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Everything posted by timo
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That is because your PhD book might define [math] \log (-5) = \log(5) + i\pi [/math], in which case then [math]\log(-5) - \log(-2) = \log(5) + i\pi - \left( \log(2) + i\pi \right) = \log(5) - \log(2) [/math]. This does indeed look like being >0. But strictly speaking you have complex numbers on the left-hand side of the inequality (even though their imaginary parts will often cancel). The relations ">" and "<" are not (canonically) defined for complex numbers. Advanced-level books often make inherit assumptions that are not explicitely mentioned; without knowing the book or the context (no need to quote, no one on sfn, safe for possibly two or three people, will understand math on a PhD level).
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Perhaps you should enlighten us and tell us a solution.
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Darn, I had surely spotted a sign-error somewhere.
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A not-so-simple but interesting example would have been my WiSci article on the derivation of the GR equation of motion which got deleted with WiSci (and the backup later being deleted with the crash of my computer's HD) . A simple example: A point-sized particle in a potential. The Lagrangian function L for this system is [math] L = T - V = mv^2/2 - qV(x) [/math] with T being kinetic energy, V being potential energy, m being the mass (so obviously classical physics, here), v the velocity, q the charge associated to the potential (e.g. mass in a gravitational potential) and V(x) the potential. The equation of motion is derived via the Euler-Lagrange equations [math] \frac{d}{dt} \frac{\partial L}{\partial \vec v} - \frac{\partial L}{\partial \vec x} = 0 [/math] (note that that the differential wrt. to a vector is just a lazy notation of mine, technically, you have to satisfy this eqn for each coordinate seperately). Plugging in the L above [math] \Rightarrow \frac{d}{dt} \left( m \vec v \right) + q\frac{\partial V}{\partial \vec x} = m \vec a + q\frac{\partial V}{\partial \vec x} = 0 [/math] [math] \Rightarrow \vec a = \frac{q}{m} \frac{-\partial V}{\partial \vec x} [/math]. A specific example: When V is a gravitational field [math] V(\vec x) = g (\vec e_z \cdot \vec x)[/math] with [math]\vec e_z [/math] being the unit vector in z-direction (well, just "up"), then [math] \vec a = -\frac{m}{m} g \vec e_z = -g \vec e_z [/math]. In other words, the object falls down with an acceleration of g, just as you would expect from the Newtonian F=ma=mg. Note that using the way shown the concept of force did not explicitely appear and was "replaced" with the concept of a potential.
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An Exceptionally Simple Theory of Everything
timo replied to D H's topic in Modern and Theoretical Physics
My question was meant a bit more specifically than "the author is not the typical sfn crackpot". And while "the author is no crackpot" probably is a necessary criterion to get interested in a paper, it certainly (not last due to the issue of limited time per day) is not a sufficient one to bother reading a paper. hep-ph alone has probably something like 5-10 papers/day, and I surely do not read them all. Just looking at the very first current entry in new, I read "TeV gravity in four dimensions?". From the topic, I see that the model might be in collider range, i.e. principially testable in collider experiments and that the model claims not to need extra dimensions. From the end of the paper (I've really only read the title and the very last sentences, I don't know the names of the authors, either) it seems that there is some rough approximation of a cross-section for whatever process. That further indicates that their idea might be testable. So if I was interested in quantum gravity or currently working on related stuff, I might go one further step and actually skim the paper because "no extra dimensions" and "possibly testable predictions" looks interesting. Ok, I hope I have roughly described what I mean by "why someone should read something". How does this fit to the paper you proposed? What is new compared to other approaches? Are there any predictions? Is the paper supposed for people with other criteria than me? In that case, who would that be (except for Martin ) and what would these criteria be (also in the case of Martin)? EDIT: Perhaps the answer is simply that the paper is not interesting for me because I am not working in quantum gravity but you probably wouldn't expect anyone working in that field to learn of it from an sfn thread if it really was an important contribution to the field. -
Lagrangian expressed with the differential form
timo replied to square173205's topic in Modern and Theoretical Physics
QED = Quantum Electrodynamics, in case that was your question. -
Gnu. But you probably wanted to know the IDE which is Code::Blocks. Dunno about its Vista compatibility. a "dynamic link library". Prossibly the library you want to link. For the executables (.exe files), it is sufficient to copy the file into a directory where windows finds it (in XP that is either the current directory or the windows/system32 directory). For linking to a program you are writing yourself: A quick google for 'linking libraries "dev-cpp"' seems to give plenty results, e.g.: http://www14.brinkster.com/aditsu/dev-cpp-faq.html http://www.thescripts.com/forum/thread129991.html You should possibly give a bit more info on your problem: - Can you run the demo? - What is the error message you get and when do you get it?
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I suppose "importing" means actually telling the linker to link the library to the project. I don't use Dev-C++ anymore, but there's certainly some option like "libraries to link" where you can and must set the library (not only the path to it). Worst case, just download some sample project that imports a library (meaning to look for a library for whatever which has tutorials written in Dev-C++) and look at the options set there.
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1) substract the 2nd addend on the left-hand side from both sides 2) divide both sides by 2
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That already clears up quite a bit. Is that what you want to show? That a cyclic interchange leaves the determinant unchanged? What =0 ? Surely not the matrix.
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Let's assume not everyone uses exactly the same notation as you do in school: There is no chance to figure out what you are trying to do and say. Slow it down, try explaining what you are talking about from the very start assuming none of the notations is obvious to the reader. As a total shot in the dark: If you are trying to prove that the determinant of a matrix is zero, then this is not the case in general.
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What are the small and capital letters with the indices?
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Ok, I think I understand your problem now. The inequality is not [math]\ln(\frac{x-1}{x+2}) > 0[/math] but [math]\ln (x-1) - \ln (x+2) > 0 [/math]. The rearranged inequality has a larger definition range than the original inequality. We hopefully agree that the original inequality is not defined for x<1. As a related example: f(x)=x/x is not the same function as g(x)=1.
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Sure. I have a lot of doubts that the logarithm of a negative number is defined. Because if you try to be smart and use a definition of log (which one would you use considering that none is given?) that maps negative numbers on complex values, then the ">" sign is not defined anymore. Put "log(-1)" in the calculator of your choice.
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The inequality isn't defined for x<-2, either. The range probably wasn't mentioned because it is "obvious" . That is correct. Do you know why you can drop the logs ? Yes. I don't understand what you are saying here. You can also do it the x-1/x+2>1 as long as you remember for which x the original equation was defined. It would also show that no x>1 satisfies the inequality.
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Actually, I wonder how you want to get any solution without violating the range, too. The original inequality obviously is defined for x>1, only. I spontaneously see no x>1 for which (x-1)/(x+2) > 1.
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I'd propose filling the gaps with sensible values.
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You supposedly meant [math] y = \frac{x^2-11}{x+\sqrt{11}} [/math] for [math] x \in R, x \neq -\sqrt{11}[/math] since the function is obviously undefined at x = -sqrt(11). If the value of the function approaches the same value when approaching -sqrt(11) from the right and from the left, you could just define a new function which is equal to the original one for all [math]x \neq -\sqrt{11}[/math] and equals that limit at [math]x = -\sqrt{11}[/math]. I suppose that's what's meant.
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An Exceptionally Simple Theory of Everything
timo replied to D H's topic in Modern and Theoretical Physics
I've not read the paper; neither am I completely sure what the purpose of this thread is. Anyways, to get some discussion started: What is a "free" parameter? Isn't it a parameter that is not fixed by the theory itself but must be measured experimentally? Examples: The speed of light, the mass of an electron, the QED coupling strength. As an alternative question (for the case we roughly agree on the understanding of the term "free parameter"): How does this approach manage to reproduce Standard Model physics without free parameters, particularly in the light of the non-gauge interactions (with non-gauge interactions having arbitrary coupling constants almost by definition) required for the electroweak symmetry breaking? Does this model reproduce the SM at all? A more general question (not to be mistaken as being pejorative): Why should anyone bother reading this paper and who would this "anyone" be? -
The year is coordinate time, not eigentime of the lightpulse.
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Luminal: You learn the stuff much faster than your lecture, thus having a lot of spare time to either learn additional stuff or do something completely different. You get top grades. And I don't think you couldn't read that "additional stuff" during the lectures you seemingly have to attend (from your post I assume there is something like an attendance-duty for lectures in your university). What exactly is your problem? That the lecture given for 30-50 people isn't tailored to your personal demands? That there's attendance-duty in your classes? Sidenote: I strongly doubt that in your country six billion people are going to university.
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An explicit (calculational) example for the statement "general covariance allows the freedom of movement forwards and backwards in time" would probably help (same goes for "accelerating forwards and backwards in time"). Particles, by relativity definition, have a path in spacetime with timelike tangent. In non-curved spacetime, this fixes the space position to exactly one position for each time position - regardless of coordinate system (except for time-coordinate being time-like and space-coordinates being space-like).