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# Vector Fields and Potential Functions

We will prove:

(1) For any twice-differentiable vector field $$\vec{A}$$ in $$\mathbb{R}^3$$:

$$\vec{\nabla} \cdot (\vec{\nabla} \times \vec{A}) = 0$$

(2) If $$\exists$$ an irrotational vector field $$\vec{F}$$ in $$\mathbb{R}^3$$ such that $$\vec{F} = \vec{\nabla}P$$ for some scalar function $$P$$, then $$P$$ must satisfy Poisson's Equation:

$$\nabla^2 P = f(x,y,z)$$

for some continuous $$f(x,y,z)$$.

Proof:

(1) Before we prove (1) directly, we need to prove something else:

Lemma: $$\forall$$ twice-differentiable multivariable functions $$a$$ in $$\mathbb{R}^3$$:

$$\frac{\partial^2 a}{\partial x \partial y} = \frac{\partial^2 a}{\partial y \partial x}$$

Subproof:

Suppose $$\partial_x\partial_y a \neq \partial_y \partial_x a$$. Then, by the definition of partial differentiation, we have:

$$\displaystyle \lim_{h\to 0} \frac{\partial_y a |_{x=x+h} - \partial_y a |_{x=x}}{h} \neq \displaystyle \lim_{h\to 0} \frac{\partial_x a |_{y=y+h} - \partial_x a |_{y=y}}{h}$$

$$\Rightarrow$$ $$\displaystyle \lim_{h\to 0} \partial_y a |_{x=x+h} - \partial_x a |_{y=y+h} \neq \partial_y a |_{x=x} - \partial_x a |_{y=y}$$

Now since we're dealing with functions in $$\mathbb{R}^3$$, and $$\mathbb{R}^3$$ is a closed metric space, we observe that this implies at least one of the following:

$$\partial_y a$$ is not differentiable with respect to x at some point in the plane $$x=x$$

$$\partial_x a$$ is not differentiable with respect to y at some point in the plane $$y=y$$

Originally, our Lemma is restricted to twice differentiable multivariable functions, and this, by definition, requires that the aforementioned derivatives exist in all planes of $$\mathbb{R}^3$$. Hence, we have arrived at a contradiction. We are then forced to conclude that:

$$\forall$$ twice-differentiable multivariable functions in $$\mathbb{R}^3$$:

$$\frac{\partial^2 a}{\partial x \partial y} = \frac{\partial^2 a}{\partial y \partial x}$$

This proves our Lemma and we may move on with (1).

The proof for (1) is straightforward now that we've established our Lemma:

Let $$\vec{A} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$$. Then:

$$\vec{\nabla} \times \vec{A} = \left[ \partial_y a_z - \partial_z a_y\right]\hat{i} +\left[-(\partial_x a_z - \partial_z a_x)\right]\hat{j} +\left[ \partial_x a_y - \partial_y a_x\right]\hat{k}$$

Then:

$$\vec{\nabla} \cdot (\vec{\nabla} \times \vec{A}) = \partial_x \partial_y a_z - \partial_x \partial_z a_y + \partial_y \partial_z a_x - \partial_y \partial_x a_z + \partial_z \partial_x a_y - \partial_z \partial_y a_x$$

And by our Lemma:

$$\vec{\nabla} \cdot (\vec{\nabla} \times \vec{A}) = 0$$

This proves (1).

(2) If $$\vec{F}$$ is irrotational and $$\vec{F} = \vec{\nabla} P$$, then:

$$\vec{\nabla} \times \vec{F} = \vec{0}$$

$$\Rightarrow$$ $$\vec{\nabla} \times \vec{\nabla} P =\vec{0}$$

$$\Rightarrow$$ $$\partial_y \partial_z P = \partial_z \partial_y P$$ and

$$\partial_x \partial_z P = \partial_z \partial_x P$$ and

$$\partial_x \partial_y P = \partial_y \partial_x P$$

Then this implies that $$P$$ is twice differentiable. Then:

$$\nabla^2 P = \partial_x^2 P + \partial_y^2 P + \partial_z^2 P$$ must exist in the entirety of $$\mathbb{R}^3$$. Then $$\exists$$ a continuous function $$f(x,y,z)$$ such that:

$$\nabla^2 P = f(x,y,z)$$

This is Poisson's Equation and thus we've proved (2).

QED

(1) can be expressed in words as "the divergence of the vector curl of any twice differentiable vector field in $$\mathbb{R}^3$$ is always zero."

(2) is a statement about potential functions. Our proof for (2) implies that all potential functions are solutions of Poisson's equation. In general, many potential functions satisfy Laplace's equation, which is the special case of Poisson's equation when $$f=0$$.

Note by Ethan Robinett
3 years, 1 month ago