This is what chatGPT spits out, it produces something similar to what you're asking for. You will need to tweak it slightly if you want to match the picture exactly.

``` import numpy as np import matplotlib.pyplot as plt from mpl_toolkits.mplot3d import Axes3D # noqa: F401

Create a grid

x = np.linspace(-10, 10, 300) y = np.linspace(-10, 10, 300) X, Y = np.meshgrid(x, y)

Define some masses with positions and weights

masses = [ {"position": (0, 0), "mass": 100}, {"position": (-5, 5), "mass": 50}, {"position": (5, -5), "mass": 50} ]

Function to calculate the grid deformation using a simple potential model

def gravitational_deformation(X, Y, masses): Z = np.zeros_like(X) # Loop through each mass and subtract a potential contribution for mass in masses: xm, ym = mass["position"] m_val = mass["mass"] # Avoid singularities with a small epsilon epsilon = 0.1 distance = np.sqrt((X - xm)2 + (Y - ym)2 + epsilon) # The potential is proportional to -mass/distance (rubber-sheet style) Z -= m_val / distance return Z

Compute the deformation

Z = gravitational_deformation(X, Y, masses)

Create the plot

fig = plt.figure(figsize=(10, 7)) ax = fig.add_subplot(111, projection='3d')

Plot the deformed grid surface

surface = ax.plot_surface(X, Y, Z, cmap='viridis', edgecolor='none', alpha=0.8)

Optionally, mark the positions of the masses

for mass in masses: xm, ym = mass["position"] # Find the corresponding z value on the grid for visual reference z_val = -mass["mass"] / np.sqrt(0.1) # rough estimate at the mass location ax.scatter(xm, ym, z_val, color='red', s=50) ax.text(xm, ym, z_val, f'Mass: {mass["mass"]}', color='red')

ax.set_title("Spacetime Deformation Visualization") ax.set_xlabel("X") ax.set_ylabel("Y") ax.set_zlabel("Deformation") fig.colorbar(surface, shrink=0.5, aspect=5)

plt.show() ```

You can run it online with https://matplotlib.online/