ecm

well, it depends on the problem, of course. In this case, since we don't have any information about Sy and Sz, and the problem is completely symmetric for the y and z axis it couldn't be any other way. But I guess in general it depends on the problem.

Yea, I think it would be ==0. But we wanted to calculate and . And sadly the average of the square is not the square of the average......

Hi, papucisse, how is it going? :) To answer your question, the longer the spring is the easier it gets to elongate it. Imagine you have a very very long spring. Wouldn't you say that it should be a lot easier to pull from the end of it, that if you take the scissors and cut just a tiny piece from the end? I do think so.... And if so, then k should be greater for the little spring. I hope this helps. Good luck! ecm

Define "move". :p The distance between us and some galaxies does, indeed, increase faster than if the galaxies were moving from us at the speed of light, but that's because space itself is expanding between them and us. Nothing moves faster than light.

I don't know about the electroweak force.....

I don't think it is, 'cause neutrinos do not feel the electromagnetic force. If you have a process which involves photons it has to happen by electromagnetic force. But if you have a process with neutrinos, then it has to be by the weak force. So I don't see how it could be by both at the same time......

Hi again, Happy new year to all! Im glad to see people are interested in this topic. :) It's too bad none of us understands too much... ;):p Yevgeny, when you talk about energy in particles physics you are always referring to relativistic energy: mc^2 if the particle is at rest, and gamma times that if it isn't. There is no potential energy, there are no fields, only interaction between particles. (At least if you don't include gravitational forces. I don't know what happens with gravitation. (particle physics can't explain it). Anyway, the energy is always positive (well, at least if you are using quantum field theory, as I said in another post. Quantum field theory is the most advanced theory we have so far, although it still isnt able to explain many things). I've been studying a bit more about virtual particles lately, so perhaps I should let you know what Ive found out. (I still don't understand anything about black holes, though) First, it is equivalent to talk about energy not being conserved during a short interval of time and virtual particles that have the momentum/energy you would expect (as vish said), as to talk about virtual particles created that are off the mass-shell, but energy is strictly conserved all the time (as I said). You can think of it whichever way that works for you. When you study processes in particles physics you describe them by means of Feynman diagrams. I don't know if you have used Feynman diagrams (they are pretty cool![banana]), but they describe interactions between particles, that happen by means of one of the three fundamental forces (I'm not including gravity, cause gravity is not understood). The basic components of these diagrams are the vertices. In each vertex, you have 3 particles being created/destroyed. For example, in the case of the electromagnetic force, you could have a vertex with an electron and a positron being destroyed and a photon being created. Or you may have an electron being destroyed, and a photon and an electron being created (in that case you could perhaps say that the initial electron has lost energy and given off a photon). And so on. Now, none of these vertices describe an actual even that can happen, because none of these vertices allow for the conservation of energy and momentum. But by coupling several of these vertices you can make real processes. Now, the particles that connect two vertices of a "possible" process is what we call virtual particles. Picture you have a process described by a Feynman diagram. Here is your choice: you may want to consider that energy and momentum are strictly conserved in each vertex, or you may consider that energy and momentum only need to be conserved in the total process. a) Energy conserved in each vertex: As I said, for real particles this can't happen. If you say energy is conserved, then your virtual particle is going to need to be very weird, that its momentum and its energy are whatever are needed to conserve the total energy/momentum. In this case you can't use the formula E=gamma*mc^2. Energy is whatever it is. And momentum is whatever it is. We say the particle is off the mass-shell. (By the way, the more off the mass-shell the particle is, the less probability there is that the process is really going to happen.) So that's how you can find negative energies. Or positive. Whatever. b) Energy is only conserved in the total process: Now you can consider that all particles, both real and virtual, are on the mass-shell, and that the exchange of the virtual particle happens fast enough that we are within Heisenberg's uncertainty principle. (Personally, I like the first approach better, and I think it is more widely used). So what does all this have to do with black holes? No idea, but I thought it was interesting..... :drunk: So keep it up, perhaps we'll figure it out yet! [:o)]

[banana][banana][banana] WOOOOOOHOOOOOO!!!!!!!!!!! Yevgeny, you're the best!!!!!! Congratulations!!!!! [bounce] [banana][banana][banana]

Yevgeny, about virtual particles with negative energy, I've read that in several different physics books, so I'm pretty sure it is so. Actually, one of those books was written by a nobel winner, who was an expert in the field, so he probably knows what he is talking about. About the Dirac's sea... well, the need for it comes up in relativistic quantum mechanics, but the thing is that rel. quant. mec. is not a theory that works, i.e. it is incomplete, and it runs into trouble to explain several things, and also people didn't like the idea of the negative energies.... so a new theory was worked out, called Quantum Field Theory, that worked much better. It, among other things, eliminates the need for the electron sea... it says that real particles (and antiparticles) always have positive energies. All the explanation I posted before of the creation of the virtual photons, I had read in a book on General Relativity. If you want I can email you the name of the book and author. Your explanation sounds ok as well, but the thing is that neither of us knows enough about this to be able to either confirm or deny any of these theories. You did not use any math to show how the gravitational field creates the pair, even if you did I probably wouldnt understand it, so we cant really argue the physics of the process at all. Actually, we are dealing here with quantum effects due to gravity, and NOBODY understands that anyway. I guess that Hawking must have proved something in his article, so it really might be a good idea to check it out. One question: supposed that a electron/positron pair is created near the hole, by whatever mechanism, such as a photon kindly providing the required energy. And then it just happens that the positron falls inside while the electron speeds away. Does that increase or decrease the energy of the hole? Im not sure, black holes are quite curious objects..... By the way, merry xmas to anyone that celebrates it! [:o)]:drunk:

First of all, let me say the Dirac's idea of the electron sea, as far as I know, is only an interesting theory, but there is no experimental evidence supporting it. Ok, now, let's see. I did not talk in my first post of the creation of the pair electron/positron. Only of the creation of a couple of virtual photons. I don't know if Hawking's radiation refers only to radiation itself (emission of photons) or also to other particles, but I would tend to think that it could happen with other particles as well. Now, the creation of the electron-positron pair has to be understood, within the theory of the sea of electrons with negative energy, as a negative electron, becoming excited (which implies the absortion of a photon) and thus abandoning the negative-energy state: becoming a "visible" electron, with positive energy, and leaving behind a hole, that behaves like a positron. Of course the electron would quickly lose its aditional energy and fall back to the "hole", which would mean the anihilation of the pair. Ok, now both the electron and the positron would be observed as particles of positive energy, but you have to understand that in this process there is a photon involved (actually, there would have to be at least two photons, because of the conservation of energy/momentum, unless there is an external field acting on the vacuum or something, but never mind that now). This is the reason why it is said that the minimal energy of a photon, to give raise to a pair electron/positron needs to be 2mc^2, (m the rest mass of the electron), so that energy is conserved. So this process is different from the one with the photons (after all photons are bosons, not fermions, so Pauli is not gonna give much trouble here). But anyway I can picture a positron created this way falling into the black hole and causing trouble as well. Conservation of charge, like that of energy, is always strictly enforced by nature. I'm sorry, yevgeny, I can't give you any more detail, 'cause I just don't know much more. Actually, I believe that if you find any "expert" in the field and you start asking questions eventually the expert would not know either. After all black holes are not well understood at all, and nobody knows yet what the relation is between gravity and quantum physics. Anyway, why don't you try to find Hawking's article of black hole radiation and check what it is talking about? Then you could let us all know........

Energy is strictly conserved, yevgeny, always. Together with all relevant quantum numbers. If you want particles created from nothing, besides having opposite quantum numbers (that is, being a particle and its antiparticle) they need to have opposite energy. These particles are called virtual for a reason: they are off the mass-shell, that is, their energy is not related to their momentum by the usual relativistic expression: E^2 = p^2 c^2 + m^2 c^4 Their energy can be whatever, both positive or negative, up to infinite. The thing is, the more off the mass-shell the particles are, the shorter their lifetime. Anyway, now that you mention it, electrons with negative energy come up when you study electrons from a relativistic quantum setting. That is, if you solve the Dirac equation (which is the relativistic quantum equation for spin 1/2 particles, equivalent to the schrodinger equation) you will get both positive and negative energies. Physicists were surprised by this, and at first assumed that only positive energy had physical meaning. But the fact remains that if you only consider positive energy you wont be able to have a base (with a closure relation). Then people started wondering that if negative energies were possible, how was it that normal "positive" electrons didn't start losing energy, getting more stable the more negative their energy was, until infinite. And Dirac suggested the "electron sea": that what we call vacuum is realy a sea of negative electrons, taking up all possible negative energies, so that the next free level corresponded to positive energies.... So, you see, it is not so clear that there are only positive energies. Anyway, this a different story. I don't really know much about it. But it doesn't matter in this case, cause we are dealing with virtual particles and they can do whatever they please.

Anyway, you could consider the creation of an electro/antielectron pair, if it makes you happier. It comes down to the same thing

If you want to consider them so..... photons are the same as antiphotons.

Hi there, yevgeny! Yes, as far as I know, Hawking's radiation involves the creation of particle-antiparticles pairs near the horizon. As you know, the "vacuum" is in a state of constant activity because of the creation/annihilation of particle-antiparticle pairs. Suppose two virtual photons are created near the horizon. Because conservation of energy, one of them would have positive energy, the other one negative. Of course photons with negative energy can't exist, but it is really not a problem 'cause these photons are virtual. Well, now imagine that it is just happens that the photon with negative energy crosses the horizon. If you know anything about general relativity and the space-time metric of black hole you will know that inside a black hole the spacetime metric is inverted: space-like components become time-like, and time-like components become space-like. As a result, the negative energy of your photon turns into a certain (and acceptable) spatial momentum, while its former momentum turns into positive energy. So now both particles can exist, the are no longer "virtual". One travels outside, the other one falls into the hole. The hole loses energy because of this "radiation". (This was the proof that there could be no "micro-black holes", created shortly after the big bang, because the would have radiated themselves away by now). So that is all I know about this. I hope it answers more questions that it brings up..... ;)

Good job, papu! [bounce] I'm glad to hear you did ok, and I hope that we were able to help a bit! :) Now good luck with the college admission process! [heartbeat] (Let us know how it goes!)

No answers? :( Ok, Ill just answer myself: What I consider a possible magnetic field is any field that obeys Maxwell's laws. See? That's the problem: the "magnetic field" you calculated in your post (B = br/r^3) is not divergenceless, and thus does not abide by Maxwell's law, and thus, as I see it, it is not a magnetic field at all. There is no possible magnetic field corresponding to a magnetic monopole, any such field would have a non-zero divergence. So I still don't understand your original question. If what you want to obtain is a vector field A such that, if you define B=rot(A) you get B = br/r^3, well, then you seem to have managed it in your post. (I assume the math you did is correct, I haven't checked it). What Im saying is that that field B does not correspond to any magnetic field, and that the field A does not correspond to any magnetic potential vector, and that the whole thing does not have any physical meaning. Now, it would be different if you are willing to accept that Maxwell's laws might not be right; especifically, that B does not need to be divergenceless. Well, ok, but then what is the meaning of a potential vector?? Give me a new set of equations or something. Define the meaning of A. If I don't know the rules of the game, I just can't play! ;) By the way, you might be wondering just where, in the process you followed to obtain your "magnetic monopole field", you "broke" from Maxwell's laws. Why is it you got a field that can not possibly be magnetic? The answer is that you used an expression for A that you can't use in your problem, namely: A = (mu x r') / r'^3 Each piece of coil may create such a vector potential, but to obtain the total potential you cannot integrate it. That expression is only valid for non-infinite currents. In the process of obtaining it, you have to accept that for your problem the flux of the magnetic field B goes to zero at infinite. That, clearly, is not the case in the situation you presented, where you have a current extending to infinite. Actually, in your situation, the flux of B at infinite has to be such that it compensates for the apparent non-zero divergence of B around the edge of your current, so that, if you integrate over all space, you actually get div(B)=0. So, let me know what you think. :)

What is a magnetic field, yevgeny? Let me rephrase that question: If you are given a vector field, how can you check if it is possible for it to represent a magnetic field?

Hey, Im not gone yet, relax! :p Sorry I didnt answer before, Ive been talking the TOEFL this morning (very easy :cool:). Since it was already paid for.... I'll PM you both today and will tell you about my offer, Id love to know your opinions. :) By the way, papucisse, might I know your news as well?

Hey, good luck to both of you on your applications! I just wanted to let everybody know I might not be taking the GRE after all. Something has come up, and as a result I might not go to the US to get my PhD, but stay here in Spain. Ive been contacted by a group of my own country, and they've made me a very tempting offer. The truth is I don't know what to do. :shy::( (Why do these things happen when I believe Ive already made a decission??) It just looks like too good a chance to let it pass... Im confused........:crazy:

Way to go, yevgeny!!!!!! [w00t][banana][bounce] Congratulations, I was sure you'd do great!! It must be nice, the feeling now that its all over... :cool::) Thanks for the detailed description of your exam-experience! Im relieved to hear you found it easier than the practise one.... and that you had enough time! I hope you still hang around the forum a bit, we were having a few interesting discussions, werent we? ;) Im very very VERY happy for you! :D

Thanks papucisse, I thought I was going to be left all alone! :) So, you think you did all right? Good job! [bounce] Was it really more difficult than the practice test? :shy::([bomb][xx(] But, were the questions more advanced, or it just took longer to solve them? By the way, the difficulty of the exam should not alter the grade you get (at least in theory), so don't worry if it was harder. If everybody does worse, then the number of question required for each grade should be lower also...

yevgeny! How did it go?? Im guessing you did good, since you seem to be preparing now for your qualifying exams! :p ;) Now, for your question. The electromagnetic force can be obtained from a potential, that Im gonna call G: G = q V - q/c v*A ok, here q is the charge of your particle, V is the scalar potential, A is the vector potential, and v is the speed of your particle (c the speed of light), * is the scalar product. The electromagnetic force is given by: F = d/dt of the partial of G with respect to the speed vector - partial of G with respect to the position vector. Ok, now, if you are in a rotating system, the position vector of the particle is the same, but the speed vector changes so: v' = v + w^r w is the angular velocity of your system, r the vector of position, ^ denotes the vectorial product. So, if you substitute v = v' - w^r in the expression of G, and then calculate the electromagnetic force (using the formula I gave before) you'll find your new electromagnetic force, with any additional term that come up. I just don't have time to work out the math right now! :yuck: By the way, the Coriolis and the centrifugal force you can get in a similar fashion, substituting v = v' - w^r in your expression of the kinetic energy, and calculating Lagrange's equations.

Hi, n0madic1, welcome to the forum! Are you taking the exam tomorrow too? If so good luck. :) Regarding your question, if your system has some sort of symmetry and you have a bit of experience with these type of problems it is usually not too hard to figure out the frequencies of the normal modes. What you have to do is to imagine the normal modes. You can only do this if your system is symmetric. For example, think of a problem where you have 3 masses aligned and connected by springs of the same constant k. The two end masses are the same. Then it isnt hard to guess what the normal modes are going to be: a) The mass in the middle doesnt move, the end masses oscillate 180º out of phase: o--> o b) The two end-masses move in phase, and the middle mass has to move the other way so that the CM remains fixed. Ok, once you can figure out what the normal modes are, you need to calculate the total potential energy of your system for each normal mode. It should look like: 1/2 K_ef x^2 K_ef, whatever it turns out to be in each case, is your "efective spring constant". x is the coordinate that is oscillating. Next you calculate the kinetic energy of your system, and you write it down like this: 1/2 m_ef (x´)^2 And yes, as you just guessed, m_ef is your "effective mass". The frequency of the normal mode will be the square root of (k_ef / m_ef). This process may look as it will take too long as well, but it is actually longer to explain it than to try it out. Work a few problems and you'll quickly see how it goes. I hope this helps. [:o)]

Good luck, yevgeny!! You can do it!!!!! [banana] Also, good luck to everybody taking the test tomorrow! Let me know how you did!

Thanks so much papucisse!! I understand it now. :)