timo

Add an "a" instead of a carriage return, then: <number of numbers>a<1st number>a<2nd number>a... Or represent each standard digit d with "0d" and use "10" as spacing between the entries and 11 as decimal point and 12 as *10^... (and interpret that stream of digits as a number if you want), so e.g. the tuple (1, 2, 3.14) would read: 03100110021003110104

Assuming it is reals that can be written down in decimal representation then it becomes almost trivial: 1st line: N. 2nd line: 1st number. 3rd line: 2nd number. . . . N+1th line: Nth number. Otherwise: A representation system with a finite number of digits can (I think) only represent a countable number of objects. Since the reals are non-countable, you cannot represent each real number in any such system. That would be your "it's impossible" proof, I think.

Higher dimensions: Seperation of variables Extending the particle-in-a-box example to two dimensions, the Hamiltonian for this problem reads [math]H(x,y) = -\frac{\hbar^2}{2m} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) + V(x,y),\ V(x,y) = \left\{ \begin{array}{ccl} 0 &:& 0 \leq x \leq L_x \text{ and } 0 \leq y \leq L_y \\ \infty & : & \text{else} \end{array} \right. [/math] with [math]L_x, L_y[/math] being the lengths of the rectangular box. Defining the two helper-operators [math] H_x(x) = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V_x(x), V_x(x) = \left\{ \begin{array}{ccl} 0 &:& 0 \leq x \leq L_x \\ \infty & : & \text{else} \end{array} \right. [/math] [math] H_y(y) = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial y^2} + V_y(y), \ V_y(y) = \left\{ \begin{array}{ccl} 0 &:& 0 \leq y \leq L_y \\ \infty & : & \text{else} \end{array} \right. [/math] the Hamiltonian can be formally rewritten as [math] H(x,y) = H_x(x) + H_y(y)[/math]. Now, these helper-operators are simply Hamiltonians for particles in a 1D box. Particularly, the possible energy levels [math]E_x, E_y[/math] and the wave-functions [math]\phi_x(x), \phi_y(y)[/math] solving [math] H_x(x)\phi_x(x) = E_x \phi_x(x)[/math] and [math] H_y(y)\phi_y(y) = E_y \phi_y(y)[/math] are known. Because of that, the multi-dimensional problem can be solved very quickly: Theorem: Assume an Hamiltonian [math]H(x,y)[/math] depending on two coordinates x and y can be rewritten as a sum of two Hamiltonians [math]H_x(x), H_y(y)[/math] each depending on only one of the coordinates. Further assume the functions [math]\phi_x(x), \phi_y(y)[/math] solve the time-independent Schrödinger equations for the reduces problems [math]H_x(x) \phi_x(x) = E_x \phi_x(x)[/math] and [math]H_y(y) \phi_y(y) = E_y \phi_y(y)[/math]. Then, the wave functions [math]\psi(x,y) = \phi_x(x) \phi_y(y)[/math] solves the full problems [math]H(x,y)\psi(x,y) = E \psi(x,y)[/math] with [math] E = E_x + E_y[/math]. Proof: [math] H(x,y) \psi(x,y) = \left( H_x(x) + H_y(y) \right) \phi_x(x) \phi_y(x) = \underbrace{H_x(x)\phi_x(x)}_{=E_x \phi_x(x)} \phi_y(x) + H_y(y) \phi_x(x) \phi_y(y)[/math] [math] = E_x \phi_x(x) \phi_y(x) + \phi_x(x) \underbrace{H_y(y) \phi_y(y)}_{=E_y \phi_y(y)} = E_x \psi(x,y) + E_y \psi(x,y)[/math] [math] =\underbrace{(E_x + E_y)}_{=:E} \psi(x,y) = E\psi(x,y) [/math] This separation of variables straightforwardly extends to particles in higher-dimensional rectangular boxes. So what does that have to do with the original question? A lot. It tells you why and how you can reduce the question (3D case) to the case you are somewhat familiar with (1D case). Of course, I only explained things up to the point of knowledge that the the question seems to require as being known. Random additional comments: - An energy state is degenerate exactly if there is more than one combination of (Ex, Ey, Ez) ( or equivalently (nx,ny.nz) ) that leads to the same energy. Note that e.g. (1,2,1) and (2,1,1) are different combinations. - You originally did some approximation. Do not do this, yet. - You can probably find all the states fulfilling the max-energy criterion by hand. Start from the state with the lowest energy, then continue looking for the state(s) with the next-lowest energy level until you exceed the maximum allowed. - The square-case is trivially included in the rectangular case. - Above mathematical steps may be non-trivial if you are not familiar with QM or differential equations. So don't be afraid if you do not immediately understand a seemingly small step. Sit down and try to understand it. Don't be afraid to ask if you still don't get it after that.

The easiest way to see how latex can be used is hitting the quote button on posts that use it and see what the poster typed in. There's a latex tutorial on sfn and also some test-bed too, I think.

You are given the total energy and the size of the box in the three directions/dimensions. No, there's not much difference between the lengths being equal or not (there is, but not much). Hm, you probably mean that if you increase the size of one side then you have to increase the size of the others to keep them being same-length. That is fine but not very exciting. What makes this statement seem a bit odd is that dimension is often understood as the maximum number of independent coordinates. Yes. Two natural questions: 1) Why? 2) Why would the 3D case with 3 equal-length sides not be Enx,ny,ny = h^2/8m ( (nx/L)^2 + (ny/L)^2 + (nz/L)^2 ), then? Btw.: Consider leaning TeX. I think learning to use the sfn math commands is a good way to do so. You can make your equations more readable. In "real world" you are likely to use TeX at some point (~100% of the physics student do, dunno for chemists) so learning the TeX math commands (used on sfn) is not a waste of time. That is pretty much what the question asked from you: You are given a maximum energy -> which combinations (nx,ny,nz) are possible so that the total energy stays below the maximum allowed. I wrote it with chemists in mind. I happen to have a chemistry book (Atkins "Physical Chemistry") at home so I think I approximately know what is demanded from chemists. Also, I once attended a chemistry course that dealt with QM (which is why I have a chemistry book at home). I am not too sure that QM is really the same or at least not taught in the same way to physicists and chemists. Explanations for chemists (and laymen) seem to focus a lot on calculus and the wave function while explanations for physicists are usually a bit more abstract talking about states, vector spaces and algebras.

Well, a short sketch for the 1D part I had prepared already yesterday, so here it is (note: (E) refers to the time-independent Schrödinger equation, (N) is the normalization criterion). The 1D particle in a box A very simple toy example is that of a particle in a 1D box. This system is described by the Hamiltonian [math] H(x) = \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x), \ V(x) = \left\{ \begin{array}{ccl} 0 &:& 0 \leq x \leq L \\ \infty & : & \text{else} \end{array} \right. [/math] The wave-function shall be zero where the potential is infinite (which means the particle is inside the box) which due to the continuity of the wave function means that [math]\psi(0) = \psi(L)=0[/math]. Inside the box (E) for the wave function [math]\psi(x)[/math] reads [math]\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \psi(x) = E \psi(x)[/math] [math]\Rightarrow \frac{\partial^2}{\partial x^2} \psi(x) = \frac{-2mE}{\hbar^2} \psi(x) [/math] [math]\Rightarrow \psi(x) = A \exp \left( i \sqrt{\frac{2mE}{\hbar^2}} x \right) + B \exp \left( -i \sqrt{\frac{2mE}{\hbar^2}} x \right)[/math] From boundary condition [math]\psi(0)=0[/math] if follows that A=-B which allows rewriting the term as [math]\psi(x) = 2A \sin \sqrt{\frac{ 2mE}{\hbar^2}} x[/math]. Now if A was zero, condition (N) would be violated, so the only way to get the solution in agreement with the other boundary condition [math]\psi(L)=0[/math] is to make the Sine zero there (which happens when the argument is a multiple of Pi). This means that [math] \sqrt{\frac{ 2mE}{\hbar^2}} L = n \pi \Rightarrow E = n^2 \frac{\hbar^2 \pi^2}{2mL^2}, \ n > 0[/math]. The solution with n=0 would again violate (N), n<0 gives the same energies. Btw.: Note how the boundary conditions create the "quantization" of energies. Now, writing the possible energies with an index n labeling the remaining degree of freedom (the multiple of Pi in the argument of the Sine), the possible energies for the particle in a 1D box are [math]E_n = n^2 \frac{\hbar^2 \pi^2}{2mL^2}[/math] (1) For the more-dimensional case the trick is that sometimes (e.g. for the particle in a box) you can rewrite [math] H(\vec x) = H_x(x) + H_y(y) + H_z(z)[/math], i.e. cast the Hamiltonian into a sum of sub-Hamiltonians each depending on less coordinates. I'll write a bit more on that over the weekend but today it's my no-physics day so it'll be tomorrow.

I wonder why you are given questions about 3D particle-in-a-box when it seems you haven't even discussed the 1D case, yet. I'll see if I can construct a short walkthrough for the problem but that'll take a few hours (during which I should better not fall asleep). EDIT: Won't get that done today; will update sometime on the weekend.

No I think you fundamentally messed it up. Fine till here except that - I think n=3 would also be fine (or why not?), - the question does not ask for a particle in 1D, - you should ask yourself what happens for n=0, - I did not check the numbers or the term for the energy, only the algebra. Where do you get that from? The degeneracy for a 1D particle in a box without additional degrees of freedom is 1 for all energies. You should look up the 2-dimensional particle in a box and understand things like [math]H = H_x + H_y \Rightarrow \psi(x,y) = \psi_x(x) \psi_y(y)[/math] (H being the hamiltonian, psi being the time-independent wave-function and x and y being the respective coordinates) and what that means for the quantum numbers (particularly their number) and the total energy and what [math]\psi_x(x)[/math] and [math]\psi_y(y)[/math] are. ???

What's the difference between time-intervals being "affected" and clocks being affected? - If "clock" refers to any generic keeping track of time then it looks pretty equivalent to me. - If clock really refers to say specifically an atomic clock then it makes me wonder: calculating planetary orbits in GR gives nice results. But the related (same spacetime geometry but simpler calculation) question of gravitational time dilatation does not work but instead accidentally happens to have the same value as the real effect: a yet-unknown influence of the gravitational field on atomic clocks? Disclaimer: There's a very prominent way to introduce relativity with clocks. I never understood those introductions. I find them silly (because I do not understand them ). So perhaps I just don't get the point here. Sidenote: Has anyone ever noticed that on internet discussions about relativity people always talk about if and whose clock goes slower and why. But seemingly no one asks how much? One additional comment now that I read the original "article": - There is no unique time that it takes to get from one event to another. Similarly to the length of a path from city A to B depends on the road in relativity the time it takes from one event to another (note that event is a fixed space and time position) depends on the path taken.

I think it is clear what you mean. That's perhaps the most important thing. Now that is completely unclear. I'll ignore it. Absolutely. What does that equation look like? Now where did you suddenly get a 2nd equation from? There is exactly one equation in the question, not two. This is (almost certainly) the point where you screw up. Finding the solution is pretty easy. Perhaps it helps you if you try to understand what the equation given sais/means.

Hm sorry, did see your question only now. C++: OpenGL support for c++ is pretty much standard for all compilers, so you could just check if you find the headers <gl/gl.h>, <gl/glu.h> and possibly even <gl/glut.h>. Higher-level libraries exist, too. Java: I do know there is jogl as an Opengl binding. I do know that there's more than those but no names at the moment. Higher-level libraries for Java exist, too. The real question is: What do you want to do?

If you take the standard model of cosmology and tune back the common time parameter then at some finite point you will find that the distances between all points in space become zero. You usually call this point or the times shortly after it (say till the the formation of the first elements) Big Bang. Therefore, the answer is "no" by definition of the term. It could be a lot, as soon as you leave mainstream science. Anything, if you just ignore common meaning of terms. There is no fifth dimension in the most generally accepted theory of gravitation (General Relativity) so there the answer is "no". Theories with additional dimensions that I met (admittedly only one rather general one) do not consider the additional dimension as a cause for gravity. In fact, it is probably more common to make the dimensions so small that there is no difference to be seen in the currently-observed (energy-)regime. I do not know exactly to what extent you can embed 4D spacetime in a higher-dimensional flat space (particularly not if 5 dimensions are sufficient). But having a mathematical apparatus that does not require you to do so is usually considered as a very strong point of General Relativity, I think. If you believe that the past dictates the future and you mean current expansion then it is probably a "yes" but with a lot of caveats. Too many to list them all; make your own thoughts about to what extent you believe a point found by extrapolation of a physical model which predicts the very same physical model not to work in this point to exist, or what something being the result of something else means.

In the exam at the moment ?!? The forum rules say something like "we do not do your homework for you". While not strictly speaking, in the sense of that message that naturally means "we will not help you cheat in an exam".

You do not have detectors telling you "there was a pion here". You measure different properties of the stuff coming out of the collision and reconstruct the event from that. For different detector types see for example Wikipedia. Experiments like ATLAS consist of different types of detectors that work together. http://www.physicsmasterclasses.org/exercises/hands-on-cern/ani/det_atlas/endview.swf is an interactive point&click adventure which looks like a nice toy to play around and try to understand it a bit.

The names of the computers in our institute are Star Trek characters (with the exception of the admin's computer which is named "Enterprise")

Try it out. Newtonian Gravity is school physics. I would not know about a generally-agreed-upon scale - it will definitely not be a hard point like in phase transitions but a gradual failure of the approximations used. I would imagine you might run into problems when spacetime considerably changes over the distance of a wavelength, whatever "considerably changes" might effectively mean in this context. You might run into problems defining photons. In short, yes: You approximate that earth accelerates with a=0 which is a good approximation for an attracting feather or an attracting car.

How would your spacetron or timetron differ from a graviton except for the name?

Unlikely. The point is: The most primitive form for an equation of motion of a particle through a gravitational field (more precisely: through a space-time) is the geodesic equation. In this equation for the case of light, the energy of the light does not come in. However, the application of the geodesic equation will certainly be limited. The geodesic equation inherently comes with a few assumptions that might break down at some point, like the object being point-like and not having any influence in space-time itself. It is a bit like the statement that in vacuum all objects fall equally fast towards the center of a an attracting planet (provided an equal starting distance, of course). If you verify this the it will work great for a feather, a block of lead and a car. It will not work well anymore for a sun because the inherent assumption that the attraction of the falling object on the attracting planet is negligible does not hold. 400 GeV is less than the mass of a single polymer so you can probably forget the influence on the gravitational field of a planet. 3 eV is a point-sized wavelength compared to stellar scales.

On typical scales (visible light in the influence of a planet or star) the path is independent of the energy.

There are OpenGL bindings and OpenGL-based environments/toolboxes/graphics engines/whatever for c++ and java. So yes.

List of sfn user titles (possibly not complete at the upper end): http://www.scienceforums.net/forum/showthread.php?t=127

Perhaps my all-time forum favorite which I posted here a few times already: <imagine disclaimer about sports activities being more respectable than arguing with strangers about topics not relevant to either participant here>

Wouldn't tachyons seem like an obvious way to break causality (unless you put the additional constraint that tachyons do not interact with anything )?

That clearly disqualifies them as "a new state of matter that is way below any temperature anybody has ever created or experienced"

Thoughts sometimes cause a sudden attraction of my head towards the nearest wall, so yes.