# Tag Info

2

The issue here is that your front wheels are turned/steered by the same angle. When you try to find the instantaneous centre of curvature, you may first want to assume the wheels won't slip from side to side, like you may get if you drive around a corner on a slippy road. As there is no slip, the velocity of each wheel must occur in the direction the ...

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N-body simulations in full general relativity are difficult because gravity is a field theory and because it is non-linear. Let's deal with the field theory part first. In Newtonian mechanics gravity is static. The field itself has no energy or momentum, no degrees of freedom at all. It is simply an instantaneous force law between all matter. Remove the ...

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If you really do not want to simulate (despite your tag), you need to program the elliptic function and the Jacobian amplitude. This is done in the Numerical Recipes. You probably need to adapt something to C#. If, on the other hand, This is about simulation, I suggest you go for the solutions provided in the comments to your question, i.e. some Runge-Kutta ...

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I had thought that if you placed two electrons in a vacuum that they would accelerate away from each other without limit. However after speaking to shminux I realised that although they accelerate forever they reach a limiting velocity which depends on their potential energy. So the particles in my sim will reach a limited v. Problem was v was too large ...

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In my experience (PhD student working on laser plasma interaction experiments) these PIC codes are pretty closely guarded by their creators. There are a few out there, for example OSIRIS, VORPAL, and TurboWave in addition to VLPL. Of these I think only VORPAL is commercially available (through Tech-X) and the others you would need to contact the groups ...

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While lemon's answer is of course correct, it is not the only way to calculate the temperature from a molecular dynamics simulation: it can also be obtained from the configurations, that is, the particle coordinates, of the system. This is called "configurational temperature" (see, e.g., this article; pay-walled). The key identity is (for a canonical ...

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The instantaneous temperature of a system of $N$ particles of masses $m_i$ and velocities $v_i$ is $$T(t) = \sum_{i=1}^N \frac{m_iv_i^2(t)}{k_BN_f}$$ where $N_f$ equals the number of degrees of freedom, typically $N_f=3N-3$ for fixed total momentum. Note that the instantaneous temperature will fluctuate 5-10% about the true (thermodynamic) temperature ...

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The premise of your question is flawed. It is not possible to know everything with perfect accuracy, even in a non-quantum fully deterministic system. (In a quantum system, even if you do know its state with perfect accuracy, you can't predict it accurately.) Ever hear of chaos? the butterfly effect? the three-body problem? No matter how well you know the ...

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We don't know is probably the right answer to this question. With current knowledge of quantum mechanics it is believed that randomness is inherent in quantum mechanics. Or we may simply do not yet know the mechanics behind. In conclusion if randomness is real than there is no way for a universe to produce exactly the same result when started multiple ...

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To my taste, the most coherent way to think about it is to realize that the whole concept of "randomness" is strongly related do lack of information. It only makes sense to talk about randomness when we have a small "agent", who is a part of a large "universe" -- thus he cannot have a complete knowledge about that universe. But he needs to make some ...

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By the sounds of it you have made a mistake with the units. In fact, you should not be using SI units at all in your simulation; astronomical values in SI units vary by such huge orders of magnitude that they are often a source of floating point errors that can destroy trajectories. You should instead use the astronomical system of units. Specifically, ...

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There is nothing to do. If you force your system strongly, energy will build up and your hydrodynamic flow will become more and more intricate. At the same time, the dissipation becomes more and more efficient since it is proportional to the spatial derivative of the velocity field. You reach a steady state when both effects compensate each other. The ...

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