⁶⁰The term $𝑐𝑢𝑟𝑙$ is mostly used within the US, instead of rotation. The $𝑟𝑜𝑡$ -operation gives how and how much a vector field rotates in space; the $𝑑𝑖𝑣$ -operation takes the derivative of a vector field. In mathematical form: $𝑟𝑜𝑡 (𝐅) \equiv \nabla \times 𝐅$ , $𝑑𝑖𝑣 (𝐅) \equiv \nabla \cdot 𝐅$ and in 3-dimensions with $𝐅 = [F_{x} F_{y} F_{z}]$ :

\nabla \times 𝐅 = | \begin{matrix} e_{x} e_{y} e_{z} \\ \frac{\partial}{∂𝑥} \frac{\partial}{∂𝑦} \frac{\partial}{∂𝑧} \\ F_{x} F_{y} F_{z} \end{matrix} |

\nabla \cdot 𝐅 = \frac{\partial F_{x}}{∂𝑥} + \frac{\partial F_{y}}{∂𝑦} + \frac{\partial F_{z}}{∂𝑧}

where

e_{x}

e_{y}

e_{z}

are unit vectors in x, y and z direction. Tricky, indeed, which is why I explain as less as possible in this reader.

The “ $\times$ ”-operator means the vector product, giving a vector which is perpendicular to the original two vectors: something with the right-hand rule. To transmit energy with an EM-field, we need a 3-dimensional world. Therefore, a 2-dimensional world would be completely dark, since light is a special form of an EM-wave.