably be "homework" for, and such classes don't assign that kind of "homework". Clearly not a homework question, and - even if it were - the original questioner will not be in any way enlightened by learning that it "sounds like homework" to a Wikipedia user. And no, a request for information, no matter how specific, is not a request for advice. - Nunh-huh 17:11, 17 October 2008 (UTC)[reply]

This statement: I've been through every course it could conceivably be "homework" for, and such classes don't assign that kind of "homework" is remarkable, considering the number of professions (medical, nursing, graduate programs...), cultures, etc that are "conceivable". Wow. --Scray (talk) 01:21, 18 October 2008 (UTC)[reply]

No matter the profession, the class would be "clinical pathology (human)", "oncology", or some variant thereof. Wow indeed. - Nunh-huh 01:42, 18 October 2008 (UTC)[reply]

Try Metastasis for the mechanics of it. Julia Rossi (talk) 07:21, 17 October 2008 (UTC)[reply]

Yes. See this article. Axl ¤ [Talk] 08:26, 17 October 2008 (UTC)[reply]

Thanks, I didn't want to post the answer without a reference. Great work! - Nunh-huh 17:11, 17 October 2008 (UTC)[reply]

The article this readers wants to read is Transitional cell carcinoma, which is a minimal stub. The question is encyclopedic in nature (just ignore the stated reason for asking; for our purposes it is irrelevant). --Una Smith (talk) 04:46, 21 October 2008 (UTC)[reply]