Is Del (or Nabla) an operator or a vector?

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Solution 1

First, let's say that $\nabla$ and $\vec \nabla$ are two equivalent notations for the same "object". This notation is used in the representation of three important vector operators: gradient, curl and divergence.

The gradient operator acts on a scalar differentiable function $f(\vec x)$, where $\vec x \in \mathbb R^n$, and returns a vector:

$$\text{grad} \ f(\vec x) = \nabla f(\vec x) \equiv \sum_{i=1}^n \frac{\partial f (\vec x)}{\partial x_i} \vec e_i $$

where $\{\vec e_i \dots\vec e_n\}$ is an orthogonal basis of $\mathbb R^n$.

The divergence operator acts on a vector field $\vec F(\vec x)$, where $\vec x,\vec F \in \mathbb R^n$, and returns a scalar function:

$$\text{div} \ \vec F(\vec x) = \nabla \cdot \vec F(\vec x) \equiv \sum_{i=1}^n \frac{\partial F_i (\vec x)}{\partial x_i} $$

The curl operator acts on a vector field $\vec F(\vec x)$, where $\vec x,\vec F \in \mathbb R^3$, and returns a vector field:

$$\text{curl} \ \vec F(\vec x) = \nabla \times \vec F(\vec x) = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \hat i+\left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \hat j+ \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) \hat k$$

where $\hat i, \hat j, \hat k$ are the unit vectors of the three Cartesian axes.

Notice that, unlike the gradient and divergence, the curl operator does not generalize simply in $n$ dimensions. Also, the notation $\nabla \times \vec F$ is only a mnemonic device useful when we work in cartesian coordinates: in other coordinate systems, applying $\nabla \times \vec F$ will hold the wrong result.

We should probably also mention the laplacian operator, which is the divergence of the gradient:

$$\nabla^2 f(\vec x) \equiv \text{div} \ (\text{grad} \ f(\vec x)) = \nabla \cdot (\nabla f(\vec x))$$


So, to sum up, $\nabla$ is just a useful notation that is used in the representation of three different vector operators. It turns out that we can often formally manipulate $\nabla$ as if it was a vector, but it is not a vector in the usual sense: $\nabla$ alone is meaningless.

To see this, just consider one of the fundamental properties of vector spaces: if $v,w$ are elements of the vector space $V$, then $v+w$ is also an element of $V$.

Let's consider the vector space $\mathbb R^n$: what meaning should we give to an expression such as

$$\nabla + \vec x \ ?$$

the answer is: no meaning at all, because $\nabla$ is not a vector.

Solution 2

Both. It's an operator that transforms as a covector under rotations. What this means is that if you rotate the coordinate system the gradient in the new coordinate system, $\nabla'$, can be written as:$$\nabla'_i = \sum_{j} R^{-1}_{ij} \nabla_j,$$ where $R^{-1}$ is the inverse of the rotation matrix, $\nabla$ is the gradient in the original coordinate system, and $\nabla'$ is the gradient in the rotated coordinate system.

Solution 3

I hate to play this card, but it depends on the object it acts on (and sometimes who you ask.) Example: many (professors, collegues, etc.) will insist on differentiating between writing $\vec{\nabla}$ and $\nabla$ (consider obliging if your grade/ income depends on it.) In reality, however $\nabla$ is NOT a specific operator, but a convenient mathematical notation. For instance, one may write $\vec{\nabla}\cdot\vec{j}$ or $\nabla\cdot \vec{j}$ and it "should" be obvious from the notation that the meaning of $\nabla$ in this case is a vector operation whether or not the vector symbol is included over it. Another example: one may write $(\vec{v}\cdot\vec\nabla) \vec{j}$ or $\vec{v}\cdot\nabla{\vec {j}}$. In ether case the same quantity is produced. I appreciate the latter notation, however, because it highlights the freedom to act the $\nabla$ upon $\vec{j}$ first (producing a matrix) and then act on $\vec{v}$ to get a vector, or to act the $\vec{v}$ on $\nabla$ first (producing a scalar operator) and then act on $\vec{j}$ producing an identical vector.

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Updated on January 28, 2020

Comments

  • Sofiane
    Sofiane almost 4 years

    Is Del (or Nabla, $\nabla$) an operator or a vector ?

    \begin{equation*} \nabla\equiv\frac{\partial}{\partial x}\vec{i}+\frac{\partial}{\partial y}\vec{j}+\frac{\partial}{\partial z}\vec{k} \end{equation*}

    In some references of vector analysis and electromagnetism, it is considered as an operator (and noted as $\nabla$), and in other ones, it is considered as a vector (and noted as $\vec\nabla$).

    • anon01
      anon01 about 7 years
      it is a vector operator. Sometimes the vector part is not noted explicitly, or is dotted (eg \nabla^2 = \nabla \cdot \nabla), but it is definitely a vector operator.
    • Selene Routley
      Selene Routley about 7 years
      To add to Sean's answer below: It's also a vector in the usual mathematician's sense: a member of the the vector space defined by the set of all linear combinations of the basis vectors $\partial_i$, with the field of scalars being either $\mathbb{R}$ or $\mathbb{C}$
    • Admin
      Admin about 7 years
      also, notice that $\nabla$ (without the arrow) is usually used to refer to the vector operator (a function from vector to the reals) called the divergence en.wikipedia.org/wiki/Divergence
    • Sean E. Lake
      Sean E. Lake about 7 years
      Disagree, @AlbertAspect. In the usage I've seen $\nabla$ is the gradient, $\nabla\cdot$ is the divergence and $\nabla\times$ is the curl.
  • Sofiane
    Sofiane about 7 years
    Sorry but can't understand "It's an operator that transforms as a vector under rotations".
  • Sean E. Lake
    Sean E. Lake about 7 years
    Added elaboration
  • Kartik
    Kartik about 7 years
    Actually it's a covector . . . The matrix should be the inverse of the rotation matrix . . .
  • Sofiane
    Sofiane about 7 years
    How can your rotation matrix-based explanation leads to that $\nabla$ can be both (vector and operator) ? What is the relation between them ? I think that it misses me some basics.
  • Selene Routley
    Selene Routley about 7 years
    It's also a vector in the usual mathematician's sense: a member of the the vector space defined by the set of all linear combinations of the basis vectors $\partial_i$, with the field of scalars being either $\mathbb{R}$ or $\mathbb{C}$
  • Sean E. Lake
    Sean E. Lake about 7 years
    It's a linear operator because it takes a function and does something to it. So, if I have a scalar function $\Phi(\mathbf{x})$ then $\nabla \Phi$ is a different, related, function that assigns a vector to every point in space. The linear operator part comes because $\nabla[ a \Phi + b \Psi] = a \nabla \Phi + b \nabla \Psi$ for constant scalars $a$ and $b$, and scalar functions $\Phi$ and $\Psi$.
  • valerio
    valerio about 7 years
    @WetSavannaAnimalakaRodVance $\partial_{x_i} \vec e_i$ (I guess you meant to write it with the basis vector) are not vectors, they are differential operators. Also, if a $\nabla$ was an element of the vector space you describe, then an expression like $\nabla \cdot \vec F$ would be meaningless because $\nabla$ and $\vec F$ would belong to different vector spaces.
  • Sean E. Lake
    Sean E. Lake about 7 years
    @valerio92 It's standard practice in differential geometry to define the basis vectors of the tangent vector spaces to the manifold as $\operatorname{d}x_i$, and the basis vectors of the covector spaces as $\frac{\partial}{\partial x_i}$. This is done because it makes the directional differential, $\sum_i \operatorname{d}x_i \frac{\partial}{\partial x_i}$, a scalar operator.
  • valerio
    valerio about 7 years
    Didn't know about this kind of formalism. I'm reading something, looks like maybe I was wrong and there is a way to interpret $\nabla$ as a (co)vector.
  • Sean E. Lake
    Sean E. Lake almost 5 years
    @Sofiane Ah, overlooked your comment. There are two different, but related, definitions of a vector. The mathematical definition is that a vector is an element of a vector space, and a vector space is a set with a bunch of properties. Another definition often used in physics is that a (co)vector is an object that transforms under the application of a single (inverse) physical rotation matrix. This is tied in more closely to group theory, and provides a slightly different understanding.
  • Apoorv Potnis
    Apoorv Potnis over 4 years
    Can you please elaborate on the definition of vectors in physics or provide a resource explaining it rigorously? I recently had one course on linear algebra so I understand the meaning of vectors in that context but I have trouble understanding in the physics sense. And how are the two definitions in maths and physics related?
  • Sean E. Lake
    Sean E. Lake over 4 years
    @ApoorvPotnis That is a new question, not a comment. Recommended reading: en.wikipedia.org/wiki/Linear_form amazon.com/Mathematics-Classical-Quantum-Physics-Dover/dp/…
  • Sean E. Lake
    Sean E. Lake over 4 years
  • Peruz
    Peruz over 3 years
    Good answer, in line with the following from Wikipedia: Strictly speaking, del is not a specific operator, but rather a convenient mathematical notation for those three operators, that makes many equations easier to write and remember.