Vector product - operator's a vector?

[Oblio]

Greetings all!
So,

$$\nabla X \vec{F}$$ is confusing me.

I understand that it can be used to tell whether a force is conservative in that, if the curl is 0 then the work done all all paths are the same... that's fine.

However,
I was looking at it, for example, in the context of the gravitational field. When drawing it out, one can see that the curl is indeed 0, and I've been told that it is proven by $$\nabla X \vec{F}$$, BUT from what I understand, and from what I've been told, the operator $$\nabla$$ has no direction, and thereby not a vector...

yet used in a vector product?

How can it even by setup in the first place if the operator isn't a vector?

 

[Oblio]

This got moved I see? Yet its from my classical mechanics class. lol

 

[Oblio]

this post can probably be disregarded unless people with a classical mechanics context can answer my question, and not a purely mathematical one; which would probably mean very little to me.

 

[arildno]

The operator is defined in such a clever way that you MAY regard it as a vector, so that the curl comes out properly.

Of course, it is not really a vector, but how we define it enables us to treat it as a vector in our calculations.

 

[CompuChip]

Actually, the curl operator is curl, rot or $\vec\nabla\times$. You can apply it to a vector $\vec V = (V_1, V_2, V_3)$ so you get $\operatorname{curl} \vec V$, $\operatorname{rot} \vec V$ or $\vec\nabla \times \vec V$, which is defined by

$$\operatorname{curl} \vec V = \operatorname{rot} \vec V = \nabla \times V \stackrel{\mathrm{def}}{=} \begin{pmatrix}<br /> \frac{\partial V_3}{\partial y} - \frac{\partial V_2}{\partial z} \\<br /> \frac{\partial V_1}{\partial z} - \frac{\partial V_3}{\partial x} \\<br /> \frac{\partial V_2}{\partial x} - \frac{\partial V_1}{\partial y} \\<br /> \end{pmatrix}$$.

By way of mnemonic, you can remember
$$\nabla = \begin{pmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{pmatrix}$$
and you can multiply it in the cross product like it is an actual vector and it will give you the correct result.
But you should consider this as a coincidence, and keep in mind that although $\nabla$ behaves in most ways like a vector, the operators that work on the vector are really things like $\nabla\cdot, \nabla\times$, etc. and it is not really a vector product in the sense of $\vec V \times \vec W$ or $\vec V \cdot \vec W$. If you want, also consider it a coincidence that the same symbol $\nabla$ occurs in all these notations, though of course this is because we think of $\nabla$ as something with partial derivatives, which is in all these operators.

Sorry, it became a mathematical explanation after all. But for most practical purposes (actually, all I have encountered in classical mechanics) you can pretend that it is a vector (but again, it is not).

 

[robphy]

$$\nabla$$ can be thought of like a vector only in cartesian coordinates.
Note the warning at the bottom of http://users.aber.ac.uk/ruw/teach/260/260del.html .

So, in starting out, it might be a useful mnemonic... but one should quickly free oneself from it. (Similarly, since evaluating determinants with the diagonal lines only work with 2x2 and 3x3, one should quickly learn the more general method and its interpretation.)

 

[Meir Achuz]

$$\nabla$$ can be thought of like a vector only in cartesian coordinates.

Del is a vector differential operator.
It's components only have that simple form in Cartesian coordinates, but it is still a vector.
All its operations can be carried out independently of any coordinate system.

 

[arildno]

$$\nabla$$ can be thought of like a vector only in cartesian coordinates.
Note the warning at the bottom of http://users.aber.ac.uk/ruw/teach/260/260del.html .

So, in starting out, it might be a useful mnemonic... but one should quickly free oneself from it. (Similarly, since evaluating determinants with the diagonal lines only work with 2x2 and 3x3, one should quickly learn the more general method and its interpretation.)

The correct application of the del operator will always work out properly, irrespective of your choice of coordinates.

Here's how the correct application goes, using cylindrical coordinates for convenience, calculating the curl.
The trick lies in applying the differential operator BEFORE making cross-products between vectors proper:
$$\nabla\times\vec{F}=\vec{i}_{r}\times\frac{\partial\vec{F}}{\partial{r}}+\vec{i}_{\theta}\times\frac{\partial\vec{F}}{r\partial\theta}+\vec{i}_{z}\times\frac{\partial\vec{F}}{\partial{z}},\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}+\vec{i}_{z}\frac{\partial}{\partial{z}}, \vec{F}=F_{r}\vec{i}_{r}+{F}_{\theta}\vec{i}_{\theta}+F_{z}\vec{i}_{z}$$

If you calculate this, you'll end up with the correct expression for the curl of F

So, there is no need to free yourself from that mnemonic, as long as you remember it IS a mnemonic..