Tag Info

New answers tagged

0

As a first step you can use a simple method, which is for every small time step $\Delta t$ approximate the acceleration as constant and use $\Delta v=a\Delta t$ for each direction, and then $\Delta x = v\Delta t$. These equations apply separately for each dimension, so calculate the x and y velocities first and then the resulting changes in position


1

An exponentially decaying tail is almost like having no tail for all practical reasons. For example consider the yukawa potential for interaction through exchange of a massive particle, it is $\propto e^{-\mu r}/r$ which is even a stronger tail than the asymptotic behavior of the hyperbolic secant. There we say that the interaction has the the effective ...


1

Short explanation: Physical systems are usually dissipative systems. Dissipative systems can be modelled with a 1st order ODE system. If you do your perturbation analysis on this system at the Hopf bifurcation and take higher orders into account, you end up with the Stuart-Landau equation. Hence by derivation, it describes the dynamics of a system ...


-1

There are also conservative and non-conservative forms of the Burgers' Equation. The non-conservative form you presented will only solve for smooth shocks. You need the conservative form if there are likely to be any sharp changes in the solution. This might also help: http://people.maths.ox.ac.uk/trefethen/pdectb/burgers2.pdf


3

In the action formalism a linear Euler-Lagrange (EL) equation corresponds to a quadratic action, i.e. an action which is quadratic in the dynamical field variables of the theory. On the other hand, self-coupling or interaction terms in the action correspond to cubic or higher terms. Such terms leads to non-linear EL equations. See also this related Phys.SE ...


0

What you are asking about is a broader topic than you probably think it is, so I can only give you some directions here: First of all, what you call phase space is usually called parameter space, while phase space is used for something else. Given that you only have two observables, one usually wouldn’t call different types of results phases, but rather ...


1

Regular quantum mechanics is linear, and as the linked question makes clear, that linearity is essential to prevent signaling given the rest of quantum mechanics: entanglement, non commuting operators, projection onto eigenspaces by strong measurements, etcetera. So if you break the linearity and want to avoid signalling, then you have to change at least ...



Top 50 recent answers are included