# How does one read $\Delta \mathbf{E} = (\mathbf{d}\cdot \nabla ) \mathbf{E}$

I'm reading a book on Electrodynamics and came across this formula:

$$\Delta \mathbf{E} = (\mathbf{d}\cdot \nabla ) \mathbf{E}$$

where $\Delta \mathbf{E}$ represents the difference (delta) in an electric field from end to end (bold letters being vectors).

It was stated as being a more compact way of showing the three equations: $$\Delta E_x \equiv ( \nabla E_x) \cdot\mathbf{d}$$ $$\Delta E_y \equiv ( \nabla E_y) \cdot\mathbf{d}$$ $$\Delta E_z \equiv ( \nabla E_z) \cdot\mathbf{d}$$ I read these as the gradient of $E_x$ (for example), which is a vector, dotted with $\mathbf{d}$, a distance vector.
But $(\mathbf{d}\cdot \nabla )$ to me seems to be dotting $\mathbf{d}$ with the del operator $\nabla$!

How is one supposed to read $(\mathbf{d}\cdot \nabla ) \mathbf{E}$? Later the author implies that $\nabla(\mathbf{d}\cdot \mathbf{E})$ is "a more convenient way" to write it.

• Do you want the words you're supposed to say when you read it or the meaning of the dot product of a vector and the del operator? – user137731 May 21 '15 at 2:30
• I guess I'm asking for both as one would support the other. Most specifically I'm interested in the meaning of the dot product of a vector and the del operator! – photogo May 21 '15 at 2:40
• I've seen the dot product of the del operator and a vector (i.e. in the other order then shown in my question). – photogo May 21 '15 at 2:42

## 1 Answer

$\mathbf d \cdot \nabla$ is just an operator. Specifically it's the operator $$d_1\frac {\partial}{\partial x} + d_2\frac {\partial}{\partial y} + d_3\frac {\partial}{\partial z}$$ where $\mathbf d = d_1\mathbf e_1 + d_2\mathbf e_2 + d_3\mathbf e_3$. As you can probably tell from the three equations you wrote, this operator is related to the directional derivative.

Note that this is completely different than the divergence of $\mathbf d$, which is often denoted $\nabla \cdot \mathbf d$.

As for how to say it, IDK. I normally read it as "'d' dot del", but that's just because that's what it looks like -- I really don't know if there is a standard way to say it or not.

• Thanks. That seems pretty obvious now when $\nabla$ is viewed as a "false vector" ... so it's just the normal dot product of two "vectors"! Is my reasoning sound? – photogo May 21 '15 at 2:51
• Yes. Basically, as long as you're careful, you can usually treat $\nabla$ as an ordinary vector. The (few) exceptions come from the fact that it doesn't have a couple of the nice properties of regular vectors. For instance it doesn't commute (or anticommute) under the dot or cross products (as seen above). – user137731 May 21 '15 at 2:53
• BTW, as I see that you're new here, you should know that if I answered your question you should click the green checkmark to the left of my answer. That way others will know that this question has already been satisfactorily answered. And if I haven't answered your question completely, let me know what you're still struggling with. – user137731 May 21 '15 at 2:58
• Thanks again very much for the answer. The check mark was grey and turned green when I clicked it. Also thanks to Ken for editing my title line as I couldn't figure out how to get the TeX to work up there! – photogo May 21 '15 at 3:03