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Circular vortex with spin vectors

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Sateesh Kandukuri
Sateesh Kandukuri el 24 de Jul. de 2023
Editada: Bruno Luong el 31 de Jul. de 2023
I need help to create a circular vortex with different polarizations like converging, diverging and clockwise etc.,
I attached an image for reference.
  2 comentarios
Dyuman Joshi
Dyuman Joshi el 24 de Jul. de 2023
Please show what you have attempted yet.
Sateesh Kandukuri
Sateesh Kandukuri el 24 de Jul. de 2023
% Parameters
numPoints = 100; % Number of points in the vortex
spinMagnitude = 0.5; % Magnitude of the spin vectors
radius = 1; % Radius of the vortex
% Generate theta values
theta = linspace(0, 2*pi, numPoints);
% Generate x and y coordinates
x = radius * cos(theta);
y = radius * sin(theta);
% Generate spin vectors
spinVectors = spinMagnitude * ones(size(x));
% Plot the vortex
figure;
quiver(x, y, spinVectors.*cos(theta), spinVectors.*sin(theta), 'b');
axis equal;
title('Circular Diverging Vortex');
xlabel('X');
ylabel('Y');

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Respuestas (2)

Dyuman Joshi
Dyuman Joshi el 24 de Jul. de 2023
% Parameters
numPoints = 50; % Number of points in the vortex
spinMagnitude = 0.5; % Magnitude of the spin vectors
r1 = 1; % Radius of the outer vortex
r2 = 0.5; %Radius of the inner vortex
% Generate theta values
theta = linspace(0, 2*pi, numPoints);
% Generate x and y coordinates
x = cos(theta);
y = sin(theta);
% Generate spin vectors
spinVectors = spinMagnitude * ones(size(x));
%%Radially outward arrows
figure;
quiver(r1*x, r1*y, spinVectors.*x, spinVectors.*y, 'b');
hold on
quiver(r2*x, r2*y, spinVectors.*x, spinVectors.*y, 'b');
axis equal;
xlabel('X');
ylabel('Y');
%%Radially outward arrows leaning in a counter clockwise direction
figure
quiver(r1*(x+y), r1*(y-x), spinVectors.*x, spinVectors.*y, 'b');
hold on
quiver(r2*(x+y), r2*(y-x), spinVectors.*x, spinVectors.*y, 'b');
axis equal;
xlabel('X');
ylabel('Y');
  4 comentarios
Dyuman Joshi
Dyuman Joshi el 31 de Jul. de 2023
Do you only have these images to work with? or do you have any data or any other piece of information?
Sateesh Kandukuri
Sateesh Kandukuri el 31 de Jul. de 2023
Actually, the polarization of the system is defined by P = (cosφ, sinφ, 0), where φ = tan-1(y/x) +Ψ and (x,y) are the spatial coordinates in the system plane with the origin at the centre. The angle Ψ determines the polarization orientation. For Ψ = 0, it gives diverging vortex. I hope this piece of information helpful.

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Bruno Luong
Bruno Luong el 31 de Jul. de 2023
Editada: Bruno Luong el 31 de Jul. de 2023
[x,y] = ndgrid(linspace(-1,1,10));
x = x(:)';
y = y(:)';
xy = [x; y];
r = vecnorm(xy, 2, 1);
r(r > 1) = NaN;
xyn = xy ./ r;
for k=1:12
Psi = 2*pi*rand();
R = [cos(Psi), -sin(Psi);
sin(Psi), cos(Psi)];
V = R * xyn;
vx = V(1,:);
vy = V(2,:);
subplot(3,4,k);
quiver(x, y, vx, vy, 'linewidth', 2);
set(gca, 'visible', 'off')
end
  2 comentarios
Sateesh Kandukuri
Sateesh Kandukuri el 31 de Jul. de 2023
Dear @Bruno Luong, I want to define these normalized spin vectors on a 200-unit diameter circular geometry with a unit vector spacing, and I need it in a separate image based on the angle Psi.
Bruno Luong
Bruno Luong el 31 de Jul. de 2023
Editada: Bruno Luong el 31 de Jul. de 2023
Feel free to adapt my code to your need.
I gave you a recipe of the cake, if you want strawberry flavor, you need to adapt my recipe and make your own cake.

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