# Tag Info

0

If you solve for the second derivatives, you won't have to use "EquationSimplification" -> "Residual" and things will work ok. Solving for the second derivatives be faster if you start with exact coefficients. Also, if you clear l, solve for the derivatives, and then substitute a value for l, Solve won't choke on the algebra. The long time it takes is ...

5

You probably entered ClebschGordan[{1/2,1/2},{0,0},{1/2,-1/2}] and got the warning message. This happens when you use angular-momentum quantum numbers that don't satisfy the conservation laws. In this case, the rule $m_1+m_2=m$ is not met, see the documentation for ClebschGordan. However, Mathematica still produces the correct result, i.e., 0. This just ...

8

Setup The region to solve the PDEs Clear["Global*"] xMax = 453.595 - Sqrt[450.05^2 - 3^2]; regA = Rectangle[{0, 0}, {xMax, 6}]; regB = ImplicitRegion[{453.595 - Sqrt[450.05^2 - (y - 3)^2] - x < 0}, {{x, 0, xMax}, {y, 0, 6}}]; reg = RegionDifference[regA, regB]; RegionPlot[reg, AspectRatio -> 1/10] Solve the PDEs using NDSolve solV = ...

1

The problem with your system of equation is that the "zero" far field solution is unstable, hence extremely sensitive to the initial conditions. Posing the problem as an initial value problem, with the "known" conditions from the successful solution: max = 50; Pr = .72; a = 0.7172594734816521 b = -0.4344414944896132` pohl = NDSolve[{f'''[\[Eta]] + 3 f[\[...

7

The problem is with the default starting initial conditions used by the shooting method in NDSolve. The shooting method is where FindRoot is being used internally, so the OP's error message is a strong hint that this is the problem. Getting convergence in a nonlinear system can depend greatly on the starting conditions. Having luckily solved the system ...

5

As @yarchik pointed out, this is strictly speaking not a two-electron problem, but a problem of "distinguishable particles," since the antisymmetry postulate (and spin) is ignored. Moreover, the 1D Coulomb potential in the initial equations can't be just $1/x$ -- it has to have an absolute sign because otherwise it would be attractive on one side and ...

8

I think this is what you're looking for: dostot = Union[dos1, dos2][[All, 3]]; {emin, emax} = MinMax[dostot]; de = .25; Histogram[dostot, {emin, emax, de}, "ProbabilityDensity"] The definition of dostot differs from yours in that I discard all the coordinate values that would only be needed to make 3D plots. The relevant energy is stored as the third ...

3

Your code cannot work because you defined the potential and electric field with incorrect syntax. Here is a corrected version: Clear[x, y, z]; PhiLineSegment[x_, y_, z_] = (segCharge/(4 Pi e0 segLength)) Log[(segLength/2 + z + Sqrt[(x + segOffset)^2 + y^2 + (z + segLength/2)^2])/(-(segLength/2) + z + Sqrt[(x + segOffset)^2 + ...

3

Summary We give the solution of the problem by first calculating the potential, setting up the equation of motion and solving it numerically assuming some typical inital conditions. The results are shown as plots of the time dependence of the components of the position vector and as a 3D-plot and an animation of the ion trajectory. This code gives a ...

3

Here is a way to do it based on Jens' answer. sphericalToCartesian = Thread[{r, θ, ϕ} -> {Sqrt[x^2 + y^2 + z^2], ArcCos[z/Sqrt[x^2 + y^2 + z^2]], Arg[x + I y]}]; (*Atomic Orbitals*) Ψ[n_, l_, m_][r_, θ_, ϕ_] := Sqrt[(n - l - 1)!/(n + l)!] E^(-(r/n)) ((2 r)/n)^l 2/n^2 LaguerreL[ n - l - 1, 2 l + 1, (2 r)/n] ...

2

This computation can be carried out using s1 = NDSolve[{(((γ[x] - 1)/2)*(1 - vr[θ]^2 - (vr'[θ])^2)*(2*vr[θ] + (vr'[θ]* N[Cot[θ*Degree]]) + (vr''[θ]))) - (vr'[θ]*(vr[θ]*vr'[θ] + vr'[θ] v''[θ])) == 0, vr[θ1] == vri, vr'[θ1] == vti, WhenEvent[vr[θ] < 0, "StopIntegration"]}, vr, {θ, θc, θs}] Note that the ODE is singular where Denominator[...

3

If you were to be more careful in your use and scoping of variables, everything would be fine. {m, k, g, r0} = {1, 1, 9.8, 2}; With[{ Q = 0, coord = {r[t], θ[t]}, T = 1/2 m (r'[t]^2 + r[t]^2 θ'[t]^2), V = 1/2 k (r[t] - r0)^2 - m g r[t] Cos[θ[t]]}, Lagrangiana[T_, V_, Q_, coords_List] := Module[{L = T - V}, (D[D[L, D[#, t]],...

Top 50 recent answers are included