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October 28

Ice cube capilliary free energy?

In a control experiment, two ice cubes are created in a freezer and they are placed in a dish of water. They are well separated and stay where they are until they melt. X amount of energy was used to pump water out the dish into an ice cube mould. In the second experiment they are placed closer together and capilary action makes them move together. If they stayed solid it would take just as much energy to move them apart. But if they melt and the same process as in the control experiment takes place, have we "separated" them for nothing, gaining free energy in the process? How can the freezer retroactively "know" the fate of the ice cubes (or the pump) to satisfy the second law of thermodynamics? (Trevor Loughlin) — Preceding unsigned comment added by 91.224.27.231 (talk) 13:11, 28 October 2013 (UTC)[reply]

Touching or no, how could they stay solid after being placed in water? ←Baseball Bugs ^{What's up, Doc?} carrots→ 14:19, 28 October 2013 (UTC)[reply]

You are failing to account for the work you performed in moving the ice cubes. If there is a potential energy gradient that depends on the separation distance (and there is a tiny one, due to the effects of surface tension and capillary action), then you will feel a force along the direction of separation, when you are moving and placing the ice cubes in the tray of water.

But, that is a tiny effect. It is so small that it is dwarfed by the air resistance acting on the ice cube as you lift it and place it in the diah of water; the air resistance is yet another non-zero effect that we can neglect because its magnitude is so small.

Were we to set up a controlled experiment to measure such an effect, we would need to manage all other sources of error to ensure that our tiny, non-zero physical process - the potential energy gradient related to the surface mechanics of capillary action - does not get lost in noise. Nimur (talk) 16:12, 28 October 2013 (UTC)[reply]

Touching a black dwarf

I know they probably do not exist yet, but they are predicted to exist if a white dwarf cools down. I was wondering, is it possible for a person to actually touch it, or would they be killed some other force? What would happen if hydrogen or any other peice of matter touches it? It's supposedly pure electrons, I'm just wondering what a solid object made of electrons might be like and if it can actually be touched like any other solid. Would some process occur that causes things to rapidly heat up if other matter touches a black dwarf? ScienceApe (talk) 17:31, 28 October 2013 (UTC)[reply]

It should feel just like a solid material, the only problem that I can see is that to touch it you have to go down the surface where the gravitational acceleration will be about 5000 g. Count Iblis (talk) 17:54, 28 October 2013 (UTC)[reply]

Note that we do have a black dwarf article. StuRat (talk) 18:29, 28 October 2013 (UTC)[reply]

The incredible surface gravity would crush the person against the surface and the atoms of the person's body would be incorporated into the star. Resistance would be futile... Dauto (talk) 19:00, 28 October 2013 (UTC)[reply]

Only if it was a superconductor. :-) StuRat (talk) 19:02, 28 October 2013 (UTC) [reply]

I believe it would inherently be so, being electron-degenerate matter (even though the article doesn't mention this). — Quondum 03:11, 29 October 2013 (UTC)[reply]

What would be the drifting velocity of an electron at those densities? Plasmic Physics (talk) 02:10, 29 October 2013 (UTC)[reply]

@ScienceApe: "It's supposedly pure electrons": Not quite, it is a plasma of nuclei and electrons.

@Plasmic Physics: Since the electrons are predominantly (or almost entirely) in the degenerate state, I understand that this implies that they will be superconducting. Not sure this answers your question? — Quondum 03:11, 29 October 2013 (UTC)[reply]

No matter what kind of gravity the outside of an object is under (even a neutron star - look at the structure image) the pressure will be very low in the outer atomic layer, and it will therefore not be degenerate matter. Wnt (talk) 16:42, 29 October 2013 (UTC)[reply]

True. The outer layer would probably be normal matter. In further response to "or would they be killed some other force", and assuming we are not concerned about the effect of 5000 g ("It's not the fall that kills you, it's the sudden stop at the end"), there is another factor to consider: the tidal forces, which I suspect will rip a person apart before they get near the surface, though I do not have figures. — Quondum 22:22, 29 October 2013 (UTC)[reply]

Infrared and microwave radiation

When directing equally strong infrared or microwave radiation towards a human, why is microwave radiation more harmful even though its wavelength is longer? Th4n3r (talk) 18:16, 28 October 2013 (UTC)[reply]

Infrared creates heat right on the surface of the skin or clothes, where it can quickly dissipate into the air, while microwaves of the frequency used in microwave ovens penetrate maybe an inch, and that heat doesn't dissipate nearly so quickly. Also, microwaves cause arcing when they hit metal parts, such as in a pacemaker. StuRat (talk) 18:22, 28 October 2013 (UTC)[reply]

Thank you. Th4n3r (talk) 19:46, 28 October 2013 (UTC)[reply]

I think most of the difference is with resonant absorption peaks of liquid water. I believe arcing is only a problem within the cavity and standing waves (a microwave oven cavity supports a finite set of TE modes and metals disrupt that and induce high currents and voltages. I suppose a wire of a pacemaker might have a microwave resonance but it seems unlikely.) I don't know if it's true that all frequencies considered "microwave" would be more harmful than all frequencies considered "infrared." It would depend on how the human body and it's constituents react. Throughout the EM spectrum the body is transparent, reflective and absorbent of different frequencies depending on the chemical. Pulse oximetry is one example of how spectral differences exist and can be used. Microwave ovens cook through a specific resonance with water but a microwave speed measuring gun doesn't have the same effect. --DHeyward (talk) 11:15, 30 October 2013 (UTC)[reply]

Yes, that's why I said "microwaves of the frequency used in microwave ovens". StuRat (talk) 21:58, 30 October 2013 (UTC)[reply]

Sorry if I misunderstood. The frequency heating water would be independant of the cavity. The arcing of a pacemeker (or any metal) is very dependent on the cavity so is not likely the same issue as a person standing near a radiating source vs. a piece of metal with a gradient against the electric field in a cavity. The cavity wants to have a static field while the metal wants an equipotential surface (and a 0 electric field at the cavity boundary). This can cause arcing. Outside the cavity, metal isn't really an issue. Think of a piece of aluminum that spans from the TE10 peak field to near he cavity edge. The aluminum will want to have an equipotential surface but the charge will be moved the the edge closest to the cavity edge. This can create heating currents and arcing voltages. But outside the cavity it just reflects the energy. --DHeyward (talk) 07:05, 31 October 2013 (UTC)[reply]

How to research symbols?

If you find some formula or symbol that you don't understand and doesn't even know the name of it, how would you look it up by yourself?

Take for example:

\sum _{k=1}^{n}k

or

{n \choose k}

Where would you go to know what they mean? Is there some tool where you could proceed in a tree-like manner until you find what you are looking for? Or maybe some tool where you draw them and it outputs the name?OsmanRF34 (talk) 19:12, 28 October 2013 (UTC)[reply]

List of mathematical symbols gives many of the more common maths symbols. There is a separate maths reference desk, for symbols in science it would depend on what area you were interested in. Dmcq (talk) 20:04, 28 October 2013 (UTC)[reply]

TinEye supposedly lets you supply a picture and scans the web for similar pics. I've never had much luck using it myself, though, so I can't recommend it. StuRat (talk) 21:42, 28 October 2013 (UTC)[reply]

Google Images has the same functionality - and works a lot better, IMHO. However, it wouldn't help with things like:

{n \choose k}

where a simple parenthesis symbol is being used for (in this case) notating a column vector rather than the more common use of grouping arithmetic operations (eg. 4x(3+1)=16). SteveBaker (talk) 03:45, 29 October 2013 (UTC)[reply]

It would depend on context, but that's almost certainly a binomial coefficient, not a column vector. Red Act (talk) 08:05, 29 October 2013 (UTC)[reply]

And, of course, this is the ideal place (either the Math or Science Desk). Draw a symbol, upload it here, and there's an excellent chance somebody will tell you what it is. StuRat (talk) 21:44, 28 October 2013 (UTC)[reply]

Yeah - that's a sure-fire way to find out - but it typically takes a few hours to get an answer. 03:45, 29 October 2013 (UTC)

Symbols.com has the great idea of an index of symbols searchable by graphic form but has a woefully small database. They have some math and logic symbols but seem more aimed at mystic users. Spinning Spark 19:02, 29 October 2013 (UTC)[reply]