# Units of vector differential operator del ($\nabla$)

My book says that $\left[\nabla \cdot (\vec E \times \vec H)\right] = \mathrm{W/m^3}$. I see that $\vec E$ is in $\mathrm{V/m}$ and $\vec H$ is $\mathrm{A/m}$, so these multiplied is $\mathrm{W/m^2}$, but how does dotting with $\nabla$ give another $\mathrm{m^{-1}}$?

Note on notation: I use $[\cdot]$ do denote the units of the quantity in brackets.

Derivatives always have units of $1/[\text{differentiation variable}]$. This can be clearly seen from the definition of the derivative in terms of difference qutionts: $$\partial_x f(x) := \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.$$ So if $x$ has some unit $[x]$ then $\partial_x f$ will have units $[f]/[x]$. (As the limit does obviously not change the units.)

As $$\nabla = \begin{pmatrix} \partial_x \\ \partial_y \\ \partial_z \end{pmatrix}$$ and the coordinates in space carry the unit $\mathrm{m}$, you have that $$[\nabla \cdot \vec V] = [\partial_x V_x + \partial_y V_y + \partial_z V_z] = [\partial_x V_x] = [V_x]/[x] = [V_x]/\mathrm{m}.$$ (Where I arbitrarily chose the $x$-component of the vector, as all components have the same units.)

• $\nabla$ can be used for co-ordinates that needn't be 3D spatial co-ordinates with dimension $L$, which is why wiki doesn't mention units. just as one wouldn't mention the units of $d/dx$... – innisfree Jun 2 '15 at 23:10
• +1, I wasn't suggesting that anything in your answer was misleading or wrong, just elaborating – innisfree Jun 2 '15 at 23:11
• This is a pretty awesome answer – Cicero Jun 2 '15 at 23:21

The $\nabla$ operator is a spatial derivative of the $\frac{\delta}{\delta x}$ etc kind. This has units of $1/m$

• Would you explain this a little bit more? In all my math classes where we used this no one ever told us that it had units, actually I can't remember learning about any operator that had units. Also I found this confirmed in a couple of forums, but even the wikipedia page doesn't mention it. – Austin Jun 2 '15 at 22:59
• @AustinMW - are you familiar with the idea that a derivative (slope) has the dimensions of "what is plotted on Y" divided by "what is plotted on X" (sometimes called "rise over run")? This is really all I am saying: in the equation you quoted, the $\nabla$ operator is just a spatial derivative of sorts, and as such has dimensions of 1 over length. I would have thought that Sebastian's answer made that clear. I don't know how else to put it, sorry. – Floris Jun 2 '15 at 23:16
• Yep I got it now, just never saw it before so it seemed weird at first, thank you – Austin Jun 2 '15 at 23:19