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## Hot answers tagged models

39

You are right, the planetary model of the atom does not make sense when one considers the electromagnetic forces involved. The electron in an orbit is accelerating continuously and would thus radiate away its energy and fall into the nucleus. One of the reasons for "inventing" quantum mechanics was exactly this conundrum. The Bohr model was proposed to ...

29

I can tell you why I don't believe in it. I think my reasons are different from most physicists' reasons, however. Regular quantum mechanics implies the existence of quantum computation. If you believe in the difficulty of factoring (and a number of other classical problems), then a deterministic underpinning for quantum mechanics would seem to imply one of ...

17

There is a great paper from the group of Howard Stone on this subject: Wetting of flexible fibre arrays (freely available here, but for some reason I am not allowed to link to it normally: http://211.144.68.84:9998/91keshi/Public/File/34/482-7386/pdf/nature10779.pdf) They specifically study when 2 closely positioned parallel fibers (i.e. hairs) clump ...

15

Let us try to rewrite the equation in approximate form of finite differences: $$\frac{A(x,t+\Delta t)-A(x,t)}{\Delta t} = C_3\frac{A(x+h,t)+A(x-h,t)-2A(x,t)}{h^2} +$$ $$+ C_2 \frac{v(x+h,t)A(x+h,t)-v(x-h,t)A(x-h,t)}{2h} + C_1 A(x,t)+C_0$$ Where $\Delta t$ -- is a time step, and $h$ -- space step. The expression becomes your PDE, in the limit $\Delta t\to0, ... 14 This could have been a comment, but as it actually anwers the question asked in the title, I'll post it as such: As far as I can tell there's no rational reason to dismiss these models out of hand - it's just that quantum mechanics (QM) has set the bar awfully high: So far, there's no experimental evidence that QM is wrong, and no one has come up with a ... 11 Theoretically, yes it should be possible to derive the boiling point of diatomic nitrogen from fundamental forces. In fact, you don't even need to involve the strong force or weak force (or the strong nuclear force, which is sort of different). The strong forces bind the quarks together into nucleons and the nucleons together into nuclei, but they have ... 9 [ text I had put here is moved to the original question, but I prefer not to erase the comments that were posted here. ] 9 Apologies in advance if the first part of this comes off a bit argumentative, but I think there is an important point about physical theory that should be made. This point is also implicit in David Zaslavsky's answer as well. Rant on effective theories Actually trying to calculate macroscopic properties like "chemistry" from fundamental theories like QCD ... 9 Actually a paper recently came out, and highlighted in Popular Science, discussing using fermionic field concepts to model crowd avoidance at Netflix. You can imagine that the same concept could be used to consider in any situation where there are large numbers of people competing for limited preferred items. Update Now that we have a few minutes, ... 8 Check out Mark Smith's PhD thesis titled Cellular automata methods in mathematical physics, specifically Chapter 4: Lorentz Invariance in Cellular Automata. The conclusion part of the chapter: Symmetry is an important aspect of physical laws, and it is therefore desirable to identify analogous symmetry in CA rules. Furthermore, the most important ... 8 At present, the Navier-Stokes equations for the dynamics of water haven't yet been derived from microscopic principles. 7 No. There is nothing wrong with perturbation theory, or with theories with known, restricted accuracy. The point of theory is to explain the results of observation from as simple an initial theoretical standpoint as possible. Therefore: Since experiment always has a finite uncertainty, one can only ask that theory match the experimental value within its ... 7 This is what I think the first bit of the calculation does. Suppose you start with a spherical eye with a hole in it (e.g. the pupil in the human eye): The radius of the eye is$ER$and the radius of the hole is$AR$, and with the length$DA$these form a right angled triangle. Pythagoras' theorem tells us: $$DA^2 + AR^2 = ER^2$$ so: $$DA = ... 6 Briefly, The Bohr--planetary model doesn't really address these issues. Bohr, a genius, just asserted that the phenomena at the atomic level were a combination of stationarity while being in an orbit, and discrete quantum jumps between the orbits. It was a postulate that yielded some agreement with experiment and was very helpful for the future ... 5 This is an example from hydrodynamics. When the effects of viscosity can be ignored (inviscid flow), a uniform incident flow can exert on immersed bodies only lift forces perpendicular to the asymptotic flow velocity. However, there exist an infinite number of solutions of the flow equations of motion satisfying the asymptotic conditions at infinity and the ... 5 No. Notoriously, supersymmetric string theory has to be formulated in 10 dimensions in order to be consistent. Another example is supergravity, which can be formulated in a maximum of 11 dimensions otherwise it predicts particles with spins higher than two. 5 Classically emission is continuous and the electron would need to occupy a "in between" energy level for a while, and that is forbidden in Bohr's scheme, so the emission can't be allowed to happen. This doesn't really explain why it can't happen, but that's phenomenology for you: you line keep lining up facts until your kludge (1) gets the right answer and ... 5 Here's my quantitative attempt at 4. and 1.: The Coandă effect here is the tendency of the airflow to adhere to the surface of the ball. This means that near the surface of the ball, the streamlines are curved with a radius of curvature approximately equal to the radius of the ball R; this curvature results in a pressure gradient just as it does in ... 5 I can't see how a negatively charged electron can stay in "orbit" around a positively charged nucleus. Even if the electron actually orbits the nucleus, wouldn't that orbit eventually decay? Yes. What you've given is a proof that the classical, planetary model of the atom fails. I can't reconcile the rapidly moving electrons required by the ... 4 There are many physical intuitions often presented in various texts on fluid dynamics. I won't mention those here. I will, however, mention that mathematically the passage from a particle point of view to a continuum point of view is still a largely un-resolved problem. (With suitable interpretation, this problem was already posed by Hilbert as his 6th of 23 ... 4 As an alternative to Christian Blatter's heat interpretation, A might describe the concentration of particles adsorbed onto a one-dimensional substrate surface (or a two-dimensional one, where we ignore one of the dimensions). New particles are adsorbed at rate C_0 per unit length. Adsorbed particles detach from the surface at rate -C_1 per particle. ... 4 You can estimate that someone swimming in water has a Reynolds number of about 10^6 - 10^7; what counts is that this number is \gg 1. In that case, you're dealing with the drag equation: http://en.wikipedia.org/wiki/Drag_(physics). If we assume that our swimmer has the same power P on the moon and on Jupiter, his velocity v scales as$$v \propto ... 4 In cellular automata I do know there is explicit dependence on step/time. In quantum mechanics (and other many other theories) it is natural to write local evolution with respect to time. On the contrary, in 'pure' relativity, time is not that different from position. And thus there is no such natural interpretation like 'the next step is the next time'. ... 4 Ok, I'm still not sure on what level you want to do this, but I will start you off with some basics. The most important factor is probably the solar elevation angle,$\theta$. As described on the wiki-page it can be calculated using this formula: $$\sin\theta=\cos h\cos\delta\cos\Phi+\sin\delta\cos\Phi$$ where$h$is the hour angle,$\delta$is the solar ... 4 If you are interested just in the direct irradiance you can neglect the emmision and scattering terms in the Radiative Transfer Equation RTE wich can in this case be simplified to allow only for absorbtion and is known under the name Beer-Bougert-Lambert's law of absorption $$\cos\theta_{0}\,\frac{\partial}{\partial p}S^{i} = -\frac{\kappa^i}{g}\,S^{i}$$ ... 4 Let me give the naivest possible estimate, so that people have something to criticize. Assuming that the most of the jet interacts with the ball and is deflected at a substantial angle then the force on the ball is roughly the momentum flow through the pen. In your units this is$\rho_{air} Q^2/(\pi d^2)$. Saying the force to levitate a ball is$1\times ...

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The article looks indeed wrong. In fact, there are two mistakes. First, you're right that the acceptance-rejection method has to be applied to $\rho(r)$, and not to $m(r)$. To understand how this idea works, suppose we want to generate a one-dimensional normalized distribution function $p(y)$. Now, let's assume we can rewrite this distribution function in ...

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