studiot

Well I will actually add the +1 for that little epithet. But for the other condition I am unsure about invoking superposition. That surely is about states, which are energy states not times or places, and combinations of these states (linear or otherwise) form the 'superposition', rather as combinations of particular solutions of differential equations form more general solutions of that equation.

That was no answer, don't mess about.

Please rephrase this in English.

Does this exclude from the Universe the empty space inside atoms, between atoms, between the Earth and Mars, between the galaxies and so on?

First it is necessary to distinguish between compressors installed as part of the nuclear power plant and those which are on site for running tools and equipment. So installed compressors. What does nuclear power produce? Large quantities of heat; it does not produce electricity. So the heat has to be used in some form of conventional electricity generator using conventional thermodynamic machinery. So steam or other gas turbines to drive the generators. The steam or gas is pumped to its destination. A compressor is a pump for gas. Compressed gas can also be used to drive machinery (more of the safety aspects of this in a moment). The steam or gas is used to remove heat (cool) the reactor. In some reactors the reactor itself is liquid cooled and the heat transferred to steam/gas via a heat exchanger. Either way the immediate ractor coolant is separated from the output hot gas by at least one stage of heat exchanger to avoid transferring radiation with the heat. Now the plant, as a whole, generates electricity. But electricity itself can't be stored in significant quantities so if the generators break down then there is no electricity to run electricl pumps. So for safety reasons the machinery is mechanically driven and this usually means indirectly via a compressor and / or turbine. It is particularly poignant to realise that the mechanisms to damped the nuclear reactions should not depend upon electricity to operated and compressed gas is the ideal answer. Compressed gas can be stored Offers no fire or spark risk like electricity Can be instantly available Offers no shock risk like electricity For this reason air tools are used in hazardous atmospheres, in preference to electric tools. Air tools are usually lighter and harder wearing than their electric equivalents as well. Roger, if you had offered a better (more detailed) question in the first place you might have received this sort of response earlier.

It is often forgotten that a 'body' can execute six types of motion. 1) Rigid body translational motion 2) Rigid body rotational motion 3) Non-rigid body translational motion 4) Non-rigid body rotational motion 5) Some form of oscillatory or repetitive motion. 6) Some combination of these So when anyone says The Universe is in motion, the first thing to ask them is what sort of motion they mean and also what they mean or include in 'the Universe.' My definition of the Universe roughly corresponds to 'everything there is including empty space' So that immediately rules out any form of translational motion as there is nowhere else for the Universe to translate to. But that still allows the universe to be constantly churning like a washing machine or tumble drier.

I'm sorry to tell you that you have missed the point I was trying to make. So I will try to explain more clearly, working from everday experience. The full answer to the OP question of what warps space is founded partly in a pure mathematics theorem by Gauss and partly in a principle of physics brought to the fore by Einstein. It also conforms to Studiot's law that fact is stranger than the most fanciful fiction. So much so that we overlook the implications of everday experience we are so familiar wth. So to start with that, look at this picture. The object is instantly recognisable, despite the unusual orientation. So how do we recognise it? Well we unconsciously use Gauss so called 'remarkable theorem' The version I am using here states that objects look the same regardless of their orientation in a coordinate system, because they have some uniquely identifiable properties that depend only on the object and not on the coordinate system. Alternatively these properties are the same in all coordinate systems. (Does that phrase sound a bit familiar?) We can either consider different orientations or the object or the background grid or both as in the next sketch. Note carefully that I have shifted and rotated the grid in each of the three examples. The petal shape is simpler than the boat but adequate for our purposes because it has varying curvature. This curvature is called gaussian curvature and it is a fundamental property that is called an intrinsic property because it only depends upon the petal. Because it only depends upon the petal we can in fact take away the coordinate system altogether and just be left with the one dimensional object I call the petal. It is one dimensional because I am only concerned with the thick black outline not the interior and I can calculate the curvature at any point along the petal just by knowing how far along the black line I am. This is called using arc length as a parameter. So that is the maths part of the conundrum. The key to the physics part to to realise that the statements about all coordiante systems being equivalent is echoed from the maths into the physics. But basically Gauss remarkable theorem translates to there is no absolute frame of reference We can discuss this next time, when you are sure you have got this idea sorted.

These were to illustrate point in post 8. Why can't you apply your epsilon-delta definition to them, apart from the fact that your statement is a bit loose, since you should note that both epsilon and delta have to be strictly greater than zero. Also 'an infinitely small epsilon' is an arguably poor way to say 'for all epsilon greater than zero'. A correct version of the definition is The function f(x) approcaches a limit L as x approaches and stated [math]\mathop {\lim }\limits_{x \to a} f(x) = L[/math] If and only if for every [math]\varepsilon > 0[/math] there is some [math]\delta > 0[/math] such that, for all x, if [math]0 < \left| {x - a} \right| < \delta [/math] then [math]\left| {f(x) - L} \right| < \varepsilon [/math] Why does this not apply to show that the limit as x tends to zero of my first function is zero? Can you see what the second function is designed to show and why I gave the value of f(0) = 1?

Rogers you might like to consider the following situation:- A container (on Earth and subject to gravity) is partly filled with liquid and moving in a straight line. It is subject to a constant acceleration. Can you describe and explain the forces acting and the shape of the surface of the liquid?

Sadly, not soon enough.

Yes, unfortunately. A space has more structure than a manifold as it not only includes the stage or region where the activity takes place but also the sets of quantities involved ie the actors and the relationships between them ie the text of the play. A manifold is little more than the stage.

Space is not warped. It follows its true and proper form. The very statement 'warps space' implies there is some unwarped absolute coordinate system to compare against. You will struggle with Relativity until you can abandon the idea of absolute space or absolute time. Draw a circle or part of one. What warps the line you are drawing? It is not a straight line - would that be somehow preferable? Is the circle a warped straight line and is the straightline absolutely perfect? Or is the circle correct for a circle?

The problem is What I am referring to in emboldened sentence is the thermodynamic idea that heat is the 'lowest grade of energy'. That energetic processess degrade higher forms like electricity or potential or motion energy to become heat. And the problem is that recovering light is trying to reverse this. Of course it can be done, but it costs you a lot in energy efficiency terms. Perhaps swansont knows more about this than I do but can't you contain laser light between mirrors and open on demand with a kerr cell shutter?

What is the battery in your car? Do you think you get the same electrons out that you put in when charging? So how to store photons as I outlined in post#4. That is photons in and later photons out. All afternoon my car has been sitting on the drive absorbing all those nice yellow photons. Tonight, after dark, members of the local furry fraternity will sunbath by basking in the infra red photons radiated under my car. There ya go, photons in photons stored, photons out. I never said they were the same photons, any more than Mr Chloride said the electrons from the car battery are the same.

No the question was Why can't light be stored? (which, of course, contains an incorrect assumption) not how can it be stored?

Wave v corpuscular doesn't really matter for the purpose of light storage.

I didn't say you could store light in a battery, I just said it could be stored.

None of the classical or semi classical theories of metals and semiconductors (Drude, Free electron, Simple band) can explain the so called 'anomalous' Hall effect coefficients. You need the full 3D quantum mehcanics combined with crystallogrphic lattice theory to get anywhere near it.

How about these functions? [math]\left\{ \begin{array}{l}f(x) = 0,x \ne 0 \\f(x) = 1,x = 0 \\\end{array} \right\}[/math] [math]\left\{ \begin{array}{l}f(x) = 0,x < 0 \\f(x) = 1,x = 0 \\f(x) = 2,x > 0 \\ \end{array} \right\}[/math]

Holes are formed when an electron is promoted into a higher level band, leaving behind a hole. The electrons we are discussing are already in the band. So there are no holes.

1) Yes this is true of monovalent metals. The monovalent metals enjoy a single electron in the outer s orbital in the free atom. Thus the orbital is half full. When N such atoms combine to form a metallic solid these combine to form the conduction band each original s orbital contributes 2 spaces to the band. So the band has "N spaces and N electrons. So it too is half full. The next (empty band) is formed from the empty outer p orbitals. This has a definitely higher energy than the s conduction band. So there is a definite gap between the upper half full band and the next empty band. 2) So why does the resulting monovalent metal sold conduct? Well the conduction band is half full. This is ideal to accept an electron moving in from somewhere else. If fact the maximum number of moving electrons that can flow is N. N are moving in and N are moving out of a given zone. To achieve this you need N filled and N vacant spaces. An insulator has full band and cannot accept further electrons. It is possible to develop this simplified explanation to cover pretty well all cases and consider the roles played by the various gaps, overlaps and abutments between bands.

So is it abandoned? Here is a better reference. Clark E S , Acta Cryst, 1958, 11, 284

It was a couple of days ago. I'm almost sure DrP gave the only answer. I was going to post a fuller one, but just can't find it. Can't remember the title but the OP was interested in auric compounds in general and auric chloride in particular.

I could have sworn there was a question recently about this but I can't find it. Can anyone help?

That's the day Marvel Comics allows Batman to take the post#3 hotrod for a spin.