Recall Gauss’ theorem, $ \int\int\int_V (\vec{\nabla} \cdot \vec{F}) dV = \int\int_{S} (\vec{F} \cdot \vec{e}_{n}) dS $. This theorem can be written more precisely. The following statement of the Divergence theorem is a copy from reference [1].Â Definitions are provided first. Volume $ V$: Define $ V$ as a region comprising three spatial dimensions. The volume has no holes in it. Boundary $ \partial V$: Define $ \partial V$ as a once-differentiable surface surrounding the volume $ V$. The boundary has a thickness of zero and no holes in it. Points not in $ V$ are not in $ \partial V,$ […]

# Category: Gauss

It turns out that deriving Gauss’ Law is easier said than done. There are several steps according to a StackExchange post [1]. The first of these steps is understanding Gauss’ Theorem. Hmm. Perhaps Gauss used his own theorem to derive his electrostatics law. After a quick online search, it is clear that Gauss’ Theorem is just another name for the Divergence theorem [2]. The Divergence Theorem is [3] $ \int\int\int_V (\vec{\nabla} \cdot \vec{F}) dV = \int\int_{S} (\vec{F} \cdot \vec{e}_{n}) dS $. Oof. That is a lot of symbolism to break down. Fortunately I am able to break this down;

Overview of Gauss’ Theorem and Basics of IntegrationRead More »

In Derivation #3, the expression, $ \frac{\partial \beta(t)}{\partial t}$, was written. This is an expression for a derivative of a function $ \beta(t)$. Now that a derivative has been introduced, Maxwell’s equations can be investigated. I start with Gauss’ Law. But first, slightly more information about derivatives is needed. I can consider the pieces of $ \frac{\partial \beta(t)}{\partial t}$ separately. These pieces are $ \beta(t)$ and $ \frac{\partial}{\partial t}$. The $ \beta(t)$ is, of course, a function while $ \frac{\partial}{\partial t}$ is an operator called a differential operator that forms $ \frac{\partial \beta(t)}{\partial t}$ if $ \frac{\partial}{\partial t}$ is