As described in the lecture a *charge distribution* \(\rho\colon \mathrm M \to \mathbb R\) in a uniformly conducting surface \(M\) induces an *electric field* \(E\), which satisfies *Gauss’s* and *Faraday’s law*\[\mathrm{div}\,E = \rho, \quad \mathrm{curl}\, E = 0.\]In particular, on a simply connected surface there is an *electric potential* \(u\colon \mathrm M \to \mathbb R\) such that\[ E =-\mathrm{grad} \,u.\]If we then apply the divergence operator on both sides we find that \(u\) solves the following *Poisson problem*\[\Delta u = -\rho.\]The potential is determined up to an additive constant thus the field is completely determined by the equation above. Since we have a discrete Laplace operator on a triangulated surface we can easily compute the electric field of a given charge:

- Compute \(\Delta\) and solve for the electric potential \(u\).
- Compute its gradient and set \(E = -\mathrm{grad}\, u\).

Here the resulting field for a distribution of positive and negative charge on the sphere:

Though there are some issues. First, we need first paint the density \(\rho\) onto \(\mathrm M\). Therefore we can just use a *paint node* – it has an *override color flag* where one can specify which attribute to write to.

A bit more serious problem is that not every density distribution leads to a solvable Poisson problem. Let us look a bit closer at this. We know that\[\Delta = A^{-1} L,\]where \(A\) is the *mass matrix* containing vertex areas and \(L\) contains the *cotangent weights* – as described in a previous post. We know that \[\mathrm{ker}\, \Delta = \{f\colon V \to \mathbb R \mid f = const\}\]and that \(\Delta\) is self-adjoint with respect to the inner product induced by \(A\) on \(\mathbb R^V\), \(\langle\!\langle f,g\rangle\!\rangle = f^t A g\). In this situation we know from linear algebra that\[\mathbb R^V = \mathrm{im}\, \Delta \oplus_\perp \mathrm{ker}\, \Delta.\]Hence the Poisson equation \(\Delta u = -\rho\) is solvable if and only if \(\rho\) is perpendicular to the constant functions, in other words\[\langle\!\langle \rho , 1 \rangle\!\rangle = \sum_{i\in V} \rho_i A_i =0.\]This is easily achieved by subtracting the mean value of \(\rho\) (orthogonal projection):\[\rho_i \leftarrow \rho_i -\tfrac{ \sum_{k\in V} \rho_k A_k}{\sum_{k\in V} A_k}.\]Then the Poisson equation \(\Delta u = – \rho\) is solvable and defines an electric field.

Actually symmetric systems are also better for the numerics. Thus we solve instead\[L u = – A \rho.\]Since the matrix \(L\) is symmetric we can use the cg-solver which is quite fast and even provides a solution if \(L\) has a non-zero kernel – provided that \(-A\rho\) lies in the image of \(L\).

Below the electric potential on the buddy bear where we have placed a positive charge on its left upper paw and a negative charge under its right lower paw.

Actually the bear on the picture does not consist of triangles – we used a *divide node* to visualize the dual cells the piecewise constant functions live on.

The graph of the function \(u\) (which is stored as a point attribute `@u`

) can then be visualized by a *polyextrude node.* Therefore one selects *Individual Elements* in the *Divide Into* field and specifies the attribute `u`

in the *Z scale *field.

The dual surface also provides a way to visualize the electric field: Once one has computed the gradient (on the primitives of the triangle mesh) the *divide node* turns it into an attribute at the points of the dual surface (corresponding to the actual triangles). This can then be used to set up attributes `@pscale`

(length of gradient) and `@orient`

(quaternion that rotates say \((1,0,0)\) to the direction of the gradient) and use them in a copy node to place arrows. Reading out the gradient at the vertices of the dual mesh needs a bit care:

1 2 3 |
vector g = v@grad; p@orient = dihedral(set(1,0,0),g); @pscale = length(g); |

Here how the end of the network might look like:

And here the resulting image:

Since the potential varies quite smoothly the coloring from blue (minimum) to red (maximum) does not reveal much of the function. Another way to visualize \(u\) is by a periodic texture. This we can do using the method `hsvtorgb`

, where we can stick in a scaled version of `u`

in the first slot (hue).

**Homework (due 19/21 July).** *Build a network that allows to specify the charge on a given geometry and computes the corresponding electric field and potential.*