Changing the basis vectors of a 2D density plot

Question

I have a 2D projection of a spherical object in the 3D space as given in the image below. The global orthonormal basis vectors are given as $\boldsymbol{x}$, $\boldsymbol{y}$, $\boldsymbol{z}$ and the local orthonormal basis vectors on the projection plane is given as $\boldsymbol{a}$, $\boldsymbol{b}$, $\boldsymbol{c}$ as in the image below.

For an object at $(-15,0,0)$ in the global coordinate system, the projection along the line $AD$ ends up at a point D with the coordinates $(-30,0,0)$ in the local coordinate system. When I create the 2D density plot of the projected object, it assumes the standard 2D basis as default and the projection ends up on the left side as in the image above.

With the local basis that I have set, I require the projection to end up on the right side or in other words I require the density plot with the same basis as the local basis ($\boldsymbol{a}$, $\boldsymbol{b}$, $\boldsymbol{c}$) of my projection geometry. For that, I want to flip the horizontal-axis of density plot so that the negative horizontal-axis values are on the right side of zero and the positive on the left of the zero. How do I do it?

The equations and the code are as follows:

FSR =   2. 2.71828^(-(( 0.0121267 (14.5272 + 0.34202 a)^2 (300. + 10. a)^2)/(-14.5272 - 0.34202 a)^2) - (0.25 (14.5272 + 0.34202 a)^2 b^2)/(-14.5272 - 0.34202 a)^2)

FSR2 = 2. 2.71828^(-((0.25 (9.39693 + 0.34202 a)^2 a^2)/(-9.39693 - 0.34202 a)^2) - (0.25 (9.39693 + 0.34202 a)^2 b^2)/(-9.39693 - 0.34202 a)^2)

DensityPlot[Max[FSR, FSR2], {a, -60, 60}, {b, -60, 60}, Mesh -> False, PlotPoints -> 200, PlotRange -> All, Exclusions -> None]

What I am trying to accomplish here is:

Here is the Matlab code that does the job in MATLAB

   [a,b] = meshgrid(-50:1:50); 
   FSR = 2*2.71828.^(-((0.012126707038032768.*((14.527228357744116 + 0.3420201433256687.* a).^2).*(300 + 10 .*a).^2)./(-14.527228357744116 - 0.3420201433256687.*a).^2) - (0.25000000000000006.*((14.527228357744116 + 0.3420201433256687* a).^2).* b.^2)./(-14.527228357744116 - 0.3420201433256687.*a).^2); 
   surf(a,b,FSR);
   set(gca,'Xdir','reverse');

Answer 1

Change your range and step-size accordingly.

DensityPlot[Sin[x] Sin[y], {x, -3 Pi, 3 Pi}, {y, -5 Pi, 5 Pi}, 
 FrameTicks -> {{Range[-5 Pi, 5 Pi, Pi], 
    None}, {Transpose@{Range[-3 Pi, 3 Pi, Pi], 
      Reverse@Range[-3 Pi, 3 Pi, Pi]}, None}}]

Edit: Here is a dirty way to do it.

   FSR = 2. 2.71828^(-((0.0121267 (14.5272 + 0.34202 a)^2 (300. + 
              10. a)^2)/(-14.5272 - 0.34202 a)^2) - (0.25 (14.5272 + 
            0.34202 a)^2 b^2)/(-14.5272 - 0.34202 a)^2);

FSR2 = 2. 2.71828^(-((0.25 (9.39693 + 0.34202 a)^2 a^2)/(-9.39693 - 
            0.34202 a)^2) - (0.25 (9.39693 + 
            0.34202 a)^2 b^2)/(-9.39693 - 0.34202 a)^2);
p1 = Rotate[
   DensityPlot[Max[FSR, FSR2], {a, -60, 60}, {b, -60, 60}, 
    Mesh -> False, PlotPoints -> 200, PlotRange -> All, 
    Exclusions -> None, Frame -> False, ImagePadding -> All], 
   180 Degree];




p2 = DensityPlot[1, {a, -60, 60}, {b, -60, 60}, 
   ColorFunction -> GrayLevel, ColorFunctionScaling -> False, 
   FrameTicks -> {{Range[-60, 60, 20], 
      None}, {Transpose@{Range[-60, 60, 20], 
        Reverse@Range[-60, 60, 20]}, None}}, ImageSize -> 390];

Overlay[{p2, p1}, Alignment -> Right]

Edit 2:

FSR[a_, b_] := 
  2. Exp[-((0.0121267 (14.5272 + 0.34202 a)^2 (300. + 
             10. a)^2)/(-14.5272 - 0.34202 a)^2) - (0.25 (14.5272 + 
           0.34202 a)^2 b^2)/(-14.5272 - 0.34202 a)^2];

FSR2[a_, b_] := 
  2. Exp[-((0.25 (9.39693 + 0.34202 a)^2 a^2)/(-9.39693 - 
           0.34202 a)^2) - (0.25 (9.39693 + 
           0.34202 a)^2 b^2)/(-9.39693 - 0.34202 a)^2];


data = Reverse /@ Tuples[Range[50, -50, -1], 2];

data2 = Max /@ Transpose@{FSR @@@ data, FSR2 @@@ data};

ListDensityPlot[Partition[data2, 101], MaxPlotPoints -> Infinity, 
 PlotRange -> MinMax@data2]