timo

Surprisingly, many people are. Sadly, some people think that solving equations is more important than the initial step of transfering a problem to equations. This is not true. You will seldomly be welcomed by two equations on a train station while the question whether a group ticket might be a better bargain than individual tickets -even if two of your group only go for half the distance- might indeed arise. I´m not imposting that you are one of those "text exercises suck"-people but if you´re bad at them, you should really try to improve your skill here. The trick actually is to translate the information given into mathematical expressions (here it´s equations). Sadly, I wouldn´t know of any cooking recipe how to do it - I do it rather intuitively (experience plays a big role, though). I don´t think giving a step-by-step guide is much of a help for you, here. So I´ll tell you what I see when I look at above problem, instead: - There is two parameters/variables/answers that´s asked for, namely the working speed of Tom and the working speed of Huck. - There is two informations given, namely their combined working speed ("They paint Aunt polly's fence and find that they can paint a 200 square foot fence in 40 minutes if they work together") and the ratio of their working speeds ("huck works 4 times faster than tom"). If this is not obvious to you, take some time and think about it. Extracting informations from a text is one of the keys to solve text exercises. - These two informations can each be cast into a mathematical form - an equation in this case. - Now you have a "two equations/two unknowns" problem in mathematical form. Solving it is only technicality, now - and since you are given that exercise I assume you learned how to do that recently.

base -> Quantum Mechanics. Next is "Darmstadt".

I think the question has already been beaten to death, resurrected and twisted into some weird abnomination (it´s only one equation to solve @Cluod and there is only one velocity that remains constant over time @Evon). It would be nice to have some feedback from Krz to know if he found the answer he was looking for (ideally with a description of what he understood and how he solved it) or if there´s some additional questions. Afterwards, there´s absolutely no problem if someone who knows how to solve the problem presents the full solution (approach, solving the solutions and plugging in the numbers - and please don´t supress the units!).

Don´t spend too much time on it if you´re busy Dave. It´s not too important. I could also have looked for all posts of Severian and recognizing those in which I contributed by the arrow-symbol on them; it´s only a few more pages to browse. I simply thought that maybe allowing for multiple username searches is only a switch or so. @Severian: Let´s see: ^^ That´s found searching for the keyword "Blubbalutsch". And not found by searching for the keyword "xyz8hb".

I am pretty sure the username is unique in here. I actually WAS talking about the advanced search where there´s a field for searching for usernames, in case that didn´t become clear. What I was asking for was an option to show all threads in which all of the posters mentioned in the username search field contributed. Adding the names in the keyword field only shows of the threads if the name of the poster was explicitely mentioned; the poster having posted there is not sufficient.

Because the frame of reference you have chosen is one in which he moves faster than you (zero <-> near c). Depends on whose frame you use. If you use the frame in which the rocket is at rest then your time passes slower. Strictly speaking, you can only compare aging when you manage to meet your twin a 2nd time because otherwise you are comparing your age to an arbitrary age of your twin. I´ve tried to visualize the problem in the attached image with two lines in R² which have a relative angle of 45°. On the left-hand side, the coordinate system is chosen such that the red line lies on the y-axis. If you go one unit along the red line, the length of the green one will increase by 1.4. So the length of the green line increases faster than that of the red line (that statement doesn´t make much sense which is what I actually want to show with the example). On the right-hand side, there is exactly the same scenario except that this time I´ve chosen the coordinate system such that the green line lies on the y-axis now. One step along the green line will increase the length of the red one by 1.4. This apparent paradox is the one you have with your problem above (with the exceptoin that the metric of spacetime is dy²-dx² instead of dy²+dx² so that the length of the line not on the axis will increase slower). Effectively, what I did in above R²-example was comparing the length from the origin to two points which have nothing in common other than having the same y-coordinate in the currently chosen (arbitrary) coordinate system. You do. Oh, and "time passed for someone" is the length of his/her/its path through spacetime, in case you didn´t know that and wondered why I´m talking about the length on a line all the time.

Is this supposed to be a running gag?

By using the ansatz Position_Speeder(time) = Position_Police(time) and solving for time.

Assuming you mean "does 1 mol Pu take up roughly 22.4 l of volume?" then the answer is no. The 22.4 l/mol are an approximation for little-interacting gases at standard pressure and standard temperature. Just try it out for water. Hopefully, you´ll quickly see that a water density of ~2 g/l is not a realistic value.

Not completely sure what you mean with "this". General Relativity in general or how the movement equation is derived in it? Keywords you could look for to read up on the movement equation are "geodesic equation", "christoffel symbols" and "connectivity". I have a the derivation online on WiSci (http://www.wisci.org/wiki/Geodesic_equation) but I´m not sure if it´s understandable without knowing some basics about relativity.

I remember having done an experiment on "Fourier Optics" in my advanced labs. It only was an ~5 hour experiment but it was a standard one (that is, we didn´t have to care about which lightsource/lenses/... to use and the tasks were given and known to be feasible within that time). It was not really the most exciting or most ambitious experiment I ever did but it was quite nice being able to physically realize and visualize some properties of the Fourier Transformation.

Yes, there´s something fundamental you haven´t considered: If you have any gravitational field G (<- Newtonian Gravity) or equivalently any spacetime g (<- General Relativity), all particles with a given starting position and a given starting velocity will have the same movement in this field. In other words: The mass of a particle does not (directly) influence it´s movement in a gravitational field. In Newtonian Gravitiy this is because a = F/m, F = G*m => a = G. The mass doesn´t enter the movement equation (a=G). You do have problems with massless particles here but you can´t consistently describe massless particles in Newtonian Physics, anyways. In Relativity, you don´t have forces. Instead you make an assumption which usually is refered to as "particles take the shortest path through spacetime". This also leads to a movement equation which is independent of the particle´s mass.

In a similar situation I did an experiment with two masses hanging on springs as shown in the sketch I attached to this message. What we did was measuring the movenent of m2 for different starting conditions and compare it with out calculated predictions. However, seeing that you bother about the grade you get and that you only have two days left, it´s possibly not a good choice for you: a) The experiment might look very simple but the components we used were far from optimal. It took us three hours to construct the experiment in a way that we could at least take some data. b) The math involved is very interesting but it might be a bit too much to learn within two days. c) In practice, things you just didn´t take into account happen. Your predictions just don´t hit the data very well (the first two or three oscillations went fine, afterwards only the shape seemed similar). You´ll need a discussion of how good your predictions are. So why did I say above? Well, I told you what experiment I did so that maybe it inspires you. I told you about a few problems we had to warn you not to underestimate simple-looking experiments.

True. Had the initial post been something like "can anyone help me writing a program that generates primes? I´ve tried this: [add code here]" I´d even feel sorry about it. Even though it´s called "tutorial" for me it seems more like a quick-reference. I´d think there are better tutorials out there. What exactly is a password program? There is two ideas that spontaneously come to my mind (don´t know why): 1) Generating all prime numbers up to a given number. 2) Check if a given number is a palindrome or not. Other than that: Well, what are you interested in? Math tools? Physics simulations? Graphics? Sound? Networking? Artificial intelligence? Something else? Except for articifial intelligence (which I know nothing about) I think there´s easy programs that even a beginner could write.

I assume your confusion about "how can anything have momentum if it doesn´t have mass" comes from the momentum in classical newtonian mechanics which is given as [math] \vec p = m \vec v[/math], there. The terminoligy in classical newtonian mechanics normally uses the parameter [math] \vec v [/math] to describe kinematics. The big advantage is that the formulas become somewhat suggestive. From a more modern standpoint it is better to consider momentum being the parameter which describes the kinematics: - In contrast to [math] \vec v [/math], for which [math] | \vec v |<c [/math], the momentum [math] \vec p [/math] can, in general, have any value. - Momentum is conserved under any processes (like the collision of two cars), velocity isn´t. - You don´t run into the problem with massless particles Using the relativistic equations, energy and velocity as a function of momentum are calculated as: [math] E = \sqrt{ m^2c^4 + | \vec p |^2 c^2 } [/math] [math] \vec v = \frac{\vec p c^2}{E} [/math] Using these, you can easily see that massless particles have a velocity of c (regardless of their momentum) and that particles with m>0 have a velocity <c (also regardless of their momentum). IN SHORT: Think of momentum as the fundamental property, not velocity. You cannot change nature (the masses of the particles are given by nature, not chosen by man) to make it fit your equations. It´s done the other way round. Neutrinos almost certainly have a mass. It is not known if photons have a mass; they are treatened as massless in all cases I know of. The equations in which photons and neutrinos are massless are good aproximations as long as the momenta of the particles are large compared to their masses. For photons, this has always been the case, so far. See above for why only particles with zero mass can have a velocity of c. This "mass increases with velocity"-thing is a very bad habit. I don´t want to go into detail here, I´ve seen too many discussion about it. If you´re interested in this, the terms "rest mass" and "relativistic mass" are the one to look for. I use "rest mass" in this post.

I´m lazy so I´ll ask you instead of looking it up from the page you linked to: What does "it´s free until Novemver" mean? Does it mean "get it till November, afterwards you´ll have to pay for it" or does it mean "when you get it now, you can use it till November for free, afterwards you´ll have to register"?

Dev-C++ is not a compiler, it´s a developement enviroment. Actually, I thought Dev-C++ uses gcc as compiler but I´m not sure about that. EDIT:

Well, it´s simply a definition; although it´s suited to get in contact with the entropy defined in (nonstatistical-) thermodynamics. Some properties which make it look like a good choice to me: 1) As long as the map from the number of microstates to entropy is a monotonous rising one (A > B => log(A) > log(B) ), the very important "the system will be in the macrostate with the most associated microstates"-axiom still translates to "entropy will be at maximum". 2) It seems like a practical definition: I can imagine you often encounter problems where you multiply numbers of microstates. Since log(A*B) = log(A)+log(B), entropy is an additive number for those problems. Sry for being so vague in point 2 but I don´t have a good example in mind right now; perhaps someone else has. Either way: From the physics-side, it doesn´t really matter if you take the log or not; it changes the equations but it´s still the same physical entity. Perhaps it´s comparable to measuring temperature in Fahrenheit or Kelvin, only a tick more sophisticated.

Telling if a number is prime and obtaining the prime factors of a number is pretty trivial (I´ve posted some metacode for that some time ago, I think). The problem is doing it quickly.

What I meant is that the value R is unaffected by the rotation. Possibly a superfluous statement, but it didn´t seem obvious to me, at least. EDIT2: k, I understand why you were talking about length contraction, now. I was actually going over rest-mass which is an invariant (my density rho would have been a "rest-mass density" not a zero-component of a momentum tensor and therefore invariant under moving <-> nonmoving) and then got it moving.

Now you got me curious: Why isn´t this a flat minkowsky metric? For me, the problem looks quite simple: I start with a wheel which has a mass distribution [math] \rho = \frac{M}{2\pi} \delta(R-r) [/math]. Then I start rotating it with an angular velocity [math] \dot \phi [/math]. For all points on the circle, the velocity will be perpendicular to the radius, therefore I wouldn´t expect any length contraction. After integrating over the whole space, the total kinetic energy of the rotating ring will then be [math] E(\dot \phi) = (\gamma-1) M = \frac{M}{\sqrt{1- (R \dot \phi)^2}} - M [/math] which of course diverges for [math] \dot \phi \to 1/R = c/R [/math] which is equivalent to a rotation frequency of [math]f \to \frac{c}{2\pi R} [/math]. Or in other words: As the frequency approaches that at which the particles in the ring had speed of light, the energy diverges. For any finite energy the system has, the particles will move slower than lightspeed.

Dunno what myspace is but it sounds like a webspace provider. I´d assume most scientists use the webspace provided by their institution (university, research facility).

http://www.scienceforums.net/forums/showthread.php?t=16896&highlight=strange+matter

I think another issue would be the energy required to get the wheel spinning so fast. You should ultimatively run into the "no matter how much energy I put in, the velocity will always be <c" problem.