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Everything posted by Royston

  1. True, I was being a nit-pick on how you worded being in control of your anger, you then went on to say 'I feel pretty good actually', so I just needed some clarification on precisely what you meant. Of course, if I was *calm as a Hindu cow, it's possible I would not of even made that post. However, logic would of superseded that emotion, but that's beside the point. It was really in response to the OP, where he/she described anger as destructive, so I just illustrated how such an emotion can be used productively. Maybe, I did make it clear that anger could be a spread of emotion, from irritability to blind rage, but point taken. Yes, in retrospect, calm could also be a general term, from quite relaxed, to practically unconscious. I'm not sure if there's a strict psychological definition. In fear of arguing of semantics, I'm pretty sure were in agreement with each other. *excuse the Fight Club quote, it just seemed appropriate.
  2. I've had quite a bit of spare time recently, plus the weather has been pretty awful the last couple of weeks, so I've been watching a few TV series, back to back (i.e not waiting a week for the next episode). At the moment, I'm glued to Breaking Bad. I think Bryan Cranston is brilliant in this series, (he plays Nobel laureate Walter White). There's been some very original consequences, that he has directly and indirectly caused, due to his change in ummm lifestyle. I've just started season 4, where the first few episodes have been a little dissapointing, but I've been told it gets a lot more interesting. I've also been watching Game of Thrones, which is ludicrously good in my opinion, and I'll be reading the books once my study finishes this year. Tyrion, played by Peter Dinklage is, hands down, my favourite character in the show. I was wondering what TV series anyone has been following, and if there's anything in particular that makes it good / compulsive viewing. EDIT: remember spoilers
  3. However, you can overcome any shortcomings in your ability, just through pure interest. During my degree, there have been areas that I've found rather dull (e.g optics...yawn), and this was reflected in my grades. When it came to the areas I was passionate about (astrophysics and cosmology), I was obtaining very good grades. Unfortunately, understanding the dull areas was necessary to get a better understanding of the exciting areas, so I had to put a lot more work in to achieve this, which didn't really bother me. My point is, if you're passionate about the subject, you'll want to learn and invest lots of time into it, regardless of your ability. That can be just as beneficial as having a natural ability in the subject.
  4. I can't see how it's possible to be angry and calm simultaneously. Anger is a general term that could mean slightly peeved, to blind rage. If you're irritated (slightly peeved) by something, you're not calm by definition. You may have knowledge that your government is collapsing, but for the time being, it doesn't irritate you. To address the OP. Our government decided to do away with heritage sites, which could increase development in rural and protected areas. When I first heard about this, I was furious, so I decided to post this to my friends and acquaintances, to encourage them to sign a petition. If I had made the post during my initial outrage, it would of read something like 'those f**king money hungry Tory s**t cakes, with their short-sighted w**ky polices, do they have the slightest idea what a f**king ecosystem IS ?!?' However, with a bit of time to relax and compose myself, I just posted their plans, mentioned the effect on ecosystems, and asked that anyone who opposes this move, please sign the petition. Anger (in this instance) is a necessary precursor to resolve a conflict. It's an initial kick-start for someone to take action, so if that's taken away, I think we'd be in a bit of a mess.
  5. Precisely, and this seems to be what's causing confusion in this thread. You can't make the transition from classical physics to quantum physics without covering wave functions, and it's natural for a student (like myself) to insist on an interpretation when making that transition e.g wave-particle duality. You need to understand wave functions and the Schrodinger equation, before you move onto the Dirac equation, which you need to understand before moving onto path integrals, and so on and so forth. There's been a number of times whilst I've been studying, where an equation pops up, that may have some quantum property e.g spin. I remember learning about chemical potential in the context of helium burning, and spin was described as angular momentum around a centre of mass. That isn't how spin should be interpreted as, but, it would be a massive diversion branching off into the true properties of spin. It is inevitable when learning science, that there will be concepts that need to abandoned as you advance, but those concepts are a crucial part of reaching those advanced stages.
  6. What wikipedia article ? Have you quoted the wrong post ? That was written using my text books for reference. What do you mean it doesn't seem? I've just shown you explicitly what an operator is.
  7. I think you could do with a few preliminaries, because you're not making any sense. This is the basics, (we'll stick with one dimension) but hopefully you'll get the idea. Operators An operator which is denoted with a hat i.e [math]\hat{O}[/math] transforms a function, say [math]f(x)[/math] to another function. So if the function is [math]f(x)=2x^2[/math] and the operator is [math]\frac {d}{dx}[/math] then [math]\hat{O}f(x) = \frac{d}{dx}2x^2 = 4x[/math] There's a good tutorial on differentiation here, if the above doesn't make sense to you. Eigenfunctions and eigenvalues Considering the role of the operator above, we can move on to eigenvalue equations. These have the relationship... [math]\hat{O}f(x) = \lambda f(x)[/math] where [math]\lambda[/math] is a complex constant. So you can see, in order for it to be an eigenvalue equation, the operator has to return the same function multiplied by a constant. The constant is the eigenvalue, and [math]f(x)[/math] is the eigenfunction. So for instance [math]\hat{O}f(x) = \frac{d^2}{dx^2}(sin(bx)) = \frac{d}{dx}(b\, cos(bx)) = -b^2 sin(bx)[/math]. In this case, [math]-b^2[/math] is the eigenvalue, and the function has not changed, so [math]sin(bx)[/math] is an eigenfunction of the operator. The de Broglie wavefunction The de Broglie relationship for an electron in an isolated system i.e free from any disturbances (forces) is [math]\lambda_{dB} = \frac{h}{p}[/math]. Where [math]\lambda_{dB}[/math] is the de Broglie wavelength, h is Planck's constant, and p is momentum. You've probably seen a wave equation before, such as [math]f(x,t) = A\,cos(kx-wt)[/math]. Where A is the amplitude of the wave, k is the wave number, [math]k = \frac{2\pi}{\lambda}[/math] and w is the angular frequency [math]w = 2\pi f[/math] (f is the frequency i.e 1/T where T is the period). The x in the equation, just denotes the direction the wave is travelling. Now, wave functions are complex, so we can write [math]\Psi_{dB}(x,t) = A[cos(kx-wt) + i\,sin(kx-wt)]=A\,e^{i(kx-wt)}[/math] (see Euler's formula) Remembering the de Broglie relationship, this implies that [math]k = \frac{2\pi\,p}{h} = \frac{p}{\hbar}[/math] where [math]\hbar = \frac{h}{2\pi}[/math] Also the photon energy is [math]E=\hbar w[/math] so [math]w = \frac{E}{\hbar}[/math] The Hamiltonian function You're probably familiar with Newtons equation [math]F=ma[/math]. This can be reformulated, by noting that F relates to the gradient of the potential energy function V(x). So for an object moving in the x direction, we have [math]F_x = -\frac{\partial V}{\partial x}[/math] Look up partial differentiation, if you're not sure about what the RHS means. Further, the derivative of velocity is acceleration, and p=mv (i.e momentum equals mass times velocity) so [math]ma_x = \frac{d(mv_x)}{dt} = \frac{dp_x}{dt}[/math] therefore we have... [math]\frac{dp_x}{dt} = -\frac{\partial V}{\partial x}[/math] Now, kinetic energy in terms of the momentum of a particle is [math]E_k = \frac{p^2}{2m}[/math] The Hamiltonian function, is the sum of the kinetic energy and potential energy of a system, with a particle of mass m, and potential energy V(x), so... [math]H = \frac{p^2_x}{2m} + V(x)[/math]. Now [math]\frac{p^2_x}{2m}[/math] is not explicitly dependant on x, so [math]\frac{\partial H}{\partial x} = \frac{\partial V}{\partial x}[/math] So, recalling the reformulated Newtonian equation, we have [math]\frac{dp_x}{dt} = -\frac{\partial H}{\partial x}[/math] The Hamiltonian operator Now we need to convert an observable (in this case the energy of the system) into an operator. Recalling the de Broglie wave function, if we partially differentiate it twice, we get (1) [math]\frac{\partial}{\partial x}\Psi (x,t) = ik\Psi(x,t)[/math] (2) [math]\frac{\partial^2}{\partial x^2}\Psi (x,t) = (ik)^2\Psi(x,t) = -k^2 \Psi(x,t)[/math] Now to obtain the eigenvalues for kinetic energy and momentum, we multiply eq (1) with [math]-i\hbar[/math] and eq (2) with [math]-\hbar^2 / 2m[/math] so (1.1) [math]-i\hbar\frac{\partial}{\partial x}\Psi (x,t) = \hbar k\Psi(x,t)[/math] (2.2) [math]-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi (x,t) = -\frac{(\hbar k)^2}{2m} \Psi(x,t)[/math] So, [math]p_x \Longrightarrow \hat{p_x} = -i\hbar\frac{\partial}{\partial x}[/math] and [math] E_k \Longrightarrow \hat{E}_k = -\frac{\hbar^2}{2m}\frac{\partial ^2}{\partial x^2}[/math] The arrow just indicates we're going from a classical variable to a quantum operator. Therefore the kinetic energy operator is simply [math]\frac{\hat p^2_x}{2m} = \frac{1}{2m} \left(-i\hbar \frac{\partial}{\partial x} \right)^2 = - \frac{\hbar^2}{2m}\frac{\partial ^2}{\partial x^2}[/math] So the Hamiltonian operator is (remembering the Hamiltonian function) [math]\hat{H} = \frac{\hat{p^2_x}}{2m} + \hat{V}(x) = -\frac{\hbar^2}{2m}\frac{\partial ^2}{\partial x^2} + V(x)[/math] The Schrodinger equation So finally, we're in a position to state the Schrodinger equation, which for a free particle could take the form [math]i \hbar \frac{\partial}{\partial t}\Psi (x,t) = \hat{E}_k \Psi(x,t)[/math] But for a bound particle, i.e there is a potential energy term, we require that [math]i \hbar \frac{\partial}{\partial t}\Psi (x,t) = (\hat{E}_k + \hat{V}(x)) \Psi(x,t)[/math] So using the kinetic energy operator, and the potential operator, this gives... [math]i \hbar \frac{\partial}{\partial t}\Psi (x,t) = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi(x,t)}{\partial x^2} + V(x) \Psi (x,t)[/math] more compactly... [math]i \hbar \frac{\partial}{\partial t}\Psi (x,t) = \hat{H}\Psi(x,t)[/math] Hopefully, this should give you a better understanding of operators, the Hamiltonian and wave functions (it's just a basic treatment). I'm not sure what your maths level is, but you can always ask if there's anything that you find confusing.
  8. I think we may have cross-posted while I was editing (see edit 2). I'm assuming you mean the old posts in the LaTex tutorial when you say OP ? If so, that is doubly weird, because the backslash is absent when I tried replying to an old post (as per edit 2) and so also absent when quoting. I was looking into a problem with Tex on Moodle in another forum, and issues can be browser dependent...especially IE (mainly due to microsoft crossing their arms to certain languages, e.g MathML). In any case, maybe an admin knows what's going on. I'm not particularly savvy in this area.
  9. You seem to be stating this as fact, and then post your own work from 'free for all' vixra. DM seems to do a much better job, when it comes to large scale structure, not just galaxy rotation. One example, MOND and the GR alternative TeVeS predicts gravitational lensing deflection should coincide with visible baryons in clusters. This has been tested, using the Bullet cluster, which shows there is a large discrepancy between the masses found from X-ray sources and gravitational lensing. A DM component remedies this. EDIT: deleted link due to mod note
  10. 2006 I'm guessing the renderer has changed on this site since then, I think it now uses Mathjax IIRC. However, the backslash (used for all symbols, operators, accents et.c) seems to be absent when you quote the old Tex, so [math]\Delta[/math] with backslash and [math]Delta[/math] without. Unrelated, but just need to check all is well (in case I need to ask any QM related stuff) [math]\langle p^2_x \rangle = \frac {\hbar^2}{2a^2} \int^{\infty}_{-\infty}\psi^*_n(x) (\hat{A}^{\dagger}\hat{A}+\hat{A}\hat{A}^{\dagger})\psi_n(x) dx[/math] EDIT: Now to quote the above Seeing as your post rendered properly, when quoted, that's not really surprising. EDIT 2: I just clicked reply on one of the old LaTex posts, and the backslash is absent, so not sure if this is related to quoting the old Tex. For example, the fourth post of the LaTex tutorial, looks like this when hitting reply... frac{a''}{a}= -frac{4pi G}{3}(rho + 3p) with math tags removed.
  11. I appreciate the response questionposter, but (perhaps impossible to not sound offensive) I really need an answer from somebody who know's what they're talking about. There are several *flaws in what you've stated, but addressing them, would be veering off topic. Again, no offense intended. *It is more than likely there is a flaw with what I've stated, for instance the commutation of the operator with the Hamiltonian isn't really relevant here...albeit I thought I had deleted that part, when I posted. Thanks, swansont. The last two points have made it a lot clearer. I got caught up with information in a system can only be obtained with making an interaction with the system...however if there is no time evolution, then the same information can be extracted due to a measurement made immediately before. Extracting information is still a measurement, albeit it seems slightly abstract in this instance.
  12. I did a search, but could not find anything on this specific problem, so apologies if this has already been answered. I just received my marked QM assignment (got a distinction, thought I'd slip that in ). In one of the essay style questions, I deliberately added a bugbear to prompt tutor feedback. However, it was not particularly satisfactory, and no different to what I have already read. Supposing I made a measurement of position on a system, that caused collapse of the wave function to a given eigenstate. If I make a measurement immediately after the collapse, (i.e there is no time evolution due to the Schrodinger equation) I should obtain the same value, say x0. However, the very act of measurement, perturbs the system. Regardless of whether the Hamiltonian commutes with the operator or not, I can't see how the act of measurement does not throw up a spread of values, due to perturbation. I'm sure this is a classic question, and I was wondering if this fell into the 'interpretation of measurement' camp. I've spent the last hour trying to come up with a mathematical treatment to the problem (this may even come up later on in the course) but it seems more of a conceptual one, perhaps. I would rather a mathematical answer over a conceptual one, because I find that a lot easier to see where the problem lies, rather than an interpretational approach. EDIT: Dirac notation is fine BTW
  13. Right, but your statement needed more information. It could easily be misinterpreted as black holes, stars or any other astronomical bodies don't merge. The system tar gave in his example were three black holes, that could only be gravitationally bound (pulling on each other). Depending on initial conditions, this could mean a merger of two black holes or all three.
  14. Err, what ? Because of dark energy, you do get galaxy mergers, but they're gravitationally bound e.g our galaxy and Andromeda. Then the lab you should be using is the Universe itself...there are a number of handy measuring tools at your disposal, e.g the CMB
  15. I think we probably need more information, before we come to any conclusions. Rule of thumb, if a claim includes a mention of Tesla, it's...
  16. IsraelUnoone, nobody can take your idea seriously, if you provide a rather confused description of your 'device' (I realise English is your second language), and expect us to take your word for it that it works. At the very least provide a diagram with dimensions, and a detailed description on how this generates electricity. For instance, I'm baffled how two tuning forks and a rotating ball, can generate usable electricity from a satellites' transmit power. What materials were used ? Why is this an improvement on other so-called renewable sources of energy ? What data can you provide e.g power output ? What is the mathematical model you used, that led to this idea ? It's impossible to provide any constructive criticism, if there is simply nothing to go on.
  17. ...tough, now available in iTunes, and all good retailers.
  18. I'll have your guts for garters. You big girls blouse. Minger. Mush (e.g 'oi mush, get down from there !'). Square (not the shape, obviously). Having a shin dig, or alternatively, have a right old knees up. Bad (to mean good). Nancy boy.
  19. Thanks for the detailed response, Klaynos That makes sense. There's no way, with my current workload, I'll be learning an entire language. However, when I do choose a programming course, once my degree is finished, I just wanted to make sure I was making the right choice before taking the plunge, so... ...is the type of information I need. It's funny you mention R, I was going to use this in an astrophysics project last year for statistics (can't remember why I changed my mind). Though you've raised a good point, i.e if somebody has already done the leg work, then build from there. IOW, looking into modifications of environments that have already been developed. Albeit I'd still need to fully understand the language, but can gloss over any complications e.g complex numbers, like you said. With your final point, that's precisely why I was looking at CUDA. It's, for me, a case of what computing power is accessible for a home set-up. So, this seemed like an obvious choice, especially as GPU's are massively multi core. So I'll have to look at platforms that have been developed for CUDA, that take away the pain of having to code complicated maths in C/C++. No problem, you've narrowed down what I should be looking towards, which is the main thing. EDIT: Yikes, I mistakenly pressed the minus button on the post rating. I'm not really into that rep stuff, but if a mod can reverse that please, thanks.
  20. There are several questions embedded in this post, plus, I was unsure whether to put this in computer science...so maybe more appropriate there. For my dissertation, I'm looking at the effects of astrophysical jets within radio galaxies...there is a clear link between galaxy evolution and the role of jets (specifically star formation rates). I think to get a clearer understanding of such an environment, you need to model that environment. I've been looking at modelling environments. The first one I came across was RAMSES code, which is built on Fortran. So my first question is, I thought Fortran was more or less redundant, so is it worth trying to understand Fortran ? I thought C/C++ would be more far reaching..(I used to programme with machine code BTW, so learning isn't a huge issue) I looked around, and found MPI is doable through SLI set up's i.e you can use multiple GPU's and a CPU to boot. So I'm now running CODE::BLOCKS as an IDE, and have Cuda, which is a development tool to start programming. So, second question, has anyone had a good experience with Cuda, and is it a viable option for simulating astrophysical situations? It opens up multiple GPU use, which is clearly a good thing ? I am a newb to this area, any feedback is welcome
  21. In the vain attempt that you read this immediately, fancy jumping on chat if you're not busy ? It's Snail if you don't remember.

  22. Royston

    GR question

    No, you don't...I think this is perfectly clear judging by your responses. Personally I think the seminal field theory is Maxwell's equations, which is really a collection of equations already proved by others. It is a blinding result of vector calculus, and I'm stumped by how such a bunch of simple equations can explain (through application) the workings of EM, but they do, and they do it very well...it gets more complicated when studying materials et.c With all due respect lemur, you are asking 'big' questions, without the background to understand the responses. This is fine to a point, but it is increasingly annoying when you try to convince your audience that you do understand these concepts. I am not being rude I am more than happy to run you through the basics of GR, if you fancy it...QFT is not my field (ha!) but I'm happy to explain what a field theory actually is...nobody knows what a quantum field theory actually 'is', and anyone who claims otherwise, is talking bullshit.
  23. Royston

    GR question

    Right, 'now' I understand what you're asking, we got there eventually What you are asking is how can a particle not only define but create the space in which it resides (sounds contradictory doesn't it), yes ? It maybe best to think of gravitons as quantified bits of curvature, in the same way as a photon could be considered as a quantified bit of a light wave (I suppose)...but that would be muddying the issue (but at this level it will do as an explanation!) It is a very good question, and I'm simply not qualified to give a thorough answer. I do know someone who specializes in QFT, if you would like me to submit this question to him ?
  24. Royston

    GR question

    Yes, qualitatively it's not particularly hard, quantitatively it becomes a nightmare ! Even with simple metrics, solving the geodesics ends up with a whole page of calculations. Lemur, I'm not sure where to start...you stated one thing in your OP (which didn't make much sense) and now you're talking about gravitons !? Please stick to distinct, and manageable questions, rather than shifting about, thanks
  25. Royston

    GR question

    Get a good book on electromagnetism, then move on to special relativity. Then wrap your head around point set / general topology, then get a good book on differential geometery ! I'm not sure what you mean, light can loose energy while passing through a gravitional well, therefore it's frequency changes. The geodesics, or the path of light is shaped by the space in which it is defined.You really need to get a grasp of the basics, before you get into GR....it is an incredibly hard subject
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