1. Jul 25, 2010

### noon0788

No, this isn't for homework. I'm trying to brush up on gradients for the Physics GRE. Let's say I have vector V.

In english, if I multiply the vector to the gradient squared of the same vector, what is that? I'm thinking that it'd just be the gradient of V, but I can't seem to prove it.

Thanks!!!

2. Jul 26, 2010

### HallsofIvy

Do you mean
$$\vec{V}\nabla^2\vec{V}$$? That is commonly called "del squared". "gradient" is specifically $\nabla f$ where f is a scalar valued function. "curl f" is $\nabla\times \vec{f}$ where f is a vector valued function, and "div f" is $\nabla\cdot \vec{f}$.

Now, $\nabla^2= \frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}+ \frac{\partial }{\partial z^2}$ (in Cartesian coordinates) is also, strictly speaking, applied to a scalar function but it is not uncommon to see it applied to a vector meaning $\nabla^2$ applied to each component.

That is, if
$$\vec{v(x,y,z)}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$$ then

$$\nabla^2 \vec{v}(x,y,z)= \nabla^2f \vec{i}$$$$+ \nabla^2g\vec{j}$$$$+ \nabla^2h \vec{k}$$

a vector. But to say what "$\vec{v}\nabla^2\vec{v}$" means we would still have to know what kind of vector product that is- dot product or cross product?

If that is dot product and $\vec{v}(x,y,z)= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$ then
$$\vec{v}\cdot \nabla^2\vec{v}= f(x,y,z)\nabla^2 f(x,y,z)+ g(x,y,z)\nabla^2 g(x,y,z)+ h(x,y,z)\nabla^2h(x, y, z)$$

If, instead, it is the cross product then
$$\vec{v}\times\nabla^2\vec{v}= \left(g(x,y,z)\nabla^2h(x,y,z)- h(x,y,z)\nabla^2g(x,y,z)\right)\vec{i}- \left(f(x,y,z)\nabla^2h(x,y,z)- h(x,y,z)\nabla^2f(x,y,z)\right)\vec{i}+ \left(f(x,y,z)\nabla^2g(x,y,z)- g(x,y,z)\nabla^2f(x,y,z)\right)\vec{k}$$

Last edited by a moderator: Jul 26, 2010
3. Jul 26, 2010

### arildno

It might also, Halls, be an outer product, i.e a matrix

4. Jul 26, 2010

### HallsofIvy

Good point. In that case it would be
$$\begin{bmatrix}f(x,y,z)\nabla^2f(x,y,z) & f(x,y,z)\nabla^2g(x,y,z) & f(x,y,z)\nabla^2h(x,y,z) \\ g(x,y,z)\nabla^2f(x,y,z) & g(x,y,z)\nabla^2g(x,y,z) & g(x,y,z)\nabla^2h(x,y,z) \\ h(x,y,z)\nabla^2f(x,y,z) & h(x,y,z)\nabla^2g(x,y,z) & h(x,y,z)\nabla^2h(x,y,z)\end{bmatrix}$$
in the x, y, z, coordinate system.