timo

Wouldn´t it be best to go to the game companies' websites and check out the requirements for their open positions? Also, be aware that creating a computer game needs much more than a few programmers. I´d bet that the numer of level designers, story writers, website admins, model designers, artists, ... is much higher than the number of actual programmers (you can probably check that out in the credits of your favorite game). I think i even read that the actual game engine (the graphics engine and the network code) which is the part that requires skills you probably learn in computer sciences are sometimes bought from other companies to reduce development costs and the number of "real" programmers required. Two more things I´d like to mention: - Be aware that there´s really a lot of very talented young people around that want to go into the game industry. I´ve been programming (with varying enthusiasm) for 15 years now but most of the 16 year old hobby programmers I´ve seen are much better coders than me. - A very good ressource for game programming infos is http://www.gamedev.net. if I remember correctly there was also a forum about game industry. You might want to go there and browse the forums for information.

@calbiterol: I´m not really sure what part you didn´t understand. Perhaps it´s best if you say what you understood (or at least what you think you understood) and what parts were problematic for you. As a sidenote: Originally my post included some variables, equations and one inequality. They seem to have disappeared somehow (at least they don´t show up for me right now). Is the missing math the problem? @swansont: EDIT: I´ve even edited my post before posting it. Atm I also think that conservation of momentum is not going to be a problem. However, if you´re interested in knowing were my original concerns lied, then go ahead reading... Yes, of course there´s recoil. And indeed for free atoms it seems quite easy to find parameters that make conservation of energy and momentum (and thus the whole process) possible. What I had in mind, however, was the process to take place in a crystal. For crystals, I think things become a bit more complicated: - You need to find a a phonon state (or a combination of those, perhaps) with a fitting pair of (E, p) to ensure conservation of energy and momentum. - Atm I cannot see if the accoustical branch of the phonons is going to be much of a help, because for small momentums E(k) it is also linear as for the photons. Intuitively Id think this might be a problem. EDIT: On 2nd though this should be no problem since you can simply make the recoil happen in opposite direction as the released photon´s momentum. If you then increase the ammount of recoil beginning from zero you reduce the emiited photon´s energy and also increase it´s momentum. You can do that until conservation of energy and momentum are true. Only problem might be the quantization of the phonon spectrum but for a macroscopic crystal this should be small enough. - The optical branches can easily absorb energy with no or little momentum being absorbed. This would make them suited for compensating the excess of energy. However, if the optical branch is narrow and does not include the energy required then it doesn´t really help. - Combinations with more than one phonon involved are going to reduce the probability for the process to happen. - A last possibility that might actually work would be giving the recoil to the crystal as a whole. Since E=P²/2m is pobably neglectible you can simply chose a recoil momentum in a way that energy and momentum are conserved. However, I remember something like "the possible momentums a crystal can take over are quantized to hbar*G", where G is a vector in the reciprocal lattice. I don´t really know why this should be because I can hardly imagine the momentums for a free particle -regardless of it´s size- to be quantized but it seems to be the case (this quantzation leads to the Moessbauer effect if I remember that correctly). So if the quanta are large enough this might also be a problem for the conservation laws. Huh, didn´t want to write that much on this, originally. Well, once you start with it ... . However, I´d like to emphasize again that I said "conservation of momentum might be a problem" not that the process is impossible due to conservation of momentum. It was also just a sidenote that came to my mind when I wrote my post and I didn´t bother (as I also didn´t do now) to do any calculations on it.

The pages are not too informative anyways. It was just a suggested start to browse. The "glossary of frequently used terms" are in english (though, they are quite short) and the links go out to english pages. I´m also not completely sure if you actually will find usefull information on the linked pages because I didn´t visit them to check it out.

My momentum conservation concerns are the following: Assume the two original photons have energy and momentum of and respectivey (c=1 for simplicity). The released photon has energy and momentum of when you assume that there´s no recoil. But as (note that the photons´ momentums are explicitely not colinear), the simple process "2 photons -> excited state -> 1 photon" is not possible in the simple form my post suggested. You have to store excess energy in the cystal or consider recoil, for example.

Given the brain has finite size and assuming there is a finite ammount of information a single cell (or a neuron or whatever stores data) can store I´d say yes.

Is the answer supposed to be a function of the resulting volume V? In that case (and if I understood your question correctly - I´m not a native english speaker) you could calculate the resulting volume V as a function of minor axis lenght a and major axis length b (V=f(a,b)) and solve the resulting equation for the length of the major axis (b=g(V,a)), then.

>> An explanation of why Rootie was wrong would have been enough to enlighten though. ^^ Well, that´s quite hard because as far as I can see it Rootje is completely right. Within the approximation of the Schwarzschild metric every particle of the star will reach the singularity within finite Eigentime and will not ever pass the event horizont for an outside observer. However, there are three points in this statement I´d like to mention: 1) It is not possible to make a distinction between a spherical symmetric mass with r=rs and a spherical symmetric mass with r=0 for an outside observer. But that was also allready said by Rootje. 2) It is said (and way beyond my understanding of GR) that not only a spherical symmetric mass distribution like in the Schwarzschild metric but a great range of conditions will lead to a sapcetime singularity. I would not bet that all these conditions involve infinite observer-time for the forming. 3) The perhaps most important point: The outer Schwarzschild metric is not a valid coordinate system for the whole spacetime of a spherical symmetric mass. While it is (at least to a good approx) the system that we observe in it might be a bit of a philosophical question whether things that we never observe -like the mass crunshing to a singluarity- do exist or not. From out point of view nothing ever passes the event horizont. Every particle of the star, however, will -from it´s point of view- fall into the forming singulariy within finite Eigentime. This reaching of the singularity within finite Eigentime is a frame-independant statement so we (the physicists) say that´s what happens. The statement that nothing ever passes the event horizont is frame-dependant and thus not phyiscally meaningfull (though it might have some meaning for you, but phyiscs isn´t very concerned about your needs ). To sum it up: The non-collapsing of a star as seen from an outside observer in the Schwarzschild metric is an effect of an invalid/incomplete map of the universe.

I don´t want to formulate a text atm so I´ll just throw in a few thoughts of mine: - As said by The Rebel photons (light beams) do not interact. - The effect of creating reactions (like emmiting light) at a cetain point within a substance by pointing two lasers at this point should be possible in theory. I´d think about the two photons in the IR-range being absorbed to create an excited state. This excited state could decay back to the ground state by absorbing a single photon in the visible range. Conservation of momentum might be a problem here. - I remember that some years ago some experimentalists at my former university showed me their labs. They were experimenting with holographic crystals which were supposed to be used as the new generation of memory chips (they allready were able to store and reread data back then -6 years ago- dunno why it didn´t come out as a new technology, yet). They used two lasers to create interference patterns in a crystal. The link to the group is here: http://www.physik.tu-darmstadt.de/lto/pro/index.html . Be aware that from my experience the pages of the TUD are not very informative, but you could read around in the "Glossary of frequently used terms" or maybe check their links. - The idea to cast rays of something from different directions to create some effect at a certain point has at least one practical application I know of: The GSI ("Gesellschaft fuer Schwerionenforschung"="Corporation for heavy ion research", also, located in Darmstadt) in cooperation with the physicians of the university of Heidelberg are using neutron beams to destroy cancer tumors. Neutrons in matter have the very pleasant feature that they give away most of their energy within a certain range. If you now adjust your beam so that this range is within the tumor and shoot beams from different directions you have a very good ration of "energy deposited in the tumor / energy deposited in healthy cells". Or in other words: You can destroy the tumor with minimal damage to the patient.

They do get absorbed in matter, yes. The absorbtion was accounted for in the exp(-kd) term in my 1st post. It´s just an approximation that assumes that for each distance interval traveled a certain percentage of the photons are absorbed. However, there will allways be a certain percentage of photons that aren´t absorbed, no matter how far the wave has travelled. This is in analogy of radioactive decay. The number of objects N(t) that didn´t decay till time t is N(t) = N(t=0)*exp(-kt). At no time there will be a certain chance that all objects decayed. However, since N(t=0) is a finite number there will be a time T where N(T)<1 which means (a bit losely speaking) that you cannot be sure to have any object left. This is what I meant with the chance to miss the wave. The wave will disappear due to absorbtion but for all distances there is a finite chance that it doesn´t.

Clasically: Yes. But at some point the intensity will be so small that quantization of radiation will play a role (I did not take into account this in above). So when the intensity of the signal falls below a certain level there is a chance to miss the wave (or part of) - regardless of the quality of the reciever. EDIT: But I´d like to add that this quantum mechanical limit is probably only a theoretical one. I would be very surprised if today´s recievers came even close to that limit.

You can put brackets wherever you want, it won´t change anything on the equation.

R = rho * l / A where rho is the specific resitance (resistivity? I don´t know the english term) of the material used, l is the length of your wire and A it´s cross-section.

Yes there is (or at least was). See this link: http://www.scienceforums.net/forums/showthread.php?t=4236 There was a problem with the TeX code some weeks ago, I don´t know if it has been fixed. Test: [math] i\hbar \partial_t \psi = H\psi [/math] EDIT: ^^ no, it still doesn´t work.

There is no general limit on how far electromagnetic waves (radio waves) travel. However, due to absorption in the air and the fact that radio waves do not travel in a straight line but spread out in all directions the intensity of the signal will be reduced with increasing distance. The intensity I(d) at distance d should in a 1st approximation behave something like I(d) = C/d² * exp(-kd), where C is a constant that depends on the power of your sender and k is a constant that describes the absorbtion in the air. so as the intesisty of your signal is reduced with increasing distance you have two options to still read the signal: a) A better reciever. The better your reciever the smaller signal intesities you can still recieve. Due to background noise there shoudl however be a lower limit to the intensities you can recieve. I´d guess that today´s recievers are able to reach that lower limit. Below that level you can still detect the signal by filtering if you have some information on it before (like knowing it´s a periodic signal) but that´s another topic. b) A more powerful sender. Given an unlimited power source you could simply increase the original power of your signal so much that it still has sufficient intesity when it reaches the reciever. Theoretically you can send any distance given a sufficient power source. To sum it up: The signal your one computer sends does not completely dissapear but it´s intensity will be so low that it becomes undetectable due to background noise.

Yes, a very famous theory (General Relativity) is based on the assumption that there is no "natural" system of reference and thus no "natural" passing of time.

Fake memory I´d guess. By the time you have a tube in your throat you are almost dead. No breathing, no conciousness, heart activity stabilized by Atropine. Imho it´s very unlikely your brain realized this, not to speak of the fact that i can hardly imagine how you recognize a tube in your throat without being able to move your hands or to feel pain.

No, the example with measuring an electrons velocity and position with a photon (ligh) is just an attempt to visualize uncertainty.

http://www.gamedev.net would be your place to go for such questions.

That´t not correct Sayonara. The Schwarzschild solution which is usually used for spherical symmetrical mass distributions is only valid outside the distribution or, if the total mass distributions end up behind the evnet horizont, outside the event horizont. The Schwarzschild coordinates have a coordinate singularity at the event horizont. You can however find coordinate systems (Kuskal coordinates) that do not have a singularity at the event horizont. They are nessecary to show that a particle falling in a black hole actually does reach the singularity and does not stop at the event horizont as is would appear in a Schwarzschild coordinate system. Due to the lack of a coordinate singularity one could see the Kruskal coordinates as the better coordinate system for describing the spacetime of a spherical symmetrical mass distribution. In Kruskal coordinates you suddenly map areas of spacetime you didn´t even see in Schwarzschild coordinates. You get two causally not connected areas of free space (parallel universes) and two regions inside an event horizont. One of those regions inside an event horizont has the strage behavior that all particles inside it must leave the area. So it´s quite the contradiction of the common behavior of a black hole where all particles inside must end up in the singularity. Because of this this area is called a White Hole. I found a picture of this on the bottom of this page: http://scholar.uwinnipeg.ca/courses/38/4500.6-001/Cosmology/Black_Holes.htm The area called "time reversal black hole" is the white hole, The areas called "outside black hole" and "inside black hole" are the areas covered by the Schwarzschild coordinates. The unnamed area is the parallel universe. The geodesic equations for light is given by eq (2.18) as u+-v = const. A few of the light cones where the time-like geodesics must lie in are also shown. As for us-2u´s question: No, afaik there is no observations of or hint towards the existence of a White Hole.

Didn´t bother to check your relativistic calculation but the classical one is flawed, at least: 0.5*m*v² = 0.5*(1 g)*(0.5 c)² = 0.5*(1 g)*0.25*c² = (1/8 g)*(3*10^8 m/s)² = (1/8000 kg)*9*10^16 m²/s² = 9/8*10^13 J = 90/8 TJ = 11.25 TJ < 13 TJ

If I see it correctly, the number of possible solutions should become obvious when you solve the problem as you get a system of linear equations with six real-valued unknowns (four in A and two in v) and six equations (one for each coordinate mapped).

>> Would it be possible to create matter from energy? Like a reversed nuclear reaction? Generally: Yes. >> Has it been done? How would it work? Extremely short-lived heavy particles are produced in particle accelerators by shooting particles with very high kinetic energy at each other, so I´d say that´s transfering energy into matter. Though, I´m not completely certain if the net rest mass after the collision is higher than the rest mass of the original particles as they are probably destroyed in the collision. But it should be possible to get a win in rest mass, theoretically. The nuclear physicists here should know that. Another process that actually happens and is sometimes taken into account for calculations is that photons (~light) can decay into an electron and an anti-electron. But I don´t know if this process is actually used on purpose (like for creating anti-electrons). >> I can only assume it would require massive amounts of energy ... The minimum energy required would be given by E=m*c² with c being the speed of light, m being the (rest) mass you want to create and E being the required energy. Calculate that out for 1 kg and see if you consider it much. >> ... and would not have any practial near-term applications Trying to create the particles predicted by theory is actually a current application. >> Create matter from energy -> gravity pulls spaceship -> Convert matter back to energy -> back to step one I don´t have a good answer on this so I´ll try this one: There is probably no point in creating matter to get more kinetic energy from falling towards a planet rather than using the energy to increase your ships kinetic energy (accellerate) without taking that detour.

Thx for the answer Pmb. However, I either didn´t understand your answer or there has been a misunderstanding because your answer actually backed up what I said: That it´s not possible to tell space from time in spacetime. I´ll also assume a flat spacetime: - >>"You measure "space" with a ruler and "time" with a clock." Yes. And the time I measure is the length of my path through spacetime. So assuming we both start from the same point and since we are obviously heading in different directions the direction in spacetime you call time is a different one. In fact, I would say that what you call a time-interval is indeed a mixture of time and space. There is no objective separation of time and space in spacetime. Guess you were speaking of the coordinates (of our both frames of rest in this case), then. - Your elevator example is a good example for my claim that you cannot tell if a trajectory is curved or not: I hope we can agree that the path of the light is completely independent of what the elevator does. As you said the beam will be a line for an observer in the elevator if it isn´t accellerating but will appear curved if it accellerates. So since both reference systems are equivalent there´s no way telling if the trajectory is curved or not. Same as above: A judgement like "this path is curved" can only be made within a coordinate system and is dependant on it. Frame-dependant statements, however, are quite useless. That´s the point in the whole tensor stuff (so you can tell if a trajectory is a geodesic or not because that´s a tensorial statement).

Partly agreed. I rather see a straight line as a special case of a geodesic for a flat space. Well, I can live with your statement but I think it cries for new confusion (keep in mind that most people here don´t know GR). But the thing that raised my attention in your post was the following statement: Is it really possible to tell space from time in spacetime in general? Or do you speak about the coordinates of your coordinate system (which are rather meaningless due to your freedom in chosing a coordinate system)? Or in other words: Is it possible to tell if a trajectory is curved or non-curved without using a coordinate system other than saying "they are all non-curved because they are geodesics"?

That´s absolutely correct. Those "straight lines" are called Geodesics which should sound familiar to you. In fact, not only light but all particles move on Geodesics.