Vector Operators

Gradient $$\nabla T = \left( \frac{\partial T}{\partial x} \hat{x} + \frac{\partial T}{\partial y}\hat{y} + \frac{\partial T}{\partial z}\hat{z} \right)$$

where $T (x, y, z)$ is any arbitrary scalar function in 3 dimensions.

It is more simply represented as $$\vec{\nabla}T = (\dd T)\cdot(\vec{\dd l})$$where $$\dd T = \frac{\partial T}{\partial x}\dd x + \frac{\partial T}{\partial y} \dd y + \frac{\partial T}{\partial z} \dd z$$
Note that $\dd T$ is also a vector, otherwise dot product wouldn't have been defined.
It is sufficient to leave $\dd T = \vec{\nabla} T$ , with appropriate unit vectors in the direction.

In different coordinate systems $$\begin{align} (\rho, \phi, z)& \ \vec{\nabla} &= \left( \frac{\partial }{\partial \rho} \hat{\rho} + \frac{1}{\rho}\frac{\partial }{\partial \phi} \hat{\phi} + \frac{\partial }{\partial z} \hat{z} \right) \ \ \ (r, \theta, \phi)& \ \vec{\nabla} &= \left( \frac{\partial}{\partial r}\hat{r} + \frac{1}{r}\frac{\partial}{\partial \theta}\hat{\theta} + \frac{1}{r\sin \theta}\frac{\partial}{\partial \phi}\hat{\phi} \right) \end{align}$$

The above two represent gradient only, the operator does not remain same for divergence

Note

Gradient, $\vec{\nabla} T$ points in the direction of maximum increase of $T$ .

Important

We obtained a vector field, $\vec{\nabla} T$ from a scalar field, $T$ .

Vector Field

$\vec{F} (x, y) = M (x, y) \hat{x} + N (x, y) \hat{y}$

where $M$ and $N$ are arbitrary functions.
The vector field is characterized as having a vector for every $(x, y)$ point.
Can be extrapolated to three dimensions.

Divergence $$\vec{\nabla} \cdot \vec{F} = \left( \frac{\partial F_{x}}{\partial x} + \frac{\partial F_{y}}{\partial y} + \frac{\partial F_{z}}{\partial z} \right)$$

Note that divergence of a vector field is scalar.

The operator $\vec{\nabla}$ is the same for gradient and divergence (when taken in cartesian system), just the operand is different, and that gives a different result.

Geometrical Interpretation : The spreading out of a field at a point, $(x, y, z)$ . It signifies flow through a closed surface.

Curl $$\vec{\nabla} \times \vec{F} = \begin{pmatrix}\hat{x} & \hat{y} & \hat{z} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ F_{x} & F_{y} & F_{z}\end{pmatrix}$$

A vector cross product between the $\vec{\nabla}$ and the vector field.

Geometrical Interpretation : How much the vector $\vec{F}$ curls around a point $(x, y, z)$ . Measure of counter-clockwise rotation.
Why counter-clockwise?
Because positive value of curl of $\vec{F}$ suggests rotation with $\vec{ω}$ pointing in the $+ z$ direction. Hence counter-clockwise is positive curl.

The operator $\vec{\nabla}$ gets transformed by coordinate system change.

Divergence and Curl in Cylindrical system

\begin{aligned} A (ρ, ϕ, z) & = (A_{ρ} \cdot e_{ρ} + A_{ϕ} \cdot e_{ϕ} + A_{z} \cdot z) \\ \nabla \cdot A & = (\frac{1}{ρ} \frac{\partial}{\partial ρ} (ρ A_{ρ}) + \frac{1}{ρ} \frac{\partial A_{ϕ}}{\partial ϕ} + \frac{\partial A_{z}}{\partial z}) \end{aligned}

\nabla \times A = \frac{1}{ρ} | \begin{matrix} e_{ρ} & ρ e_{ϕ} & \hat{z} \\ \frac{\partial}{\partial ρ} & \frac{\partial}{\partial ϕ} & \frac{\partial}{\partial z} \\ A_{ρ} & ρ A_{ϕ} & A_{z} \end{matrix} |

Divergence and curl in spherical system

\begin{aligned} A (r, θ, ϕ) \\ \vec{\nabla} \cdot A & = (\frac{1}{r^{2}} \frac{\partial (r^{2} A_{r})}{\partial r} + \frac{1}{r \sin θ} \frac{\partial (A_{θ} \sin θ)}{\partial θ} + \frac{1}{r \sin θ} \frac{\partial A_{ϕ}}{\partial ϕ}) \end{aligned}

\vec{\nabla} \times A = \frac{1}{r^{2} \sin θ} | \begin{matrix} e_{r} & r e_{θ} & \sin θ e_{ϕ} \\ \frac{\partial}{\partial r} & \frac{\partial}{\partial θ} & \frac{\partial}{\partial ϕ} \\ A_{r} & r A_{θ} & \sin θ A_{ϕ} \end{matrix} |

\vec{\nabla} \times A = e_{r} \frac{1}{r \sin θ} [\frac{\partial}{\partial θ} (A_{ϕ} \sin θ) - \frac{\partial A_{θ}}{\partial ϕ}] + e_{θ} \frac{1}{r} [\frac{1}{\sin θ} \frac{\partial A_{r}}{\partial ϕ} - \frac{\partial}{\partial r} (r A_{ϕ})] + e_{ϕ} \frac{1}{r} [\frac{\partial}{\partial r} (r A_{θ}) - \frac{\partial A_{r}}{\partial θ}]

Laplacian $$\vec{\nabla}^2f = \vec{\nabla} \cdot(\vec{\nabla}f)$$

It is a scalar operator, ie, works on scalar fields.