# Dot and Cross Products (of Gradients)

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

## Answers and Replies

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" [Broken] on what's called the "triple product."

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

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

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.

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alphysicist
Homework Helper
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
Science Advisor
Homework Helper
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. 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
Homework Helper
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.

That is incorrect; it is the negative of that.

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