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timo

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Everything posted by timo

  1. Yes, that´s both correct. Note, however, that "photons are massless" refers to their rest-mass (a term which is a bit misleading because photons cannot be at rest). The fact that the term "mass" can either mean the rest-mass or the total energy (relativisic mass) can be a bit confusing if you´re not used to keep that in mind. But that´s not really important for your question, just wanted to say it. Simple answer: The movement of a particle (due to gravity) is independant of it´s mass. Example: A feather will fall as fast as a cannon-ball if there´s no air-resistance. This is even partly true in Newtonian Mechanics where the "heavy mass" and the "inert mass" are experimentally verified to be proportional and hence cancel each other out in the equations.
  2. I´m not a native english speaker so I don´t really know the english term. By "unit" I meant the one-element and I used the term because Meital used it. Well possible that it´s called identity (would indeed even make more sense).
  3. >> What is the probablility equation that all these dates are numerologically >> tied to the above subset of numbers? Given that those events were chosen because they do have an association the probablity is 100%. >> 2nd question: specifically regarding the most prominent of dates/numbers, >> the number "11", which occurs at least on 10 major dates of global >> history, what is the probability equation and result for that? That really depends on the total number of major global events (EDIT: Not the equation of course, but the probability). Since I suspect this number to be very high so would be the probability for 10 of them being associated to number 11.
  4. Actually it´s quite ok. Just two remarks: - Might be a good idea to explicitely mention that there is only one unit in a ring. Because you just show that f(1) is a unit. Ok, that´s nitpicking perhaps. - Your remark "since R' is commutative" is a bit disturbing. Simple reason: You said that nowhere before. Also, this proof is usually made for non-commutative rings. So I´d guess you´ve got a flaw there. The proof which does not use commutativity is quite similar to your attempt, though.
  5. The biggest problem with this would be to recognize which points on both pictures do correspond. I wouldn´t know how to do that. You could try an approach with comparing colors (I´d compare then on similarity not on total equality). This might work for objects where distinct faces have an own color each. For a human face I´d not expect it to work (it might do so though but I doubt it). Also keep in mind that certain points are only seen by one of the cameras. You could either ignore them for your 3D pic or extrapolate them. Assuming you have identified points that are seen by both cameras constructing the 3D position of them is fairly easy. For each camera you know that the points lie on a certain line in 3D (a ray cast into space). Take the intersection of the lines (or the point of closest approach to avoid problems with inaccurancies) and you have the corresponding 3D point. While the reconstruction is fairly easy and can be programmed by anyone with a little skill in programming I´d expect the recognition to be very complex. So unless you´re a really good programmer (well, you probably wouldn´t ask for advice if you were) or have fun with experimenting on code and being happy with non-optimal progs I think you´d best search the net for informations on that. Alternatively you could try a programming forum since to my experience there are a lot of competent people around there. I used to post on www.gamedev.net some time ago which is a pretty good forum. They are dedicated to game developement, however, so it´s possible they won´t like your question there.
  6. Great. I know it´s possible to write an english text with the required length, then. Could have guessed so before. No, that´s what we are debating about, I think. Of course it´s possible to get the correct key but it´s impossible to know it when you have it.
  7. I wonder if simply calling the police is considered "none" (as that´s the force I´d personally use) or "lethal" (as that´s how it can turn out for the burglar if he´s very stupid) or anything inbetween (minimum required perhaps).
  8. I´m not completely certain about the size of a "word". However (from what I´ve learned): A Byte is 8 bits. An Integer (the data type called "int" or "integer" in most programming languages) is the size of a processor register which would be 32bit for today´s standards (though, 64bit processors are allready available). The latter would be what I´d consider a "word". Dave´s statement is correct in the way that you need to load the memory adress in a register to address it. Hence, with 32bit registers you can address 2^32 bytes of memory. EDIT: OH, and Dave: The max memory is one byte more than you said, because 0 is also a valid address
  9. @5614: I do very well understand what you say, I think. I am however afraid that you dont get my point which probably is because I was too lazy to write a good post explaining my point in detail and because you should really take more time than 5 minutes before writing a response (honestly: 5 minutes does not make the impression you really tried to understand). Let´s agree that the message is coded with a random key of the same lenght by an XOR-operation so we have at least some basis to discuss upon. What you are saying is that I just have to check each key and you will have the original message sooner or later. That´s absolutely correct. What you also say is that the problem is to know when you got the right key. That´s in fact the point. Try to switch message and key. The transmitted message will of course remain the same because K XOR M = M XOR K. Now try each possible message until you get the correct key. I hope we can agree that you are absolutely not able to tell which one of the keys was the original one (because it was random by definition). I hope that clearifies the problem a bit. While I said "you can´t get the message because you don´t know which of the possible keys was used" you said "you can get the key if you know which one of the possible messages was transmitted". Both statement are correct.
  10. We seem to have different views on crackable. If I take the set of all possible messages (it´s even a countable one) and say "your message is in there, I cracked your code" then I have cracked each code even before the actual message is written. EDIT: To clearify on my example with the 100 Bytes. You do know the message is definitely one of those of the set of all messages containing 100 Bytes (size of this set is 8^100). However, you have absolutely no way to know which one it is because they all do have the same probability (1/(8^100)) to be the right one.
  11. >> and the computer would need to recognise when it had cracked the encryption. That´s the major point. If I send 100 Bytes that you don´t have any additional infos on, every of the 8^100 possibilities could be the message. You could as well just intercept the length of the transmission and make guesses what it could be or not intercept anything and simply guess.
  12. Wouldn´t it have been a better idea to apply this comment on the statement my remark referred to? That statement makes even less sense, then. However: [smart-assing] We are talking about Black Holes. The only known solutions of the Einstein equations for that imply that there is no mass except for the singularity.You´ll have a hard time finding light that doesn´t travel at c there [/smart-assing]
  13. EDIT: All you said is completely correct. The more energy you transfer the more far the ball is going to fly (within reasonable circumstances). Only the last sentence could be a bit misleading. The force you applied to the ball while it was in your hand of course also determines it´s movement. Actually, the energy you transfered to the ball can to some point be considered as a measure of the force that worked on the ball while it was in your hand. Maybe you have heard that energy = force * distance. If, for example, you apply a force of 10 Newton on the ball over a distance of 1 metre, the kinetic energy you store into the ball by doing so is 10 Newton * 1 metre = 10 Newton-metres = 10 Joules. Remark: A surprisingly well-formed question.
  14. It´s another level of unbreakable, at least. I can write a program that factorizes large numbers within 5-20 minutes if the numbers fit into an integer supplied by that language. Methods of factorizing large numbers can be rather trivial. The method only works because my program would probably run a few years before giving a result. The other one is unbreakable even in theory. Unbreakable, however, -by how I use the term- does not mean that there´s absolutely no chance of getting the information. If, for example, you intercept the message "sd4a" which was sent by above means, just guessig the original message was your favorite four letter word has at least a finite chance of being correct. But that´s not breaking the encryption and your chances drastically decrease with the length of the message.
  15. Oh, if you agree that it´ll work if the numbers are really random, then I can save my time making thoughts about how to explain it (I had to look up a few things so I´d prefer not doing so). Indeed, computers only create pseudo-randoms but above encryption method simply assumed you take real randoms. They don´t have to be made with a computer. Of course I could think of few (=no) methods for creating large ammounts of (pseudo-) randoms other than using a computer so the problem would indeed lie in the random generator (apart from other problems like that I could simply steal the information from any of the computers). As for real randoms: This somehow rings a bell. I think it´s one of the advantages of Quantum Computers that they are able to produce real randoms. Assume you prepare a state [math] \frac{1}{\sqrt{2}} \left( |0> + |1> \right) [/math] with |0> and |1> being perpendicular states (polarization of a photon for example). If you now measure on |0> it´ll give you a count 50% of the time without any correlations between the different measurements. Well, that´s all idealized. The real problems lie in the actual building of Quantum Computers. The theoretical background and some algorithms are allready there.
  16. May strongly depend on if that guy called a file of random numbers a code I think (that´s all you have left when you apply a random key of the same length to a message). Have to leave now, but I´ll try to go into more detail later.
  17. That´s actually the reason why no light escapes the area within the event horizon. Spacetime is curved in a way that none of the possible light-paths (and no "matter"-paths also) lead out of it. Do you know that, did you hear that somewhere or is it simply what you think? If case 1 or 3: Could you please explain what you mean with that statement? I don´t understand that, especially not when I consider that light allways travels at lightspeed. @Herme3: I hardly doubt that anything exceeds the speed of light when falling into a Black Hole although I have to look up the formulas to be certain. If I´m right, then your idea is simply wrong. Otherwise I might have met a very interesting new phenomenon today.
  18. Current encryption methods work because it´s very hard to find the prime factors of the product P of two large primes. Don´t ask me how that actually works - I´m just repeating what I´ve heard everywhere and I don´t know the details. The length of this prime, however, is small compared with the length of the message. If you are able to break down the publicly known P into it´s primes it seems like you can easily decode the message. Normal computers are quite slow doing this. On Quantum Computers, however, algorithms have been proposed that would run much more efficiently and would do the job in a reasonable time. If you for example chose a key with the same length as the message, a simple XOR-operation will destroy all information stored in the message except it´s size. This can be considered as completely unbreakable (test: Try to find the message of 1MB of random numbers). The problem here is that you have to transport your key to the reviever as well as the actual message. That´s where Quantum-Cryptography comes into play: Because it is possible to determine if someone intercepted your message sent by QM-ways you can simply send the key to the reciever. If someone intercepted the key you will notice that and simply send another one untill it hasn´t been intercepted. Now if you successfully sent your key without anyone intercepting it you can simply decode your message with is, send it per E-Mail and let the other one decode it with the key only the two of you have. Above method of QCryptography was referred to as the BB 84 Protocol by the lecturer of my Quantum Information lecture if that helps anyone. To sum iit up: It is impossible to break a message that was encrypted by above means of Quantum-Cryptography, regardless of what kind of computer you use.
  19. [1] Didn´t get your time definition. Especially the statement "there is no time, since all the units move at the speed of light" confused me. [2] So if the units react individually under collision what is your rule to tell which block later on belongs to which object? Let´s use the notation (x,y) for an object where x is the number of >´s and y the number of <'s.. So my example about the photons would read (2,0) + (0,1) --> (0,1) + (2,0) (note that the summands have an ordering). That works easy with (x,0)+(0,b) also. But what happens to (3,1)+(1,2), for example? Where do I distinguish between the particles? If I thread every block with it´s individual "collisions" I´d get (0,3)+(4,0) afterwise. Total matter-anihilation for every physical process? Simply asked: What is the result of (x,y) + (a,b) ? [3] I´m really interested how you take the relativistic mass-change into account. If you manage to do this and still have energy-conservation you might have a first result. I see little point in enhancing your "this all is described by my theory"-list if there´s an obwious weakness at this basic part. [4] If you allways thought that energy in general in quantized (I´ll try to remember the correct word ) then you now learned that it isn´t, at least in the sense I was talking about (free photons). Funnily you gave a formula showing this yourself (E=h*f). Think about it a while to see what I mean. Remember what I said about in-depth insights about current theories? Well, this one is not really a big secret but I´d expect a lot of users here to give the same answer when asked why the energy of a free photon is quantized. [5] How do you derive the number 10^30 ? Also, do you think there is any experiment being able to measure to an accurancy of 10^-30 so the difference between your quantized units and non-quantized mass and velocity can ever be observed? That´s my thoughts for now. Especially the points [2] and [3] would interest me as they would actually give at least some results if you make them work.
  20. I´ve spent too much time on this post allready so I´ll make this short: 1) Never post a reply while being angry/frustrated. 2) Learn to deal with criticism. ------------------------------------------------------- Now for your (now surprisingly understandable) approach - the actual reason that I even bothered posting here again: For your assumptions: - What is speed, what is the universe? Are you assuming a universe as in Newtonian Mechanics with time being a quantity existing independently from space? - Thinking about elastic collision even helps that much that I can ask my 1st question here: What happens when two different photons >> and < meet each other? According to your elastic collision that would result in << and > after collision, respectively. Photons, however, do not interact. So it´s still >> and < in reality (or < and >> depending on your view). For your clearifications in the follow-up post: - Are you aware that you are quantitizing speed? An object having a mass of 10 units for example can only have 11 different speeds (I´ll call them velocities from now on, that´s probably what you mean). If you counter this with "I make the unit so small you won´t recognize it" then I´ll be very interested where the whole point in integers of units is going to be. - Mass - Energy equivalence: Splitting up the mc² term into two is ok. But if you consider the kinetic term you have a problem: Even if you can account for the relatvistic mass-change correctly and can give an explanation how and why you do that, it´ll most certainly involve the lorentz-factor. The lorentz-factor, however, is the inverse of a square root. I´d be very surprised if you can model this with integers. - Energy-quantization: I hope you are aware that energy in general is neither predicted nor observed to be quantitized. If it really is, your energy quantum "building block" must be smaller than 1/3rd of the tiny mass of a neutrino. If they, however, travel close to lightspeed you need a lot of >´s to make up for the one < that makes the neutrino-mass nonzero. That´ll make the energy of your building block even more tiny. Are you aware what numbers of units we´re thus talking about when even considering particles as small as a proton? There´s a lot more questions that could be asked but I think that´s enough for now. Note that I consider any response that took less than three hours of thinking as an insult (yes, really).
  21. I wouldn´t interpret his statement as a hostile attitude. Being told "write down your ideas as formulas" is the standard answer you´ll get from any physicist when you present weird (new) ideas to him. I´ve been told that very often. Sometimes I indeed took the time to try to write it down (even made a seminary talk about one of those ideas). In the end I allways recognized why my idea was bad or useless at least by doing so. But it´s a great exercise doing so as you sometimes get some in-depth insight into current theories that others who simply accept it might miss.
  22. timo

    cause of gravity

    yep, and no curvature in the sense of GR (or at least I wouldn´t know where extrinsic curvature plays a role in GR).
  23. Good question. But not a valid one in GR. Time dilatation and length contraction can be considered as an effect of changing your base-vectors when switching from one coordinate system to another. But in GR vectors are only defined in a point of spacetime. So you -usually- can compare only vectors in the same point. And the effect of graviational time dilatation (whatever that´s supposed to be) is of course the same for all observers in one point. Also note that the statement "in a gravitational field, time slows down" is vague at best. It sais that Eigentime increases slower with the time-coordinate of your coordinate system. The time-coordinate of your coordinate system, however, has little to no physical meaning.
  24. timo

    Men Not Included!

    Good post Aardvark and there is not much in it I could refuse in general. Just a few remarks: - If I come to the doctor with bad general health I´ll certainly be given priority over an older person being in better shape unless the doc is a complete idiot. Not much point in debating over this point, however. I know what you mean and I think you also know what I mean. - I simply feel like the importance of "one should know his roots" is generally overrated. I certainly do not deny that some people have a strong interest in knowing. Maybe you don´t know your biological parents and know what you´re talking of. But the point that practically noone takes a gene-test on his parents if he´s told that they are his biological ones seems as a hint to me that it just isn´t of fundamental importance for one´s life (and honestly: Who would like to find out his parent´s actually aren´t the biological ones?). Perhaps I´m biased in this matter as I personally know someone who doesn´t know her biological parents and has not raised any interest in them so far although meeting them -at least the mother- would be possible. One last thing: I would expect that people who are considering using the mennotincluded-service think about the point that their children will never meet their father more than both of us do (more than me at least).
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