# Dot product and divergence

1,882

It is pretty much simply a short way to notate both vector field operations by looking at $\nabla$ as a vector operator by writing \begin{equation} \nabla=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right) \end{equation} in $\mathbb{R}^3$, or equivalently \begin{equation} \nabla=\frac{\partial}{\partial x}\hat{\imath}+\frac{\partial}{\partial y}\hat{\jmath}+\frac{\partial}{\partial z}\hat{k}. \end{equation} Performing this vector operator on a scalar field gives you the expression for that field's gradient, whereas applying it to a vector field via a dot product gives you the vector field's divergence (analogoulsy for the cross product, which gives you the field's curl instead).

It is important to note, however, that unlike with regular three-vectors, this expression for divergence is not commutative, as the $\nabla$ operator is not a vector in $\mathbb{R}^3$.

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### nelson ningombam

Updated on January 24, 2020

• nelson ningombam almost 3 years

Divergence is represented by dot product. How is the divergence related to dot product? And curl is represented by cross product. How is the curl related to cross product?

• $\operatorname{div} F = \nabla \cdot F$ , $\operatorname{curl} F = \nabla \times F$
• Suriya almost 7 years
Its just the dot/cross product of the Del operator with the vector
• ACuriousMind almost 7 years
I'm voting to close this question as off-topic because it's not a physics question and I consider it of too low quality to migrate it.
• dmckee --- ex-moderator kitten almost 7 years
Thinking about the meaning of these notations (and that for the gradient) will start you on the road to understanding the algebra of differential operators.
• nelson ningombam almost 7 years
it is related to physics...without this knowledge how 1 can understand the maxwell`s equations @ACuriousMind
• nelson ningombam almost 7 years
i know that the dot product represents the divergence, but why the dot product represents the divergence?