studiot

I know that , but any Engineer would (should) shy away from describing the failure of concrete in compression as cracking. It doesn't. It suffers crushing failure if at all. But don't forget how difficult it is for any material to fail in compression. In order for that to happen it has to be in triaxial compression, otherwise some other form of failure occurs. The 'compression' failure of concrete cubes, for instance is (uniaxial compression) is a diagonal shear failure, and the BS rules out cube failures which depart significantly from this pattern.

Thanks to all who replied, particularly Mordred and Uncool (+1 apiece). I really hadn't considered the time aspect which makes it impossible in space time for the same coordinate points to be traversed, even if you take the scanario I proposed which was simultaneous viewing from both ends. I suppose we will have to rephrase it so that the timescale of observation is short compared to that of any material changes. I really don't know the answer to my question which is why I asked it. This is all part of my attempt to reconcile theoretical and observational / practical Physics.

Please explain why concrete in such heavy compression as you describe is 'cracked' ?

Gosh this really is a piece of string question. What country are you in? This is relevant because in some places they build to the codes, in some places they ignore them. You need to provide a lot more information about what you really want to do, and, as Bender says, about the ceiling, walls and so on.

Thank you for posting this information. +1 It will take some time to digest it. Aberdeen huh? I've just collected someone from Aberdeen at Bristol airport.

1) I don't think I said I was only concerned with solution for a single particle translating freely in space. I think I said this was the simplest solution. It is the simplest because of the second part of your statement. 2) Thank you for bringing to my attention that I missed a (very important) variable. Here is the correct equation, including Energy. [math]\Psi = A\sin \sqrt {\left( {\frac{{8{\pi ^2}mE}}{{{h^2}}}} \right)} [/math] Bearing in mind that the wave function itself, psi, does not translate - that is the activity of the 'particle', The question then arises, How can the energy of that particle change if it is (your words) not subject to any forces or (my words) not subject to any potential? Which brings me back to my question you have not answered What does your input have to do with a particular interpretation of QM as that is the OP? @swansont - any chance you could put a pointer on my previous incorrect post as I can't now alter it?

I remember doing a material failure investigation on a german super alloy (steel). The emergency repair on the bridge bearing cost £1/4 million, leaving 63 more bearings at risk on the viaduct.

Yes that's right, SJ. The problem is that light travels along geodesics. Surveyors are used to the fact that if you set up a theodolite at some point A on the Earth's surface, sight on B, and set out a row of pegs leading to B, and then move to B and set out a row of pegs from B to A there will be two rows of pegs, not one. This due to the fact that the direction of gravity at A and B will point to slightly different centres. I am just trying to get my head round the implications extending this to the universal spacetime manifold which is more irregular than the Earth. So thanks guys for responding.

I'm not entirely sure why you posted a derivation of Schroedinger, or where you are going with it? Your Schroedinger equation is only true because the potential term is zero, otherwise it is incomplete. The solution I was thinking about is [math]\Psi = A\sin \sqrt {\left( {\frac{{8{\pi ^2}m}}{{{h^2}}}} \right)} x[/math] Which I think is more useful as you can plug numbers into it. As I understand it, the method of finding this solution is 'inspired guess', which means guessing solutions, based on experience, and testing to see if they satisfy the DE or PDE, which they do in this case.

Does General Relativity imply that the Principle of Reversibility of Light does not hold in our Universe?

Coriolis only affects horizontal motion, it does not affect vertical motion. Tides are a vertical movement of water so are unaffected directly. However in order for the water surface to rise and fall, water must advect in and out horizontally. These tidals streams are affected by Coriolis, though to a lesser amount than ocean currents. This is because the biggest effects are close-to-shore effects. In the open ocean the tidal rise and fall is small as are the oceanic tidals streams. Here the local flow is small but spread over such a vast area that the volume is substantial. Close-to shore effect demonstate resonance and/or forced oscillation effects such as the amphidromic points in the North Sea. The size and timing of the vertical water movements and the direction of the horizontal ones are strongly influenced by the topography.

Well I see that you have exhibited a Schroedinger equation, but I don't see you exhibiting a solution in explicit form [math]\Psi \left( {x,t} \right) = ???[/math] That would be what I call a solution.

And yet spatial displacement was the originally proposed test of GR and the actual one which provided the first practical verification. I am aware of your longitude analogy, and am hoping to offer a fuller answer to that and your other questions, but in turn, I would also be grateful for some discussion of the OP.

Except that I was not questioning your point about the sky or its blueness. I was questioning the appropriateness of the use of the word why. Do little children ever ask any question other than why (not).

Really? Why do you say that? And how does that pertain to the OP? Or, if you like, how does any bending in the ct direction allow light to go round a solid obstacle?

What a refreshing attitude. +1 You should go far. Welcome indeed.

Why is a rather overused word, which sometimes oversteps its boundaries. Let me change your question slightly to Why is your car blue? Answer 1) Because I Iike blue best. Answer 2) Because it reflects the non blue light present in sunlight. Answer (1) is a motivational answer and not Physics. Answer (2) is a Physics answer but would not be more correct to ask "By what mechanism...?" IOW how (come) ?

One of the earliest questions asked was How do I light a fire? That is a Physics question. Should I light a fire?, why light a fire? and many more are not Physics questions since Physics does not deal with motivation (the why question) or morality .

Well I don't think that is a poor analogy, though obviously every analogy is imperfect. Classically if a third person places each face (without noting which is which) in a separate sealed envelope, that surely is as near as a classical model can get?

The curvature is a local phenomenon, which means it varies from point to point in spacetime and can be determined with reference to nearby points in spacetime. However remember the fourth axis of spacetime is ct not time. This has the dimension of length, and units of metres. So, unlike Strange's example where the radius arm points to a common point in 3D spacetime, there is no such common centre in 5D spacetime. The radius arm at every point in our 4D manifold extends in a different direction in the mathematical 5D spactime. So long as this doesn't give the impression that the radius is somehow in 'time'. Which brings me back to my OP calculation (in metres). Still no takers?

Suppose you made a coin by pressing together a tail face and a head face. You have effectively entangled these two faces. Now suppose you split the faces and took one to the other end of the universe, leaving the other face behind with a friend, but sealed in an envelope. If you friend then opens the envelope he will instantly know which face you have taken, but he will not know until he opens the envelope. This is a model of how entanglement works.

Unless that purpose was less than straightforward. +1

You didn't make clear if these motors are part of the vehicle system (eg hydraulic pump) or extra equipment you plug in (eg mounted camera rotator). In aircraft a 400Hz supply is often used because the higher the frequency the smaller and lighter the motors and transformers. Frequency drives? Is this to vary the speed of the motor? You did mention safety not filtering? Filtering? Does this mean your drive circuitry is square or pulsed wave? How about incorporating some better filtering? Gyrators? High value capacitances also readily available these days. You also need to consider the possibility of fault conditions energising exposed metal parts at supply voltage and the possibility of sparks.

To get entangled, those particles must have been together on the same side of the universe at some time. So how long did it take to get separated by a whole universe? What are the chances of something happening to disentangle them in that time? So how practical is this proposition?

Whilst I have some reservations helping you with this, as you should not be working on 230 volt circuitry unless you or someone in your team has the competence, I will offer a couple of pointers. The Earth is not part of the working 230V AC circuitry and is, in fact not used in mobile equipment. Instead you should employ double insulation methods throughout. What is not clear is the purpose of the 230V circuitry. Is it a permanent installation supplying some on board equipment or is it providing a connection point for portable equipment?