studiot

Thank you for answering my question. That helps to get what we say at the right level. And yes the person in the video said pretty much what we have been telling you here, although he did not mention the Lennard -Jones force by name, he talked about it.

Apart from being rude enough so as not to respond to my comments, What was your point in posting the video in post#31?

Mike, these extracts from 'Bergmann :Introduction to the Theory of Relativity' might help you sort out what you want to ask us. Particularly the part in italics I have indicated at the sides.

John I think I would need much better scientific convincing than that article before I swapped OPC construction for talcum powder, although I have always liked this Further, and as a matter of interest, carbonation by absorbtion of atmospheric carbon dioxide has long been known as a degrading factor for OPC concrete. Sensei, thank you for your thoughts, but those installations are not the sort of 'plant' I have in my garden. I really had hoped some bio/eco scientist here might know the % carbon in kg / kg for soft plant material such as leaves and harder plant material such as roots and wood. Nor is it true that plant material, especially woody material, returns to atmospheric carbon dioxide in any rush. I have considerably increased the heart my soil by a vigorous composting programme that has greatly increased the organic content of my soil over the last 30 years. Perhaps in 300 years time i will have generated some high quality Somerset peat.

I'm not convinced that neutron interaction is electromagnetic. Thinking further about the question of macro v atomic and smaller, there is a whole science of what is called contact mechanics or tribology.

Not until all soul's eve. To choirs of ghastly cackling laughter

Since the distance between two touching objects is defined as zero the question has no meaning. You can't half-touch.

Strange was answering your question about wavefunctions. Mordred has already councilled (as I do) not mixing up wavefunction and discrete models. For quantum models 'touching' has no meaning since wavefunctions extend to infinity, even through other atoms etc. So either use the particulate model or the quantum wave model but not both at the same time.

Atoms (and other particles) can and do touch each other. For instance neutrons and alpha particles are fired into the nucleus in particle accelerators. You can't get much more intimate than that. Take heart, there is nothing wrong with the bullet model, so long as you realise it is just that. A model which does not display all the characteristics of the real thing. Yes electrons do repel each other. This is a force model and the correct full model of this is known as the Lennard Jones model, if you would like to know more. What level are you studying this at?

Not really. QM and einstinian relativity address different matters, although there is some overlap in that there are versions of the equations of QM that incorporate relativistic mechanics, rather than classical mechanics. I believe that Paul Dirac was the first to offer this. There are yet remaining areas of Physics that are (fully) addressed by neither QM or einstinian relativity, for example electric Charge and number (of molecules or particles). You may also not be aware that the 'uncertainty principle' also applies in classical mathematics and even in the rules of simple arithmetic of ordinary numbers, which is why we have adopted certain conventions about arithmetic to resolve this. Edit +1 to Strange. I have trying to make this point for some time.

1) Yes 2) No The choice of the word 'Event' is an unfortunate one given the usual english use of the word. It simply means that whoever chose it (or translated the German?) chose to add an unnecessary word to the vocabulary of mechanics. The word 'point', as in its usual geometric meaning of a collection of definite coordinate measured against some system of axes, has served faithfully for some centuries and that is the whole meaning of the word 'event' in this context.

I think that teaching proportion or proportionality (nouns which name the phenomenon) and proportionate (old fashioned) or proportional (modern) - the adjectives used to describe some noun has sadly fallen out of fashion. I find that for most folks using proportion is actually easier and more natural to understand and use than more explicit mathematics. Certainly in Newton's day pretty well all work was stated in terms of proportion so if it was good enough for that genius it is more than good enough for me. Most of the discussion about the 1/2 has centered on propelling a body to a specific speed. A much easier derivation of the KE is found by considering instead the constant force needed to bring a body to rest from some velocity v. Consider a body of mass m moving with steady velocity, that is brought to rest by the application of a constant force P. P produces a constant retardation, f so that P = mf, in a distance x Applying the simple equation (final velocity)2 = (initial velocity)2 - 2 (acceleration) (distance) and noting that the final velocity is zero at rest. 0 = v2 - 2fx Thus fx = 1/2 v2 But the work done by the particle against force P is Px = mfx Therefore mfx = 1/2 mv2 This quantity is defined as the kinetic energy

ESRI have a good long standing reputation for all sorts of GIS (geographic information systems), though it is nearly twenty years since I last used their products. It is good to be brought up to date.

OK so have you developed the contours themselves or do you just have a grid of number values?

This is a difficult and modern subject to hit me with on a Sunday morning, especially with so little information about what your goal is, although I understand what you have told us so far. Is this an exercise in geomorphology, terrain analysis, ground feature analysis or what? It looks as though I have guessed correctly about the need for recognising features described by your contours and that you are looking for a way to best partition your 'map' into zones that best show these features?

I guess you are attempting something like this?

I particularly liked this description, as well as tour previous post. +1 But here is a question for consideration. Is the fluid in an eddy flowing ? After all it is going nowhere.

Careful. You might get to like some of this stuff.

The table only contains the very simple, I'm sure you can cope with [math]y = a{x^n}[/math] [math]\frac{{dy}}{{dx}} = na{x^{\left( {n - 1} \right)}}[/math] u and v are universally used to show an intermediate function of x. This is also known as function of a function or the chain rule, but both are fancy names for substitution. Watch [math]y = \sqrt {3{x^2}} [/math] [math]Let\;u = 3{x^2}[/math] [math]y = \sqrt u = {u^{\frac{1}{2}}}[/math] to differentiate split the derivative into two parts [math]\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}*\frac{{du}}{{dx}}[/math] [math]\frac{{du}}{{dx}} = 6x[/math] [math]\frac{{dy}}{{du}} = \frac{d}{{du}}\left( {{u^{\frac{1}{2}}}} \right) = \frac{1}{2}{u^{\left( {\frac{1}{2} - 1} \right)}} = \frac{1}{2}{u^{ - \frac{1}{2}}} = \frac{1}{{2\sqrt u }} = \frac{1}{{2\sqrt {3{x^2}} }}[/math] [math]\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}*\frac{{du}}{{dx}} = \frac{1}{{2\sqrt {3{x^2}} }}*6x = \frac{{3x}}{{\sqrt {3{x^2}} }} = \frac{3}{{\sqrt 3 }} = \sqrt 3 [/math]

Careful here. Both the sets [0,1] and (0,1) are infinite bounded subsets of R. The question is surely fairly trivial since R is unbounded (1) So partitioning R into subsets S and S' where S' is the null set and S is R is contradicts (1) Sheff, Is this homework??? I will leave you to finish off formally, moderators may move this to the homework section.

The expression in the picture (This site has difficulty with my latex generator for matrices) is one stage in the calculation of the fixed end moments for a beam spanning between two walls and built into each at its ends, by matrix methods. Restraining the beam by building the ends in considerably stiffens the beam compared to simply resting it on top of the wall because the end restraints apply additional moments to the beam called fixed end moments. Stiffening means the same beam deflects less under load when built in as opposed to resting on. The point is that if you look carefully it illustrates my comment about linear combinations of non linear expressions nicely. The 3 x 3 matrix is called the stiffness matrix and it premultiplies the deflection vector, which has 3 elements. So by the usual operation of matrix multiplication the output is another column vector of applied loads (moments in this case), which also has 3 elements. So the first row is equivalent to [math]\frac{{4EI}}{a}*{\theta _1} + 0 + \frac{{6EI}}{{{a^2}}}*\Delta = - M[/math] Where theta is a rotation and delta is the linear deflection and M the applied moment. Other letters are constants (for that beam)

It is normal to have a table of standard derivatives available in exams, though it is quicker if you know them.

I really liked and agree with the underlined sentences. Too many words which already have sound meanings are redefined, causing much confusion, in my opinion. This also applies to 'curvature' when what is meant is 'the curvature tensor'. The curvature tensor reduces, as do many properties such as tangents, to a single choice in 2 dimensions - the scalar radius. There are even models of relativity that introduce a the extra dimension, such as Kaluza's theory. https://www.google.co.uk/search?q=kaluza%27s+theory&rlz=1C1AVNG_enGB673GB673&oq=kaluza%27s+theory&aqs=chrome..69i57.6791j0j8&sourceid=chrome&ie=UTF-8 But we do not need curvature at all if we restrict our analysis to the points of the manifold plus some conditions of constraint. That is the whole message in my surveyor's example.

What you may be interested in the the following property of determinants. The sum of the elements in one row, i, by the corresponding cofactors, in another row. m, is zero. [math]\sum\limits_{j = 1}^n {{S_{ij}}C{o_{mj}}} = 0[/math] [math]i \ne m[/math] This leads to Laplaces expansion of determinants https://www.google.co.uk/search?hl=en-GB&source=hp&biw=&bih=&q=laplaces+expansion+of+determinants&gbv=2&oq=laplaces+expansion+of+determinants&gs_l=heirloom-hp.3...1437.8656.0.8859.34.9.0.25.1.0.140.827.3j5.8.0....0...1ac.1.34.heirloom-hp..26.8.811.h5aMLZkxu6w As regards combinations of matrix element here is a matrix equaltion from structural enfgineering illustrating the point I made in post#5

The mistake in post#15 was entirely mine. I copy/pasted it from the post immediately above it, but didn't look properly. My apologies to Tim and Geordie and anyone else inconvenienced. Thank you Geordie for putting me right so pleasantly. The point I was making still stands however. Let us take mass density [math]\rho = \mathop {\lim }\limits_{\delta v \to 0} \frac{{\delta m}}{{\delta v}}[/math] Since we are saying delta v tends to zero, it makes no sense to talk of compressed space or compressed anything else. You can't compress below zero! I did wonder if the article was referring to what is called tensor density, which is quite a different thing. Perhaps Mordred will enlighten us since that is his field. Finally if spacetime has any sort of 'density' then how can it not be physical?