timo

Oh, that´s easy: If he´s a lazy guy he´ll use his own reference frame as the targetting is trivial in it. If, on the other hand, he is a frequent visitor of http://www.scienceforums.net then he will most certainly find a more complicated way of targeting. During his completely incomrehensive calculations he will mixup classical and relativistic view. This will result in a miscalculation. The laser will miss. The captain will post this here as an experiment that clearly disproves relativity.

From what I know: No. A friend of mine did his Diploma Thesis in Set Theory (if that really is what I´m thinking of, atm - I´m not entirely familiar with the english terms for the different branches of math). From what I´ve seen there, Set Theory is really, really abstract. I once took a look in one of his books and shyed away when I read the first Theorem which startet with something like "Let T be a complete theory ...". Well, the definition for a complete theory wasn´t that hard to understand when he explained it to me (but I forgot it anyways) but this sentence still left an impression. Group Theory is much more illustrative in comparison to that as it deals with symmetries and their representations a lot.

I have no idea to be honest. Main reason I proposed this was that you seemed like you were going to buy a laser anyways. And this is the part that I´d intuitively consider the most expensive anyways. However, I´m not completely certain if you can produce good holograms without the conditions you have in a lab. Obtaining it´s information from an inerference pattern (that is: structures of size in the micro- to nanometer range) a hologram is very sensitive to outside disturbances (you better not breathe nor move when creating one ). On the other hand, I´d be very surprised if noone has allready posted his/her experiences with doing holograms at home at some internet page, so I´d say you best google around a bit for more information about the feasibility of this project.

It´s hard to tell what you´d like. However, apart from the standard answers that would be topology or differential geometry I´d say you might want to take a look at general algebra and perhaps group theory in particular. In fact, that´s what I (personally) consider "real math" in contrast to stuff like differential equations or linear algebra which every engineer or natural scientist learns.

How about holography? You´ll need some equipment (laser, beamsplitter, a table that doesn´t vibrate, sensitive photo plates) but I think it can be an interesting (although perhaps a bit ambitious) project.

"Permanently Banned" doesn´t sounds like it if you ask me.

I wouldn´t know of any other (sensible) definition of division othen than multiplication by the inverse, so "yes".

The only one I quoted. And to emphasize the line on Wolfram referring to this:

Shame on me but I´ve never heard about the "axioms of real numbers". What are they supposed to be? The definition of a field? If "yes", then your statement isn´t true (see http://mathworld.wolfram.com/FieldAxioms.html)

the two of you should probably take a look at the dates of the first six posts and the (nonexisting) heat of the debate that was going on back then to judge the relevance of this thread. To add something constructive: http://redwing.hutman.net/~mreed/warriorshtm/necromancer.htm

That is because you checked

Not much time to answer right now since I have to go to work (damn, I´m allready quite late). But I drew a sketch for you that you might like. I´ll check back here later this evening.

If you have a point P=(x, y, z) then it´s squared distance from the origin is x²+y²+z². Whether these x,y,z depend on another parameter t or doesn´t matter. So why did you use "|OD|² = (x-5)² + (y+1)² + (z+1)²" instead of "|OD|² = x²+y²+z²" ? The rest of your math seems ok, so try it again with |OD|² = x²+y²+z².

It´s often usefull to write complex numbers a+bi in a polar coordinate way as r*exp(ip) with r and p both being reals. If you draw the complex plane in the usual way, r will be the distance of a+bi from the origin and p will be the counter-clockwise angle of this connection line to the positive Re-axis. The multiplication rule in this representation is z1 * z2 = r1*exp(ip1) *r2*exp(ip2) = (r1*r2) * exp(i(p1+p2) or in other words: Multiply the distances, add the angles. So for the square root of a complex number z=d*exp(ip) you are looking for a number z' =d' *exp(ip') such that d'*d' = d and p'+p' = p (you have to pay attention that the addition is modulo 2pi, so p/2 is as well a solution for p' as p/2+pi is). The imaginary number can be written as i = 1*exp(i*pi/2) so d=1 and p=pi/2. From d=1 it follows directly that d'=1 (negative distances are not possible). From p=pi/2 => p' = pi/4 or pi/4+pi.

Starting to write down the equation for the line that D lies on (your r = A + t(B-A)) is a good start. But why you then proceed with calculating anything on |AD| is completely unclear to me. Unless you made a typo somewhere, it´s |OD| = |OA| that you have as 2nd condition to fix the point D, not a condition involving |AD|. The straightforward solution is to start from |OD| = |OA|. |OA| is known because you know the coordinates of A, |OD| can be expressed by the line r(t) you allready got. Solve for t. You´ll get two possible results for t. One will be such that r(t)=A which was excluded in the question. The other one is the one you´re looking for, then. As a hint here: As distances are allways positive, the condition |OA|² = |OD|² is the same as |OA| = |OD| and eleminates the annoying sqare root. Differentiation (what you did) is often used to find extremal values (the differential is zero, then). So if you use d(|AD|²)/dt = 0 and solve it, you find the point D that closest to A. Not much of a surprise that you´d get t=0 and the point closest to A would be A itself. But that´s not what is asked for. There´s an alternative way to solve the question which is quite elegant in my opinion but needs a bit of understanding of vector math and might be a bit more work to actually do. But since it might help you understand the problem and you might also find it interesting, I´ll sketch it here: Given an abritrary vector k, any vector v can be split up into a part parallel to k and a part perpendicular to k in a unique way. In your case, you could split up the vector a which is the coordinates of A into a part a_par parallel to the line through AB and a part a_perp which is perpendicular to that line. a_perp will hit the line at a point P. This is the point on the line which is closest to the origin. If you add a_par to that point, you´ll end up on point A, of course. But if you substract a_par from there, you´ll end up on another point which has the same distance from the origin as A (because |a_perp|² + |a_par|² = |a_perp|² + |-a_par|² <-- phytagoras). As a_par is parallel to the line so is -a_par. The point you got by D = a_perp - a_par is the one you´re looking for. This method is probably best understood if you draw it on a piece of paper, so you might do that after you solved the question the straightforward way proposed above using the correct coordinates, then.

Or to extent your question: What is it supposed to be? What's anti-electromagnetism? Things are especially hard to understand if you don't even know what they are. No, there's another force called "strong force" that's responsible for nuclei to be stable. Yes. Nice idea, but a) this would still be normal gravity only that antimatter had negative mass and b) it just isn't like that. Still "yes, it's there and it's proven".

It´ll be rather hard to know someone well enough to trust him if he´s living on the other end of the planet and is only known to you by a fake name in an internet forum . What´s a german high school student, btw ?

>> does it mean that k varies with the number of atoms in the lattice? Not really. The k-value is a quantum number characterizing the quantum state. For phonons, there´s as many possible k-values as there are primitive crystal cells (but still an infinite number of vibrational states since the number of phonons in the crystal is arbitrary). For electrons, it´s almost the same, although I wouldn´t know the reason for it atm (it pobably comes from periodic boundary conditions). >> and is the probability of finding an electron near an atom is greater than that of >> between the atoms? That entirely depends on the state the electron is in. But as a rule of thumb I´d guess that states with the electron density closer to the ions have lower energies than states with electron density farther away from the ions. Reason: The potential energy is lower if the electrons are closer to the positively charged ions. >> If so, is it because the atoms (or should I say the positive ions) attract the >> electrons so it changed the wave function of it? There is something similar to that idea in solid state phyiscs, indeed. As you might or might not know, there are so called band gaps at the end of the brillouin zone. That is: The energy of the lowest band and the 2nd lowest band do not have the same value at the end of the Brillouin zone which they had if the electrons weren´t influenced by the ions. Due to reasons I can´t remember atm (the experimentalists' reasoning as being shown in Kittel seemed a bit weird to me - I once had a better one but I forgot it) you can assume the Bloch waves for k near the Brillouin zone to be standing sine- or cosine-waves. Both those waves have different electron densities near the ion. This results in a difference of the potential energy which results in the band gap.

Addendum on how the different energy bands appear: Writing the wavefunction as above has the disadvantage that this notation is not unique. Any factor exp(iKx) with K being a vector of the reciprocal lattice can be pulled out of the exp(i kx) term and applied to the Bloch wave, since exp(i Kx) is also lattice periodic. You can get a unique representation by restricting k to the 1st Brillouin zone (that´s the set of vectors in the reciprocal space for which the origin is the closest reciprocal lattice point, in case you don´t know the Brillouin zones). So for any k in the exp(i kx) not being in the 1st Brillouin zone you substract a suited reciprocal lattice point K so that the resulting k' is in the first Brillouin zone: exp( i kx) = exp(i (k' + K)x) = exp(i k'x) * exp(i Kx). The exp(i Kx) is then considered part of the new Bloch wave. This new Bloch wave can result in a different energy than the old one. Hence, for any given restricted k you have multiple solutions (because of having multiple Bloch waves) and multiple energies. That´s the energy bands (in 1D, each band would correspond to a certain K being substracted).

If I remember correctly the Bloch wave is the part of the wavefunction that is periodic with the lattice (we are talking about wave functions in a crystal lattice, are we?). Or in other words: psi_k(x) = exp(i kx) B_k (x) with psi being the wavefunction and B_k being the lattice-periodic Bloch function. Different k -their number is limited to the number of elemental crystal cells but tends towards infinity for macroscopic crystals- lead to different energies which is what you plot in the E(k) diagrams. Dunno about how the energy bands appear in this picture atm, but I'll look it up later. I'm at work atm so I can't really give any more detailed answers but I'll give a more detailed answer when I'm at home later (isn't THAT long since I had my final exams in solid state physics so some knowledge should still be there ...).

Oscillations per coordinate time.

About Eigentime: ---------------- Generally, the position of a paticle is given as position as a function of time P(t). In Relativity this is replaced by a trajectory in spacetime. Let A and B both lie on this trajectory and B be in the future of A: - In any coordinate system A and B are defined at a certain time t(A) and t(B), respectively. The coordinate time it takes for the particle to travel from A to B is dt=t(B)-t(A), then. This is a quantity that depends on the chosen coordinate system. - But there´s also a quantity that´s independent on any chosen coordinate system: The length of the trajectory between A and B (draw a line on a piece of paper, the lenght of the line won´t change if you rotate the paper). This quantity is called Eigentime because it is the coordinate time difference you´d get if you were measuring the coordinate time in the particle´s frame of rest. - Now, because the metric in Minkowski-space is not positive definite (in fact, calling it metric isn´t even correct - it should be called pseudo-metric but physicists are lazy people) you can have connecting lines between two non-identical points A and B that have a length of zero. The trajectories of massless particles are such trajectories. That´s why their Eigentime between any two points is zero. btw.: One thing that just came to my mind: Perhaps it helps you to know that Eigentime is a german word (but it´s also the term used in english) which means "[the particle´s] own time" or simply the time measured by the particle itself (as said above). About the freezing time and the "cannot oscillate in no time": ---------------------------------------------------------- I find this rather hard to explain in detail so I´ll try the "why this argument is not valid"-method. There is no frame in which the photon is at rest. Furthermore, there is no coordinate system in which the coordiante time between any to nonidentical points on the photon´s trajectory is zero. This leads to the result that the "time freezes" scenario can not happen when you speak about coordinate time. And to come back to my initial statement: Frequency is given in units that do not lead to problems. EDIT: Oh, and what you call "time" is the coordinate time - the time measured in the coordinate system you are using.

Not really a book, but a link I bookmarked some time ago that might keep you occupied for some time: http://www.cs.berkeley.edu/~russell/ai.html

Most 26 year old men know that getting an answer in an internet forum can take longer than 3 minutes. On topic: The frequency is given in units of coordinate time (which directly leads to frequency being dependant on the coordinate system) not in units of the Eigentime (which would be the time measured by the photon and which would indeed be kind of freezed).