timo

No, that´s not what my argument is. My claim is that most people were already respected or at least not being considered crackpots prior to becoming famous. No I don´t. I claim that a failed math test has no relevance for your argument. For me sufficient proof would be looking at the people´s bio. Einstein, for example, was a trained physicist (so it seems he managed to finish a PhD in physics, despite of that very important failed math test in school) when he became famous for his Relativity. That does not exactly fit the "complete outsider rocks the world of science"-image that is usually attributed to him. And considering he wrote his SR stuff around the time that he got his PhD and also the time he wrote the paper that later brought him the nobel prize, I see no reason to believe he was considered a crackpot by the scientific community. Disagreeing with past/current believes does not necessarily make you being considered a crackpot. It´s probably a mandatory but not a sufficient criterion for being considered a crackpot. Where is your problem in understanding the statement that your point of view might just be a romantic myth about rebels/outsiders changing the world? EDIT: And please stick to what I said and don´t say something else and claim I said it in the future - it is not a good ground to start a discussion on. I did not say "historical figures were never considered anything similar to crackpots". I said that I doubt that MOST were considered crackpots. Yes, I´d limit it to natural scientists for a start. Politicians, philosophers and human right activists are a different issue because there´s less possibilities (well, actually none) to objetively tell right from wrong in these matters.

There´s no fixed scale at which quantum effects become evident, less a scale at which you can see them. You can easily make the interference pattern resulting from a double-slit experiment as large as a meter or more. Typically, at the scale of a few atomic diameters (roughly 10^-9 m) you cannot neglect quantum mechanical effects anymore but you cannot reverse that statement: Many properties of solid objects -even large ones- are of quantum mechanical origin, like magnetism and the mechanism of valence and conduction bands in metals and semi-metals.

I thought that is just an urban myth. I´d see a point had he gotten an F in his physics exam at university for claiming lightspeed is constant. But this way it looks like an attempt to desperately see connections where none exist. Which famous intellectual was considered a crackpot prior to becoming famous? Newton? Goethe? Leibnitz? Maxwell? Gauss? Darwin? Feynman? I´m no historian and I don´t know it. But I bet with assuming they were all highly respected I get more hits than with assuming they all were considered crackpots. I don´t know if your statement about most famous intellectuals being initially considered crackpots is right or wrong. But rational thinking tells me it´s probably wrong, especially when it comes to science (might be different in arts and literature).

C´mon, be little more elaborative. How does the network work? In what respect is any bandwidth reduced? If n people want to dowload a file of size X, I am pretty sure you cannot push the total amount of data transfer below nX, so either you are reducing the overhead caused by [??] or you are outsourcing tasks from the server (what´s that?) to the clients (is it server-client based at all?) or your network has a different strcuture than O(n) networks (what´s the structure of both network types?). There´s a lot of stuff that could be explained here and I am pretty sure that quite some people might be interested to learn a bit about file transfer systems.

It does. And giving Severians post a 2nd thought you´ll see that this case was included there.

And what exactly scales with O(log(n)) ?

Having the longest at least once in his life.

If you´re talking about what I suppose, then this thread might be a starting point for you: http://www.scienceforums.net/showthread.php?t=22585&highlight=orange+band

1/sqrt(2) |true> + 1/sqrt(2) |false> You were talking about quantum computers, were you? EDIT: I should note that I was kinda joking, there. It´s a nice feature that in quantum computers, you can find a complete set of eigenstates for every operator, in this very case even one with an eigenvalue of +1. For completeness: The other eigenstate would be 1/sqrt(2) |true> - 1/sqrt(2) |false> which would be the eigenstate to the eigenvalue -1.

Since 5614 replaced the term quark with particle: Of course taus are particles but the term you were probably looking for is lepton. About tracing back to high energies: I see no initial reason why taus should be produced more than electrons. At most, I´d see a point in claiming that the energies involved are so high that one can neglect the masses of the particles and that therefore electrons behave exactly as taus or myons (because they are equal except for their mass). A little catch there is that I doubt that you can even speak of particles in the normal sense in a scenario such as early stages of the universe (normally particles are defined as free, non-interacting particles). The term generation is not related to an order of production in the universe.

Somehow I always read gf=grandfather instead of gf=girlfried at first. But this has been the first time that it cause THAT much confusion.

std::cout TwoDigitInt(int i) { if (i<10) return std::cout<<"0"<<i else return std::cout<<i; } . . . std::cout<<"The new time is"<< TwoDigitInt(newHHEndTime << newMMEndTime<<std::endl; I´d try something like that, but I´m not familiar with the std stuff so don´t pin me of the correctness of the syntax.

I don´t have the time thinking about that atm, but since I suppose others might have the same question: Are the small subscripts at the u derivatives with respect to that parameter?

Funnily, I read about programs anonymizing your browsing just yesterday on Wikipedia (I won´t go look for the link, I think it was the german version, anyways). That article also mentioned anonymized file-sharing techniques, also techniques that did not sound as if they slow down everything (t´was something about sending the data encrypted). I have a pretty split opinion about anonymous surfing. On the one hand, I usually leave my house and go shopping or whatever without wearing a mask so I see little reason to do a similar thing on the web. On the other hand I find it strange that people (machines in this case) might collect data about me for reasons they are not telling me.

I suppose the question boils down to asking what [math]\theta [/math] actually is supposed to be. So: What is it? Are you sure it´s the same as what your teacher defined it to be? EDIT: @uncool: It´s only clear if you know what [math]\theta [/math], A and B are. And that is absolutely not obvious to me.

What?

I´d suppose the european space agency (ESA or ESOC, not sure what´s the name) has a few.

What do you use it for? I mean: I seldomly have to look up the top-mass in everyday life.

I am not really sure if the collapse of the wave function is actually understood at all. For technical purposes, you can simply assume that it happens at the point at which you don´t expect anything which requires a QM-description to happen anymore. Example: If you approximate your photo plate as a binary system which either marks a dot when it's hit by an electron or doesn´t when there is no electron coming in, then that seems like a very suitable point to collapse the wave function. Making it dependent on whether you look at the plate seems a bit too esoteric to me. What if I look at it and tell you the result, for example?

The result you get is exactly the same. I already wrote "that´s the rule to raise and lower indices" but I failed to note how it looks if you raise them, sry. Raising an index: [math] \partial^{\mu} = g^{\mu \nu} \partial_{\nu} [/math] EDIT2: But I don´t understand where you get powers of diagonal elements. The term is linear in g.

What is your problem? The factor 5? Differentiation is a linear operation, therefore d/dx (5 f(x)) = 5 (d/dx f(x)). EDIT: Alternatively, use the product rule: d/dx (5 f(x)) = ( d/dx 5) f(x) + 5 (d/dx f(x)) = 0 + 5 (d/dx f(x))

In flat spacetime, points whose connection line has a velocity (=direction; velocity in 3D is a direction in 4D, roughly speaking) >c can obtain a reversed time-ordering under Lorentz transformations. Take two points (t1, x1) and (t2,x2) with t2>t1. If their connection line is <=c, then t2>t1 in all frames of reference. If v>c, then there are frames of reference in which t1>t2. So if an event at (t2,x2) is caused by an event at (t1,x1) you have the puzzling effect that some people (in a suitable frame of reference) might see the effect happen before the cause. That´s usually considered as impossible, hence the statement that all dependencies should propagate <=c so that the effect happens after the cause independently of the frame of reference (except for explicit inversion of the time axis, of course). Now what does this have to do with black holes? I have no idea. Black holes are defenitely not flat space time, hence the argument cannot remain unchanged. I sometimes have the feeling that results from flat spacetime are just carried over to curved spacetime where they, not surprisingly, result in puzzling effects - but it´s also well possible that the argumentation is not related to the causality problem in SR I mentioned above. Two notes on black holes and attempted time-travel scenarios: 1) You can double the spacetime of the black hole solution resulting in a "white hole" on the other side. It´s just a dirty mathematical trick (allowing negative values for a variable whose physical interpretation is a radius) but does not violate any rules other than that it doesn´t seem to make sense. Now, someone had the glorious idea that this new spacetime is just the old one and that you have a double-mapping of spacetime (like if you desribe a circle by letting your angle run from 0° to 720°). In that case, the white hole could be somewhere else in space and you might (after some small modifications to the metric which are not compatible to any known physical processes) find a way to pass from the black side to the white one and effectively taking a shortcut. Taking the shortcut and then completely ignoring that there is a black hole, you could then apply the causality statement from above (which is rather cynical because you just got to your destination so fast because it doesn´t apply) you can then can construct a statement of violating causality. 2) In above, I spoke of nonrotating, noncharged black holes which is the simplemost case. Rotating black holes have a slightly different spacetime, notably additional singularities called ring singularities. They are slightly different from the main singularity (event horizont). I think ajb somewhere gave a link to a paper about how these singularities might be used for time-travel.

Treating electric and magnetic interaction in one object also makes the fact that it is an electromagnetic interaction nicely visible. The notation I used more or less simply is the relativistic notation (often called covariant notation) of the same. Yes, it looks more elegant to me, too. In fact, most, if not all, relativistic equations look more elegant when written in relativistic notation. I know the convention [math] \square = \partial^{\mu} \partial_{\mu} = \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} - \frac{\partial^2}{\partial z^2} [/math] or, in a notation that you sometimes see but that imho is slightly outdated, [math] \square = \partial^{\mu} \partial_{\mu} = - \frac{\partial^2}{\partial t^2} + \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} [/math] (as always, I omit constants i.e. c=1 for me) - it actually depends on the choice for the metric tensor of Minkowsky (flat) spacetime which can be chosen as g=diag(1,-1,-1,-1) or g=diag(-1,1,1,1), respectively. So yes, my □ would be your □². But I am not sure if I´ve ever seen the notation you are using and I can barely imagine that it´s very usefull for GR because with the factor i on the time-derivative you are actually hiding a very important point of GR, namely that the metric is not euclidian like g=diag(1,1,1,1) (note the signs, strictly speaking it´s not even a metric). Quick convention suggestion: [math] \partial^{\mu} = \left( \frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) [/math] [math] g_{\mu \nu} = diag(1,-1,-1-1) [/math] [math] \partial_{\mu} = g_{\mu \nu} \partial^{\nu} [/math] <- this is in fact the rule to raise and lower indices, not just for the derivative operator; summing over (contracting) two indices should only happen when one is an upper one and the other is a lower one, otherwise there´s probably an error somewhere. [math] \Rightarrow \square = \partial^{\mu} \partial_{\mu} = g^{\mu \nu} \partial_{\mu} \partial_{\nu} = \dots [/math] Oh, and of course you´ll have to add the 1/c in the definition of the partial derivative if you don´t take c=1.

I´d really like to elaborate on the topic but I had to look up too many things for a coherent response and I absolutely don´t have the time for that right now - poke me again in two weeks if there´s questions left over, I´m in principle interested in getting a bit better understanding of the stuff myself. So in short: The directions are: Field equations of GR (= Einstein equations), Energy-momentum tensor (also called Energy-momentum-stress tensor) and relativistic hydrodynamics. I am no fan of relativistic mass myself. I don´t see how it is any good and especially in GR it seems a pretty useless concept. Relativistic mass tends to be used by people in the context of "gravitation is caused by relativistic mass" which seems a questionable statement to me. Let´s take a look of what I remember off my head: The source of electromagnetism is the electrical current which, in relativistic notation, is a four-vector (in the following just called vector, add it to your list of things to look up if you don´t know that concept). For an electric charge density C and no electrical current it´s simply (C,0,0,0) up to a constant factor. For a charge density C and a current [math] \vec j [/math] it is something like (C,jx,jy,jz). In a suitable gauge the field equations for the electromagnetic field are [math] \square A = j [/math] where [math]\square[/math] is the D´alembert operator (a 2nd derivative) and A is the four-potential that describes the field (if you want the E-field and the B-field used in classical electrodynamics you can take suitable derivatives of A to obtain them). Electromagnetism has the nice property that it is compatible with SR which loosely speaking is because that notation uses four-vectors. The natural attempt for a gravitational theory compatible with SR therefore is to try the same trick: Take the matter-density and the matter current (whatever that shall be) and try to find a vector potential A describing the gravitational field For reasons I cannot remember out of my head this does not work - the reason was that some objects have a different behavior under coordiante transformation that that what would be required from a four-vector field equation. It turns out that the field equations for GR are tensor equations of rank two which, for example can be written in terms of a matrix. The question is: How is the matter which should be the cause the source of the gravitational field described in terms of such an object? It is described by the energy-momentum tensor. Sadly, I cannot tell you too much about that object without looking it up but let´s take some simplified version which might already show why I don´t like the "gravitation is caused by relativistic mass" statement: For an energy density D, no movement of the masses and vanishing pressure (pressure can add significant terms, especially for photons), the energy-momentum tensor is [math]\left( \begin{array}{cccc} D & & & \\ & 0 & & \\ & & 0 & \\ & & & 0 \end{array}\right) [/math] where I have omitted any constants that might appear (I always omit constants). From that I called D an energy-density you might think "so it is relativistic mass that gravitation couples to". Well, imho that´s not really true. I have assumed "no movement, no pressure" so it´s just the same as mass-density. So let´s see what happens when you go to another coordinate system (the trick about relativistic field equation is that they hold true in any coordinate system!). I am too lazy to really calculate it and I already typed much more that I originally intended to, so just an heuristic argument: energy-density is energy divided by volume. Applying a Lorentz transformation on that, the energy gets a factor gamma. But: Due to what one usually calls length contraction, the volume gets a factor 1/gamma at the same time. Therefore, D -> gamma² D (thinking of it, I think that transformation rule was exactly the reason why a gravitational field equation must be tensorial and not vectorial as in electrodynamics). That´s not really what I´d expect from a coupling to relativistic mass which only gets a factor gamma^1 under transformations. There´s two catches here: 1) I have completely omitted the other components of the energy-momentum tensor. 2) I have not even said what gravity is supposed to be . In fact, I don´t really know it myself. I tend to see gravitation as a distortion from flat spacetime but that really seems just a non-saying simplification once you start giving the statement a few second thoughts. Ok, time to finish that post, a short summary: - The important keywords you might want to look up are mentioned above. - Internal energies of a system (heat and the pressure can play a significant role) also contribute to the field equations. However, they are macroscopic properties (thermodynamics). A microscpopic treatment (gravitational field caused by a single elementary particle) might get along without those. - I cannot prove that the statement "gravitation is caused by relativistic mass" is wrong, because I don´t know how the microscopic treatment of GR looks like and I therefore don´t even understand the statement. But the transformation properties of the energy-momentum tensor make me think that the statement is wrong. @Spyman: One might think it is correct. For the two parallel beams you could think of asymtotically transform into a system in which the energy of the beams vanishes if they are parellel while there´s no such transformation if they are not parallel. But I am not convinced if that idea would really stand a rigorous investigation. Additions, elaborations and corrections are highly welcome!

Perhaps it isn´t exactly what you meant, but: How about using a Wiki software ?