Dot and Cross Products (of Gradients)

[jeff1evesque]

Statement:
I was wondering if the following are identical,
$$\nabla \bullet \nabla \times \vec{A} = \nabla \bullet (\nabla \times \vec{A}) ?$$ (#1)

Also, more importantly I was wondering if someone could explain to me why the following is zero for any vector,

$$\nabla \bullet \nabla \times \vec{A} = 0?$$

Reasoning:
If we look at $$\nabla \bullet \nabla \times \vec{A}$$, Isn't $$\nabla \bullet \nabla$$ the laplacian, or $$\nabla^{2}?$$ Is that the reason we cannot perform the operation in equation (#1) above- the laplacian is a function that needs a vector field to be operated on, and there is none. So $$\nabla \bullet \nabla = 0.$$Thanks, alot,JL

 

[nickmai123]

Statement:
I was wondering if the following are identical,
$$\nabla \bullet \nabla \times \vec{A} = \nabla \bullet (\nabla \times \vec{A}) ?$$ (#1)

Also, more importantly I was wondering if someone could explain to me why the following is zero for any vector,

$$\nabla \bullet \nabla \times \vec{A} = 0?$$

Reasoning:
If we look at $$\nabla \bullet \nabla \times \vec{A}$$, Isn't $$\nabla \bullet \nabla$$ the laplacian, or $$\nabla^{2}?$$ Is that the reason we cannot perform the operation in equation (#1) above- the laplacian is a function that needs a vector field to be operated on, and there is none. So $$\nabla \bullet \nabla = 0.$$


Thanks, alot,


JL

Yeah they're identical. Check out http://en.wikipedia.org/wiki/Triple_product" on what's called the "triple product."

 

[Feldoh]

$$\nabla \bullet \nabla \times \vec{A} = 0 = div(curl(A))$$

I mean if you think of it in terms of geometrical vectors, then if we're looking at the divergence of the amount of A which is contained it sort of intuitively seems like it should be zero.

If it's too hard to visualize then just work it out. Assume A is a vector function and take the curl of A then the divergence of the curl of A and by Clairaut's theorem you should see that it all cancels out.

Secondly $$(\nabla \bullet \nabla) \times \vec{A}$$ makes no sense you cannot take the cross product of a scalar operator (which the laplacian is).

Finally the laplacian doesn't equal zero. It's a linear operator, it doesn't equal anything be itself. In fact it really doesn't have any meaning by itself if it's not being applied to a function in euclidean space.

 

[Pengwuino]

It's more complicated then that. Work it out and you'll see the derivatives canceling.

 

[jeff1evesque]

It's more complicated then that. Work it out and you'll see the derivatives canceling.

I tried, for example I let $$\vec{H} = 2xy\hat{x} + x^2y^2z^2\hat{y} + x^3y^3z^3\hat{y}.$$

When I take the curl,
$$\nabla \times \vec{H} = [3x^3y^2z^3 - 2x^{2}y^{2}z]\hat{x} + [3x^{2}y^{3}z^{3} - 0]\hat{y} + [2xy^{2}z^{2} - 2x]\hat{z}$$ and let that whole thing equal $$\alpha$$

So now when I take the divergence,
$$\nabla \bullet \alpha = (9x^{2}y^{2}z^{3} - 4xy^{2}z) + 9(x^{2}y^{2}z^{3}) + (4xy^{2}z) = 18(x^{2}y^{2}z^{3}) \neq 0$$

It seems like it almost worked out, the second term above needed a sign change, but I've reviewed the calculation, and it doesn't seem that I've made a mistake.

...makes no sense you cannot take the cross product of a scalar operator (which the laplacian is).

That was along the lines I was thinking, the laplacian requires a vector to be operated on, and it doesn't have one.

 

[alphysicist]

I tried, for example I let $$\vec{H} = 2xy\hat{x} + x^2y^2z^2\hat{y} + x^3y^3z^3\hat{y}.$$

When I take the curl,
$$\nabla \times \vec{H} = [3x^3y^2z^3 - 2x^{2}y^{2}z]\hat{z} + [3x^{2}y^{3}z^{3} - 0]\hat{y} + [2xy^{2}z^{2} - 2x]\hat{z}$$ and let that whole thing equal $$\alpha$$

I think your second term (the y-hat term) is missing a minus sign here.

 

[tiny-tim]

Hi jeff1evesque!

(have a del: ∇ and use \cdot, not \bullet)

I was wondering if the following are identical,
$$\nabla \bullet \nabla \times \vec{A} = \nabla \bullet (\nabla \times \vec{A}) ?$$ (#1)

Also, more importantly I was wondering if someone could explain to me why the following is zero for any vector,

$$\nabla \bullet \nabla \times \vec{A} = 0?$$

Reasoning:
If we look at $$\nabla \bullet \nabla \times \vec{A}$$, Isn't $$\nabla \bullet \nabla$$ the laplacian, or $$\nabla^{2}?$$ Is that the reason we cannot perform the operation in equation (#1) above- the laplacian is a function that needs a vector field to be operated on, and there is none. So $$\nabla \bullet \nabla = 0.$$

You're very confused.

Learn the following:

div curl = curl grad = 0

div(curlA) = ∇.(∇xA) = 0

curl(gradφ) = ∇x(∇φ) = 0

div(gradφ) = ∇²φ = Laplacian.

∇.(∇xA) is zero for the same reason that B.(BxA) is zero.

 

[jeff1evesque]

I think your second term (the y-hat term) is missing a minus sign here.

The y-hat consists of the following,
$$[\frac{\partial}{\partial x}(x^{3}y^{3}z^{3}) - \frac{\partial}{\partial z}(2xy)]\hat{y} = [3x^{2}y^{3}z^{3} - 0]\hat{y}$$

Am I incorrect?

 

[alphysicist]

The y-hat consists of the following,
$$[\frac{\partial}{\partial x}(x^{3}y^{3}z^{3}) - \frac{\partial}{\partial z}(2xy)]\hat{y} = [3x^{2}y^{3}z^{3} - 0]\hat{y}$$

Am I incorrect?

That is incorrect; it is the negative of that.

 

[jeff1evesque]

That is incorrect; it is the negative of that.

Thanks, I totally forgot (the notation is similar to taking determinants).