The Gradient

Line Integrals and the Gradient

We looked at the relationship between

  1. $\oint_C \mathbf{F} \cdot \hat{\mathbf{t}} \\, ds = 0$ for any closed curve $C$.
  2. $\nabla \times \mathbf{F} = 0$.

Two results:

There’s a third way to express this:

  1. $\mathbf{F}$ is gradient of a scalar field $\psi$.

That is, if there’s a scalar function $\psi$ such that

$$ F_x = \frac{\partial \psi}{\partial x}, \quad F_y = \frac{\partial \psi}{\partial y}, \quad F_z = \frac{\partial \psi}{\partial z} $$

then line integral of $\mathbf{F} \cdot \hat{\mathbf{t}}$ is path-independent.

($\implies$) Proving if $\mathbf{F}$ is gradient of some scalar function $\psi$, then line integral is path-independent:

To show this, first note that using the chain rule, we have:

$$ \mathbf{F} \cdot \hat{\mathbf{t}} = \frac{\partial \psi}{\partial x} \frac{dx}{ds} + \frac{\partial \psi}{\partial y} \frac{dy}{ds} + \frac{\partial \psi}{\partial z} \frac{dz}{ds} = \frac{d\psi}{ds} $$

then using the Fundamental Theorem of Calculus, we have:

$$ \int_C \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = \int_C \frac{d\psi}{ds} \, ds = \psi(x_1, y_1, z_1) - \psi(x_0, y_0, z_0) $$

($\impliedby$) Now, we prove the converse: if line integral is path-independent, then $\mathbf{F}$ is a gradient of some scalar function $\psi$.

Fix $(x_0, y_0, z_0)$ and define $\psi(x, y, z)$ as:

$$ \psi(x, y, z) = \int_{(x_0, y_0, z_0)}^{(x, y, z)} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds $$

Since the line integral is path-independent, we can change the path to:

Then:

$$ \begin{align*} \psi(x, y, z) &= \int_{P_0}^{P_1} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds + \int_{P_1}^{P} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds \\ &= \int_{P_0}^{P_1} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds + \int_{P_1}^{P} F_x(x', y, z) \, dx' \end{align*} $$

Since the first term on the right-hand side is independent of $x$, when we differentiate $\psi$ with respect to $x$, we have:

$$ \frac{\partial \psi}{\partial x} = F_x(x, y, z) $$

Similarly, we can show that:

$$ \frac{\partial \psi}{\partial y} = F_y(x, y, z), \quad \frac{\partial \psi}{\partial z} = F_z(x, y, z) $$

This can be written as:

$$ \begin{align*} \mathbf{F} &= \mathbf{i} \frac{\partial \psi}{\partial x} + \mathbf{j} \frac{\partial \psi}{\partial y} + \mathbf{k} \frac{\partial \psi}{\partial z} \\[0.5em] &= \left( \mathbf{i} \frac{\partial}{\partial x} + \mathbf{j} \frac{\partial}{\partial y} \+ \mathbf{k} \frac{\partial}{\partial z} \right) \psi = \nabla \psi \end{align*} $$

which we read “del psi”. This operator is called the gradient.

The gradient of $\psi$ is a vector function of position.

Similarly, if $\mathbf{F} = \nabla \psi$ under suitable conditions, then $\nabla \times \mathbf{F} = 0$:

$$ \begin{align*} (\nabla \times \mathbf{F})_x &= \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = \frac{\partial}{\partial y} \left( \frac{\partial \psi}{\partial z} \right) - \frac{\partial}{\partial z} \left( \frac{\partial \psi}{\partial y} \right) \\[0.5em] &= \frac{\partial^2 \psi}{\partial y \partial z} - \frac{\partial^2 \psi}{\partial z \partial y} = 0 \end{align*} $$

The last equality holds if $\psi$ and its first and second derivatives are continuous.

The other components can be shown similarly to be zero.

The converse: If $\nabla \times \mathbf{F} = 0$ in a simply connected region, then $\mathbf{F}$ is a gradient of some scalar function $\psi$.

In normal circumstances, where our functions have continuous first derivatives and the regions are simply connected, then the three conditions are equivalent:

  1. $\oint_C \mathbf{F} \cdot \hat{\mathbf{t}} \\, ds = 0$ independent of path.
  2. $\nabla \times \mathbf{F} = 0$.
  3. $\mathbf{F} = \nabla \psi$ for some scalar function $\psi$.

Finding the Electrostatic Field

As we saw, differential form of Gauss’ Law is:

$$ \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} $$

It is not very useful, since it is one equation with three unknowns ($E_x$, $E_y$, and $E_z$).

Since $\mathbf{E}$ is a gradient of some scalar function, we can define electrostatic potential $\Phi$ such that:

$$ \mathbf{E} = - \nabla \Phi $$

Substituting this into Gauss’ Law, we have:

$$ \frac{\partial^2 \Phi}{\partial x^2} + \frac{\partial^2 \Phi}{\partial y^2} + \frac{\partial^2 \Phi}{\partial z^2} = -\frac{\rho}{\epsilon_0} \tag{IV-5} $$

We define the Laplacian operator as:

$$ \nabla^2 = \nabla \cdot \nabla = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \tag{IV-6} $$

Then (IV-5) can be written as:

$$ \nabla^2 \Phi = -\frac{\rho}{\epsilon_0} \tag{IV-7} $$

This is called the Poisson’s equation. It is a linear, second-order partial differential equation in one unknown, the scalar function $\Phi$.

At any point in space where there is no charge, $\rho = 0$, then (IV-7) becomes:

$$ \nabla^2 \Phi = 0 $$

This is called the Laplace’s equation.

Directional Derivatives and the Gradient

The rate of change of the function $f$ in the direction of a unit vector $\hat{\mathbf{u}}$ is given by the directional derivative:

$$ \frac{df}{ds} = \hat{\mathbf{u}} \cdot \nabla f $$

The book gives an argument on how to derive this using the Taylor expansion.

It follows from this that the gradient of a scalar function $f$ is the vector that points in the direction of greatest increase of $f$, and its magnitude is the rate of increase in that direction.