two variables for same coordinate

Hey,

I have this potential V=1/4*pi*epsilon*sqrt(((r1-r2)^2)+const.^2).

I solved for V =1/4*pi*epsilon*r using finite difference method but I'm unable to understand how to vary two variables for same axis. I have tried to do using different different loop for both parameters but can't find the solutions.

please help

Answer 1

[Image Analyst]

Pooja: You can use either meshgrid() or for loops. Below I show you both ways.

const = 2;
epsilon = 3;
maxR1 = 5.5;
maxR2 = 7.4;
% Define size of output matrix.
rows = 5;
columns = 4;
% Get x and y coordinates at each (y, x) location.
R1 = linspace(1, maxR1, columns); % x
R2 = linspace(1, maxR2, rows); % y

% Method 1 : vectorized using meshgrid()
[r1, r2] = meshgrid(R1, R2)
V = (1/4) * pi * epsilon * sqrt(((r1-r2).^2)+const ^ 2)

% Method 2 : for loops
V = zeros(rows, columns);
for col = 1 : columns
	r1 = R1(col);
	for row = 1 : rows
		r2 = R2(row);
		V(row, col) = (1/4) * pi * epsilon * sqrt(((r1-r2).^2)+const ^ 2);
	end
end
V
fprintf('Done running %s.m ...\n', mfilename);

[pooja sudha]

Hey,

Actually I don't want this. let me explain again briefly.

Actually what I'm trying to do is, I have [-((hbar^2)/(2*m))(d^2/dr1^2 +d^2/dr2^2) +(1/4) * pi * epsilon * sqrt(((r1-r2).^2)+const ^ 2)] psi = E*psi equation.

and i am creating two matrices one for first part and another is for second part in left side of equation using finite difference method and will find eigen values & eigen vectors using this.

and here r1 is radial coordinate for one particle and r2 is radial coordinate for another particle. i want to plot eigen vectors.

[pooja sudha]

will it have four coordinates x1,y1 & x2, y2 and corresponding to this four for loops.

[Image Analyst]

No, why would it? I have only 2 for loops, not 4. I scan the values of r1 and r2 to get every possible combination. Then I compute the V for that combination of r1 and r2. V is not really an image, it's more like a table of values where the rows represent different r2 values and the columns represent different r1 values. There is not really an x and y like you'd think of in an image. There really are no "coordinates" in this situation. Like I said, it's more like a table or chart of values for the variety of r1 and r2 you might encounter.

If this is not what you want, then perhaps you need to provide a diagram showing what's going on.

[pooja sudha]

Hey Thankyou. It worked

Answer 2

[pooja sudha]

please find the attached code.

here i computed for I-dimension & for 1-particle by using the potential V=1/sqrt(r^2+constant^2) ,now same thing I'm trying to do for 2-particle by adding the coordinate for second particle in the potential like V=1/sqrt((r1-r2)^2+constant^2).

please help if you know about this.

Thank you