How to calculate flux through a plane parallel to the XZ plane? Given a 3d vector field.

Say we have a 3D vector field, let's say the field and grid are already defined

n = 100;
rmax = 10000;
x = linspace(-rmax,rmax,n)
y = linspace(-rmax,rmax,n)
z = linspace(-rmax,rmax,n)

% Say FX,FY, and FZ are the components of the vector field and we want to calculate the flux through the plane y = 9800

Fy9800 = (FY(Y==y(n-2)))'; %Picks out the Y- component of the vector field in the y = 9800 plane

Fy9800R = reshape(Fy9800,[n,n]); 

[X2,Z2] = meshgrid(x,z)

surf(X2,Z2,Fy9800R) %Graphs the flux through the y=9800 plane

xlabel('X-Axis')
ylabel('Z-Axis')
zlabel('Flux')

Would this work? Thanks in advanced for the help.

Answer 1

[Rahul]

Hi Omar,

I understand that you are working with a 3D vector field defined on a grid and would like to extract and visualize the Y-component of the field on a specific plane, where one dimension (in this case, y) has a fixed value. The given approach involves trying to plot this as a surface to get a visual representation of the flux through that plane.

Here a few modifications that you can consider adopting in the given MATLAB script:

Fy9800 = (FY(Y == y(n-2)))';

1. Floating-point comparison: Using 'Y == y(n-2)' may not behave as intended because of floating-point precision. Even if 'y(n-2)' is exactly 9800 in the vector, comparing it directly like this might not return 'true' due to small numerical differences.
2. Indexing structure: This approach treats 'FY' and 'Y' as flat arrays, but if 'FY' is a 3D matrix (n x n x n), this kind of logical indexing can flatten and disrupt the spatial layout. A better and more structured way could be to directly use the known index for 'y = 9800'.

Since the given script utilizes 'linspace(-rmax, rmax, n)' to define the y vector, and since 'y(n-2)' is 9800, we can safely use the following indexing method:

iy = n - 2; % Index where y ≈ 9800

The relevant slice of 'FY' can then be extracted using regular 3D indexing, as shown below:

Fy9800 = squeeze(FY(:, iy, :));  % Slice of FY at y = 9800 (X-Z plane)

The 'squeeze' function has been used here to remove the singleton Y-dimension and get a 2D matrix of size (n x n) that corresponds to X and Z. Here is how you can align it properly with the plotting grid:

[X2, Z2] = meshgrid(x, z);
Fy9800R = Fy9800';  
% Transpose to match dimensions of X2 and Z2

The transpose of the 'meshgrid(x, z)' creates matrices where rows correspond to Z and columns to X, and we want 'Fy9800R(i, j)' to correspond to 'Z(i)' and 'X(j)'.

This approach should allow you to create a clear surface plot of the Y-component of your vector field on the specified plane.

For more information regarding various functions used in the above mentioned code snippet, you can refer to the following documentation links:

• Create 2-D and 3-D grids using 'meshgrid' function: https://www.mathworks.com/help/matlab/ref/meshgrid.html
• Remove dimensions of length 1 using 'squeeze' function: https://www.mathworks.com/help/matlab/ref/squeeze.html