Del should not be confused with Dell.
Del, or nabla, is an operator used in mathematics (particularly in vector calculus) as a vector differential operator, usually represented by the nabla symbol ∇. When applied to a function defined on a one-dimensional domain, it denotes the standard derivative of the function as defined in calculus. When applied to a field (a function defined on a multi-dimensional domain), it may denote any one of three operations depending on the way it is applied: the gradient or (locally) steepest slope of a scalar field (or sometimes of a vector field, as in the Navier–Stokes equations); the divergence of a vector field; or the curl (rotation) of a vector field.
Del is a very convenient mathematical notation for those three operations (gradient, divergence, and curl) that makes many equations easier to write and remember. The del symbol (or nabla) can be formally defined as a vector operator whose components are the corresponding partial derivative operators. As a vector operator, it can act on scalar and vector fields in three different ways, giving rise to three different differential operations: first, it can act on scalar fields by a formal scalar multiplication—to give a vector field called the gradient; second, it can act on vector fields by a formal dot product—to give a scalar field called the divergence; and lastly, it can act on vector fields by a formal cross product—to give a vector field called the curl. These formal products do not necessarily commute with other operators or products. These three uses, detailed below, are summarized as:
\operatorname{grad}f=\nablaf
\operatorname{div}v=\nabla ⋅ v
\operatorname{curl}v=\nabla x v
Rn
(x1,...,xn)
\{e1,...,en\}
x1,...,xn
{\partial\over\partialx1},...,{\partial\over\partialxn}
\nabla=
n | |
\sum | |
i=1 |
ei{\partial\over\partialxi}=\left({\partial\over\partialx1},\ldots,{\partial\over\partialxn}\right)
Where the expression in parentheses is a row vector. In three-dimensional Cartesian coordinate system
R3
(x,y,z)
\{ex,ey,ez\}
\nabla=ex{\partial\over\partialx}+ey{\partial\over\partialy}+ez{\partial\over\partialz}=\left({\partial\over\partialx},{\partial\over\partialy},{\partial\over\partialz}\right)
As a vector operator, del naturally acts on scalar fields via scalar multiplication, and naturally acts on vector fields via dot products and cross products.
More specifically, for any scalar field
f
F=(Fx,Fy,Fz)
\left(ei{\partial\over\partialxi}\right)f:={\partial\over\partialxi}(eif)={\partialf\over\partialxi}ei
\left(ei{\partial\over\partialxi}\right) ⋅ F:={\partial\over\partialxi}(ei ⋅ F)={\partialFi\over\partialxi}
\left(ex{\partial\over\partialx}\right) x F:={\partial\over\partialx}(ex x F)={\partial\over\partialx}(0,-Fz,Fy)
\left(ey{\partial\over\partialy}\right) x F:={\partial\over\partialy}(ey x F)={\partial\over\partialy}(Fz,0,-Fx)
\left(ez{\partial\over\partialz}\right) x F:={\partial\over\partialz}(ez x F)={\partial\over\partialz}(-Fy,Fx,0),
then using the above definition of
\nabla
\nablaf=\left(ex{\partial\over\partialx}\right)f+\left(ey{\partial\over\partialy}\right)f+\left(ez{\partial\over\partialz}\right)f={\partialf\over\partialx}ex+{\partialf\over\partialy}ey+{\partialf\over\partialz}ez
\nabla ⋅ F=\left(ex{\partial\over\partialx} ⋅ F\right)+\left(ey{\partial\over\partialy} ⋅ F\right)+\left(ez{\partial\over\partialz} ⋅ F\right)={\partialFx\over\partialx}+{\partialFy\over\partialy}+{\partialFz\over\partialz}
\begin{align} \nabla x F&=\left(ex{\partial\over\partialx} x F\right)+\left(ey{\partial\over\partialy} x F\right)+\left(ez{\partial\over\partialz} x F\right)\\ &={\partial\over\partialx}(0,-Fz,Fy)+{\partial\over\partialy}(Fz,0,-Fx)+{\partial\over\partialz}(-Fy,Fx,0)\\ &=\left({\partialFz\over\partialy}-{\partialFy\over\partialz}\right)ex+\left({\partialFx\over\partialz}-{\partialFz\over\partialx}\right)ey+\left({\partialFy\over\partialx}-{\partialFx\over\partialy}\right)ez \end{align}
Example:
f(x,y,z)=x+y+z
\nablaf=ex{\partialf\over\partialx}+ey{\partialf\over\partialy}+ez{\partialf\over\partialz}=\left(1,1,1\right)
Del can also be expressed in other coordinate systems, see for example del in cylindrical and spherical coordinates.
Del is used as a shorthand form to simplify many long mathematical expressions. It is most commonly used to simplify expressions for the gradient, divergence, curl, directional derivative, and Laplacian.
f
\operatorname{grad}f={\partialf\over\partialx}\hatx+{\partialf\over\partialy}\haty+{\partialf\over\partialz}\hatz=\nablaf
It always points in the direction of greatest increase of
f
h(x,y)
In particular, this notation is powerful because the gradient product rule looks very similar to the 1d-derivative case:
\nabla(fg)=f\nablag+g\nablaf
However, the rules for dot products do not turn out to be simple, as illustrated by:
\nabla(u ⋅ v)=(u ⋅ \nabla)v+(v ⋅ \nabla)u+u x (\nabla x v)+v x (\nabla x u)
The divergence of a vector field
v(x,y,z)=vx\hatx+vy\haty+vz\hatz
\operatorname{div}v={\partialvx\over\partialx}+{\partialvy\over\partialy}+{\partialvz\over\partialz}=\nabla ⋅ v
The divergence is roughly a measure of a vector field's increase in the direction it points; but more accurately, it is a measure of that field's tendency to converge toward or diverge from a point.
The power of the del notation is shown by the following product rule:
\nabla ⋅ (fv)=(\nablaf) ⋅ v+f(\nabla ⋅ v)
The formula for the vector product is slightly less intuitive, because this product is not commutative:
\nabla ⋅ (u x v)=(\nabla x u) ⋅ v-u ⋅ (\nabla x v)
The curl of a vector field
v(x,y,z)=vx\hatx+vy\haty+vz\hatz
\operatorname{curl}v=\left({\partialvz\over\partialy}-{\partialvy\over\partialz}\right)\hatx+\left({\partialvx\over\partialz}-{\partialvz\over\partialx}\right)\haty+\left({\partialvy\over\partialx}-{\partialvx\over\partialy}\right)\hatz=\nabla x v
The curl at a point is proportional to the on-axis torque that a tiny pinwheel would be subjected to if it were centered at that point.
The vector product operation can be visualized as a pseudo-determinant:
\nabla x v=\left|\begin{matrix}\hatx&\haty&\hatz\\[2pt]{
\partial | |
\partialx |
Again the power of the notation is shown by the product rule:
\nabla x (fv)=(\nablaf) x v+f(\nabla x v)
The rule for the vector product does not turn out to be simple:
\nabla x (u x v)=u(\nabla ⋅ v)-v(\nabla ⋅ u)+(v ⋅ \nabla)u-(u ⋅ \nabla)v
The directional derivative of a scalar field
f(x,y,z)
a(x,y,z)=ax\hatx+ay\haty+az\hatz
(a ⋅ \nabla)f=\limh{
f(x+axh,y+ayh+z+azh)-f(x,y,z) | |
h |
a ⋅ \operatorname{grad}f=ax{\partialf\over\partialx}+ay{\partialf\over\partialy}+az{\partialf\over\partialz}=a ⋅ (\nablaf)
This gives the rate of change of a field
f
a
a
Note that
(a ⋅ \nabla)
The Laplace operator is a scalar operator that can be applied to either vector or scalar fields; for cartesian coordinate systems it is defined as:
\Delta={\partial2\over\partialx2}+{\partial2\over\partialy2}+{\partial2\over\partialz2}=\nabla ⋅ \nabla=\nabla2
The Laplacian is ubiquitous throughout modern mathematical physics, appearing for example in Laplace's equation, Poisson's equation, the heat equation, the wave equation, and the Schrödinger equation.
While
\nabla2
\nabla2
\nabla
\nabla
\nabla2=\nabla ⋅ \nablaT
So whether
\nabla2
Del can also be applied to a vector field with the result being a tensor. The tensor derivative of a vector field
v
\nabla ⊗ v
⊗
For a small displacement
\deltar
\deltav=(\nabla ⊗ v)T\sdot\deltar
For vector calculus:
\begin{align} \nabla(fg)&=f\nablag+g\nablaf\\ \nabla(u ⋅ v)&=u x (\nabla x v)+v x (\nabla x u)+(u ⋅ \nabla)v+(v ⋅ \nabla)u\\ \nabla ⋅ (fv)&=f(\nabla ⋅ v)+v ⋅ (\nablaf)\\ \nabla ⋅ (u x v)&=v ⋅ (\nabla x u)-u ⋅ (\nabla x v)\\ \nabla x (fv)&=(\nablaf) x v+f(\nabla x v)\\ \nabla x (u x v)&=u(\nabla ⋅ v)-v(\nabla ⋅ u)+(v ⋅ \nabla)u-(u ⋅ \nabla)v \end{align}
For matrix calculus (for which
u ⋅ v
uTv
\begin{align} \left(A\nabla\right)Tu&=\nablaT\left(ATu\right)-\left(\nablaTAT\right)u \end{align}
Another relation of interest (see e.g. Euler equations) is the following, where
u ⊗ v
\begin{align} \nabla ⋅ (u ⊗ v)=(\nabla ⋅ u)v+(u ⋅ \nabla)v \end{align}
When del operates on a scalar or vector, either a scalar or vector is returned. Because of the diversity of vector products (scalar, dot, cross) one application of del already gives rise to three major derivatives: the gradient (scalar product), divergence (dot product), and curl (cross product). Applying these three sorts of derivatives again to each other gives five possible second derivatives, for a scalar field f or a vector field v; the use of the scalar Laplacian and vector Laplacian gives two more:
\begin{align} \operatorname{div}(\operatorname{grad}f)&=\nabla ⋅ (\nablaf)=\nabla2f\\ \operatorname{curl}(\operatorname{grad}f)&=\nabla x (\nablaf)\\ \operatorname{grad}(\operatorname{div}v)&=\nabla(\nabla ⋅ v)\\ \operatorname{div}(\operatorname{curl}v)&=\nabla ⋅ (\nabla x v)\\ \operatorname{curl}(\operatorname{curl}v)&=\nabla x (\nabla x v)\\ \Deltaf&=\nabla2f\\ \Deltav&=\nabla2v \end{align}
These are of interest principally because they are not always unique or independent of each other. As long as the functions are well-behaved (
Cinfty
\begin{align} \operatorname{curl}(\operatorname{grad}f)&=\nabla x (\nablaf)=0\\ \operatorname{div}(\operatorname{curl}v)&=\nabla ⋅ (\nabla x v)=0 \end{align}
Two of them are always equal:
\operatorname{div}(\operatorname{grad}f)=\nabla ⋅ (\nablaf)=\nabla2f=\Deltaf
The 3 remaining vector derivatives are related by the equation:
\nabla x \left(\nabla x v\right)=\nabla(\nabla ⋅ v)-\nabla2v
And one of them can even be expressed with the tensor product, if the functions are well-behaved:
\nabla(\nabla ⋅ v)=\nabla ⋅ (v ⊗ \nabla)
Most of the above vector properties (except for those that rely explicitly on del's differential properties - for example, the product rule) rely only on symbol rearrangement, and must necessarily hold if the del symbol is replaced by any other vector. This is part of the value to be gained in notationally representing this operator as a vector.
Though one can often replace del with a vector and obtain a vector identity, making those identities mnemonic, the reverse is not necessarily reliable, because del does not commute in general.
A counterexample that demonstrates the divergence (
\nabla ⋅ v
v ⋅ \nabla
\begin{align} (u ⋅ v)f&\equiv(v ⋅ u)f\\ (\nabla ⋅ v)f&=\left(
\partialvx | |
\partialx |
+
\partialvy | |
\partialy |
+
\partialvz | |
\partialz |
\right)f =
\partialvx | |
\partialx |
f+
\partialvy | |
\partialy |
f+
\partialvz | |
\partialz |
f\\ (v ⋅ \nabla)f&=\left(vx
\partial | |
\partialx |
+vy
\partial | |
\partialy |
+vz
\partial | |
\partialz |
\right)f =vx
\partialf | |
\partialx |
+vy
\partialf | |
\partialy |
+vz
\partialf | |
\partialz |
\\ ⇒ (\nabla ⋅ v)f&\ne(v ⋅ \nabla)f\\ \end{align}
A counterexample that relies on del's differential properties:
\begin{align} (\nablax) x (\nablay)&=\left(ex
\partialx | |
\partialx |
+ey
\partialx | |
\partialy |
+ez
\partialx | |
\partialz |
\right) x \left(ex
\partialy | |
\partialx |
+ey
\partialy | |
\partialy |
+ez
\partialy | |
\partialz |
\right)\\ &=(ex ⋅ 1+ey ⋅ 0+ez ⋅ 0) x (ex ⋅ 0+ey ⋅ 1+ez ⋅ 0)\\ &=ex x ey\\ &=ez\\ (ux) x (uy)&=xy(u x u)\\ &=xy0\\ &=0 \end{align}
Central to these distinctions is the fact that del is not simply a vector; it is a vector operator. Whereas a vector is an object with both a magnitude and direction, del has neither a magnitude nor a direction until it operates on a function.
For that reason, identities involving del must be derived with care, using both vector identities and differentiation identities such as the product rule.