📂Mathematical Physics

Gauss's Theorem, Divergence Theorem

Theorem¹

The following holds for a 3-dimensional vector function $\mathbf{F}$:

$$ \begin{equation} \int_{\mathcal{V}} \nabla \cdot \mathbf{F} dV = \oint_{\mathcal{S}} \mathbf{F} \cdot d \mathbf{S} \label{1} \end{equation} $$

Here, $\nabla \cdot \mathbf{F}$ is divergence, $\int_{\mathcal{V}}$ is volume integration, and $\oint_{\mathcal{S}}$ is closed surface integration.

Description

This is called Gauss’s theorem, Green’s theorem, or divergence theorem. The divergence theorem is especially used in electromagnetics.

Mathematical Meaning

Mathematically, it means that a surface integral can be expressed as a volume integral, and a volume integral can be expressed as a surface integral. In other words, it’s possible to convert triple integrals and double integrals to each other.

Physical Meaning

Physically, the total sum of amounts coming in and out at each point (small volume) on the left side of $\big( \eqref{1}$ is the same as the total sum of amounts coming in and out on the entire surface of the volume on the right side of $\big( \eqref{1}$.

As a simple example, imagine there are people in a room. People enter and leave the room through the door. Suppose there are two observers, one looking inside the room and the other at the door. Let’s say 2 people enter the room and 3 people leave. Then, the change of people as seen by the observer inside the room² is $|2-3|=1$, and the change of people as seen by the doorkeeper³ is $|3-2|=1$. (Count as $+1$ when $1$ people open the door and leave) These two are always the same.

Proof

Let’s check if the volume integral of divergence for each face adds up to the same.

Adding up the surface integral for all faces results in:

$$ \int _{S_{1}} \mathbf{F} \cdot d \mathbf{S}_{1} + \int_ {S_2} \mathbf{F} \cdot d \mathbf{S}_2 + \int_ {S_{3}} \mathbf{F} \cdot d \mathbf{S}_ + \int_ {S_{4}} \mathbf{F} \cdot d\mathbf{S}_{4} + \int _{S_{5}} \mathbf{F} \cdot d\mathbf{S}_{5}+\int _{S_{6}} \mathbf{F} \cdot d\mathbf{S}_{6} $$

First, let’s calculate for faces $S_{1}$ and $S_2$, which are perpendicular to the $x$ axis. It’s $\mathbf{F}= F_{x} \hat{\mathbf{x}} + F_{y} \hat{\mathbf{y}} + F_{z} \hat{\mathbf{z}}$, and each surface’s direction is outward. Assume the direction of $\mathbf{F}$ is the same as $S_{1}$. Then, the two surface integrals are as follows:

$$ \begin{align*} \int _{S_{1}} \mathbf{F} \cdot d \mathbf{S}_{1} + \int _{S_2} \mathbf{F} \cdot d \mathbf{S}_2 &= \int_ {S_{1}} F_{x} dS_{1} - \int _{S_2} F_{x} dS_2 \\ &= \int_{z}^{z+\Delta z} \int_{y}^{y+\Delta y} F_{x} (x+\Delta x,y,z) dydz - \int_{z}^{z+\Delta z} \int_{y}^{y+\Delta y} F_{x} (x,y,z) dydz \\ &= \int_{z}^{z+\Delta z} \int_{y}^{y+\Delta y} \bigg[ F_{x} (x+\Delta x,y,z) - F_{x} (x,y,z) \bigg] dydz \end{align*} $$

At this point, by the Fundamental Theorem of Calculus, since $\displaystyle \int _{a} ^b \dfrac{ dF(x)}{dx}dx=F(b) - F(a)$, it can be summarized as follows:

$$ \begin{align*} & \int_{z}^{z+\Delta z} \int_{y}^{y+\Delta y} \left[ F_{x} (x+\Delta x,y,z) - F_{x} (x,y,z) \right] dydz \\ =&\ \int_{z}^{z+\Delta z} \int_{y}^{y+\Delta y} \left[ \int_{x} ^{x +\Delta x} \dfrac{ \partial F_{x}(x,y,z) }{\partial x} dx \right] dydz \\ =&\ \int_{z}^{z+\Delta z} \int_{y}^{y+\Delta y} \int_{x} ^{x +\Delta x} \dfrac{ \partial F_{x}(x,y,z) }{\partial x} dx dydz \\ =&\ \iiint \dfrac{ \partial F_{x} }{\partial x} dV \end{align*} $$

Thus, the following result is obtained:

$$ \int_ {S_{1}} \mathbf{F} \cdot d \mathbf{S}_{1} + \int_ {S_2} \mathbf{F} \cdot d \mathbf{S}_2 =\iiint \dfrac{ \partial F_{x} }{\partial x} dV $$

Similarly, if we calculate the surface integrals for $S_{3}$ and $S_{4}$, and the surface integrals for $S_{5}$ and $S_{6}$, we get:

$$ \int _{S_{3}} \mathbf{F} \cdot d \mathbf{S}_{1} + \int _{S_{4}} \mathbf{F} \cdot d \mathbf{S}_2 =\iiint \dfrac{ \partial F_{y} }{\partial y} dV $$

$$ \int_ {S_{5}} \mathbf{F} \cdot d \mathbf{S}_{5} + \int_ {S_{6}} \mathbf{F} \cdot d \mathbf{S}_2 =\iiint \dfrac{ \partial F_{z} }{\partial z} dV $$

Finally, adding all the surface integrals for the 6 faces gives:

$$ \begin{align*} \oint _\mathcal{S} \mathbf{F} \cdot d \mathbf{S} &= \iiint \dfrac{ \partial F_{x} }{\partial x} dV + \iiint \dfrac{ \partial F_{y} }{\partial y} dV +\iiint \dfrac{ \partial F_{z} }{\partial z} dV \\ &= \iiint \left[ \dfrac{ \partial F_{x} }{\partial x} + \dfrac{ \partial F_{y} }{\partial y} + \dfrac{ \partial F_{z} }{\partial z} \right] dV \\ &= \iiint \nabla \cdot \mathbf{F} dV \\ &= \int_\mathcal{V} \nabla \cdot \mathbf{F} dV \end{align*} $$

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