Nabla is a command representation for the nabla differential operator. Thus, it can be used to calculate the gradient, divergence, curl or Laplacian of a function as well.

%Nabla is the inert form of Nabla, that is: it represents the same mathematical operation while holding the operation and checking of arguments unperformed. The expansion rules defined for Nabla, however, also work (are implemented) for the inert %Nabla. To activate the operation use value.

Nabla works in a context sensitive manner. For instance, Nabla(F) returns the gradient of $F$ or the divergence of $F$ according to whether $F$ is identified as a scalar or a vector function (see ?Identify). So when called with only one argument Nabla(F) = Nabla . F. When called with two arguments, Nabla(F, Curl), Nabla(F, Divergence), Nabla(F, Gradient) and Nabla(F,Laplacian) respectively return the curl the divergence, the gradient and the Laplacian of $F$. The result is expressed in the specific coordinates and orthonormal vector basis system related to the functional dependence detected in $F$. When $F$ does not depend on the geometrical coordinates $\left[x\,y\,z\,\mathrm{\rho }\,\mathrm{\phi }\,r\,\mathrm{\theta }\right]$, Nabla(F, ...) returns 0; and when $F$ depends on coordinates of more than one system, the calculations are interrupted an error message is displayed on the screen.

A call to Nabla with one or two arguments returns the evaluation of the corresponding vector differential operation unless $F$ is a non-projected vector, in which case Nabla(F) and Nabla(F, Divergence) return unevaluated as $\mathrm{Divergence}\left(F\right)$ and Nabla(F, Curl) returns Curl(F) (see ?Divergence and ?Curl). Generally speaking, to evaluate the corresponding vector differential operation it would be necessary to know the components of $F$ in some orthonormal vector basis. It is possible, however, to make Nabla return the vector differential operation performed by representing these unknown components using the Component command; for that purpose the optional argument useComponent should be given.

Since Nabla performs differentiation using the standard diff command, all differentiation knowledge of diff is used when computing with Nabla.

For the conventions about the geometrical coordinates and vectors see Identify.

$\mathrm{with}\left(\mathrm{Physics}\left[\mathrm{Vectors}\right]\right)$

$\left[\mathrm{\&x}\,\mathrm{`+`}\,\mathrm{`.`}\,\mathrm{Assume}\,\mathrm{ChangeBasis}\,\mathrm{ChangeCoordinates}\,\mathrm{CompactDisplay}\,\mathrm{Component}\,\mathrm{Curl}\,\mathrm{DirectionalDiff}\,\mathrm{Divergence}\,\mathrm{Gradient}\,\mathrm{Identify}\,\mathrm{Laplacian}\,\nabla \,\mathrm{Norm}\,\mathrm{ParametrizeCurve}\,\mathrm{ParametrizeSurface}\,\mathrm{ParametrizeVolume}\,\mathrm{Setup}\,\mathrm{Simplify}\,\mathrm{`^`}\,\mathrm{diff}\,\mathrm{int}\right]$

$\mathrm{Setup}\left(\mathrm{mathematicalnotation}=\mathrm{true}\right)$

$\left[\mathrm{mathematicalnotation}=\mathrm{true}\right]$

$\nabla \left(f\left(x\,y\,z\right)\right)$

$\left(\frac{\partial }{\partial x}f\left(x\,y\,z\right)\right)\stackrel{\wedge }{i}+\left(\frac{\partial }{\partial y}f\left(x\,y\,z\right)\right)\stackrel{\wedge }{j}+\left(\frac{\partial }{\partial z}f\left(x\,y\,z\right)\right)\stackrel{\wedge }{k}$

$\nabla ·f\left(x\,y\,z\right)$

$\left(\frac{\partial }{\partial x}f\left(x\,y\,z\right)\right)\stackrel{\wedge }{i}+\left(\frac{\partial }{\partial y}f\left(x\,y\,z\right)\right)\stackrel{\wedge }{j}+\left(\frac{\partial }{\partial z}f\left(x\,y\,z\right)\right)\stackrel{\wedge }{k}$

$\nabla \left(f\left(x\,y\,z\right)\,\mathrm{Laplacian}\right)$

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

$\mathrm{Physics}:-\mathrm{Setup}\left(\mathrm{vectorpostfix}=\_\right)$

$\left[\mathrm{vectorpostfix}=\_\right]$

$\nabla \left(\mathrm{f\_}\left(x\,y\,z\right)\right)$

$\nabla ·\overrightarrow{f}\left(x\,y\,z\right)$

$\nabla \left(\mathrm{f\_}\left(x\,y\,z\right)\,\mathrm{Divergence}\right)$

$\nabla ·\overrightarrow{f}\left(x\,y\,z\right)$

$\nabla ·\mathrm{f\_}\left(x\,y\,z\right)$

$\nabla ·\overrightarrow{f}\left(x\,y\,z\right)$

$\nabla \left(\mathrm{f\_}\left(x\,y\,z\right)\,\mathrm{Curl}\right)$

$\nabla \times \overrightarrow{f}\left(x\,y\,z\right)$

$\nabla \left(\mathrm{f\_}\left(x\,y\,z\right)\,\mathrm{Curl}\,\mathrm{useComponent}\right)$

$\left({\left(\frac{\partial }{\partial y}\overrightarrow{f}\left(x\,y\,z\right)\right)}_3-{\left(\frac{\partial }{\partial z}\overrightarrow{f}\left(x\,y\,z\right)\right)}_2\right)\stackrel{\wedge }{i}+\left({\left(\frac{\partial }{\partial z}\overrightarrow{f}\left(x\,y\,z\right)\right)}_1-{\left(\frac{\partial }{\partial x}\overrightarrow{f}\left(x\,y\,z\right)\right)}_3\right)\stackrel{\wedge }{j}+\left({\left(\frac{\partial }{\partial x}\overrightarrow{f}\left(x\,y\,z\right)\right)}_2-{\left(\frac{\partial }{\partial y}\overrightarrow{f}\left(x\,y\,z\right)\right)}_1\right)\stackrel{\wedge }{k}$