📂Mathematical Physics

Stokes' Theorem

Theorem¹

Let’s call something a vector and an area in 3D space as $\mathbf{v}, \mathcal{S}$, respectively. The area vector of $\mathcal{S}$ is denoted as $d\mathbf{a}$, the border of $\mathcal{S}$ as $\mathcal{P}$, and the path moving along $\mathcal{P}$ as $d\mathbf{l}$. Then, the following equation holds:

$$ \int_{\mathcal{S}} (\nabla \times \mathbf{v} )\cdot d\mathbf{a} = \oint_{\mathcal{P}} \mathbf{v} \cdot d\mathbf{l} $$

This is called Stokes’ theorem or the fundamental theorem for curl.

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Incidentally, outside of theorem names, as used in the shrimp sushi restaurant, the word ‘rotation’ alone is used as ‘curl’.

Explanation

The theorem can be explained as follows:

• The total amount of rotation of vector $\mathbf{v}$ within a certain region $\mathcal{S}$ (left side) is the same as the sum of the values of vector $\mathbf{v}$ along the border $\mathcal{P}$ of that region (right side).

For those studying physics, understanding the meaning behind Stokes’ theorem is far more important than understanding its proof.

Since $d\mathbf{l}$ is a closed path, it doesn’t matter from which point or in which direction we start, the outcome remains unaffected. Thus, the direction of the area vector $d\mathbf{a}$ is determined by the right-hand rule.

Without looking at the picture, it might be hard to understand that the total amount of rotation within the area and the total amount along the path are the same. Let’s look at the picture below.

The integral value is determined solely by the boundary $\mathcal{P}$ of $\mathcal{S}$

Since Stokes’ theorem is an equation, regardless of what shape $\mathcal{S}$ is in, if its boundary is the same curve $\mathcal{P}$, the integral value always remains the same. Therefore, what the surface looks like is irrelevant. As shown in the picture below, even if there are various surfaces, if their edges are the same, then $\int (\nabla \times \mathbf{v})\cdot d\mathbf{a}$ has the same value. Therefore, the integral value is determined by the boundary of $\mathcal{S}$.

The integral value for a closed surface $\mathcal{S}$ is always $0$

In the case of a closed surface, since the length of the border is $0$, the length of the path is $0$, and the right-side closed path integral is always $0$. Hence, the following result is obtained:

$$ \int_{\mathcal{S}} (\nabla \times \mathbf{v} )\cdot d\mathbf{a} = 0 = \oint_{\mathcal{P}} \mathbf{v} \cdot d\mathbf{l} $$

If it’s hard to understand that the length of the border of a closed surface is $0$, look at the picture below.