Scalar Potential

Certain vector fields can be represented as the gradient of scalar functions. These vector fields are called gradient fields, and the associated scalar function is known as the scalar potential. The connection between scalar potentials and their gradient fields is very obvious in physics.

Definition §

Scalar Potential

Let $\vec{F}$ be a $C^1$ irrotational vector field (that is, $\nabla \times \vec{F}=\vec{0}$). Then, there exists a scalar function $\varphi$ and a constant vector $\vec{c}$ such that:

$$\vec{F}=\nabla (\varphi +\vec{c})$$

That is, $\vec{F}$ can be represented as the gradient of $\varphi +\vec{c}$.

• $\vec{F}$ is called the gradient field of $\varphi$. It is also called a conservative vector field.
• $\varphi$ is called the scalar potential of $\vec{F}$
• $\varphi$ is unique up to $\vec{c}$ (which is a constant vector) (that is, we can change $\vec{c}$)

• Note that every gradient field is irrotational!
• This uses Stokes’ Theorem

Connection to Physical Force Fields §

A scalar potential is almost always connected to a potential energy field in physics. Moreover, any conservative force has a scalar potential:

• The gravitational potential is a scalar potential for gravitational field
• The electrical potential is a scalar potential for electrical field
• Magnetic field does not have a scalar potential, because it is a non-conservative force.

Examples §

E1: Finding a scalar potential

Is $\vec{F}$ defined below a gradient field?

$$\vec{F}(x,y,z)=(2x{y}^2,2x^{2}y+2z^{2}y,2y^{2}z)$$

Solution $\vec{F}$ is irrotational. If it isn't, we can stop since $\vec{F}$ can only have a scalar potential (i.e. be a gradient field) if it is irrotational. Let $\vec{F}(x,y,z)=(u,v,w)$ where: $u=2x{y}^2,v=2x^{2}y+2z^{2}y,w=2y^{2}z$

First, we need to check if

$$\begin{array}{rl}\nabla \times \vec{F}& =\det \left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ u& v& w\end{array}\right]\\ & =\hat{i}(4zy-4zy)-\hat{j}(0-0)+\hat{k}(4xy-4xy)=\vec{0}\end{array}$$

Since the curl returns $\vec{0}$, we know that $\vec{F}$ is irrotational. To find the scalar potential, $\varphi$, of $\vec{F}$, we need to construct a series of partial differential equations:

$$(2x{y}^2,2x^{2}y+2z^{2}y,2y^{2}z)=\nabla \varphi =\left(\frac{\partial \varphi }{\partial x},\frac{\partial \varphi }{\partial y},\frac{\partial \varphi }{\partial z}\right)$$

$$\{\begin{array}{l}2x{y}^2=\frac{\partial \varphi }{\partial x}\\ 2x^{2}y+2z^{2}y=\frac{\partial \varphi }{\partial y}\\ 2y^{2}z=\frac{\partial \varphi }{\partial z}\end{array}$$

$$2x{y}^2=\frac{\partial \varphi }{\partial x}⟹\varphi =\int 2x{y}^2\text{d}x=x^2{y}^2+C_1(y,z)$$

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