DQW

The short answer : "yes"

Not to trash this party, but dear pljames : in answer to your first question, I'd have to say that you are (or were, when you wrote the first post) not close at all to understanding (the special theory of) relativity. I hope you do not take this as an insult at all. It is an honest opinion in response to your question. I only hope it spurs you on to read and learn more. Now it may be that you have a better understanding after reading 'fang's post. However, reading your response to it - especially the part that you chose to quote, I have to say, again, that you are still a good way off. What fang is talking about is one of the postulates (assumptions) that SR is based upon - not any of the amazing results that come out of it (though the validity of the postulate has been established through measurements/verifications of the results). In fact, an only slightly weaker form of that postulate was proposed about half a millenium ago, by Galileo. Much of the interesting physics in SR comes out of the Lorentz transforms that ensure the second postulate (the constancy of the speed of light in vacuum is independent of the inertial frame from which it is measured) holds, and the geometric construction of these transforms. It is this departure from the Galilean transforms that makes SR interesting, as well as highly non-intuitive (one might say, "confounding" even). Some of the interesting ideas that come out of SR are the understanding of how lengths and times measured in different inertial frames are related; and how electric and magnetic fields transform. While I claim no superior understanding of relativity, I can say for sure that if what you have learnt excites you, then there's tons more of wondrous beauty to be found in learning relativity more carefully. I hope your love for science only grows, but keep in mind that it takes careful study to unearth much of the beauty that lies hid in it..

Yes.

But surely, you see how that geometry is different from yours.

I know there are several polymers that go transparent in UV, but I have no data on my hands as of now. And then there's thinning effects that could be used to tune the reflectivity. Still haven't forgotten this one...

I second EL's suggestion there. Albert is not going to learn very much chemistry by shooting random questions at a forum (the answers to which, require an understanding of the fundamentals) - and not hardly even waiting for a response before the next question comes up. I repeat my recommendation of Atkins' Physical Chemistry. EL : I shall respond to your previous post to me later (if you so wish it). Essentially, we are both not on the same page...and I don't want to get into a lengthy argument based on our independent interpretations of an ill-defined question. What I said in my post is not what you have understood of it, so I'll be happy to clear it up, if required.

And to perform an actual FT, you must know how to integrate.

None of the alkali metals would work - even though they have plasma frequencies that are on the low side (being only monovalent). The best of them would be Cs (largest atom), which goes transparent only above 5.3 eV. Naturally, none of the transition metals will work (or that's what I think - know any large, monovalent transition metals ?). Hmmm....let me think a little longer ....

Do you care if it is transparent above 10 eV as well ?

The reason people like rectangular co-ordinates (or spherical, cylindrical, etc.) is that it is useful to have orthonormality between basis vectors, and the dot product is a convenient inner product.

Who said anything about final bond energies ? And I didn't know that "reactivity" was a physical quantity that could be described by an equation. From the reactant as well as product bond energies, one can calculate the reaction enthalpy. From a knowledge of the physical states, concentrations and other conditions, you can determine the entropies of the various species and hence the entropy change for the reaction. From the enthalpy, entropy and temperature, you can find the free energy change for the reaction (and hence the equilibrium constant at that temperature). This essentially tells you how much the reactants want to react under the given conditions. How fast the reaction proceeds depends on concentrations and the rate constant, which in turn can be determined (in theory) from the Arrhenius factor (molecular geometries, steric effects, molecularity, etc.), the activation energy (which essentially, can be found from bond energies - as well as other relevant energies, such as solvation, lattice, ionization, etc. - and a knowledge of the reaction mechanism) and the reaction temperature.

Phi, I admire your patience !!

I could not open that link. Will try again later.

Bond energies (when interpreted in the context of the reaction conditions, molecular geometry, etc.) can give you a good idea about the reactivity of a particular compound to other specified compounds under a given set of conditions (temperature, pressure, etc). Also, you can calculate bond energies from enthalpies and vice versa. So, it is possible to determine the thermicity of a reaction given all the bond energies, but that's about it.

Yes, I don't listen to myself, do I ? Okay, clearly the lower arc works (test with z=-2i). Naturally, the upper arc maps onto arg(M(z))=5pi/4

I don't recall doing this. In any case, I think it very much is a circle - and exactly the one whose equation is written in post #2. It is a circle centered on (1,-1) with radius \sqrt{2}. To the OP : An alternate approach (there's nothing wrong with what you've done) is to write z = x+iy and hence express z/(z-2) as some a+ib and find b/a.

Hmm. I've been told to keep in mind that Bi is stightly toxic. Never bothered to look it up....until now :http://www.sciencestuff.com/msds/C1309.html http://ptcl.chem.ox.ac.uk/MSDS/BI/bismuth.html The first link suggests negative effects, but the second says there's no toxicity. Perhaps the reaction with gastric acid is the problem ?

It isn't ??? I get the same locus that the OP has in the second post. Only it's a circle of radius [imath]\sqrt{2} [/imath]. The two real points on the circle are at (0,0) (for which, z/(z-2) is the origin and hence has any argument you choose to give it) and at (2,0) (which must be excluded from the domain for z/(z-2) to belong in C). Surely I'm making some stupid mistake ...

Ouch !1. Light is not "bent" in an optical fiber. It is reflected internally. 2. It does get slowed down.

I do not understand the origin of this "drag force". Could you please derive it or show me an argument for a force along the [imath] \hat {\phi} [/imath] direction ? I'm a little confused right now...

No, the farther apart you go, the closer you get to an ideal dipole-dipole interaction (and demagnetizing fields vanish too). The limiting behavior will not be x^-5 Sure, it happens where 1/x = 0

What is the direction of magnetization of the magnet ?

If you showed us the directions of the magnetizations of each of your cubes, one might be able to make a rough guess of the field geometry.

Okay, here's a completely blind guess : 540 (on the M/B search results)