Skip to main content

plotting - particle motion in 1D anharmonic well


I have set of data points but I want to draw the trajectory of particle moving inside this well either in phase space. How can I do this? I tried with VectorPlot but not come out with good and set of data point are as follows: mass of particle is 10^-24 kg and here in this data first column is position & second potential. I am very thankful to you if you can help me to do this. Thank you.


x={2.*10^-6, 1.9*10^-6, 1.8*10^-6, 1.7*10^-6, 1.6*10^-6, 1.5*10^-6, 
     1.4*10^-6, 1.3*10^-6, 1.2*10^-6, 1.1*10^-6, 1.*10^-6, 9.*10^-7, 
     8.*10^-7, 7.*10^-7, 6.*10^-7, 5.*10^-7, 4.*10^-7, 3.*10^-7, 2.*10^-7,
      1.*10^-7, 0, -1.*10^-7, -2.*10^-7, -3.*10^-7, -4.*10^-7, -5.*10^-7, \
    -6.*10^-7, -7.*10^-7, -8.*10^-7, -9.*10^-7, -1.*10^-6, -1.1*10^-6, \
    -1.2*10^-6, -1.3*10^-6, -1.4*10^-6, -1.5*10^-6, -1.6*10^-6, \
    -1.7*10^-6, -1.8*10^-6, -1.9*10^-6, -2.*10^-6}  


 Vx = {-2.39203*10^-19, -2.38026*10^-19, -2.37166*10^-19, -2.36809*10^-19, \
-2.37117*10^-19, -2.38208*10^-19, -2.40143*10^-19, -2.42925*10^-19, \
-2.46503*10^-19, -2.50782*10^-19, -2.55644*10^-19, -2.60954*10^-19, \
-2.6657*10^-19, -2.72342*10^-19, -2.78094*10^-19, -2.83618*10^-19, \
-2.88644*10^-19, -2.92835*10^-19, -2.95779*10^-19, -2.97001*10^-19, \
-2.96*10^-19, -2.92287*10^-19, -2.8545*10^-19, -2.75213*10^-19, \
-2.61492*10^-19, -2.44433*10^-19, -2.24423*10^-19, -2.0208*10^-19, \
-1.782*10^-19, -1.53693*10^-19, -1.29492*10^-19, -1.06466*10^-19, \
-8.5334*10^-20, -6.6606*10^-20, -5.0546*10^-20, -3.7171*10^-20, \
-2.6278*10^-20, -1.75*10^-20, -1.0373*10^-20, -4.409*10^-21, 

 1.*10^-21}


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more