timo

Somehow, I feel a bit uncomfortable seeing a link to a page where I´m listed as "physics expert" directly below a link to WiSci. I can´t see a point in those two links, anyways: The staff-list seems superfluous and the link to the dead-but-not-buried-yet WiSci is more scary than helpful to get people interested.

How would the test pad work, then? I mean: If we agree that the preview button is better than the submit button, then there should be something even better than the preview button. What?

Depends on what "an equation for the solar system" is for you. Force on body i is the sum of the gravitational forces exerted by the other bodies j : [math] m_i \vec a_i = \vec F_i = \sum_{j\neq i} \vec F_{ij} [/math]. Writing down the movement equation obviously isn´t a problem at all. The problem is not that you can´t write down the movement equations, the problem is that no one knows how to solve the movement equation for more than two bodies analytically. But for example doing a numerical simulation of such a system is pretty straightforward and easy (just in case anyone in here likes writing physics simulations).

Might be a good idea to copy-paste the explanation from the nominations thread to the respective vote threads.

The zero function is a function, actually a pretty important one . The constant function is a function, too. Of course the function changes for different parameters (A, k and w). Strictly speaking, it does not ask if the solutions are physically meaningfull - at least to my reading. It asks "are the functions solutions of a particular diff. eq" where the particular diff. eq is the Schroedinger eq. for a free particle. That might sound like nitpicking but there´s a catch in it. The functions [math] \psi_{\pm} [/math] Severian brought up are not solutions for a free particle, either. But as you might know, for differential equations there is something which I suppose is called the "solution space". For linear diff. eqs this solution space can be seen as a vector space (unless you refuse the zero-function to be a function - in that case you lack the zero-vector). In this case, the functions [math] \psi_{\pm} [/math] are a (mathematically) valid base for this vector space. You can construct all solutions of the diff. eq. from these base vectors - including the physically meaningfull ones. That´s why asking if a particular function is a solution for a 'physical' differential equation can make sense even if the function itself cannot represent a physical entity.

I wouldn´t see what´s wrong with your solution (I didn´t check the numbers; I trust you know how to use a pocket calculator). Is the question you posted the original wording of the exercise? Perhaps you misunderstood the question.

I don´t have any books on such advanced topics, sorry. So let´s give an example: R², cartesian coordiantes. Red curve is {(0,p) : 0<=p<=1}, green curve is {(p,p) : 0<=p<=5, (p, 0.5-p): 0.5<p<=1}. Which matrix to you apply that results in the green curve being shorter than the red one? EDIT: ^^ I knew I shouldn´t have posted this but I´m afraid it´s too late now. As a matter of fact, above example doesn´t really matter. Relativity isn´t based on LA but on differential geometry. Lengths of curves simply ARE independent of coordinate systems - the whole formalism of tensor algebra is based upon that principle.

Print it out on a piece of paper and then tell me which way to turn the piece of paper around such that the green line becomes shorter. There is none. Using scissors or streching parts of the paper is not valid, btw. That´s equivalent to modifying spacetime (the piece of paper). EDIT: To be a bit more helpful: Within SR, transformations between inertial frames of reference are really only rotations in spacetime (plus an offset, perhaps - but feel free to move the piece of paper around your desk, too). In fact, my statement that you can tell which line is longer is realized in an even stronger way in relativity. In relativity, the statement "lengths of lines remain unchanged under ANY coordinate transformations" holds true. It finds its mathematical realization in the transformation laws for covariant and contravariant indices and the transformation law for the metric. I´d even say that it´s a nessecary statement for relativity to be physically meaningfull but don´t pin me on this as I didn´t bother to formally check this.

EDIT: I misread your post first. I read "... except when they eventually meet up". As I tried to show in my first post in here, you cannot objectively tell who ages faster (which line inceases faster). But you can compare age differences when they meet. Translating this to my line-example: You can tell which connection from A to B is the shorter one:

Seems more effective than the "submit reply" button to me. After pressing "preview post", I have a preview on top with the source below it. After pressing submit, I only see the outcome but have to press edit to see the source text again.

^^ That sounds exactly like what I looked for (even though for the original problem it comes a bit late since I designed the program such that the numbers are always O(1)). Thank you!

"V(x)=0 because it´s a free particle" is ok. Using the time-dependent version sounds reasonable, too. After all, a time-parameter is given explicitely. Sure. You´re given two functions and are asked if they are solutions to a differential equation. Plug them in and see if both sides of the diff. eq. equal (and under what conditions they do so - there´s two free parameters w and k in the problem). Perhaps it helps to rewrite the sine and the cosine as cos(x) = 0.5(exp(ix) + exp(-ix)) and sin(x) = -0.5i(exp(ix)-exp(-ix)), perhaps it´s more a burden than a help - figure it out yourself.

Hopefully so (but I´d think lighting a christmas tree and illuminating a street are still different orders of magnitude in terms of required amount of light). It´s a terrible waste of energy when something like 95% (perhaps it´s even more like 99%; not sure) of the energy goes away as heat. Just out of interest: Do you have any technical data for your LEDs? Especially efficiency and power output would interest me. EDIT: I found that Wikipedia has articles on both, LEDs and lightbulbs, which basically give the information I was looking for.

What about efficiency (well, I think it´s pretty high for LEDs, so perhaps we already are at the 90%+ range) and power output? Especially in the case of power output, I would expect the range of LEDs to be limited. Another issue for LEDs would be easy, cheap mass-production of them. I don´t think that cheap, powerfull and effective LEDs are already available. At least, the streets of Dresden are still illuminated by ordinary lightbulbs.

NumberFormat works but it doesn´t help me. As far as I´ve seen, NumberFormat rounds the number to the specified number of fractional digits. I really don´t want to carry along 15 digits just to diplay 10^-15.

Doesn´t seem to be a known method but thanks for the suggestion.

Not sure. At least I didn´t find anything with that name. What would the actual command look like? What I am mainly interested in is the "e-15" part of the number; the number of decimal digits is 2nd priority (two or three would probably be ideal).

I think the main force resisting a complete gravitational collapse is due to the fact that neutrons are fermions - but I have to say that this is just a guess the name "Neutron star" suggests to me. Being fermions they must occupy a seperate state, each. As a good approx, this will be all of the lowest states. With decreasing volume, the energies of the states grow, so the total energy of the neutrons must grow. Because of that, it takes energy F*dr to compress a sphere of fermionic gas by dr (r is radius, here). You can associate this with a force F resisting compression. @[Thyco?]: I didn´t say there is no evidence for black holes. And perhaps saying there is no real experimental proof was not a good idea, too. It´s a bit up to debate where the lines between "real experimental proof", "strong evidence", "evidence" and "indirect evidence" lie.´I just wanted to say that E.g. atoms are on a more solid ground than black holes.

I´m currently trying out Java. So far, my impressions are mixed but tending towards "it´s nice". However, there is one issue that bugs me a bit: I have not yet found out how to convert a double to a string with a reasonable formatting. The toString()-metod and the string = ""+double trick format 1.3322676295501878E-15 to 1.332267 which really is more than useless (in fact, I encoutered the problem because I was looking for a bug - I knew the value had to be O(10^-10) or smaller). I have already taken a small look at google and to my surprise, I found a lot of classes converting doubles to strings. But I didn´t find what I was looking for: A native Java method (by that I mean a method/function of the core language) doing so. Is there really no native Java method converting a double to a string or did I simply not find it, yet? I can hardly believe that the developers of Java really think ommiting the exponent on a real is a smart idea.

What about skipping "most knowledgeable"? I don´t think that any member of this board can even judge E.g. Matt Grime´s skill, less compare it to that of others.

That is, if you replace it with a black hole of the sun´s mass. As you indirectly said in the previous paragraph, the gravitational effects created by a black hole and the gravitational effects created by any other spherical symmetric obejct with the same mass are the same - with the little difference that a black hole is smaller and you therefore can get closer to get really strong effects. I think gravitational lensing has been observed on our sun, not on black holes. I am not sure if Hawking Radiation is more than a handwaving hypothesis. But I don´t know much about it anyways; I´m just a bit cautious when the name Hawking is mentioned because there´s a lot of hype about him. Some random remarks for the original question: - Up to today, there is no real (experimental) proof that black holes really exist. - As mentioned above, the gravitational effects of a black hole and any other spherical symmetric mass distribution (sun, planet) are the same - with the exception that a black hole is small enough so that you can get as close as to reach the "event horizont", an area where some strange effects happen. - Therefore, as a first guess I´d say that the chance of a black hole eating of a whole galaxy is as big as the chance of a planet of equal mass attracting a galaxy to it (in the sense of "all the stars/planets bash into that planet because they are attracted by it). - There´s one difference in the analogy of the previous point. For planets/suns, there´s the tendency to release part of the attracted mass by whatever process. In the sun it´s light created by fusion; supernovae also play a role, I think. For black holes, there is no such process known except for the Hawking Radiation RyanJ mentioned.

@intothevoidx: I don´t know why you do that. After all, all I know about "x" is that it´s a letter appearing at the end of the alphabet. No, seriously: All I can offer you is leading you on the way that I would take to solve the problem. I think the steps 1) - 4) in above are really simple questions; step 4 is almost the answer to the original question. There´s not nessecarily always only a single way to solve a problem and how should I or anyone else on this board know how your teacher solved it? If you don´t want to try another way of solving the problem, then post what your teacher wrote down - you´ll surely find someone who can explain it to you.

- I must confess I was a bit hasty when I said "There is two parameters/variables/answers that´s asked for, namely the working speed of Tom and the working speed of Huck". Actually, it´s the time it takes them to paint the 500 square feet fence which is asked for. However, getting from the working speed to the time the work takes is pretty easy. And solving for their working speeds first seems easier to me. - For converting to equations: I am not sure it helps but maybe it´s worth a try: Try answering these questions: 1) If I paint 10 square feet per minute, how much fence will be painted after 4 minutes? 2) If you help me and paint 15 square feet per minute for the same 4 minutes, how much of the fence will be painted, then? 3) What if I don´t paint 10 square feet per minute but Q square feet per minute - how much square feet of fence will be painted after 4 minutes, then? 4) What if I paint Q square feet per minute and you paint Z times as fast as me?