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For a scalar field $f(\mathbf{r})$, where $\mathbf{r} = (x, y, z)$, what kind of mathematical object results from each of the operations below:

$$ \nabla f(\textbf{r}) $$

$$ \nabla \cdot f(\textbf{r}) $$

$$ \nabla \times f(\textbf{r})? $$

Adrian Keister
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Rob
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2 Answers2

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For the first, you get a vector representing how fast $f$ changes in each direction. The second two expressions are not defined, because you can only dot or cross a vector with another vector. Here is a table of the "possibilities", where $f$ is a scalar field, and $\mathbf{F}$ is a vector field: $$ \begin{array}{|c|c|}\hline \text{Operation} &\text{Result} \\ \hline \nabla f &\text{vector field} \\ \hline \nabla \cdot f & \text{undefined} \\ \hline \nabla \times f & \text{undefined} \\ \hline \nabla\mathbf{F} &\text{tensor} \\ \hline \nabla \cdot\mathbf{F} & \text{scalar} \\ \hline \nabla\times\mathbf{F} & \text{vector} \\ \hline \end{array} $$ For a reference on the expression $\nabla\mathbf{F}$, see this M.SE question, and this wiki article to which it links.

Adrian Keister
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  • So the first one is valid for nabla operator, and the other two are valid if nabla is a vector? – Rob Apr 10 '18 at 17:10
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    Nabla is always a vector. The result of nabla applied to anything is another vector. Your second two expressions are only valid when nabla is operating on a vector field. That is, you can write $\nabla\cdot\mathbf{F}$ or $\nabla\times\mathbf{E}$, where the $\mathbf{F}$ and $\mathbf{E}$ are vector fields. – Adrian Keister Apr 10 '18 at 17:12
  • OK, that makes more sense. Thanks – Rob Apr 10 '18 at 17:29
  • You're very welcome! – Adrian Keister Apr 10 '18 at 17:31
  • Nabla is not a vector (at least not the same kind as $\mathbf r$), but you can pretend that it is for many purposes. – amd Apr 10 '18 at 23:27
  • @amd: Right - it's technically a vector operator. But I find that for students, thinking of it primarily as a vector is more helpful. – Adrian Keister Apr 11 '18 at 13:18
  • @Rob Ch: I made a mistake in a previous comment: $\nabla\cdot\mathbf{F}$ is a scalar. I had said that nabla applied to anything is another vector, which is not true. – Adrian Keister Apr 11 '18 at 13:19
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The gradient operator $\nabla$ takes a scalar function $f:\mathbf{R}^3 \to \mathbf{R}$ and produces a vector field $(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}).$ The notation $\nabla \cdot F$ denotes the application of $\nabla$ componentwise on the components of a vector field $\mathbf{R}^3\to \mathbf{R}^3$, and describes the divergence. For instance, if the vector field is $$ F(x,y,z)=P(x,y,z)i+Q(x,y,z)j+R(x,y,z),$$ the divergence operator gives us $$ \nabla\cdot F(x,y,z)=\frac{\partial P}{\partial x}(x,y,z)+\frac{\partial Q}{\partial y}(x,y,z)+\frac{\partial R}{\partial z}(x,y,z).$$ In particular, this quantity lives in $\mathbf{R}$. However, the divergence operator needs to act on a vector field to be meaningful. In the case where $f:\mathbf{R}^3\to \mathbf{R}$ is a scalar field, there is not a good idea of what that would mean. For the last case, we have the standard curl operator $\nabla\times F$ where $F$ is as above. It is given by $$ \nabla\times F=\bigg(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\bigg)i+\bigg(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\bigg)j+\bigg(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\bigg)k.$$ As we can see, this eats a vector field $F:\mathbf{R}^3\to\mathbf{R}^3$ and produces a vector field in $\mathbf{R}^3$. In your case, it is not obvious how to apply $\nabla\times f$ where $f$ is a scalar field, because the curl depends on the cross product in its definition, which is a construction special to $\mathbf{R}^3$.

In summation, the only one of these expressions that is well-defined is the first.

Alekos Robotis
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