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5

Enzo is fundamentally a grid-based finite-volume hydrodynamics code. That is, the domain is divided into cells, each is assigned various fluid properties (density, velocity, etc.), and at each timestep fluxes of those quantities across the interfaces between cells are used to update the quantities in the cells. It has a choice of particular methods for ...


1

In a reference frame where the center of mass of the Earth-Sun system is at rest, we will have $$ m_\text{Sun} \vec{v}_\text{Sun} + m_\text{Earth} \vec{v}_\text{Earth} = 0 \quad \Rightarrow \quad \vec{v}_\text{Sun} = - \frac{m_\text{Earth}}{m_\text{Sun}} \vec{v}_\text{Earth}. $$ In particular, if this is true at some initial time, then it's true at all ...


5

In situations like this, it is a good idea to adjust the time step based on the gradient of the force - because the whole concept of numerical integration is that "things don't change too much from now until the next time step", and that assumption is violated when you move rapidly through a region with fast-changing force. This has a risky side-effect: if ...


0

You're getting particles that get 0 distance from each other, and therefore have infinite acceleration. The computer cannot resolve this, and so it explodes according to how close the step gets to zero. I'm not sure what "physical" behaviour would be, given that zero distance point masses are difficult to study. But if you added in a y coordinate and gave ...


1

For any PIC simulation, you are necessarily tying yourself to particles, particles that experience forces. Thus, we have the generic force law: \begin{align} m_i\frac{d\mathbf v_i}{dt}&=\mathbf F_i \tag{1a}\\ \frac{d\mathbf x_i}{dt}&=\mathbf v_i \tag{1b} \end{align} In the case of PIC, you are often considering the electromagnetic (Lorentz) force, ...


-4

I have found something that may explain why you got an envelope that was unexpected. I greatly enlarged the intercetion of the point where the curve iad envelope intersect. It is not at the peak (which would make it horziontal, like your 'wobbley' envelope) But the envelope 'mashes the sineusoid into distortion, shifting the true peak toward it's ...


0

The Hamiltonian is given by \begin{equation} H=\frac{\mathbf{p}^2}{2m}+U\left(\mathbf{r}\right), \end{equation} where $U\left(\mathbf{r}\right)$ is the potential landscape due to the crystal lattice. The Bloch theorem asserts that the solution to the problem \begin{equation} H\Psi_{\mathbf{k}}=E\left(\mathbf{k}\right)\Psi_{\mathbf{k}}, \end{equation} is to ...


0

Functions on a circle with angular frequency components up to $n$ and average 0 form a vector space of dimension $2n$ spanned by (e.g.) $\cos(2\pi nx)$ and $\sin(2\pi nx)$, so you need at least $2n$ samples to reconstruct it. In this case taking samples at regular intervals of length $1/(2n)$ of each basis element gives you a linearly independent set of $2n$ ...


0

This would most likely violate the Church Turing Thesis, namely that everything that is computable is computable by a turing machine. If we could measure this parameter, we could construct a machine that could measure it to arbitrary precision, we would have hypercomputation, which we do not believe exists.


0

You need the integration step to be much shorter than the impact time. For most real world collisions, 0.1 second will be much too long. The force will change during the collision - and given the very simplistic integration method you use, you have to integrate over sufficiently short steps during which the force doesn't change much. You could do this by ...


0

Total energy has independent contributions from kinetic energy and potential energy. The kinetic energy component is determined by temperature and is not affected by particle mass (although obviously velocities will be affected). The potential energy components also do not depend on mass (e.g. Coulombic effects). It's somewhat counter-intuitive but the ...



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