Proof: Nabla X (Nabla X a) = Nabla (Nabla · a) - Nabla^2 a

[madmike159]

Homework Statement

Prove that:

$$\nabla$$X($$\nabla$$Xa) = $$\nabla$$($$\nabla$$$$\cdot$$a) - $$\nabla^{2}$$a

where a is a vector point function.

(X is the cross product and that dot is a dot product.)

Homework Equations

curl, grad, div

The Attempt at a Solution

I have just done another question of the form curl curl A = grad div A - $$\nabla$$^2A

I'm stuck on this one as I don't know what a vector point function is. I tried it with a unit vector, but that just gave me 0 = 0.

 

[HallsofIvy]

A "vector point function" is just a "vector field"- a function that defines a vector at each point. It is, in fact, exactly the kind of functions you have been applying $\nabla\times$ or $\nabla\cdot$ to all along.

(By the way, there is no "[text]" command in LaTex (there is a \text but I don't know why you would want that here).)
You want to show that
$$\nabla\times(\nabla\times\vec{a})= \nabla(\nabla\cdot\vec{a})- \nabla^2\vec{a}$$

Okay, "just do it"- it's really just manipulation. Let $\vec{a}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$ and calculate both sides.

Of course,
$$\nabla\times\vec{a}= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \vec{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f(x,y,z) & g(x,y,z) & h(x,y,z)\end{array}\right|$$
$$= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}+ \left(\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}$$

The $\nabla(\nabla\cdot \vec{a})$ should be straight forward. $\nabla\cdot\vec{a}$ is a scalar function and then you take grad of that.

But be careful about [itex]\nabla^2\vec{a}[/tex]. Strictly speaking,<br /> $$\nabla^2= \frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}+ \frac{\partial^2}{\partial z^2}$$<br /> the "Laplacian" applies to scalar valued functions. To apply it to a vector function, apply it to each component separately:<br /> $$\nabla^2\vec{a}= \left(\frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2}+ \frac{\partial^2 f}{\partial z}^2\right)\vec{i}$$$$+ \left(\frac{\partial^2 g}{\partial x^2}+ \frac{\partial^2 g}{\partial y^2}+ \frac{\partial^2 g}{\partial z}^2\right)\vec{j}$$$$+ \left(\frac{\partial^2 h}{\partial x^2}+ \frac{\partial^2 h}{\partial y^2}+ \frac{\partial^2 h}{\partial z}^2\right)\vec{k}$$[/itex]

 

[madmike159]

I'm still having trouble with this (and my lecturer gave a strong hint that this would be in our exam).

I'm also having trouble getting LaTex to work.

I'll use d here (for $$\partial$$).

For curl curl a i get

(d/dy (dg/dx - df/dy) - d/dz (df/dz - dh/dx)) i and so on for j and k. I think I have these right. What I'm not sure on is how to expand the brackets.

I have the same problem for grad div a where i get

d/dx (df/dx + dg/dy + dh/dz) i and so on for j and k.

For d/dx x df/dx I think you would get d^2f/dx^2, but I have no idea what to do when the bottoms are different.