Skip to main content

Creating a numerical table of values and using an interpolation of them to transform pixels in an image by the ImageTransformation command


I am trying to carry out point "Implementing Map" on this paper Interstellar wormholes however I am using my own equations not those given (only the same method).


I have generated a list of values from my code ie numericalmap however would like to use the command Interpolation to interpolate these points and then use ImageTransformation to transform images from these interpolations. Note the image used here is just a random URL link, feel free to insert any test image you like.


Background



The numerical map is a list of (ϕcs,ϕ[λend]) where ϕcs is the angle in an observers view at which he sees a light ray and ϕ[λend] is the angle which the ray is deflected through from that ϕcs as it passes by a black hole.


The interpolation of these therefore gives a function so that for any given initial angle we can now how much it is deflected by as it goes around a black hole.


Nx,Ny,Nz just represent Cartesian components of the light ray in the observers sky, the point of the Interpolation is to get a transformation relationship between these light ray components in the observers sky and the deflection angle caused by the black hole (ie ϕ[λend] which is the 2nd entry when evaluating numerical map).


I have managed to create the interpolating function from my numerical map so far


numericalmap = {};
n = 100;
For[i = 0, i < n + 1, i++, \[Phi]csgen = (1.009 + (1/2) i/n)*Pi;
 M = 1; E0 = 1; \[Theta]cs = Pi/2;
 Nx = Sin[\[Theta]cs]*Cos[\[Phi]csgen];
 Ny = Sin[\[Theta]cs]*Sin[\[Phi]csgen];

 Nz = Cos[\[Theta]cs];
 nr = -Nx;
 n\[Phi] = -Ny;
 n\[Theta] = Nz;
 b = rc*Sin[\[Theta]c]*n\[Phi]/(1 - 2 M/rc)^(1/2);
 B2 = rc^2/(1 - 2 M/rc)*(n\[Phi]^2 + n\[Theta]^2);
 prinitial = ((1 - 2 M/rc)^(-1))*nr;
 p\[Theta]initial = ((1 - 2 M/rc)^(-1/2))*rc*n\[Theta];
 {rc, \[Theta]c, \[Phi]c} = {200, Pi/2, 0};
 lambdaend = -100000;

 ham = {
   t'[\[Lambda]] + E0/(1 - (2 M)/r[\[Lambda]]) == 0,
   r'[\[Lambda]] - (1 - (2 M)/r[\[Lambda]]) pr[\[Lambda]] == 0,
   \[Theta]'[\[Lambda]] - P\[Theta][\[Lambda]]/r[\[Lambda]]^2 == 0,
   \[Phi]'[\[Lambda]] - b/(r[\[Lambda]]*Sin[\[Theta][\[Lambda]]])^2 ==
     0, pr'[\[Lambda]] + 
     M/r[\[Lambda]]^2 (E0^2/(1 - (2 M)/r[\[Lambda]])^2 + 
        pr[\[Lambda]]^2) - B2/r[\[Lambda]]^3 == 0, 
   P\[Theta]'[\[Lambda]] - (b^2*
        Cos[\[Theta][\[Lambda]]])/(r[\[Lambda]]^2*

        Sin[\[Theta][\[Lambda]]]^3) == 0
   };
 haminital = {t[0] == 0, 
   r[0] == rc, \[Theta][0] == \[Theta]c, \[Phi][0] == \[Phi]c, 
   pr[0] == prinitial, P\[Theta][0] == p\[Theta]initial};
 \[Phi]2 = 
  NDSolveValue[{ham, haminital}, {t, r, \[Theta], \[Phi], pr, 
     P\[Theta]}, {\[Lambda], lambdaend, 0}][[4]];
 numericalmap = 
  Append[numericalmap, {\[Phi]csgen - Pi, Pi + \[Phi]2[lambdaend]}]]


f=Interpolation[numericalmap]
img = Import["https://i.stack.imgur.com/2W9SD.png"]

ImageForwardTransformation[img,
g[x_,y_]:=ToPolarCoordinates[{x,y}],
k=f[#] & g[x_,y_],
h[k_,y]:=FromPolarCoordinates[{h,y}]
]




Main question


I am trying to convert the pixels in the image to polar coordinates then apply the f to the radial distance and then convert it back to Cartesian form. I think I am doing this wrong and would appreciate someone deforming the radial coordinate of the image correctly using the interpolation function.


After this I also tried using ImageTransformation on this list as follows but am not entirely sure how to input my Interpolation function of the numericalmap into it to transform the pixels by this function.


What this interpolation function is meant to show is the deformation in the radial direction of a 2d image as the input angle of the image changes (Where the angle of an image should change to). I think you have to use polar coordinates on the image with the coordinate system centred on the image. Not 100% sure how to do this but resulting images should look like


Image


Note: I think the issue lies within the way I am inputting the Interpolation into ImageTransformationfunction Marked by ## in code




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