Gradient Definition: What is the Vector Operator \mathbf\nabla?

[Greg Bernhardt]

Definition/Summary

The gradient is a vector operator denoted by the symbol $\mathbf\nabla$ or grad. The gradient of a differentiable scalar function $f\left({\mathbf x}\right)$ of a vector $\mathbf{x}=\left(x_1,x_2,\ldots,x_n\right)$ is a vector field whose components are the partial derivatives of $f\left({\mathbf x}\right)$ with respect to the variables $x_1,x_2,\ldots,x_n\,.$ Explicitly,

$$\mathbf\nabla f = \left(\frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \ldots, \frac{\partial f}{\partial x_n}\right)$$

Equations

For a function of three variables in Cartesian Coordinates,

$$\nabla f\left(x,y,z\right) = \frac{\partial f}{\partial x}\hat{\mathbf{i}} + \frac{\partial f}{\partial y}\hat{\mathbf{j}} + \frac{\partial f}{\partial z}\hat{\mathbf{k}}$$


In Cylindrical Polar Coordinates,

$$\nabla f\left(r,\theta,z\right) = \frac{\partial f}{\partial r}\hat{\mathbf{e_r}} + \frac{1}{r}\frac{\partial f}{\partial \theta}\hat{\mathbf{e_\theta}} + \frac{\partial f}{\partial z}\hat{\mathbf{k}}$$

Where $\hat{\mathbf{e_r}}$ and $\hat{\mathbf{e_\theta}}$ are unit vectors in the radial and angular directions respectively.


In spherical coordinates,

$$\nabla f\left(r,\phi,\theta\right) = \frac{\partial f}{\partial r}\hat{\mathbf{e_r}} + \frac{1}{r} \ \frac{\partial f}{\partial \phi} \hat{\mathbf{e_\phi}} + \frac{1}{r \ \sin \phi} \ \frac{\partial f}{\partial \theta} \hat{\mathbf{e_\theta}}$$

where $\phi$ is the angle from the +z-axis to the point $(r, \phi, \theta )$. Also $\hat{\mathbf{e_r}}$, etc., denote unit vectors.

NOTE: this definition of $\phi, \theta$ is the one commonly used in math and engineering textbooks. PHYSICS TEXTBOOKS USUALLY HAVE $\phi, \theta$ DEFINED THE OTHER WAY ROUND.

Extended explanation

The main property of the gradient of $f$, is that it lies in the domain of the function $f$, and points in the direction in which $f$ is increasing fastest. In particular the gradient at a point $\mathbf{p}$ is perpendicular to the "level set" of $f$ through $\mathbf{p}$, where $f$ is constantly equal to $f(\mathbf{p})$.

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[fresh_42]

See also: https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/