# Electrodynamics: Vector Calculus Question

#### majormuss

Gold Member
Problem Statement
Please see attached screenshot. Why are the red circled Del operators not combining to give a net result of 0?
Relevant Equations
Vector Calculus
Why are the red circled Del operators not combining to become 'Del-squared' to cancel out the second term to give a net result of 0?

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#### Abhishek11235

Because $$\nabla(\nabla. ) \neq (\nabla.\nabla = \nabla^2)$$

Alternatively,the red circled termed is vector whereas $\nabla^2$ is scalar called Laplacian.

This thing you can find in almost all Basic Vector Calculus books.

#### majormuss

Gold Member
Because $$\nabla(\nabla. ) \neq (\nabla.\nabla = \nabla^2)$$

Alternatively,the red circled termed is vector whereas $\nabla^2$ is scalar called Laplacian.

This thing you can find in almost all Basic Vector Calculus books.
Thanks!

#### JD_PM

$$\nabla(\nabla \cdot \vec v)$$
Is the OP pointing at the gradient of the divergence operation?

#### majormuss

Gold Member
$$\nabla(\nabla \cdot \vec v)$$
Is the OP pointing at the gradient of the divergence operation?
yea. Sorry, I only just saw this.

#### majormuss

Gold Member
Because $$\nabla(\nabla. ) \neq (\nabla.\nabla = \nabla^2)$$

Alternatively,the red circled termed is vector whereas $\nabla^2$ is scalar called Laplacian.

This thing you can find in almost all Basic Vector Calculus books.
Could you please cite a page number or section from an online page where that distinction is fully explained? I thought I understood it what you meant but I am still getting tripped up.

#### kuruman

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There is a basic difference between the two, one operates on a vector and the result is a vector,
$$\vec \nabla(\vec \nabla \cdot \vec A)=\left(\frac {\partial}{\partial x} ~\hat x+\frac {\partial}{\partial y} ~\hat y+\frac {\partial}{\partial z} ~\hat z \right )\left(\frac {\partial A_x}{\partial x} +\frac {\partial A_y}{\partial y}+\frac {\partial A_z}{\partial z} \right ).$$The other operates on a scalar and the result is a scalar,
$$\vec \nabla \cdot \vec \nabla \varphi =\left(\frac {\partial ^2}{\partial x^2} +\frac {\partial ^2}{\partial y^2}+\frac {\partial ^2}{\partial z^2} \right )\varphi.$$

#### PeroK

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There is a basic difference between the two, one operates on a vector and the result is a vector,
$$\vec \nabla(\vec \nabla \cdot \vec A)=\left(\frac {\partial}{\partial x} ~\hat x+\frac {\partial}{\partial y} ~\hat y+\frac {\partial}{\partial z} ~\hat z \right )\left(\frac {\partial A_x}{\partial x} +\frac {\partial A_y}{\partial y}+\frac {\partial A_z}{\partial z} \right ).$$The other operates on a scalar and the result is a scalar,
$$\vec \nabla \cdot \vec \nabla \varphi =\left(\frac {\partial ^2}{\partial x^2} +\frac {\partial ^2}{\partial y^2}+\frac {\partial ^2}{\partial z^2} \right )\varphi.$$
To add to this. In the attached example, $\nabla^2$ is actually acting on a vector. In this case, by definition:
$$\nabla^2 \varphi = \vec \nabla \cdot \vec \nabla \varphi =\left(\frac {\partial ^2}{\partial x^2} +\frac {\partial ^2}{\partial y^2}+\frac {\partial ^2}{\partial z^2} \right )\varphi.$$
And, also by definition:
$$\nabla^2 \vec{A} = (\nabla^2 A_x) \hat{x} + (\nabla^2 A_y) \hat{y} + (\nabla^2 A_z) \hat{z}\ne \vec \nabla(\vec \nabla \cdot \vec A)$$

#### kuruman

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To add to this. In the attached example, $\nabla^2$ is actually acting on a vector. In this case, by definition:
$$\nabla^2 \varphi = \vec \nabla \cdot \vec \nabla \varphi =\left(\frac {\partial ^2}{\partial x^2} +\frac {\partial ^2}{\partial y^2}+\frac {\partial ^2}{\partial z^2} \right )\varphi.$$
And, also by definition:
$$\nabla^2 \vec{A} = (\nabla^2 A_x) \hat{x} + (\nabla^2 A_y) \hat{y} + (\nabla^2 A_z) \hat{z}\ne \vec \nabla(\vec \nabla \cdot \vec A)$$
And the distinction between the two is illustrated by the vector calculus identity
$$\vec \nabla \times (\vec \nabla \times \vec A)=\vec \nabla(\vec \nabla \cdot \vec A)-\nabla^2 \vec A.$$

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