The symbol $\nabla$ (nabla) represents the del operator.



$$\nabla = \nail{\fracp{}{x_1}, \dots, \fracp{}{x_n}}$$

For example, using the standard basis $\hat i, \hat j, \hat k$ of $\RR^3$, we get $$\nabla = \nail{\fracp{}{x}, \fracp{}{y}, \fracp{}{z}} = \hat i\fracp{}{x} + \hat j\fracp{}{y} + \hat k\fracp{}{z}$$


If $f:\RR^3\to\RR$ is a scalar field, $$\Mr{grad} f = \nabla f = \nail{\fracp{f}{x}, \fracp{f}{y}, \fracp{f}{z}} \in \RR^3$$ The gradient is the "slope" direction and magnitude.


If $f:\RR^3\to\RR^3$ is a vector field and $f(x,y,z) = (f_x,f_y,f_z)$, $$\Mr{div} f = \nabla\cdot f = \fracp{f_x}{x} + \fracp{f_y}{y} + \fracp{f_z}{z} \in \RR$$ The divergence measures how much field diverges from the given point.


If $f:\RR^3\to\RR^3$ is a vector field and $f(x,y,z) = (f_x,f_y,f_z)$, $$\Mr{curl} f = \nabla\times f = \abs{\begin{matrix} \fracp{}{x} & \fracp{}{y} & \fracp{}{z} \\ f_x & f_y & f_z \\ \hat i & \hat j & \hat k \end{matrix}}\in\RR^3$$ The curl is the torque at a given point.

Directional Derivative

If $f:\RR^3\to\RR$ is a scalar field and $a(x,y,z) = (a_x,a_y,a_z)$, $$a\cdot\Mr{grad} f = (a\cdot\nabla) f = a_x\fracp{f}{x} + a_y\fracp{f}{y} + a_z\fracp{f}{z} \in \RR$$

Hessian / Laplacian

This one is confusing. ML people use $\nabla^2$ to denote the Hessian matrix. For $f:\RR^3\to\RR$, $$\nabla^2 f = \matx{ \fracp{^2}{x^2} f & \fracp{^2}{x\partial y} f & \fracp{^2}{x\partial z} f \\ \fracp{^2}{y\partial x} f & \fracp{^2}{y^2} f & \fracp{^2}{y\partial z} f \\ \fracp{^2}{z\partial x} f & \fracp{^2}{z\partial y} f & \fracp{^2}{z^2} f \\ }\in\RR^3$$

However, in physics, $\nabla^2$ denotes the Laplacian operator $$\Delta f = \nabla^2 f = \nabla\cdot\nabla f = \fracp{^2}{x^2} f + \fracp{^2}{y^2} f + \fracp{^2}{z^2} f \in \RR$$

Both operators can also be applied on $f:\RR^3\to\RR^\text{higher}$, but the results will have more dimensions.


Exported: 2016-07-13T01:31:43.734581